\(\int \frac {\sec ^4(c+d x) (B \sec (c+d x)+C \sec ^2(c+d x))}{(a+a \sec (c+d x))^{5/2}} \, dx\) [401]

Optimal result

Integrand size = 42, antiderivative size = 261 \[ \int \frac {\sec ^4(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx=\frac {(163 B-283 C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{16 \sqrt {2} a^{5/2} d}+\frac {(B-C) \sec ^4(c+d x) \tan (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}+\frac {(13 B-21 C) \sec ^3(c+d x) \tan (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}-\frac {(985 B-1729 C) \tan (c+d x)}{120 a^2 d \sqrt {a+a \sec (c+d x)}}-\frac {(85 B-157 C) \sec ^2(c+d x) \tan (c+d x)}{80 a^2 d \sqrt {a+a \sec (c+d x)}}+\frac {(475 B-787 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{240 a^3 d} \] Output:

1/32*(163*B-283*C)*arctan(1/2*a^(1/2)*tan(d*x+c)*2^(1/2)/(a+a*sec(d*x+c))^ 
(1/2))*2^(1/2)/a^(5/2)/d+1/4*(B-C)*sec(d*x+c)^4*tan(d*x+c)/d/(a+a*sec(d*x+ 
c))^(5/2)+1/16*(13*B-21*C)*sec(d*x+c)^3*tan(d*x+c)/a/d/(a+a*sec(d*x+c))^(3 
/2)-1/120*(985*B-1729*C)*tan(d*x+c)/a^2/d/(a+a*sec(d*x+c))^(1/2)-1/80*(85* 
B-157*C)*sec(d*x+c)^2*tan(d*x+c)/a^2/d/(a+a*sec(d*x+c))^(1/2)+1/240*(475*B 
-787*C)*(a+a*sec(d*x+c))^(1/2)*tan(d*x+c)/a^3/d
 

Mathematica [A] (warning: unable to verify)

Time = 14.95 (sec) , antiderivative size = 316, normalized size of antiderivative = 1.21 \[ \int \frac {\sec ^4(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx=\frac {(163 B-283 C) \arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ) \cos ^4\left (\frac {1}{2} (c+d x)\right ) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sec ^{\frac {5}{2}}(c+d x) \sqrt {1+\sec (c+d x)}}{4 d \sqrt {\sec ^2\left (\frac {1}{2} (c+d x)\right )} (a (1+\sec (c+d x)))^{5/2}}+\frac {\cos ^5\left (\frac {1}{2} (c+d x)\right ) \sec ^3(c+d x) \left (\frac {1}{30} (-1495 B+2671 C) \sin \left (\frac {1}{2} (c+d x)\right )+\frac {16}{5} C \sec ^2(c+d x) \sin \left (\frac {1}{2} (c+d x)\right )+\frac {1}{4} \sec ^2\left (\frac {1}{2} (c+d x)\right ) \left (21 B \sin \left (\frac {1}{2} (c+d x)\right )-29 C \sin \left (\frac {1}{2} (c+d x)\right )\right )+\frac {1}{2} \sec ^4\left (\frac {1}{2} (c+d x)\right ) \left (B \sin \left (\frac {1}{2} (c+d x)\right )-C \sin \left (\frac {1}{2} (c+d x)\right )\right )-\frac {16}{15} \sec (c+d x) \left (-5 B \sin \left (\frac {1}{2} (c+d x)\right )+11 C \sin \left (\frac {1}{2} (c+d x)\right )\right )\right )}{d (a (1+\sec (c+d x)))^{5/2}} \] Input:

Integrate[(Sec[c + d*x]^4*(B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + a*Sec[ 
c + d*x])^(5/2),x]
 

Output:

((163*B - 283*C)*ArcSin[Tan[(c + d*x)/2]]*Cos[(c + d*x)/2]^4*Sqrt[Cos[c + 
d*x]/(1 + Cos[c + d*x])]*Sec[c + d*x]^(5/2)*Sqrt[1 + Sec[c + d*x]])/(4*d*S 
qrt[Sec[(c + d*x)/2]^2]*(a*(1 + Sec[c + d*x]))^(5/2)) + (Cos[(c + d*x)/2]^ 
5*Sec[c + d*x]^3*(((-1495*B + 2671*C)*Sin[(c + d*x)/2])/30 + (16*C*Sec[c + 
 d*x]^2*Sin[(c + d*x)/2])/5 + (Sec[(c + d*x)/2]^2*(21*B*Sin[(c + d*x)/2] - 
 29*C*Sin[(c + d*x)/2]))/4 + (Sec[(c + d*x)/2]^4*(B*Sin[(c + d*x)/2] - C*S 
in[(c + d*x)/2]))/2 - (16*Sec[c + d*x]*(-5*B*Sin[(c + d*x)/2] + 11*C*Sin[( 
c + d*x)/2]))/15))/(d*(a*(1 + Sec[c + d*x]))^(5/2))
 

Rubi [A] (verified)

Time = 1.92 (sec) , antiderivative size = 287, normalized size of antiderivative = 1.10, number of steps used = 20, number of rules used = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.452, Rules used = {3042, 4560, 3042, 4507, 27, 3042, 4507, 27, 3042, 4509, 27, 3042, 4498, 27, 3042, 4489, 3042, 4282, 216}

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.

\(\displaystyle \int \frac {\sec ^4(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a \sec (c+d x)+a)^{5/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^4 \left (B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^{5/2}}dx\)

\(\Big \downarrow \) 4560

\(\displaystyle \int \frac {\sec ^5(c+d x) (B+C \sec (c+d x))}{(a \sec (c+d x)+a)^{5/2}}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^5 \left (B+C \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^{5/2}}dx\)

\(\Big \downarrow \) 4507

\(\displaystyle \frac {\int \frac {\sec ^4(c+d x) (8 a (B-C)-a (5 B-13 C) \sec (c+d x))}{2 (\sec (c+d x) a+a)^{3/2}}dx}{4 a^2}+\frac {(B-C) \tan (c+d x) \sec ^4(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {\sec ^4(c+d x) (8 a (B-C)-a (5 B-13 C) \sec (c+d x))}{(\sec (c+d x) a+a)^{3/2}}dx}{8 a^2}+\frac {(B-C) \tan (c+d x) \sec ^4(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^4 \left (8 a (B-C)-a (5 B-13 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2}}dx}{8 a^2}+\frac {(B-C) \tan (c+d x) \sec ^4(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 4507

\(\displaystyle \frac {\frac {\int \frac {\sec ^3(c+d x) \left (6 a^2 (13 B-21 C)-a^2 (85 B-157 C) \sec (c+d x)\right )}{2 \sqrt {\sec (c+d x) a+a}}dx}{2 a^2}+\frac {a (13 B-21 C) \tan (c+d x) \sec ^3(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}+\frac {(B-C) \tan (c+d x) \sec ^4(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\int \frac {\sec ^3(c+d x) \left (6 a^2 (13 B-21 C)-a^2 (85 B-157 C) \sec (c+d x)\right )}{\sqrt {\sec (c+d x) a+a}}dx}{4 a^2}+\frac {a (13 B-21 C) \tan (c+d x) \sec ^3(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}+\frac {(B-C) \tan (c+d x) \sec ^4(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^3 \left (6 a^2 (13 B-21 C)-a^2 (85 B-157 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{4 a^2}+\frac {a (13 B-21 C) \tan (c+d x) \sec ^3(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}+\frac {(B-C) \tan (c+d x) \sec ^4(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 4509

\(\displaystyle \frac {\frac {\frac {2 \int -\frac {\sec ^2(c+d x) \left (4 a^3 (85 B-157 C)-a^3 (475 B-787 C) \sec (c+d x)\right )}{2 \sqrt {\sec (c+d x) a+a}}dx}{5 a}-\frac {2 a^2 (85 B-157 C) \tan (c+d x) \sec ^2(c+d x)}{5 d \sqrt {a \sec (c+d x)+a}}}{4 a^2}+\frac {a (13 B-21 C) \tan (c+d x) \sec ^3(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}+\frac {(B-C) \tan (c+d x) \sec ^4(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {-\frac {\int \frac {\sec ^2(c+d x) \left (4 a^3 (85 B-157 C)-a^3 (475 B-787 C) \sec (c+d x)\right )}{\sqrt {\sec (c+d x) a+a}}dx}{5 a}-\frac {2 a^2 (85 B-157 C) \tan (c+d x) \sec ^2(c+d x)}{5 d \sqrt {a \sec (c+d x)+a}}}{4 a^2}+\frac {a (13 B-21 C) \tan (c+d x) \sec ^3(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}+\frac {(B-C) \tan (c+d x) \sec ^4(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {-\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^2 \left (4 a^3 (85 B-157 C)-a^3 (475 B-787 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{5 a}-\frac {2 a^2 (85 B-157 C) \tan (c+d x) \sec ^2(c+d x)}{5 d \sqrt {a \sec (c+d x)+a}}}{4 a^2}+\frac {a (13 B-21 C) \tan (c+d x) \sec ^3(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}+\frac {(B-C) \tan (c+d x) \sec ^4(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 4498

\(\displaystyle \frac {\frac {-\frac {\frac {2 \int -\frac {\sec (c+d x) \left (a^4 (475 B-787 C)-2 a^4 (985 B-1729 C) \sec (c+d x)\right )}{2 \sqrt {\sec (c+d x) a+a}}dx}{3 a}-\frac {2 a^2 (475 B-787 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{5 a}-\frac {2 a^2 (85 B-157 C) \tan (c+d x) \sec ^2(c+d x)}{5 d \sqrt {a \sec (c+d x)+a}}}{4 a^2}+\frac {a (13 B-21 C) \tan (c+d x) \sec ^3(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}+\frac {(B-C) \tan (c+d x) \sec ^4(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {-\frac {-\frac {\int \frac {\sec (c+d x) \left (a^4 (475 B-787 C)-2 a^4 (985 B-1729 C) \sec (c+d x)\right )}{\sqrt {\sec (c+d x) a+a}}dx}{3 a}-\frac {2 a^2 (475 B-787 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{5 a}-\frac {2 a^2 (85 B-157 C) \tan (c+d x) \sec ^2(c+d x)}{5 d \sqrt {a \sec (c+d x)+a}}}{4 a^2}+\frac {a (13 B-21 C) \tan (c+d x) \sec ^3(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}+\frac {(B-C) \tan (c+d x) \sec ^4(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {-\frac {-\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (a^4 (475 B-787 C)-2 a^4 (985 B-1729 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{3 a}-\frac {2 a^2 (475 B-787 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{5 a}-\frac {2 a^2 (85 B-157 C) \tan (c+d x) \sec ^2(c+d x)}{5 d \sqrt {a \sec (c+d x)+a}}}{4 a^2}+\frac {a (13 B-21 C) \tan (c+d x) \sec ^3(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}+\frac {(B-C) \tan (c+d x) \sec ^4(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 4489

\(\displaystyle \frac {\frac {-\frac {-\frac {15 a^4 (163 B-283 C) \int \frac {\sec (c+d x)}{\sqrt {\sec (c+d x) a+a}}dx-\frac {4 a^4 (985 B-1729 C) \tan (c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{3 a}-\frac {2 a^2 (475 B-787 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{5 a}-\frac {2 a^2 (85 B-157 C) \tan (c+d x) \sec ^2(c+d x)}{5 d \sqrt {a \sec (c+d x)+a}}}{4 a^2}+\frac {a (13 B-21 C) \tan (c+d x) \sec ^3(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}+\frac {(B-C) \tan (c+d x) \sec ^4(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {-\frac {-\frac {15 a^4 (163 B-283 C) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx-\frac {4 a^4 (985 B-1729 C) \tan (c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{3 a}-\frac {2 a^2 (475 B-787 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{5 a}-\frac {2 a^2 (85 B-157 C) \tan (c+d x) \sec ^2(c+d x)}{5 d \sqrt {a \sec (c+d x)+a}}}{4 a^2}+\frac {a (13 B-21 C) \tan (c+d x) \sec ^3(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}+\frac {(B-C) \tan (c+d x) \sec ^4(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 4282

\(\displaystyle \frac {\frac {-\frac {-\frac {-\frac {30 a^4 (163 B-283 C) \int \frac {1}{\frac {a^2 \tan ^2(c+d x)}{\sec (c+d x) a+a}+2 a}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}-\frac {4 a^4 (985 B-1729 C) \tan (c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{3 a}-\frac {2 a^2 (475 B-787 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{5 a}-\frac {2 a^2 (85 B-157 C) \tan (c+d x) \sec ^2(c+d x)}{5 d \sqrt {a \sec (c+d x)+a}}}{4 a^2}+\frac {a (13 B-21 C) \tan (c+d x) \sec ^3(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}+\frac {(B-C) \tan (c+d x) \sec ^4(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {\frac {-\frac {2 a^2 (85 B-157 C) \tan (c+d x) \sec ^2(c+d x)}{5 d \sqrt {a \sec (c+d x)+a}}-\frac {-\frac {2 a^2 (475 B-787 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}-\frac {\frac {15 \sqrt {2} a^{7/2} (163 B-283 C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{d}-\frac {4 a^4 (985 B-1729 C) \tan (c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{3 a}}{5 a}}{4 a^2}+\frac {a (13 B-21 C) \tan (c+d x) \sec ^3(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}+\frac {(B-C) \tan (c+d x) \sec ^4(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

Input:

Int[(Sec[c + d*x]^4*(B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + a*Sec[c + d* 
x])^(5/2),x]
 

Output:

((B - C)*Sec[c + d*x]^4*Tan[c + d*x])/(4*d*(a + a*Sec[c + d*x])^(5/2)) + ( 
(a*(13*B - 21*C)*Sec[c + d*x]^3*Tan[c + d*x])/(2*d*(a + a*Sec[c + d*x])^(3 
/2)) + ((-2*a^2*(85*B - 157*C)*Sec[c + d*x]^2*Tan[c + d*x])/(5*d*Sqrt[a + 
a*Sec[c + d*x]]) - ((-2*a^2*(475*B - 787*C)*Sqrt[a + a*Sec[c + d*x]]*Tan[c 
 + d*x])/(3*d) - ((15*Sqrt[2]*a^(7/2)*(163*B - 283*C)*ArcTan[(Sqrt[a]*Tan[ 
c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/d - (4*a^4*(985*B - 1729*C) 
*Tan[c + d*x])/(d*Sqrt[a + a*Sec[c + d*x]]))/(3*a))/(5*a))/(4*a^2))/(8*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 216

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

rule 3042

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

rule 4282

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

rule 4489

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

rule 4498

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

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 4560

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

Maple [A] (warning: unable to verify)

Time = 2.00 (sec) , antiderivative size = 326, normalized size of antiderivative = 1.25

method	result	size
default	\(-\frac {\sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (\left (-2445 \cos \left (d x +c \right )^{3}-7335 \cos \left (d x +c \right )^{2}-7335 \cos \left (d x +c \right )-2445\right ) \sqrt {2}\, B \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \ln \left (\sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )+\left (4245 \cos \left (d x +c \right )^{3}+12735 \cos \left (d x +c \right )^{2}+12735 \cos \left (d x +c \right )+4245\right ) \sqrt {2}\, C \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \ln \left (\sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )+\left (2990 \cos \left (d x +c \right )^{3}+5030 \cos \left (d x +c \right )^{2}+1600 \cos \left (d x +c \right )-320\right ) B \tan \left (d x +c \right )+\left (-5342 \cos \left (d x +c \right )^{4}-9110 \cos \left (d x +c \right )^{3}-3136 \cos \left (d x +c \right )^{2}+320 \cos \left (d x +c \right )-192\right ) C \sec \left (d x +c \right ) \tan \left (d x +c \right )\right )}{480 d \,a^{3} \left (\cos \left (d x +c \right )^{3}+3 \cos \left (d x +c \right )^{2}+3 \cos \left (d x +c \right )+1\right )}\)	\(326\)
parts	\(\frac {B \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (\left (-598 \cos \left (d x +c \right )^{3}-1006 \cos \left (d x +c \right )^{2}-320 \cos \left (d x +c \right )+64\right ) \tan \left (d x +c \right )+\left (489 \cos \left (d x +c \right )^{3}+1467 \cos \left (d x +c \right )^{2}+1467 \cos \left (d x +c \right )+489\right ) \sqrt {2}\, \ln \left (\sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\right )}{96 d \,a^{3} \left (\cos \left (d x +c \right )^{3}+3 \cos \left (d x +c \right )^{2}+3 \cos \left (d x +c \right )+1\right )}-\frac {C \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (\left (-5342 \cos \left (d x +c \right )^{4}-9110 \cos \left (d x +c \right )^{3}-3136 \cos \left (d x +c \right )^{2}+320 \cos \left (d x +c \right )-192\right ) \sec \left (d x +c \right ) \tan \left (d x +c \right )+\left (4245 \cos \left (d x +c \right )^{3}+12735 \cos \left (d x +c \right )^{2}+12735 \cos \left (d x +c \right )+4245\right ) \sqrt {2}\, \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \ln \left (\sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )\right )}{480 d \,a^{3} \left (\cos \left (d x +c \right )^{3}+3 \cos \left (d x +c \right )^{2}+3 \cos \left (d x +c \right )+1\right )}\)	\(376\)

Input:

int(sec(d*x+c)^4*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(5/2),x,me 
thod=_RETURNVERBOSE)
 

Output:

-1/480/d/a^3*(a*(1+sec(d*x+c)))^(1/2)/(cos(d*x+c)^3+3*cos(d*x+c)^2+3*cos(d 
*x+c)+1)*((-2445*cos(d*x+c)^3-7335*cos(d*x+c)^2-7335*cos(d*x+c)-2445)*2^(1 
/2)*B*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*ln((-2*cos(d*x+c)/(cos(d*x+c)+1)) 
^(1/2)-cot(d*x+c)+csc(d*x+c))+(4245*cos(d*x+c)^3+12735*cos(d*x+c)^2+12735* 
cos(d*x+c)+4245)*2^(1/2)*C*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*ln((-2*cos(d 
*x+c)/(cos(d*x+c)+1))^(1/2)-cot(d*x+c)+csc(d*x+c))+(2990*cos(d*x+c)^3+5030 
*cos(d*x+c)^2+1600*cos(d*x+c)-320)*B*tan(d*x+c)+(-5342*cos(d*x+c)^4-9110*c 
os(d*x+c)^3-3136*cos(d*x+c)^2+320*cos(d*x+c)-192)*C*sec(d*x+c)*tan(d*x+c))
 

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 596, normalized size of antiderivative = 2.28 \[ \int \frac {\sec ^4(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx =\text {Too large to display} \] Input:

integrate(sec(d*x+c)^4*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(5/2 
),x, algorithm="fricas")
 

Output:

[1/960*(15*sqrt(2)*((163*B - 283*C)*cos(d*x + c)^5 + 3*(163*B - 283*C)*cos 
(d*x + c)^4 + 3*(163*B - 283*C)*cos(d*x + c)^3 + (163*B - 283*C)*cos(d*x + 
 c)^2)*sqrt(-a)*log(-(2*sqrt(2)*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x 
 + c))*cos(d*x + c)*sin(d*x + c) - 3*a*cos(d*x + c)^2 - 2*a*cos(d*x + c) + 
 a)/(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)) - 4*((1495*B - 2671*C)*cos(d*x 
+ c)^4 + 5*(503*B - 911*C)*cos(d*x + c)^3 + 32*(25*B - 49*C)*cos(d*x + c)^ 
2 - 160*(B - C)*cos(d*x + c) - 96*C)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c 
))*sin(d*x + c))/(a^3*d*cos(d*x + c)^5 + 3*a^3*d*cos(d*x + c)^4 + 3*a^3*d* 
cos(d*x + c)^3 + a^3*d*cos(d*x + c)^2), -1/480*(15*sqrt(2)*((163*B - 283*C 
)*cos(d*x + c)^5 + 3*(163*B - 283*C)*cos(d*x + c)^4 + 3*(163*B - 283*C)*co 
s(d*x + c)^3 + (163*B - 283*C)*cos(d*x + c)^2)*sqrt(a)*arctan(sqrt(2)*sqrt 
((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)/(sqrt(a)*sin(d*x + c))) + 
 2*((1495*B - 2671*C)*cos(d*x + c)^4 + 5*(503*B - 911*C)*cos(d*x + c)^3 + 
32*(25*B - 49*C)*cos(d*x + c)^2 - 160*(B - C)*cos(d*x + c) - 96*C)*sqrt((a 
*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(a^3*d*cos(d*x + c)^5 + 3*a 
^3*d*cos(d*x + c)^4 + 3*a^3*d*cos(d*x + c)^3 + a^3*d*cos(d*x + c)^2)]
 

Sympy [F]

\[ \int \frac {\sec ^4(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx=\int \frac {\left (B + C \sec {\left (c + d x \right )}\right ) \sec ^{5}{\left (c + d x \right )}}{\left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{\frac {5}{2}}}\, dx \] Input:

integrate(sec(d*x+c)**4*(B*sec(d*x+c)+C*sec(d*x+c)**2)/(a+a*sec(d*x+c))**( 
5/2),x)
 

Output:

Integral((B + C*sec(c + d*x))*sec(c + d*x)**5/(a*(sec(c + d*x) + 1))**(5/2 
), x)
 

Maxima [F(-1)]

Timed out. \[ \int \frac {\sec ^4(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx=\text {Timed out} \] Input:

integrate(sec(d*x+c)^4*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(5/2 
),x, algorithm="maxima")
 

Output:

Timed out
 

Giac [A] (verification not implemented)

Time = 1.26 (sec) , antiderivative size = 362, normalized size of antiderivative = 1.39 \[ \int \frac {\sec ^4(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx=\frac {\frac {{\left ({\left ({\left (15 \, {\left (\frac {2 \, {\left (\sqrt {2} B a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - \sqrt {2} C a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}{a^{2}} + \frac {21 \, \sqrt {2} B a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - 29 \, \sqrt {2} C a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}{a^{2}}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - \frac {3685 \, \sqrt {2} B a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - 6733 \, \sqrt {2} C a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}{a^{2}}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + \frac {5 \, {\left (1133 \, \sqrt {2} B a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - 1973 \, \sqrt {2} C a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}}{a^{2}}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - \frac {15 \, {\left (155 \, \sqrt {2} B a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - 291 \, \sqrt {2} C a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}}{a^{2}}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{2} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}} - \frac {15 \, {\left (163 \, \sqrt {2} B - 283 \, \sqrt {2} C\right )} \log \left ({\left | -\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} \right |}\right )}{\sqrt {-a} a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}}{480 \, d} \] Input:

integrate(sec(d*x+c)^4*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(5/2 
),x, algorithm="giac")
 

Output:

1/480*((((15*(2*(sqrt(2)*B*a^2*sgn(cos(d*x + c)) - sqrt(2)*C*a^2*sgn(cos(d 
*x + c)))*tan(1/2*d*x + 1/2*c)^2/a^2 + (21*sqrt(2)*B*a^2*sgn(cos(d*x + c)) 
 - 29*sqrt(2)*C*a^2*sgn(cos(d*x + c)))/a^2)*tan(1/2*d*x + 1/2*c)^2 - (3685 
*sqrt(2)*B*a^2*sgn(cos(d*x + c)) - 6733*sqrt(2)*C*a^2*sgn(cos(d*x + c)))/a 
^2)*tan(1/2*d*x + 1/2*c)^2 + 5*(1133*sqrt(2)*B*a^2*sgn(cos(d*x + c)) - 197 
3*sqrt(2)*C*a^2*sgn(cos(d*x + c)))/a^2)*tan(1/2*d*x + 1/2*c)^2 - 15*(155*s 
qrt(2)*B*a^2*sgn(cos(d*x + c)) - 291*sqrt(2)*C*a^2*sgn(cos(d*x + c)))/a^2) 
*tan(1/2*d*x + 1/2*c)/((a*tan(1/2*d*x + 1/2*c)^2 - a)^2*sqrt(-a*tan(1/2*d* 
x + 1/2*c)^2 + a)) - 15*(163*sqrt(2)*B - 283*sqrt(2)*C)*log(abs(-sqrt(-a)* 
tan(1/2*d*x + 1/2*c) + sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)))/(sqrt(-a)*a^2 
*sgn(cos(d*x + c))))/d
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\sec ^4(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx=\int \frac {\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}}{{\cos \left (c+d\,x\right )}^4\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \] Input:

int((B/cos(c + d*x) + C/cos(c + d*x)^2)/(cos(c + d*x)^4*(a + a/cos(c + d*x 
))^(5/2)),x)
 

Output:

int((B/cos(c + d*x) + C/cos(c + d*x)^2)/(cos(c + d*x)^4*(a + a/cos(c + d*x 
))^(5/2)), x)
 

Reduce [F]

\[ \int \frac {\sec ^4(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx=\frac {\sqrt {a}\, \left (\left (\int \frac {\sqrt {\sec \left (d x +c \right )+1}\, \sec \left (d x +c \right )^{6}}{\sec \left (d x +c \right )^{3}+3 \sec \left (d x +c \right )^{2}+3 \sec \left (d x +c \right )+1}d x \right ) c +\left (\int \frac {\sqrt {\sec \left (d x +c \right )+1}\, \sec \left (d x +c \right )^{5}}{\sec \left (d x +c \right )^{3}+3 \sec \left (d x +c \right )^{2}+3 \sec \left (d x +c \right )+1}d x \right ) b \right )}{a^{3}} \] Input:

int(sec(d*x+c)^4*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(5/2),x)
 

Output:

(sqrt(a)*(int((sqrt(sec(c + d*x) + 1)*sec(c + d*x)**6)/(sec(c + d*x)**3 + 
3*sec(c + d*x)**2 + 3*sec(c + d*x) + 1),x)*c + int((sqrt(sec(c + d*x) + 1) 
*sec(c + d*x)**5)/(sec(c + d*x)**3 + 3*sec(c + d*x)**2 + 3*sec(c + d*x) + 
1),x)*b))/a**3