\(\int \frac {(a+a \sec (c+d x))^2}{(e \tan (c+d x))^{5/2}} \, dx\) [116]

Optimal result

Integrand size = 25, antiderivative size = 258 \[ \int \frac {(a+a \sec (c+d x))^2}{(e \tan (c+d x))^{5/2}} \, dx=\frac {a^2 \arctan \left (1-\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} d e^{5/2}}-\frac {a^2 \arctan \left (1+\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} d e^{5/2}}-\frac {a^2 \text {arctanh}\left (\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}+\sqrt {e} \tan (c+d x)}\right )}{\sqrt {2} d e^{5/2}}-\frac {4 a^2}{3 d e (e \tan (c+d x))^{3/2}}-\frac {4 a^2 \sec (c+d x)}{3 d e (e \tan (c+d x))^{3/2}}-\frac {2 a^2 \operatorname {EllipticF}\left (c-\frac {\pi }{4}+d x,2\right ) \sec (c+d x) \sqrt {\sin (2 c+2 d x)}}{3 d e^2 \sqrt {e \tan (c+d x)}} \] Output:

1/2*a^2*arctan(1-2^(1/2)*(e*tan(d*x+c))^(1/2)/e^(1/2))*2^(1/2)/d/e^(5/2)-1 
/2*a^2*arctan(1+2^(1/2)*(e*tan(d*x+c))^(1/2)/e^(1/2))*2^(1/2)/d/e^(5/2)-1/ 
2*a^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)/d/e^(5/2)-4/3*a^2/d/e/(e*tan(d*x+c))^(3/2)-4/3*a^2*sec(d*x+c)/d/e/( 
e*tan(d*x+c))^(3/2)-2/3*a^2*InverseJacobiAM(c-1/4*Pi+d*x,2^(1/2))*sec(d*x+ 
c)*sin(2*d*x+2*c)^(1/2)/d/e^2/(e*tan(d*x+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 = 14.21 (sec) , antiderivative size = 224, normalized size of antiderivative = 0.87 \[ \int \frac {(a+a \sec (c+d x))^2}{(e \tan (c+d x))^{5/2}} \, dx=-\frac {a^2 \cos ^2\left (\frac {1}{2} (c+d x)\right ) \cos (c+d x) \cot \left (\frac {1}{2} (c+d x)\right ) \sec ^4\left (\frac {1}{2} \arctan (\tan (c+d x))\right ) \left (16 \operatorname {Hypergeometric2F1}\left (-\frac {3}{4},\frac {1}{2},\frac {1}{4},-\tan ^2(c+d x)\right )+16 \operatorname {Hypergeometric2F1}\left (-\frac {3}{4},1,\frac {1}{4},-\tan ^2(c+d x)\right )+3 \sqrt {2} \left (2 \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )-2 \arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )+\log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )-\log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )\right ) \tan ^{\frac {3}{2}}(c+d x)\right )}{24 d e^2 \sqrt {e \tan (c+d x)}} \] Input:

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

Output:

-1/24*(a^2*Cos[(c + d*x)/2]^2*Cos[c + d*x]*Cot[(c + d*x)/2]*Sec[ArcTan[Tan 
[c + d*x]]/2]^4*(16*Hypergeometric2F1[-3/4, 1/2, 1/4, -Tan[c + d*x]^2] + 1 
6*Hypergeometric2F1[-3/4, 1, 1/4, -Tan[c + d*x]^2] + 3*Sqrt[2]*(2*ArcTan[1 
 - Sqrt[2]*Sqrt[Tan[c + d*x]]] - 2*ArcTan[1 + Sqrt[2]*Sqrt[Tan[c + d*x]]] 
+ Log[1 - Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]] - Log[1 + Sqrt[2]*Sqr 
t[Tan[c + d*x]] + Tan[c + d*x]])*Tan[c + d*x]^(3/2)))/(d*e^2*Sqrt[e*Tan[c 
+ d*x]])
 

Rubi [A] (verified)

Time = 0.68 (sec) , antiderivative size = 316, normalized size of antiderivative = 1.22, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4374

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

(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*Tan[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]])
 

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]
 

Maple [A] (verified)

Time = 1.82 (sec) , antiderivative size = 300, normalized size of antiderivative = 1.16

method	result	size
parts	\(\frac {2 a^{2} e \left (-\frac {\left (e^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {e \tan \left (d x +c \right )+\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \tan \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}{e \tan \left (d x +c \right )-\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \tan \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {e \tan \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {e \tan \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 e^{4}}-\frac {1}{3 e^{2} \left (e \tan \left (d x +c \right )\right )^{\frac {3}{2}}}\right )}{d}-\frac {2 a^{2}}{3 d e \left (e \tan \left (d x +c \right )\right )^{\frac {3}{2}}}-\frac {2 a^{2} \left (\sqrt {-\csc \left (d x +c \right )+\cot \left (d x +c \right )}\, \sqrt {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}\, \sqrt {2 \cot \left (d x +c \right )-2 \csc \left (d x +c \right )+2}\, \left (1+\sec \left (d x +c \right )\right ) \operatorname {EllipticF}\left (\sqrt {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )+2 \csc \left (d x +c \right )\right )}{3 d \sqrt {e \tan \left (d x +c \right )}\, e^{2}}\)	\(300\)
default	\(-\frac {a^{2} \left (1+\cos \left (d x +c \right )\right ) \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 ) \sin \left (d x +c \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 ) \sin \left (d x +c \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 ) \sin \left (d x +c \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 ) \sin \left (d x +c \right )-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 )}\, \operatorname {EllipticF}\left (\sqrt {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right ) \sin \left (d x +c \right )+8 \cos \left (d x +c \right )\right ) \sec \left (d x +c \right ) \csc \left (d x +c \right )}{6 d \sqrt {e \tan \left (d x +c \right )}\, e^{2}}\)	\(500\)

Input:

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

Output:

2*a^2/d*e*(-1/8/e^4*(e^2)^(1/4)*2^(1/2)*(ln((e*tan(d*x+c)+(e^2)^(1/4)*(e*t 
an(d*x+c))^(1/2)*2^(1/2)+(e^2)^(1/2))/(e*tan(d*x+c)-(e^2)^(1/4)*(e*tan(d*x 
+c))^(1/2)*2^(1/2)+(e^2)^(1/2)))+2*arctan(2^(1/2)/(e^2)^(1/4)*(e*tan(d*x+c 
))^(1/2)+1)-2*arctan(-2^(1/2)/(e^2)^(1/4)*(e*tan(d*x+c))^(1/2)+1))-1/3/e^2 
/(e*tan(d*x+c))^(3/2))-2/3*a^2/d/e/(e*tan(d*x+c))^(3/2)-2/3*a^2/d/(e*tan(d 
*x+c))^(1/2)/e^2*((-csc(d*x+c)+cot(d*x+c))^(1/2)*(-cot(d*x+c)+csc(d*x+c)+1 
)^(1/2)*(2*cot(d*x+c)-2*csc(d*x+c)+2)^(1/2)*(1+sec(d*x+c))*EllipticF((-cot 
(d*x+c)+csc(d*x+c)+1)^(1/2),1/2*2^(1/2))+2*csc(d*x+c))
 

Fricas [F(-1)]

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

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

Output:

Timed out
 

Sympy [F]

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

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

Output:

a**2*(Integral((e*tan(c + d*x))**(-5/2), x) + Integral(2*sec(c + d*x)/(e*t 
an(c + d*x))**(5/2), x) + Integral(sec(c + d*x)**2/(e*tan(c + d*x))**(5/2) 
, x))
 

Maxima [F(-2)]

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

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

Output:

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(e>0)', see `assume?` for more de 
tails)Is e
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {(a+a \sec (c+d x))^2}{(e \tan (c+d x))^{5/2}} \, dx=\frac {\sqrt {e}\, a^{2} \left (\int \frac {\sqrt {\tan \left (d x +c \right )}}{\tan \left (d x +c \right )^{3}}d x +\int \frac {\sqrt {\tan \left (d x +c \right )}\, \sec \left (d x +c \right )^{2}}{\tan \left (d x +c \right )^{3}}d x +2 \left (\int \frac {\sqrt {\tan \left (d x +c \right )}\, \sec \left (d x +c \right )}{\tan \left (d x +c \right )^{3}}d x \right )\right )}{e^{3}} \] Input:

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

Output:

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