\(\int \frac {\sec ^{\frac {7}{2}}(c+d x) (A+B \sec (c+d x))}{a+a \sec (c+d x)} \, dx\) [201]

Optimal result

Integrand size = 33, antiderivative size = 229 \[ \int \frac {\sec ^{\frac {7}{2}}(c+d x) (A+B \sec (c+d x))}{a+a \sec (c+d x)} \, dx=\frac {3 (5 A-7 B) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 a d}+\frac {5 (A-B) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 a d}-\frac {3 (5 A-7 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 a d}+\frac {5 (A-B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 a d}-\frac {(5 A-7 B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 a d}+\frac {(A-B) \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{d (a+a \sec (c+d x))} \] Output:

3/5*(5*A-7*B)*cos(d*x+c)^(1/2)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))*sec(d 
*x+c)^(1/2)/a/d+5/3*(A-B)*cos(d*x+c)^(1/2)*InverseJacobiAM(1/2*d*x+1/2*c,2 
^(1/2))*sec(d*x+c)^(1/2)/a/d-3/5*(5*A-7*B)*sec(d*x+c)^(1/2)*sin(d*x+c)/a/d 
+5/3*(A-B)*sec(d*x+c)^(3/2)*sin(d*x+c)/a/d-1/5*(5*A-7*B)*sec(d*x+c)^(5/2)* 
sin(d*x+c)/a/d+(A-B)*sec(d*x+c)^(7/2)*sin(d*x+c)/d/(a+a*sec(d*x+c))
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 7.05 (sec) , antiderivative size = 461, normalized size of antiderivative = 2.01 \[ \int \frac {\sec ^{\frac {7}{2}}(c+d x) (A+B \sec (c+d x))}{a+a \sec (c+d x)} \, dx=\frac {\cos ^2\left (\frac {1}{2} (c+d x)\right ) (A+B \sec (c+d x)) \left (-30 \sqrt {2} A e^{-i d x} \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt {1+e^{2 i (c+d x)}} \csc (c) \left (-3 \sqrt {1+e^{2 i (c+d x)}}+e^{2 i d x} \left (-1+e^{2 i c}\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-e^{2 i (c+d x)}\right )\right )+42 \sqrt {2} B e^{-i d x} \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt {1+e^{2 i (c+d x)}} \csc (c) \left (-3 \sqrt {1+e^{2 i (c+d x)}}+e^{2 i d x} \left (-1+e^{2 i c}\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-e^{2 i (c+d x)}\right )\right )+100 A \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}-100 B \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}+2 \sqrt {\sec (c+d x)} \left (-18 (5 A-7 B) \cos (d x) \csc (c)+(25 A-13 B+4 (5 A-2 B) \cos (c+d x)+25 (A-B) \cos (2 (c+d x))) \sec ^2(c+d x) \tan \left (\frac {1}{2} (c+d x)\right )\right )\right )}{30 a d (B+A \cos (c+d x)) (1+\sec (c+d x))} \] Input:

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

Output:

(Cos[(c + d*x)/2]^2*(A + B*Sec[c + d*x])*((-30*Sqrt[2]*A*Sqrt[E^(I*(c + d* 
x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x))]*Csc[c]*(-3*Sq 
rt[1 + E^((2*I)*(c + d*x))] + E^((2*I)*d*x)*(-1 + E^((2*I)*c))*Hypergeomet 
ric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))]))/E^(I*d*x) + (42*Sqrt[2]*B*Sq 
rt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x)) 
]*Csc[c]*(-3*Sqrt[1 + E^((2*I)*(c + d*x))] + E^((2*I)*d*x)*(-1 + E^((2*I)* 
c))*Hypergeometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))]))/E^(I*d*x) + 1 
00*A*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]] - 100 
*B*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]] + 2*Sqr 
t[Sec[c + d*x]]*(-18*(5*A - 7*B)*Cos[d*x]*Csc[c] + (25*A - 13*B + 4*(5*A - 
 2*B)*Cos[c + d*x] + 25*(A - B)*Cos[2*(c + d*x)])*Sec[c + d*x]^2*Tan[(c + 
d*x)/2])))/(30*a*d*(B + A*Cos[c + d*x])*(1 + Sec[c + d*x]))
 

Rubi [A] (verified)

Time = 1.03 (sec) , antiderivative size = 212, normalized size of antiderivative = 0.93, number of steps used = 14, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.424, Rules used = {3042, 4507, 27, 3042, 4274, 3042, 4255, 3042, 4255, 3042, 4258, 3042, 3119, 3120}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4507

\(\displaystyle \frac {\int \frac {1}{2} \sec ^{\frac {5}{2}}(c+d x) (5 a (A-B)-a (5 A-7 B) \sec (c+d x))dx}{a^2}+\frac {(A-B) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{d (a \sec (c+d x)+a)}\)

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4274

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4255

\(\displaystyle \frac {5 a (A-B) \left (\frac {1}{3} \int \sqrt {\sec (c+d x)}dx+\frac {2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}\right )-a (5 A-7 B) \left (\frac {3}{5} \int \sec ^{\frac {3}{2}}(c+d x)dx+\frac {2 \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{5 d}\right )}{2 a^2}+\frac {(A-B) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{d (a \sec (c+d x)+a)}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4255

\(\displaystyle \frac {5 a (A-B) \left (\frac {1}{3} \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}\right )-a (5 A-7 B) \left (\frac {3}{5} \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\int \frac {1}{\sqrt {\sec (c+d x)}}dx\right )+\frac {2 \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{5 d}\right )}{2 a^2}+\frac {(A-B) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{d (a \sec (c+d x)+a)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {5 a (A-B) \left (\frac {1}{3} \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}\right )-a (5 A-7 B) \left (\frac {3}{5} \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\int \frac {1}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )+\frac {2 \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{5 d}\right )}{2 a^2}+\frac {(A-B) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{d (a \sec (c+d x)+a)}\)

\(\Big \downarrow \) 4258

\(\displaystyle \frac {5 a (A-B) \left (\frac {1}{3} \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx+\frac {2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}\right )-a (5 A-7 B) \left (\frac {3}{5} \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\cos (c+d x)}dx\right )+\frac {2 \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{5 d}\right )}{2 a^2}+\frac {(A-B) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{d (a \sec (c+d x)+a)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {5 a (A-B) \left (\frac {1}{3} \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}\right )-a (5 A-7 B) \left (\frac {3}{5} \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx\right )+\frac {2 \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{5 d}\right )}{2 a^2}+\frac {(A-B) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{d (a \sec (c+d x)+a)}\)

\(\Big \downarrow \) 3119

\(\displaystyle \frac {5 a (A-B) \left (\frac {1}{3} \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}\right )-a (5 A-7 B) \left (\frac {2 \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{5 d}+\frac {3}{5} \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )\right )}{2 a^2}+\frac {(A-B) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{d (a \sec (c+d x)+a)}\)

\(\Big \downarrow \) 3120

\(\displaystyle \frac {5 a (A-B) \left (\frac {2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}+\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}\right )-a (5 A-7 B) \left (\frac {2 \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{5 d}+\frac {3}{5} \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )\right )}{2 a^2}+\frac {(A-B) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{d (a \sec (c+d x)+a)}\)

Input:

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

Output:

((A - B)*Sec[c + d*x]^(7/2)*Sin[c + d*x])/(d*(a + a*Sec[c + d*x])) + (5*a* 
(A - B)*((2*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x] 
])/(3*d) + (2*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(3*d)) - a*(5*A - 7*B)*((2* 
Sec[c + d*x]^(5/2)*Sin[c + d*x])/(5*d) + (3*((-2*Sqrt[Cos[c + d*x]]*Ellipt 
icE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/d + (2*Sqrt[Sec[c + d*x]]*Sin[c + 
d*x])/d))/5))/(2*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 3042

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

rule 3119

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)* 
(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
 

rule 3120

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2 
)*(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
 

rule 4255

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* 
x]*((b*Csc[c + d*x])^(n - 1)/(d*(n - 1))), x] + Simp[b^2*((n - 2)/(n - 1)) 
  Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] 
&& IntegerQ[2*n]
 

rule 4258

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(b*Csc[c + d*x] 
)^n*Sin[c + d*x]^n   Int[1/Sin[c + d*x]^n, x], x] /; FreeQ[{b, c, d}, x] && 
 EqQ[n^2, 1/4]
 

rule 4274

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + 
(a_)), x_Symbol] :> Simp[a   Int[(d*Csc[e + f*x])^n, x], x] + Simp[b/d   In 
t[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]
 

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]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(778\) vs. \(2(206)=412\).

Time = 7.45 (sec) , antiderivative size = 779, normalized size of antiderivative = 3.40

method	result	size
default	\(\text {Expression too large to display}\)	\(779\)

Input:

int(sec(d*x+c)^(7/2)*(A+B*sec(d*x+c))/(a+a*sec(d*x+c)),x,method=_RETURNVER 
BOSE)
 

Output:

-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)/a*((-2*A+2*B)/s 
in(1/2*d*x+1/2*c)^2/(2*sin(1/2*d*x+1/2*c)^2-1)*(-2*sin(1/2*d*x+1/2*c)^4+si 
n(1/2*d*x+1/2*c)^2)^(1/2)*(2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2-Ellip 
ticE(cos(1/2*d*x+1/2*c),2^(1/2))*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d 
*x+1/2*c)^2-1)^(1/2))+(A-B)*(cos(1/2*d*x+1/2*c)*(sin(1/2*d*x+1/2*c)^2)^(1/ 
2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(EllipticF(cos(1/2*d*x+1/2*c),2^(1/2)) 
-EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x 
+1/2*c)^2)/cos(1/2*d*x+1/2*c)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^ 
2)^(1/2)+(2*A-2*B)*(-1/6*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1 
/2*d*x+1/2*c)^2)^(1/2)/(cos(1/2*d*x+1/2*c)^2-1/2)^2+1/3*(sin(1/2*d*x+1/2*c 
)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin( 
1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2)))+2/5*B/(8*si 
n(1/2*d*x+1/2*c)^6-12*sin(1/2*d*x+1/2*c)^4+6*sin(1/2*d*x+1/2*c)^2-1)/sin(1 
/2*d*x+1/2*c)^2*(24*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)-12*EllipticE(c 
os(1/2*d*x+1/2*c),2^(1/2))*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2 
*c)^2-1)^(1/2)*sin(1/2*d*x+1/2*c)^4-24*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2* 
c)^4+12*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(sin(1/2*d*x+1/2*c)^2)^(1/2) 
*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*sin(1/2*d*x+1/2*c)^2+8*cos(1/2*d*x+1/2*c 
)*sin(1/2*d*x+1/2*c)^2-3*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(sin(1/2*d* 
x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2))*(-2*sin(1/2*d*x+1/2...
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.10 (sec) , antiderivative size = 339, normalized size of antiderivative = 1.48 \[ \int \frac {\sec ^{\frac {7}{2}}(c+d x) (A+B \sec (c+d x))}{a+a \sec (c+d x)} \, dx=-\frac {25 \, {\left (\sqrt {2} {\left (i \, A - i \, B\right )} \cos \left (d x + c\right )^{3} + \sqrt {2} {\left (i \, A - i \, B\right )} \cos \left (d x + c\right )^{2}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 25 \, {\left (\sqrt {2} {\left (-i \, A + i \, B\right )} \cos \left (d x + c\right )^{3} + \sqrt {2} {\left (-i \, A + i \, B\right )} \cos \left (d x + c\right )^{2}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 9 \, {\left (\sqrt {2} {\left (-5 i \, A + 7 i \, B\right )} \cos \left (d x + c\right )^{3} + \sqrt {2} {\left (-5 i \, A + 7 i \, B\right )} \cos \left (d x + c\right )^{2}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 9 \, {\left (\sqrt {2} {\left (5 i \, A - 7 i \, B\right )} \cos \left (d x + c\right )^{3} + \sqrt {2} {\left (5 i \, A - 7 i \, B\right )} \cos \left (d x + c\right )^{2}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) + \frac {2 \, {\left (9 \, {\left (5 \, A - 7 \, B\right )} \cos \left (d x + c\right )^{3} + 2 \, {\left (10 \, A - 19 \, B\right )} \cos \left (d x + c\right )^{2} - 2 \, {\left (5 \, A - 2 \, B\right )} \cos \left (d x + c\right ) - 6 \, B\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{30 \, {\left (a d \cos \left (d x + c\right )^{3} + a d \cos \left (d x + c\right )^{2}\right )}} \] Input:

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

Output:

-1/30*(25*(sqrt(2)*(I*A - I*B)*cos(d*x + c)^3 + sqrt(2)*(I*A - I*B)*cos(d* 
x + c)^2)*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + 25*( 
sqrt(2)*(-I*A + I*B)*cos(d*x + c)^3 + sqrt(2)*(-I*A + I*B)*cos(d*x + c)^2) 
*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 9*(sqrt(2)*(- 
5*I*A + 7*I*B)*cos(d*x + c)^3 + sqrt(2)*(-5*I*A + 7*I*B)*cos(d*x + c)^2)*w 
eierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x 
+ c))) + 9*(sqrt(2)*(5*I*A - 7*I*B)*cos(d*x + c)^3 + sqrt(2)*(5*I*A - 7*I* 
B)*cos(d*x + c)^2)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d 
*x + c) - I*sin(d*x + c))) + 2*(9*(5*A - 7*B)*cos(d*x + c)^3 + 2*(10*A - 1 
9*B)*cos(d*x + c)^2 - 2*(5*A - 2*B)*cos(d*x + c) - 6*B)*sin(d*x + c)/sqrt( 
cos(d*x + c)))/(a*d*cos(d*x + c)^3 + a*d*cos(d*x + c)^2)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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