\(\int \frac {(e \tan (c+d x))^{3/2}}{(a+a \sec (c+d x))^2} \, dx\) [132]

Optimal result

Integrand size = 25, antiderivative size = 258 \[ \int \frac {(e \tan (c+d x))^{3/2}}{(a+a \sec (c+d x))^2} \, dx=\frac {e^{3/2} \arctan \left (1-\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} a^2 d}-\frac {e^{3/2} \arctan \left (1+\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} a^2 d}-\frac {e^{3/2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}+\sqrt {e} \tan (c+d x)}\right )}{\sqrt {2} a^2 d}-\frac {4 e^3}{3 a^2 d (e \tan (c+d x))^{3/2}}+\frac {4 e^3 \sec (c+d x)}{3 a^2 d (e \tan (c+d x))^{3/2}}+\frac {2 e^2 \operatorname {EllipticF}\left (c-\frac {\pi }{4}+d x,2\right ) \sec (c+d x) \sqrt {\sin (2 c+2 d x)}}{3 a^2 d \sqrt {e \tan (c+d x)}} \] Output:

1/2*e^(3/2)*arctan(1-2^(1/2)*(e*tan(d*x+c))^(1/2)/e^(1/2))*2^(1/2)/a^2/d-1 
/2*e^(3/2)*arctan(1+2^(1/2)*(e*tan(d*x+c))^(1/2)/e^(1/2))*2^(1/2)/a^2/d-1/ 
2*e^(3/2)*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-4/3*e^3/a^2/d/(e*tan(d*x+c))^(3/2)+4/3*e^3*sec(d*x+c)/a^2 
/d/(e*tan(d*x+c))^(3/2)+2/3*e^2*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)
 

Mathematica [F]

\[ \int \frac {(e \tan (c+d x))^{3/2}}{(a+a \sec (c+d x))^2} \, dx=\int \frac {(e \tan (c+d x))^{3/2}}{(a+a \sec (c+d x))^2} \, dx \] Input:

Integrate[(e*Tan[c + d*x])^(3/2)/(a + a*Sec[c + d*x])^2,x]
 

Output:

Integrate[(e*Tan[c + d*x])^(3/2)/(a + a*Sec[c + d*x])^2, x]
 

Rubi [A] (verified)

Time = 0.78 (sec) , antiderivative size = 323, normalized size of antiderivative = 1.25, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4376

\(\displaystyle \frac {e^4 \int \frac {(a-a \sec (c+d x))^2}{(e \tan (c+d x))^{5/2}}dx}{a^4}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4374

\(\displaystyle \frac {e^4 \int \left (\frac {\sec ^2(c+d x) a^2}{(e \tan (c+d x))^{5/2}}-\frac {2 \sec (c+d x) a^2}{(e \tan (c+d x))^{5/2}}+\frac {a^2}{(e \tan (c+d x))^{5/2}}\right )dx}{a^4}\)

\(\Big \downarrow \) 2009

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

Input:

Int[(e*Tan[c + d*x])^(3/2)/(a + a*Sec[c + d*x])^2,x]
 

Output:

(e^4*((a^2*ArcTan[1 - (Sqrt[2]*Sqrt[e*Tan[c + d*x]])/Sqrt[e]])/(Sqrt[2]*d* 
e^(5/2)) - (a^2*ArcTan[1 + (Sqrt[2]*Sqrt[e*Tan[c + d*x]])/Sqrt[e]])/(Sqrt[ 
2]*d*e^(5/2)) + (a^2*Log[Sqrt[e] + Sqrt[e]*Tan[c + d*x] - Sqrt[2]*Sqrt[e*T 
an[c + d*x]]])/(2*Sqrt[2]*d*e^(5/2)) - (a^2*Log[Sqrt[e] + Sqrt[e]*Tan[c + 
d*x] + Sqrt[2]*Sqrt[e*Tan[c + d*x]]])/(2*Sqrt[2]*d*e^(5/2)) - (4*a^2)/(3*d 
*e*(e*Tan[c + d*x])^(3/2)) + (4*a^2*Sec[c + d*x])/(3*d*e*(e*Tan[c + d*x])^ 
(3/2)) + (2*a^2*EllipticF[c - Pi/4 + d*x, 2]*Sec[c + d*x]*Sqrt[Sin[2*c + 2 
*d*x]])/(3*d*e^2*Sqrt[e*Tan[c + d*x]])))/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] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 1.38 (sec) , antiderivative size = 535, normalized size of antiderivative = 2.07

method	result	size
default	\(\frac {e \sqrt {e \tan \left (d x +c \right )}\, \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1\right )^{2} \left (3 i \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 )-3 i \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 )-10 \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 {EllipticF}\left (\sqrt {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )+3 \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 )+3 \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 )+4 \left (1-\cos \left (d x +c \right )\right )^{3} \csc \left (d x +c \right )^{3}-4 \csc \left (d x +c \right )+4 \cot \left (d x +c \right )\right ) \left (1+\cos \left (d x +c \right )\right )^{2} \tan \left (d x +c \right ) \sec \left (d x +c \right )}{24 a^{2} d \left (-1+\cos \left (d x +c \right )\right )}\)	\(535\)

Input:

int((e*tan(d*x+c))^(3/2)/(a+a*sec(d*x+c))^2,x,method=_RETURNVERBOSE)
 

Output:

1/24/a^2/d/(-1+cos(d*x+c))*e*(e*tan(d*x+c))^(1/2)*((1-cos(d*x+c))^2*csc(d* 
x+c)^2-1)^2*(3*I*(-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))-3*I*(-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)*Elliptic 
Pi((-cot(d*x+c)+csc(d*x+c)+1)^(1/2),1/2+1/2*I,1/2*2^(1/2))-10*(-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)*EllipticF((-cot(d*x+c)+csc(d*x+c)+1)^(1/2),1/2*2^(1/2))+3*(- 
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))+3*(-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))+4*(1-cos(d*x+c))^3*csc(d*x+c)^3-4*cs 
c(d*x+c)+4*cot(d*x+c))*(1+cos(d*x+c))^2*tan(d*x+c)*sec(d*x+c)
 

Fricas [F(-1)]

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

integrate((e*tan(d*x+c))^(3/2)/(a+a*sec(d*x+c))^2,x, algorithm="fricas")
 

Output:

Timed out
 

Sympy [F]

\[ \int \frac {(e \tan (c+d x))^{3/2}}{(a+a \sec (c+d x))^2} \, dx=\frac {\int \frac {\left (e \tan {\left (c + d x \right )}\right )^{\frac {3}{2}}}{\sec ^{2}{\left (c + d x \right )} + 2 \sec {\left (c + d x \right )} + 1}\, dx}{a^{2}} \] Input:

integrate((e*tan(d*x+c))**(3/2)/(a+a*sec(d*x+c))**2,x)
 

Output:

Integral((e*tan(c + d*x))**(3/2)/(sec(c + d*x)**2 + 2*sec(c + d*x) + 1), x 
)/a**2
 

Maxima [F(-1)]

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

integrate((e*tan(d*x+c))^(3/2)/(a+a*sec(d*x+c))^2,x, algorithm="maxima")
 

Output:

Timed out
 

Giac [F]

\[ \int \frac {(e \tan (c+d x))^{3/2}}{(a+a \sec (c+d x))^2} \, dx=\int { \frac {\left (e \tan \left (d x + c\right )\right )^{\frac {3}{2}}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{2}} \,d x } \] Input:

integrate((e*tan(d*x+c))^(3/2)/(a+a*sec(d*x+c))^2,x, algorithm="giac")
 

Output:

integrate((e*tan(d*x + c))^(3/2)/(a*sec(d*x + c) + a)^2, x)
 

Mupad [F(-1)]

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

int((e*tan(c + d*x))^(3/2)/(a + a/cos(c + d*x))^2,x)
 

Output:

int((cos(c + d*x)^2*(e*tan(c + d*x))^(3/2))/(a^2*(cos(c + d*x) + 1)^2), x)
 

Reduce [F]

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

int((e*tan(d*x+c))^(3/2)/(a+a*sec(d*x+c))^2,x)
 

Output:

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