\(\int \frac {(e \tan (c+d x))^{5/2}}{(a+a \sec (c+d x))^2} \, dx\) [131]

Optimal result

Integrand size = 25, antiderivative size = 251 \[ \int \frac {(e \tan (c+d x))^{5/2}}{(a+a \sec (c+d x))^2} \, dx=\frac {e^{5/2} \arctan \left (1-\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} a^2 d}-\frac {e^{5/2} \arctan \left (1+\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} a^2 d}+\frac {e^{5/2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}+\sqrt {e} \tan (c+d x)}\right )}{\sqrt {2} a^2 d}-\frac {4 e^3}{a^2 d \sqrt {e \tan (c+d x)}}+\frac {4 e^3 \cos (c+d x)}{a^2 d \sqrt {e \tan (c+d x)}}+\frac {4 e^2 \cos (c+d x) E\left (\left .c-\frac {\pi }{4}+d x\right |2\right ) \sqrt {e \tan (c+d x)}}{a^2 d \sqrt {\sin (2 c+2 d x)}} \] Output:

1/2*e^(5/2)*arctan(1-2^(1/2)*(e*tan(d*x+c))^(1/2)/e^(1/2))*2^(1/2)/a^2/d-1 
/2*e^(5/2)*arctan(1+2^(1/2)*(e*tan(d*x+c))^(1/2)/e^(1/2))*2^(1/2)/a^2/d+1/ 
2*e^(5/2)*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*e^3/a^2/d/(e*tan(d*x+c))^(1/2)+4*e^3*cos(d*x+c)/a^2/d/( 
e*tan(d*x+c))^(1/2)-4*e^2*cos(d*x+c)*EllipticE(cos(c+1/4*Pi+d*x),2^(1/2))* 
(e*tan(d*x+c))^(1/2)/a^2/d/sin(2*d*x+2*c)^(1/2)
 

Mathematica [C] (warning: unable to verify)

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

Time = 7.91 (sec) , antiderivative size = 812, normalized size of antiderivative = 3.24 \[ \int \frac {(e \tan (c+d x))^{5/2}}{(a+a \sec (c+d x))^2} \, dx=\frac {\cos ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) \csc ^2(c+d x) \left (\frac {32 \cos \left (\frac {c}{2}\right ) \cos (d x) \sec (2 c) \sin \left (\frac {c}{2}\right )}{d}+\frac {16 \sec \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}+\frac {d x}{2}\right ) \sin \left (\frac {d x}{2}\right )}{d}-\frac {16 \cos (c) \sec (2 c) \sin (d x)}{d}+\frac {16 \tan \left (\frac {c}{2}\right )}{d}\right ) (e \tan (c+d x))^{5/2}}{(a+a \sec (c+d x))^2}+\frac {e^{-2 i c} \left (-e^{4 i c} \sqrt {-1+e^{4 i (c+d x)}} \arctan \left (\sqrt {-1+e^{4 i (c+d x)}}\right )+2 \sqrt {-1+e^{2 i (c+d x)}} \sqrt {1+e^{2 i (c+d x)}} \text {arctanh}\left (\sqrt {\frac {-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}}\right )\right ) \cos ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) \sec (2 c) \sec ^2(c+d x) (e \tan (c+d x))^{5/2}}{d \sqrt {-\frac {i \left (-1+e^{2 i (c+d x)}\right )}{1+e^{2 i (c+d x)}}} \left (1+e^{2 i (c+d x)}\right ) (a+a \sec (c+d x))^2 \tan ^{\frac {5}{2}}(c+d x)}-\frac {e^{-2 i c} \left (\sqrt {-1+e^{4 i (c+d x)}} \arctan \left (\sqrt {-1+e^{4 i (c+d x)}}\right )-2 e^{4 i c} \sqrt {-1+e^{2 i (c+d x)}} \sqrt {1+e^{2 i (c+d x)}} \text {arctanh}\left (\sqrt {\frac {-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}}\right )\right ) \cos ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) \sec (2 c) \sec ^2(c+d x) (e \tan (c+d x))^{5/2}}{d \sqrt {-\frac {i \left (-1+e^{2 i (c+d x)}\right )}{1+e^{2 i (c+d x)}}} \left (1+e^{2 i (c+d x)}\right ) (a+a \sec (c+d x))^2 \tan ^{\frac {5}{2}}(c+d x)}-\frac {8 e^{i (c-d x)} \cos ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) \left (3-3 e^{4 i (c+d x)}+e^{4 i d x} \left (1+e^{4 i c}\right ) \sqrt {1-e^{4 i (c+d x)}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},e^{4 i (c+d x)}\right )\right ) \sec (2 c) \sec ^2(c+d x) (e \tan (c+d x))^{5/2}}{3 d \sqrt {-\frac {i \left (-1+e^{2 i (c+d x)}\right )}{1+e^{2 i (c+d x)}}} \left (1+e^{2 i (c+d x)}\right ) (a+a \sec (c+d x))^2 \tan ^{\frac {5}{2}}(c+d x)} \] Input:

Integrate[(e*Tan[c + d*x])^(5/2)/(a + a*Sec[c + d*x])^2,x]
 

Output:

(Cos[c/2 + (d*x)/2]^4*Csc[c + d*x]^2*((32*Cos[c/2]*Cos[d*x]*Sec[2*c]*Sin[c 
/2])/d + (16*Sec[c/2]*Sec[c/2 + (d*x)/2]*Sin[(d*x)/2])/d - (16*Cos[c]*Sec[ 
2*c]*Sin[d*x])/d + (16*Tan[c/2])/d)*(e*Tan[c + d*x])^(5/2))/(a + a*Sec[c + 
 d*x])^2 + ((-(E^((4*I)*c)*Sqrt[-1 + E^((4*I)*(c + d*x))]*ArcTan[Sqrt[-1 + 
 E^((4*I)*(c + d*x))]]) + 2*Sqrt[-1 + E^((2*I)*(c + d*x))]*Sqrt[1 + E^((2* 
I)*(c + d*x))]*ArcTanh[Sqrt[(-1 + E^((2*I)*(c + d*x)))/(1 + E^((2*I)*(c + 
d*x)))]])*Cos[c/2 + (d*x)/2]^4*Sec[2*c]*Sec[c + d*x]^2*(e*Tan[c + d*x])^(5 
/2))/(d*E^((2*I)*c)*Sqrt[((-I)*(-1 + E^((2*I)*(c + d*x))))/(1 + E^((2*I)*( 
c + d*x)))]*(1 + E^((2*I)*(c + d*x)))*(a + a*Sec[c + d*x])^2*Tan[c + d*x]^ 
(5/2)) - ((Sqrt[-1 + E^((4*I)*(c + d*x))]*ArcTan[Sqrt[-1 + E^((4*I)*(c + d 
*x))]] - 2*E^((4*I)*c)*Sqrt[-1 + E^((2*I)*(c + d*x))]*Sqrt[1 + E^((2*I)*(c 
 + d*x))]*ArcTanh[Sqrt[(-1 + E^((2*I)*(c + d*x)))/(1 + E^((2*I)*(c + d*x)) 
)]])*Cos[c/2 + (d*x)/2]^4*Sec[2*c]*Sec[c + d*x]^2*(e*Tan[c + d*x])^(5/2))/ 
(d*E^((2*I)*c)*Sqrt[((-I)*(-1 + E^((2*I)*(c + d*x))))/(1 + E^((2*I)*(c + d 
*x)))]*(1 + E^((2*I)*(c + d*x)))*(a + a*Sec[c + d*x])^2*Tan[c + d*x]^(5/2) 
) - (8*E^(I*(c - d*x))*Cos[c/2 + (d*x)/2]^4*(3 - 3*E^((4*I)*(c + d*x)) + E 
^((4*I)*d*x)*(1 + E^((4*I)*c))*Sqrt[1 - E^((4*I)*(c + d*x))]*Hypergeometri 
c2F1[1/2, 3/4, 7/4, E^((4*I)*(c + d*x))])*Sec[2*c]*Sec[c + d*x]^2*(e*Tan[c 
 + d*x])^(5/2))/(3*d*Sqrt[((-I)*(-1 + E^((2*I)*(c + d*x))))/(1 + E^((2*I)* 
(c + d*x)))]*(1 + E^((2*I)*(c + d*x)))*(a + a*Sec[c + d*x])^2*Tan[c + d...
 

Rubi [A] (verified)

Time = 0.76 (sec) , antiderivative size = 317, normalized size of antiderivative = 1.26, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4376

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4374

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {e^4 \left (\frac {a^2 \arctan \left (1-\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} d e^{3/2}}-\frac {a^2 \arctan \left (\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}+1\right )}{\sqrt {2} d e^{3/2}}-\frac {a^2 \log \left (\sqrt {e} \tan (c+d x)-\sqrt {2} \sqrt {e \tan (c+d x)}+\sqrt {e}\right )}{2 \sqrt {2} d e^{3/2}}+\frac {a^2 \log \left (\sqrt {e} \tan (c+d x)+\sqrt {2} \sqrt {e \tan (c+d x)}+\sqrt {e}\right )}{2 \sqrt {2} d e^{3/2}}+\frac {4 a^2 \cos (c+d x) E\left (\left .c+d x-\frac {\pi }{4}\right |2\right ) \sqrt {e \tan (c+d x)}}{d e^2 \sqrt {\sin (2 c+2 d x)}}-\frac {4 a^2}{d e \sqrt {e \tan (c+d x)}}+\frac {4 a^2 \cos (c+d x)}{d e \sqrt {e \tan (c+d x)}}\right )}{a^4}\)

Input:

Int[(e*Tan[c + d*x])^(5/2)/(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^(3/2)) - (a^2*ArcTan[1 + (Sqrt[2]*Sqrt[e*Tan[c + d*x]])/Sqrt[e]])/(Sqrt[ 
2]*d*e^(3/2)) - (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^(3/2)) + (a^2*Log[Sqrt[e] + Sqrt[e]*Tan[c + 
d*x] + Sqrt[2]*Sqrt[e*Tan[c + d*x]]])/(2*Sqrt[2]*d*e^(3/2)) - (4*a^2)/(d*e 
*Sqrt[e*Tan[c + d*x]]) + (4*a^2*Cos[c + d*x])/(d*e*Sqrt[e*Tan[c + d*x]]) + 
 (4*a^2*Cos[c + d*x]*EllipticE[c - Pi/4 + d*x, 2]*Sqrt[e*Tan[c + d*x]])/(d 
*e^2*Sqrt[Sin[2*c + 2*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] (verified)

Result contains complex when optimal does not.

Time = 0.97 (sec) , antiderivative size = 256, normalized size of antiderivative = 1.02

method	result	size
default	\(\frac {\left (\frac {1}{2}-\frac {i}{2}\right ) \left (\operatorname {EllipticPi}\left (\sqrt {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}, \frac {1}{2}-\frac {i}{2}, \frac {\sqrt {2}}{2}\right )+i \operatorname {EllipticPi}\left (\sqrt {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}, \frac {1}{2}+\frac {i}{2}, \frac {\sqrt {2}}{2}\right )-4 i \operatorname {EllipticE}\left (\sqrt {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )+2 i \operatorname {EllipticF}\left (\sqrt {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )-4 \operatorname {EllipticE}\left (\sqrt {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )+2 \operatorname {EllipticF}\left (\sqrt {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )\right ) \sqrt {e \tan \left (d x +c \right )}\, e^{2} \sqrt {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}\, \sqrt {2 \cot \left (d x +c \right )-2 \csc \left (d x +c \right )+2}\, \sqrt {-\csc \left (d x +c \right )+\cot \left (d x +c \right )}\, \left (\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )}{a^{2} d}\)	\(256\)

Input:

int((e*tan(d*x+c))^(5/2)/(a+a*sec(d*x+c))^2,x,method=_RETURNVERBOSE)
 

Output:

(1/2-1/2*I)/a^2/d*(EllipticPi((-cot(d*x+c)+csc(d*x+c)+1)^(1/2),1/2-1/2*I,1 
/2*2^(1/2))+I*EllipticPi((-cot(d*x+c)+csc(d*x+c)+1)^(1/2),1/2+1/2*I,1/2*2^ 
(1/2))-4*I*EllipticE((-cot(d*x+c)+csc(d*x+c)+1)^(1/2),1/2*2^(1/2))+2*I*Ell 
ipticF((-cot(d*x+c)+csc(d*x+c)+1)^(1/2),1/2*2^(1/2))-4*EllipticE((-cot(d*x 
+c)+csc(d*x+c)+1)^(1/2),1/2*2^(1/2))+2*EllipticF((-cot(d*x+c)+csc(d*x+c)+1 
)^(1/2),1/2*2^(1/2)))*(e*tan(d*x+c))^(1/2)*e^2*(-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)*( 
csc(d*x+c)+cot(d*x+c))
 

Fricas [F(-1)]

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

integrate((e*tan(d*x+c))^(5/2)/(a+a*sec(d*x+c))^2,x, algorithm="fricas")
 

Output:

Timed out
 

Sympy [F(-1)]

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

integrate((e*tan(d*x+c))**(5/2)/(a+a*sec(d*x+c))**2,x)
 

Output:

Timed out
 

Maxima [F]

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

integrate((e*tan(d*x+c))^(5/2)/(a+a*sec(d*x+c))^2,x, algorithm="maxima")
 

Output:

integrate((e*tan(d*x + c))^(5/2)/(a*sec(d*x + c) + a)^2, x)
 

Giac [F]

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

integrate((e*tan(d*x+c))^(5/2)/(a+a*sec(d*x+c))^2,x, algorithm="giac")
 

Output:

integrate((e*tan(d*x + c))^(5/2)/(a*sec(d*x + c) + a)^2, x)
 

Mupad [F(-1)]

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

int((e*tan(c + d*x))^(5/2)/(a + a/cos(c + d*x))^2,x)
 

Output:

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

Reduce [F]

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

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

Output:

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