\(\int \frac {(e \tan (c+d x))^{11/2}}{(a+a \sec (c+d x))^2} \, dx\) [128]

Optimal result

Integrand size = 25, antiderivative size = 281 \[ \int \frac {(e \tan (c+d x))^{11/2}}{(a+a \sec (c+d x))^2} \, dx=\frac {e^{11/2} \arctan \left (1-\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} a^2 d}-\frac {e^{11/2} \arctan \left (1+\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} a^2 d}-\frac {e^{11/2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}+\sqrt {e} \tan (c+d x)}\right )}{\sqrt {2} a^2 d}+\frac {2 e^6 \operatorname {EllipticF}\left (c-\frac {\pi }{4}+d x,2\right ) \sec (c+d x) \sqrt {\sin (2 c+2 d x)}}{3 a^2 d \sqrt {e \tan (c+d x)}}+\frac {2 e^5 \sqrt {e \tan (c+d x)}}{a^2 d}-\frac {4 e^5 \sec (c+d x) \sqrt {e \tan (c+d x)}}{3 a^2 d}+\frac {2 e^3 (e \tan (c+d x))^{5/2}}{5 a^2 d} \] Output:

1/2*e^(11/2)*arctan(1-2^(1/2)*(e*tan(d*x+c))^(1/2)/e^(1/2))*2^(1/2)/a^2/d- 
1/2*e^(11/2)*arctan(1+2^(1/2)*(e*tan(d*x+c))^(1/2)/e^(1/2))*2^(1/2)/a^2/d- 
1/2*e^(11/2)*arctanh(2^(1/2)*(e*tan(d*x+c))^(1/2)/(e^(1/2)+e^(1/2)*tan(d*x 
+c)))*2^(1/2)/a^2/d+2/3*e^6*InverseJacobiAM(c-1/4*Pi+d*x,2^(1/2))*sec(d*x+ 
c)*sin(2*d*x+2*c)^(1/2)/a^2/d/(e*tan(d*x+c))^(1/2)+2*e^5*(e*tan(d*x+c))^(1 
/2)/a^2/d-4/3*e^5*sec(d*x+c)*(e*tan(d*x+c))^(1/2)/a^2/d+2/5*e^3*(e*tan(d*x 
+c))^(5/2)/a^2/d
 

Mathematica [F]

\[ \int \frac {(e \tan (c+d x))^{11/2}}{(a+a \sec (c+d x))^2} \, dx=\int \frac {(e \tan (c+d x))^{11/2}}{(a+a \sec (c+d x))^2} \, dx \] Input:

Integrate[(e*Tan[c + d*x])^(11/2)/(a + a*Sec[c + d*x])^2,x]
 

Output:

Integrate[(e*Tan[c + d*x])^(11/2)/(a + a*Sec[c + d*x])^2, x]
 

Rubi [A] (verified)

Time = 0.77 (sec) , antiderivative size = 342, normalized size of antiderivative = 1.22, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3042, 4376, 3042, 4374, 2009}

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.

\(\displaystyle \int \frac {(e \tan (c+d x))^{11/2}}{(a \sec (c+d x)+a)^2} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\left (-e \cot \left (c+d x+\frac {\pi }{2}\right )\right )^{11/2}}{\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^2}dx\)

\(\Big \downarrow \) 4376

\(\displaystyle \frac {e^4 \int (a-a \sec (c+d x))^2 (e \tan (c+d x))^{3/2}dx}{a^4}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {e^4 \int \left (-e \cot \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2} \left (a-a \csc \left (c+d x+\frac {\pi }{2}\right )\right )^2dx}{a^4}\)

\(\Big \downarrow \) 4374

\(\displaystyle \frac {e^4 \int \left (\sec ^2(c+d x) (e \tan (c+d x))^{3/2} a^2-2 \sec (c+d x) (e \tan (c+d x))^{3/2} a^2+(e \tan (c+d x))^{3/2} a^2\right )dx}{a^4}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {e^4 \left (\frac {a^2 e^{3/2} \arctan \left (1-\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} d}-\frac {a^2 e^{3/2} \arctan \left (\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}+1\right )}{\sqrt {2} d}+\frac {a^2 e^{3/2} \log \left (\sqrt {e} \tan (c+d x)-\sqrt {2} \sqrt {e \tan (c+d x)}+\sqrt {e}\right )}{2 \sqrt {2} d}-\frac {a^2 e^{3/2} \log \left (\sqrt {e} \tan (c+d x)+\sqrt {2} \sqrt {e \tan (c+d x)}+\sqrt {e}\right )}{2 \sqrt {2} d}+\frac {2 a^2 e^2 \sqrt {\sin (2 c+2 d x)} \sec (c+d x) \operatorname {EllipticF}\left (c+d x-\frac {\pi }{4},2\right )}{3 d \sqrt {e \tan (c+d x)}}+\frac {2 a^2 (e \tan (c+d x))^{5/2}}{5 d e}+\frac {2 a^2 e \sqrt {e \tan (c+d x)}}{d}-\frac {4 a^2 e \sec (c+d x) \sqrt {e \tan (c+d x)}}{3 d}\right )}{a^4}\)

Input:

Int[(e*Tan[c + d*x])^(11/2)/(a + a*Sec[c + d*x])^2,x]
 

Output:

(e^4*((a^2*e^(3/2)*ArcTan[1 - (Sqrt[2]*Sqrt[e*Tan[c + d*x]])/Sqrt[e]])/(Sq 
rt[2]*d) - (a^2*e^(3/2)*ArcTan[1 + (Sqrt[2]*Sqrt[e*Tan[c + d*x]])/Sqrt[e]] 
)/(Sqrt[2]*d) + (a^2*e^(3/2)*Log[Sqrt[e] + Sqrt[e]*Tan[c + d*x] - Sqrt[2]* 
Sqrt[e*Tan[c + d*x]]])/(2*Sqrt[2]*d) - (a^2*e^(3/2)*Log[Sqrt[e] + Sqrt[e]* 
Tan[c + d*x] + Sqrt[2]*Sqrt[e*Tan[c + d*x]]])/(2*Sqrt[2]*d) + (2*a^2*e^2*E 
llipticF[c - Pi/4 + d*x, 2]*Sec[c + d*x]*Sqrt[Sin[2*c + 2*d*x]])/(3*d*Sqrt 
[e*Tan[c + d*x]]) + (2*a^2*e*Sqrt[e*Tan[c + d*x]])/d - (4*a^2*e*Sec[c + d* 
x]*Sqrt[e*Tan[c + d*x]])/(3*d) + (2*a^2*(e*Tan[c + d*x])^(5/2))/(5*d*e)))/ 
a^4
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

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

rule 4374

Int[(cot[(c_.) + (d_.)*(x_)]*(e_.))^(m_)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + ( 
a_))^(n_), x_Symbol] :> Int[ExpandIntegrand[(e*Cot[c + d*x])^m, (a + b*Csc[ 
c + d*x])^n, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && IGtQ[n, 0]
 

rule 4376

Int[(cot[(c_.) + (d_.)*(x_)]*(e_.))^(m_)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + ( 
a_))^(n_), x_Symbol] :> Simp[a^(2*n)/e^(2*n)   Int[(e*Cot[c + d*x])^(m + 2* 
n)/(-a + b*Csc[c + d*x])^n, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && EqQ[a 
^2 - b^2, 0] && ILtQ[n, 0]
 

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 2.02 (sec) , antiderivative size = 593, normalized size of antiderivative = 2.11

method	result	size
default	\(\frac {\sqrt {2}\, e^{5} \sqrt {-\frac {2 \sin \left (d x +c \right ) \cos \left (d x +c \right )}{\left (1+\cos \left (d x +c \right )\right )^{2}}}\, \sqrt {e \tan \left (d x +c \right )}\, \left (48-40 \sec \left (d x +c \right )+12 \sec \left (d x +c \right )^{2}+50 \sqrt {2 \cot \left (d x +c \right )-2 \csc \left (d x +c \right )+2}\, \sqrt {-\csc \left (d x +c \right )+\cot \left (d x +c \right )}\, \operatorname {EllipticF}\left (\sqrt {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right ) \sqrt {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}\, \left (\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )+15 \sqrt {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}\, \sqrt {2 \cot \left (d x +c \right )-2 \csc \left (d x +c \right )+2}\, \sqrt {-\csc \left (d x +c \right )+\cot \left (d x +c \right )}\, \operatorname {EllipticPi}\left (\sqrt {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}, \frac {1}{2}-\frac {i}{2}, \frac {\sqrt {2}}{2}\right ) \left (-\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )+15 \sqrt {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}\, \sqrt {2 \cot \left (d x +c \right )-2 \csc \left (d x +c \right )+2}\, \sqrt {-\csc \left (d x +c \right )+\cot \left (d x +c \right )}\, \operatorname {EllipticPi}\left (\sqrt {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}, \frac {1}{2}+\frac {i}{2}, \frac {\sqrt {2}}{2}\right ) \left (-\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )+15 i \sqrt {2 \cot \left (d x +c \right )-2 \csc \left (d x +c \right )+2}\, \sqrt {-\csc \left (d x +c \right )+\cot \left (d x +c \right )}\, \sqrt {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}\, \operatorname {EllipticPi}\left (\sqrt {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}, \frac {1}{2}-\frac {i}{2}, \frac {\sqrt {2}}{2}\right ) \left (-\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )+15 i \sqrt {2 \cot \left (d x +c \right )-2 \csc \left (d x +c \right )+2}\, \sqrt {-\csc \left (d x +c \right )+\cot \left (d x +c \right )}\, \sqrt {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}\, \operatorname {EllipticPi}\left (\sqrt {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}, \frac {1}{2}+\frac {i}{2}, \frac {\sqrt {2}}{2}\right ) \left (\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )\right )}{60 a^{2} d \sqrt {-\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{\left (1+\cos \left (d x +c \right )\right )^{2}}}}\)	\(593\)

Input:

int((e*tan(d*x+c))^(11/2)/(a+a*sec(d*x+c))^2,x,method=_RETURNVERBOSE)
 

Output:

1/60/a^2/d*2^(1/2)*e^5*(-2*sin(d*x+c)*cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)*( 
e*tan(d*x+c))^(1/2)/(-sin(d*x+c)*cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)*(48-40 
*sec(d*x+c)+12*sec(d*x+c)^2+50*(2*cot(d*x+c)-2*csc(d*x+c)+2)^(1/2)*(-csc(d 
*x+c)+cot(d*x+c))^(1/2)*EllipticF((-cot(d*x+c)+csc(d*x+c)+1)^(1/2),1/2*2^( 
1/2))*(-cot(d*x+c)+csc(d*x+c)+1)^(1/2)*(csc(d*x+c)+cot(d*x+c))+15*(-cot(d* 
x+c)+csc(d*x+c)+1)^(1/2)*(2*cot(d*x+c)-2*csc(d*x+c)+2)^(1/2)*(-csc(d*x+c)+ 
cot(d*x+c))^(1/2)*EllipticPi((-cot(d*x+c)+csc(d*x+c)+1)^(1/2),1/2-1/2*I,1/ 
2*2^(1/2))*(-cot(d*x+c)-csc(d*x+c))+15*(-cot(d*x+c)+csc(d*x+c)+1)^(1/2)*(2 
*cot(d*x+c)-2*csc(d*x+c)+2)^(1/2)*(-csc(d*x+c)+cot(d*x+c))^(1/2)*EllipticP 
i((-cot(d*x+c)+csc(d*x+c)+1)^(1/2),1/2+1/2*I,1/2*2^(1/2))*(-cot(d*x+c)-csc 
(d*x+c))+15*I*(2*cot(d*x+c)-2*csc(d*x+c)+2)^(1/2)*(-csc(d*x+c)+cot(d*x+c)) 
^(1/2)*(-cot(d*x+c)+csc(d*x+c)+1)^(1/2)*EllipticPi((-cot(d*x+c)+csc(d*x+c) 
+1)^(1/2),1/2-1/2*I,1/2*2^(1/2))*(-cot(d*x+c)-csc(d*x+c))+15*I*(2*cot(d*x+ 
c)-2*csc(d*x+c)+2)^(1/2)*(-csc(d*x+c)+cot(d*x+c))^(1/2)*(-cot(d*x+c)+csc(d 
*x+c)+1)^(1/2)*EllipticPi((-cot(d*x+c)+csc(d*x+c)+1)^(1/2),1/2+1/2*I,1/2*2 
^(1/2))*(csc(d*x+c)+cot(d*x+c)))
 

Fricas [F(-1)]

Timed out. \[ \int \frac {(e \tan (c+d x))^{11/2}}{(a+a \sec (c+d x))^2} \, dx=\text {Timed out} \] Input:

integrate((e*tan(d*x+c))^(11/2)/(a+a*sec(d*x+c))^2,x, algorithm="fricas")
 

Output:

Timed out
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(e \tan (c+d x))^{11/2}}{(a+a \sec (c+d x))^2} \, dx=\text {Timed out} \] Input:

integrate((e*tan(d*x+c))**(11/2)/(a+a*sec(d*x+c))**2,x)
 

Output:

Timed out
 

Maxima [F(-1)]

Timed out. \[ \int \frac {(e \tan (c+d x))^{11/2}}{(a+a \sec (c+d x))^2} \, dx=\text {Timed out} \] Input:

integrate((e*tan(d*x+c))^(11/2)/(a+a*sec(d*x+c))^2,x, algorithm="maxima")
 

Output:

Timed out
 

Giac [F]

\[ \int \frac {(e \tan (c+d x))^{11/2}}{(a+a \sec (c+d x))^2} \, dx=\int { \frac {\left (e \tan \left (d x + c\right )\right )^{\frac {11}{2}}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{2}} \,d x } \] Input:

integrate((e*tan(d*x+c))^(11/2)/(a+a*sec(d*x+c))^2,x, algorithm="giac")
 

Output:

integrate((e*tan(d*x + c))^(11/2)/(a*sec(d*x + c) + a)^2, x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(e \tan (c+d x))^{11/2}}{(a+a \sec (c+d x))^2} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^2\,{\left (e\,\mathrm {tan}\left (c+d\,x\right )\right )}^{11/2}}{a^2\,{\left (\cos \left (c+d\,x\right )+1\right )}^2} \,d x \] Input:

int((e*tan(c + d*x))^(11/2)/(a + a/cos(c + d*x))^2,x)
 

Output:

int((cos(c + d*x)^2*(e*tan(c + d*x))^(11/2))/(a^2*(cos(c + d*x) + 1)^2), x 
)
 

Reduce [F]

\[ \int \frac {(e \tan (c+d x))^{11/2}}{(a+a \sec (c+d x))^2} \, dx=\frac {\sqrt {e}\, \left (\int \frac {\sqrt {\tan \left (d x +c \right )}\, \tan \left (d x +c \right )^{5}}{\sec \left (d x +c \right )^{2}+2 \sec \left (d x +c \right )+1}d x \right ) e^{5}}{a^{2}} \] Input:

int((e*tan(d*x+c))^(11/2)/(a+a*sec(d*x+c))^2,x)
 

Output:

(sqrt(e)*int((sqrt(tan(c + d*x))*tan(c + d*x)**5)/(sec(c + d*x)**2 + 2*sec 
(c + d*x) + 1),x)*e**5)/a**2