\(\int \frac {\sec ^{\frac {7}{2}}(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^{3/2}} \, dx\) [255]

Optimal result

Integrand size = 35, antiderivative size = 247 \[ \int \frac {\sec ^{\frac {7}{2}}(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^{3/2}} \, dx=-\frac {(12 A-19 B) \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{4 a^{3/2} d}+\frac {(9 A-13 B) \text {arctanh}\left (\frac {\sqrt {a} \sqrt {\sec (c+d x)} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{2 \sqrt {2} a^{3/2} d}+\frac {(A-B) \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}+\frac {(6 A-7 B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{4 a d \sqrt {a+a \sec (c+d x)}}-\frac {(A-2 B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{2 a d \sqrt {a+a \sec (c+d x)}} \] Output:

-1/4*(12*A-19*B)*arcsinh(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))/a^(3/2 
)/d+1/4*(9*A-13*B)*arctanh(1/2*a^(1/2)*sec(d*x+c)^(1/2)*sin(d*x+c)*2^(1/2) 
/(a+a*sec(d*x+c))^(1/2))*2^(1/2)/a^(3/2)/d+1/2*(A-B)*sec(d*x+c)^(7/2)*sin( 
d*x+c)/d/(a+a*sec(d*x+c))^(3/2)+1/4*(6*A-7*B)*sec(d*x+c)^(3/2)*sin(d*x+c)/ 
a/d/(a+a*sec(d*x+c))^(1/2)-1/2*(A-2*B)*sec(d*x+c)^(5/2)*sin(d*x+c)/a/d/(a+ 
a*sec(d*x+c))^(1/2)
 

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(497\) vs. \(2(247)=494\).

Time = 2.93 (sec) , antiderivative size = 497, normalized size of antiderivative = 2.01 \[ \int \frac {\sec ^{\frac {7}{2}}(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^{3/2}} \, dx=\frac {4 (6 A-7 B) \arcsin \left (\sqrt {1-\sec (c+d x)}\right ) \cos ^3\left (\frac {1}{2} (c+d x)\right ) \sec ^2(c+d x) \sin \left (\frac {1}{2} (c+d x)\right )+8 (9 A-13 B) \arcsin \left (\sqrt {\sec (c+d x)}\right ) \cos ^3\left (\frac {1}{2} (c+d x)\right ) \sec ^2(c+d x) \sin \left (\frac {1}{2} (c+d x)\right )+6 A \sqrt {1-\sec (c+d x)} \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)-7 B \sqrt {1-\sec (c+d x)} \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)+4 A \sqrt {1-\sec (c+d x)} \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)-3 B \sqrt {1-\sec (c+d x)} \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)+2 B \sqrt {1-\sec (c+d x)} \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)-9 \sqrt {2} A \arctan \left (\frac {\sqrt {2} \sqrt {\sec (c+d x)}}{\sqrt {1-\sec (c+d x)}}\right ) \tan (c+d x)+13 \sqrt {2} B \arctan \left (\frac {\sqrt {2} \sqrt {\sec (c+d x)}}{\sqrt {1-\sec (c+d x)}}\right ) \tan (c+d x)-9 \sqrt {2} A \arctan \left (\frac {\sqrt {2} \sqrt {\sec (c+d x)}}{\sqrt {1-\sec (c+d x)}}\right ) \sec (c+d x) \tan (c+d x)+13 \sqrt {2} B \arctan \left (\frac {\sqrt {2} \sqrt {\sec (c+d x)}}{\sqrt {1-\sec (c+d x)}}\right ) \sec (c+d x) \tan (c+d x)}{4 d \sqrt {1-\sec (c+d x)} (a (1+\sec (c+d x)))^{3/2}} \] Input:

Integrate[(Sec[c + d*x]^(7/2)*(A + B*Sec[c + d*x]))/(a + a*Sec[c + d*x])^( 
3/2),x]
 

Output:

(4*(6*A - 7*B)*ArcSin[Sqrt[1 - Sec[c + d*x]]]*Cos[(c + d*x)/2]^3*Sec[c + d 
*x]^2*Sin[(c + d*x)/2] + 8*(9*A - 13*B)*ArcSin[Sqrt[Sec[c + d*x]]]*Cos[(c 
+ d*x)/2]^3*Sec[c + d*x]^2*Sin[(c + d*x)/2] + 6*A*Sqrt[1 - Sec[c + d*x]]*S 
ec[c + d*x]^(3/2)*Sin[c + d*x] - 7*B*Sqrt[1 - Sec[c + d*x]]*Sec[c + d*x]^( 
3/2)*Sin[c + d*x] + 4*A*Sqrt[1 - Sec[c + d*x]]*Sec[c + d*x]^(5/2)*Sin[c + 
d*x] - 3*B*Sqrt[1 - Sec[c + d*x]]*Sec[c + d*x]^(5/2)*Sin[c + d*x] + 2*B*Sq 
rt[1 - Sec[c + d*x]]*Sec[c + d*x]^(7/2)*Sin[c + d*x] - 9*Sqrt[2]*A*ArcTan[ 
(Sqrt[2]*Sqrt[Sec[c + d*x]])/Sqrt[1 - Sec[c + d*x]]]*Tan[c + d*x] + 13*Sqr 
t[2]*B*ArcTan[(Sqrt[2]*Sqrt[Sec[c + d*x]])/Sqrt[1 - Sec[c + d*x]]]*Tan[c + 
 d*x] - 9*Sqrt[2]*A*ArcTan[(Sqrt[2]*Sqrt[Sec[c + d*x]])/Sqrt[1 - Sec[c + d 
*x]]]*Sec[c + d*x]*Tan[c + d*x] + 13*Sqrt[2]*B*ArcTan[(Sqrt[2]*Sqrt[Sec[c 
+ d*x]])/Sqrt[1 - Sec[c + d*x]]]*Sec[c + d*x]*Tan[c + d*x])/(4*d*Sqrt[1 - 
Sec[c + d*x]]*(a*(1 + Sec[c + d*x]))^(3/2))
 

Rubi [A] (verified)

Time = 1.64 (sec) , antiderivative size = 259, normalized size of antiderivative = 1.05, number of steps used = 17, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.457, Rules used = {3042, 4507, 27, 3042, 4509, 27, 3042, 4509, 27, 3042, 4511, 3042, 4288, 222, 4295, 219}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {\sec ^{\frac {7}{2}}(c+d x) (A+B \sec (c+d x))}{(a \sec (c+d x)+a)^{3/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{7/2} \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^{3/2}}dx\)

\(\Big \downarrow \) 4507

\(\displaystyle \frac {\int \frac {\sec ^{\frac {5}{2}}(c+d x) (5 a (A-B)-4 a (A-2 B) \sec (c+d x))}{2 \sqrt {\sec (c+d x) a+a}}dx}{2 a^2}+\frac {(A-B) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {\sec ^{\frac {5}{2}}(c+d x) (5 a (A-B)-4 a (A-2 B) \sec (c+d x))}{\sqrt {\sec (c+d x) a+a}}dx}{4 a^2}+\frac {(A-B) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{5/2} \left (5 a (A-B)-4 a (A-2 B) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{4 a^2}+\frac {(A-B) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 4509

\(\displaystyle \frac {\frac {\int -\frac {2 \sec ^{\frac {3}{2}}(c+d x) \left (3 a^2 (A-2 B)-a^2 (6 A-7 B) \sec (c+d x)\right )}{\sqrt {\sec (c+d x) a+a}}dx}{2 a}-\frac {2 a (A-2 B) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a^2}+\frac {(A-B) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {-\frac {\int \frac {\sec ^{\frac {3}{2}}(c+d x) \left (3 a^2 (A-2 B)-a^2 (6 A-7 B) \sec (c+d x)\right )}{\sqrt {\sec (c+d x) a+a}}dx}{a}-\frac {2 a (A-2 B) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a^2}+\frac {(A-B) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {-\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (3 a^2 (A-2 B)-a^2 (6 A-7 B) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{a}-\frac {2 a (A-2 B) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a^2}+\frac {(A-B) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 4509

\(\displaystyle \frac {-\frac {\frac {\int -\frac {\sqrt {\sec (c+d x)} \left (a^3 (6 A-7 B)-a^3 (12 A-19 B) \sec (c+d x)\right )}{2 \sqrt {\sec (c+d x) a+a}}dx}{a}-\frac {a^2 (6 A-7 B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{a}-\frac {2 a (A-2 B) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a^2}+\frac {(A-B) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {-\frac {-\frac {\int \frac {\sqrt {\sec (c+d x)} \left (a^3 (6 A-7 B)-a^3 (12 A-19 B) \sec (c+d x)\right )}{\sqrt {\sec (c+d x) a+a}}dx}{2 a}-\frac {a^2 (6 A-7 B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{a}-\frac {2 a (A-2 B) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a^2}+\frac {(A-B) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {-\frac {-\frac {\int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \left (a^3 (6 A-7 B)-a^3 (12 A-19 B) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{2 a}-\frac {a^2 (6 A-7 B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{a}-\frac {2 a (A-2 B) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a^2}+\frac {(A-B) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 4511

\(\displaystyle \frac {-\frac {-\frac {2 a^3 (9 A-13 B) \int \frac {\sqrt {\sec (c+d x)}}{\sqrt {\sec (c+d x) a+a}}dx-a^2 (12 A-19 B) \int \sqrt {\sec (c+d x)} \sqrt {\sec (c+d x) a+a}dx}{2 a}-\frac {a^2 (6 A-7 B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{a}-\frac {2 a (A-2 B) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a^2}+\frac {(A-B) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {-\frac {-\frac {2 a^3 (9 A-13 B) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx-a^2 (12 A-19 B) \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{2 a}-\frac {a^2 (6 A-7 B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{a}-\frac {2 a (A-2 B) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a^2}+\frac {(A-B) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 4288

\(\displaystyle \frac {-\frac {-\frac {2 a^3 (9 A-13 B) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx+\frac {2 a^2 (12 A-19 B) \int \frac {1}{\sqrt {\frac {a \tan ^2(c+d x)}{\sec (c+d x) a+a}+1}}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}}{2 a}-\frac {a^2 (6 A-7 B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{a}-\frac {2 a (A-2 B) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a^2}+\frac {(A-B) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 222

\(\displaystyle \frac {-\frac {-\frac {2 a^3 (9 A-13 B) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx-\frac {2 a^{5/2} (12 A-19 B) \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}}{2 a}-\frac {a^2 (6 A-7 B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{a}-\frac {2 a (A-2 B) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a^2}+\frac {(A-B) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 4295

\(\displaystyle \frac {-\frac {-\frac {-\frac {4 a^3 (9 A-13 B) \int \frac {1}{2 a-\frac {a^2 \sin (c+d x) \tan (c+d x)}{\sec (c+d x) a+a}}d\left (-\frac {a \sqrt {\sec (c+d x)} \sin (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}-\frac {2 a^{5/2} (12 A-19 B) \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}}{2 a}-\frac {a^2 (6 A-7 B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{a}-\frac {2 a (A-2 B) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a^2}+\frac {(A-B) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {-\frac {-\frac {\frac {2 \sqrt {2} a^{5/2} (9 A-13 B) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x) \sqrt {\sec (c+d x)}}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{d}-\frac {2 a^{5/2} (12 A-19 B) \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}}{2 a}-\frac {a^2 (6 A-7 B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{a}-\frac {2 a (A-2 B) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a^2}+\frac {(A-B) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\)

Input:

Int[(Sec[c + d*x]^(7/2)*(A + B*Sec[c + d*x]))/(a + a*Sec[c + d*x])^(3/2),x 
]
 

Output:

((A - B)*Sec[c + d*x]^(7/2)*Sin[c + d*x])/(2*d*(a + a*Sec[c + d*x])^(3/2)) 
 + ((-2*a*(A - 2*B)*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(d*Sqrt[a + a*Sec[c + 
 d*x]]) - (-1/2*((-2*a^(5/2)*(12*A - 19*B)*ArcSinh[(Sqrt[a]*Tan[c + d*x])/ 
Sqrt[a + a*Sec[c + d*x]]])/d + (2*Sqrt[2]*a^(5/2)*(9*A - 13*B)*ArcTanh[(Sq 
rt[a]*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])] 
)/d)/a - (a^2*(6*A - 7*B)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(d*Sqrt[a + a*S 
ec[c + d*x]]))/a)/(4*a^2)
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 222

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[Rt[b, 2]*(x/Sqrt 
[a])]/Rt[b, 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4288

Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) 
+ (a_)], x_Symbol] :> Simp[-2*(a/(b*f))*Sqrt[a*(d/b)]   Subst[Int[1/Sqrt[1 
+ x^2/a], x], x, b*(Cot[e + f*x]/Sqrt[a + b*Csc[e + f*x]])], x] /; FreeQ[{a 
, b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && GtQ[a*(d/b), 0]
 

rule 4295

Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) 
+ (a_)], x_Symbol] :> Simp[-2*b*(d/(a*f))   Subst[Int[1/(2*b - d*x^2), x], 
x, b*(Cot[e + f*x]/(Sqrt[a + b*Csc[e + f*x]]*Sqrt[d*Csc[e + f*x]]))], x] /; 
 FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0]
 

rule 4507

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( 
a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[d*(A*b 
- a*B)*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^(n - 1)/(a*f*( 
2*m + 1))), x] - Simp[1/(a*b*(2*m + 1))   Int[(a + b*Csc[e + f*x])^(m + 1)* 
(d*Csc[e + f*x])^(n - 1)*Simp[A*(a*d*(n - 1)) - B*(b*d*(n - 1)) - d*(a*B*(m 
 - n + 1) + A*b*(m + n))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, 
A, B}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)] && G 
tQ[n, 0]
 

rule 4509

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( 
a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-B)*d* 
Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^(n - 1)/(f*(m + n))), 
 x] + Simp[d/(b*(m + n))   Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n - 
 1)*Simp[b*B*(n - 1) + (A*b*(m + n) + a*B*m)*Csc[e + f*x], x], x], x] /; Fr 
eeQ[{a, b, d, e, f, A, B, m}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] 
&& GtQ[n, 1]
 

rule 4511

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( 
a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(A*b - 
a*B)/b   Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^n, x], x] + Simp[B/b 
 Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^n, x], x] /; FreeQ[{a, b 
, d, e, f, A, B, m}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0]
 

Maple [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(505\) vs. \(2(208)=416\).

Time = 4.55 (sec) , antiderivative size = 506, normalized size of antiderivative = 2.05

method	result	size
default	\(\frac {\sec \left (d x +c \right )^{\frac {7}{2}} \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (\sqrt {2}\, A \arctan \left (\frac {\sqrt {2}\, \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )}{2 \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right ) \left (18 \cos \left (d x +c \right )^{5}+18 \cos \left (d x +c \right )^{4}\right )+\sqrt {2}\, B \arctan \left (\frac {\sqrt {2}\, \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )}{2 \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right ) \left (-26 \cos \left (d x +c \right )^{5}-26 \cos \left (d x +c \right )^{4}\right )+A \arctan \left (\frac {\cot \left (d x +c \right )-\csc \left (d x +c \right )+1}{2 \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right ) \left (-12 \cos \left (d x +c \right )^{5}-12 \cos \left (d x +c \right )^{4}\right )+B \arctan \left (\frac {\cot \left (d x +c \right )-\csc \left (d x +c \right )+1}{2 \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right ) \left (19 \cos \left (d x +c \right )^{5}+19 \cos \left (d x +c \right )^{4}\right )+A \arctan \left (\frac {\cot \left (d x +c \right )-\csc \left (d x +c \right )-1}{2 \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right ) \left (-12 \cos \left (d x +c \right )^{5}-12 \cos \left (d x +c \right )^{4}\right )+B \arctan \left (\frac {\cot \left (d x +c \right )-\csc \left (d x +c \right )-1}{2 \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right ) \left (19 \cos \left (d x +c \right )^{5}+19 \cos \left (d x +c \right )^{4}\right )+\sin \left (d x +c \right ) \cos \left (d x +c \right )^{3} \left (6 \cos \left (d x +c \right )+4\right ) \sqrt {2}\, A \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}+\sin \left (d x +c \right ) \left (-7 \cos \left (d x +c \right )^{2}-3 \cos \left (d x +c \right )+2\right ) \sqrt {2}\, B \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \cos \left (d x +c \right )^{2}\right )}{8 d \,a^{2} \left (\cos \left (d x +c \right )^{2}+2 \cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\)	\(506\)
parts	\(\frac {A \cos \left (d x +c \right )^{3} \left (\sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2 \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right ) \left (-9 \cos \left (d x +c \right )^{2}-9 \cos \left (d x +c \right )\right )+\arctan \left (\frac {\cot \left (d x +c \right )-\csc \left (d x +c \right )-1}{2 \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right ) \left (-6 \cos \left (d x +c \right )^{2}-6 \cos \left (d x +c \right )\right )+\arctan \left (\frac {\cot \left (d x +c \right )-\csc \left (d x +c \right )+1}{2 \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right ) \left (-6 \cos \left (d x +c \right )^{2}-6 \cos \left (d x +c \right )\right )+\sin \left (d x +c \right ) \left (3 \cos \left (d x +c \right )+2\right ) \sqrt {2}\, \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \sec \left (d x +c \right )^{\frac {7}{2}}}{4 d \,a^{2} \left (\cos \left (d x +c \right )^{2}+2 \cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}-\frac {B \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \sec \left (d x +c \right )^{\frac {9}{2}} \left (\sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2 \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right ) \left (-26 \cos \left (d x +c \right )^{6}-26 \cos \left (d x +c \right )^{5}\right )+\arctan \left (\frac {\cot \left (d x +c \right )-\csc \left (d x +c \right )-1}{2 \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right ) \left (-19 \cos \left (d x +c \right )^{6}-19 \cos \left (d x +c \right )^{5}\right )+\arctan \left (\frac {\cot \left (d x +c \right )-\csc \left (d x +c \right )+1}{2 \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right ) \left (-19 \cos \left (d x +c \right )^{6}-19 \cos \left (d x +c \right )^{5}\right )+\sin \left (d x +c \right ) \left (7 \cos \left (d x +c \right )^{2}+3 \cos \left (d x +c \right )-2\right ) \sqrt {2}\, \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \cos \left (d x +c \right )^{3}\right )}{8 d \,a^{2} \left (\cos \left (d x +c \right )^{2}+2 \cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\)	\(558\)

Input:

int(sec(d*x+c)^(7/2)*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^(3/2),x,method=_RET 
URNVERBOSE)
 

Output:

1/8/d/a^2*sec(d*x+c)^(7/2)*(a*(1+sec(d*x+c)))^(1/2)/(cos(d*x+c)^2+2*cos(d* 
x+c)+1)/(-1/(1+cos(d*x+c)))^(1/2)*(2^(1/2)*A*arctan(1/2*2^(1/2)*(-csc(d*x+ 
c)+cot(d*x+c))/(-1/(1+cos(d*x+c)))^(1/2))*(18*cos(d*x+c)^5+18*cos(d*x+c)^4 
)+2^(1/2)*B*arctan(1/2*2^(1/2)*(-csc(d*x+c)+cot(d*x+c))/(-1/(1+cos(d*x+c)) 
)^(1/2))*(-26*cos(d*x+c)^5-26*cos(d*x+c)^4)+A*arctan(1/2*(cot(d*x+c)-csc(d 
*x+c)+1)/(-1/(1+cos(d*x+c)))^(1/2))*(-12*cos(d*x+c)^5-12*cos(d*x+c)^4)+B*a 
rctan(1/2*(cot(d*x+c)-csc(d*x+c)+1)/(-1/(1+cos(d*x+c)))^(1/2))*(19*cos(d*x 
+c)^5+19*cos(d*x+c)^4)+A*arctan(1/2*(cot(d*x+c)-csc(d*x+c)-1)/(-1/(1+cos(d 
*x+c)))^(1/2))*(-12*cos(d*x+c)^5-12*cos(d*x+c)^4)+B*arctan(1/2*(cot(d*x+c) 
-csc(d*x+c)-1)/(-1/(1+cos(d*x+c)))^(1/2))*(19*cos(d*x+c)^5+19*cos(d*x+c)^4 
)+sin(d*x+c)*cos(d*x+c)^3*(6*cos(d*x+c)+4)*2^(1/2)*A*(-2/(1+cos(d*x+c)))^( 
1/2)+sin(d*x+c)*(-7*cos(d*x+c)^2-3*cos(d*x+c)+2)*2^(1/2)*B*(-2/(1+cos(d*x+ 
c)))^(1/2)*cos(d*x+c)^2)
 

Fricas [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 758, normalized size of antiderivative = 3.07 \[ \int \frac {\sec ^{\frac {7}{2}}(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^{3/2}} \, dx =\text {Too large to display} \] Input:

integrate(sec(d*x+c)^(7/2)*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^(3/2),x, algo 
rithm="fricas")
 

Output:

[-1/16*(2*sqrt(2)*((9*A - 13*B)*cos(d*x + c)^3 + 2*(9*A - 13*B)*cos(d*x + 
c)^2 + (9*A - 13*B)*cos(d*x + c))*sqrt(a)*log(-(a*cos(d*x + c)^2 + 2*sqrt( 
2)*sqrt(a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin( 
d*x + c) - 2*a*cos(d*x + c) - 3*a)/(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)) 
+ ((12*A - 19*B)*cos(d*x + c)^3 + 2*(12*A - 19*B)*cos(d*x + c)^2 + (12*A - 
 19*B)*cos(d*x + c))*sqrt(a)*log((a*cos(d*x + c)^3 - 7*a*cos(d*x + c)^2 - 
4*(cos(d*x + c)^2 - 2*cos(d*x + c))*sqrt(a)*sqrt((a*cos(d*x + c) + a)/cos( 
d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)) + 8*a)/(cos(d*x + c)^3 + cos(d*x 
 + c)^2)) - 4*((6*A - 7*B)*cos(d*x + c)^2 + (4*A - 3*B)*cos(d*x + c) + 2*B 
)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c))) 
/(a^2*d*cos(d*x + c)^3 + 2*a^2*d*cos(d*x + c)^2 + a^2*d*cos(d*x + c)), -1/ 
8*(2*sqrt(2)*((9*A - 13*B)*cos(d*x + c)^3 + 2*(9*A - 13*B)*cos(d*x + c)^2 
+ (9*A - 13*B)*cos(d*x + c))*sqrt(-a)*arctan(sqrt(2)*sqrt(-a)*sqrt((a*cos( 
d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))/(a*sin(d*x + c))) + ((12*A 
- 19*B)*cos(d*x + c)^3 + 2*(12*A - 19*B)*cos(d*x + c)^2 + (12*A - 19*B)*co 
s(d*x + c))*sqrt(-a)*arctan(1/2*(cos(d*x + c)^2 - 2*cos(d*x + c))*sqrt(-a) 
*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))/(a*sqrt(cos(d*x + c))*sin(d*x + c 
))) - 2*((6*A - 7*B)*cos(d*x + c)^2 + (4*A - 3*B)*cos(d*x + c) + 2*B)*sqrt 
((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)))/(a^2* 
d*cos(d*x + c)^3 + 2*a^2*d*cos(d*x + c)^2 + a^2*d*cos(d*x + c))]
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\sec ^{\frac {7}{2}}(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^{3/2}} \, dx=\text {Timed out} \] Input:

integrate(sec(d*x+c)**(7/2)*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))**(3/2),x)
 

Output:

Timed out
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 13364 vs. \(2 (208) = 416\).

Time = 1.15 (sec) , antiderivative size = 13364, normalized size of antiderivative = 54.11 \[ \int \frac {\sec ^{\frac {7}{2}}(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^{3/2}} \, dx=\text {Too large to display} \] Input:

integrate(sec(d*x+c)^(7/2)*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^(3/2),x, algo 
rithm="maxima")
 

Output:

-1/16*(4*(12*(sin(4*d*x + 4*c) + 2*sin(2*d*x + 2*c) + 2*sin(3/2*arctan2(si 
n(2*d*x + 2*c), cos(2*d*x + 2*c))) + 2*sin(1/2*arctan2(sin(2*d*x + 2*c), c 
os(2*d*x + 2*c))))*cos(7/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - 
8*(sin(5/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - sin(3/4*arctan2( 
sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - 3*sin(1/4*arctan2(sin(2*d*x + 2*c), 
 cos(2*d*x + 2*c))))*cos(3/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) 
+ 4*(sin(4*d*x + 4*c) + 2*sin(2*d*x + 2*c) + 2*sin(1/2*arctan2(sin(2*d*x + 
 2*c), cos(2*d*x + 2*c))))*cos(5/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2 
*c))) - 4*(sin(4*d*x + 4*c) + 2*sin(2*d*x + 2*c) + 2*sin(1/2*arctan2(sin(2 
*d*x + 2*c), cos(2*d*x + 2*c))))*cos(3/4*arctan2(sin(2*d*x + 2*c), cos(2*d 
*x + 2*c))) - 12*(sin(4*d*x + 4*c) + 2*sin(2*d*x + 2*c))*cos(1/4*arctan2(s 
in(2*d*x + 2*c), cos(2*d*x + 2*c))) + 3*(sqrt(2)*cos(4*d*x + 4*c)^2 + 4*sq 
rt(2)*cos(2*d*x + 2*c)^2 + 4*sqrt(2)*cos(3/2*arctan2(sin(2*d*x + 2*c), cos 
(2*d*x + 2*c)))^2 + 4*sqrt(2)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x 
+ 2*c)))^2 + sqrt(2)*sin(4*d*x + 4*c)^2 + 4*sqrt(2)*sin(4*d*x + 4*c)*sin(2 
*d*x + 2*c) + 4*sqrt(2)*sin(2*d*x + 2*c)^2 + 4*sqrt(2)*sin(3/2*arctan2(sin 
(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 + 4*sqrt(2)*sin(1/2*arctan2(sin(2*d*x 
+ 2*c), cos(2*d*x + 2*c)))^2 + 2*(2*sqrt(2)*cos(2*d*x + 2*c) + sqrt(2))*co 
s(4*d*x + 4*c) + 4*(sqrt(2)*cos(4*d*x + 4*c) + 2*sqrt(2)*cos(2*d*x + 2*c) 
+ 2*sqrt(2)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + sqrt...
 

Giac [F(-2)]

Exception generated. \[ \int \frac {\sec ^{\frac {7}{2}}(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^{3/2}} \, dx=\text {Exception raised: RuntimeError} \] Input:

integrate(sec(d*x+c)^(7/2)*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^(3/2),x, algo 
rithm="giac")
 

Output:

Exception raised: RuntimeError >> an error occurred running a Giac command 
:INPUT:sage2OUTPUT:Unable to divide, perhaps due to rounding error%%%{%%{[ 
%%%{%%{[23574053482485268906770432,0]:[1,0,-2]%%},[16]%%%},0]:[1,0,%%%{-1, 
[1]%%%}]%%},
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\sec ^{\frac {7}{2}}(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^{3/2}} \, dx=\int \frac {\left (A+\frac {B}{\cos \left (c+d\,x\right )}\right )\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{7/2}}{{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{3/2}} \,d x \] Input:

int(((A + B/cos(c + d*x))*(1/cos(c + d*x))^(7/2))/(a + a/cos(c + d*x))^(3/ 
2),x)
 

Output:

int(((A + B/cos(c + d*x))*(1/cos(c + d*x))^(7/2))/(a + a/cos(c + d*x))^(3/ 
2), x)
 

Reduce [F]

\[ \int \frac {\sec ^{\frac {7}{2}}(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^{3/2}} \, dx=\frac {\sqrt {a}\, \left (\left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sqrt {\sec \left (d x +c \right )+1}\, \sec \left (d x +c \right )^{4}}{\sec \left (d x +c \right )^{2}+2 \sec \left (d x +c \right )+1}d x \right ) b +\left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sqrt {\sec \left (d x +c \right )+1}\, \sec \left (d x +c \right )^{3}}{\sec \left (d x +c \right )^{2}+2 \sec \left (d x +c \right )+1}d x \right ) a \right )}{a^{2}} \] Input:

int(sec(d*x+c)^(7/2)*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^(3/2),x)
 

Output:

(sqrt(a)*(int((sqrt(sec(c + d*x))*sqrt(sec(c + d*x) + 1)*sec(c + d*x)**4)/ 
(sec(c + d*x)**2 + 2*sec(c + d*x) + 1),x)*b + int((sqrt(sec(c + d*x))*sqrt 
(sec(c + d*x) + 1)*sec(c + d*x)**3)/(sec(c + d*x)**2 + 2*sec(c + d*x) + 1) 
,x)*a))/a**2