3.355 \(\int \frac {(a+a \tan (e+f x))^3}{(d \tan (e+f x))^{5/2}} \, dx\)

Optimal. Leaf size=117 \[ \frac {2 \sqrt {2} a^3 \tanh ^{-1}\left (\frac {\sqrt {d} \tan (e+f x)+\sqrt {d}}{\sqrt {2} \sqrt {d \tan (e+f x)}}\right )}{d^{5/2} f}-\frac {16 a^3}{3 d^2 f \sqrt {d \tan (e+f x)}}-\frac {2 \left (a^3 \tan (e+f x)+a^3\right )}{3 d f (d \tan (e+f x))^{3/2}} \]

[Out]

2*a^3*arctanh(1/2*(d^(1/2)+d^(1/2)*tan(f*x+e))*2^(1/2)/(d*tan(f*x+e))^(1/2))*2^(1/2)/d^(5/2)/f-16/3*a^3/d^2/f/
(d*tan(f*x+e))^(1/2)-2/3*(a^3+a^3*tan(f*x+e))/d/f/(d*tan(f*x+e))^(3/2)

________________________________________________________________________________________

Rubi [A]  time = 0.18, antiderivative size = 117, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {3565, 3628, 3532, 208} \[ -\frac {16 a^3}{3 d^2 f \sqrt {d \tan (e+f x)}}+\frac {2 \sqrt {2} a^3 \tanh ^{-1}\left (\frac {\sqrt {d} \tan (e+f x)+\sqrt {d}}{\sqrt {2} \sqrt {d \tan (e+f x)}}\right )}{d^{5/2} f}-\frac {2 \left (a^3 \tan (e+f x)+a^3\right )}{3 d f (d \tan (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Tan[e + f*x])^3/(d*Tan[e + f*x])^(5/2),x]

[Out]

(2*Sqrt[2]*a^3*ArcTanh[(Sqrt[d] + Sqrt[d]*Tan[e + f*x])/(Sqrt[2]*Sqrt[d*Tan[e + f*x]])])/(d^(5/2)*f) - (16*a^3
)/(3*d^2*f*Sqrt[d*Tan[e + f*x]]) - (2*(a^3 + a^3*Tan[e + f*x]))/(3*d*f*(d*Tan[e + f*x])^(3/2))

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rule 3532

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[(-2*d^2)/f,
Subst[Int[1/(2*c*d + b*x^2), x], x, (c - d*Tan[e + f*x])/Sqrt[b*Tan[e + f*x]]], x] /; FreeQ[{b, c, d, e, f}, x
] && EqQ[c^2 - d^2, 0]

Rule 3565

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[((b*c - a*d)^2*(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^(n + 1))/(d*f*(n + 1)*(c^2 + d^2)), x] - D
ist[1/(d*(n + 1)*(c^2 + d^2)), Int[(a + b*Tan[e + f*x])^(m - 3)*(c + d*Tan[e + f*x])^(n + 1)*Simp[a^2*d*(b*d*(
m - 2) - a*c*(n + 1)) + b*(b*c - 2*a*d)*(b*c*(m - 2) + a*d*(n + 1)) - d*(n + 1)*(3*a^2*b*c - b^3*c - a^3*d + 3
*a*b^2*d)*Tan[e + f*x] - b*(a*d*(2*b*c - a*d)*(m + n - 1) - b^2*(c^2*(m - 2) - d^2*(n + 1)))*Tan[e + f*x]^2, x
], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && Gt
Q[m, 2] && LtQ[n, -1] && IntegerQ[2*m]

Rule 3628

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f
_.)*(x_)]^2), x_Symbol] :> Simp[((A*b^2 - a*b*B + a^2*C)*(a + b*Tan[e + f*x])^(m + 1))/(b*f*(m + 1)*(a^2 + b^2
)), x] + Dist[1/(a^2 + b^2), Int[(a + b*Tan[e + f*x])^(m + 1)*Simp[b*B + a*(A - C) - (A*b - a*B - b*C)*Tan[e +
f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x] && NeQ[A*b^2 - a*b*B + a^2*C, 0] && LtQ[m, -1] && NeQ[a^2
+ b^2, 0]

Rubi steps

\begin {align*} \int \frac {(a+a \tan (e+f x))^3}{(d \tan (e+f x))^{5/2}} \, dx &=-\frac {2 \left (a^3+a^3 \tan (e+f x)\right )}{3 d f (d \tan (e+f x))^{3/2}}+\frac {2 \int \frac {4 a^3 d^2+3 a^3 d^2 \tan (e+f x)+a^3 d^2 \tan ^2(e+f x)}{(d \tan (e+f x))^{3/2}} \, dx}{3 d^3}\\ &=-\frac {16 a^3}{3 d^2 f \sqrt {d \tan (e+f x)}}-\frac {2 \left (a^3+a^3 \tan (e+f x)\right )}{3 d f (d \tan (e+f x))^{3/2}}+\frac {2 \int \frac {3 a^3 d^3-3 a^3 d^3 \tan (e+f x)}{\sqrt {d \tan (e+f x)}} \, dx}{3 d^5}\\ &=-\frac {16 a^3}{3 d^2 f \sqrt {d \tan (e+f x)}}-\frac {2 \left (a^3+a^3 \tan (e+f x)\right )}{3 d f (d \tan (e+f x))^{3/2}}-\frac {\left (12 a^6 d\right ) \operatorname {Subst}\left (\int \frac {1}{-18 a^6 d^6+d x^2} \, dx,x,\frac {3 a^3 d^3+3 a^3 d^3 \tan (e+f x)}{\sqrt {d \tan (e+f x)}}\right )}{f}\\ &=\frac {2 \sqrt {2} a^3 \tanh ^{-1}\left (\frac {\sqrt {d}+\sqrt {d} \tan (e+f x)}{\sqrt {2} \sqrt {d \tan (e+f x)}}\right )}{d^{5/2} f}-\frac {16 a^3}{3 d^2 f \sqrt {d \tan (e+f x)}}-\frac {2 \left (a^3+a^3 \tan (e+f x)\right )}{3 d f (d \tan (e+f x))^{3/2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 4.03, size = 272, normalized size = 2.32 \[ \frac {a^3 (\tan (e+f x)+1)^3 \left (8 \sin ^3(e+f x) \tan (e+f x) \, _2F_1\left (\frac {3}{4},1;\frac {7}{4};-\tan ^2(e+f x)\right )-8 \sin (e+f x) \cos ^2(e+f x) \, _2F_1\left (-\frac {3}{4},1;\frac {1}{4};-\tan ^2(e+f x)\right )-72 \sin ^2(e+f x) \cos (e+f x) \, _2F_1\left (-\frac {1}{4},1;\frac {3}{4};-\tan ^2(e+f x)\right )-9 \sqrt {2} \cos ^3(e+f x) \tan ^{\frac {5}{2}}(e+f x) \left (2 \tan ^{-1}\left (1-\sqrt {2} \sqrt {\tan (e+f x)}\right )-2 \tan ^{-1}\left (\sqrt {2} \sqrt {\tan (e+f x)}+1\right )+\log \left (\tan (e+f x)-\sqrt {2} \sqrt {\tan (e+f x)}+1\right )-\log \left (\tan (e+f x)+\sqrt {2} \sqrt {\tan (e+f x)}+1\right )\right )\right )}{12 f (d \tan (e+f x))^{5/2} (\sin (e+f x)+\cos (e+f x))^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Tan[e + f*x])^3/(d*Tan[e + f*x])^(5/2),x]

[Out]

(a^3*(1 + Tan[e + f*x])^3*(-8*Cos[e + f*x]^2*Hypergeometric2F1[-3/4, 1, 1/4, -Tan[e + f*x]^2]*Sin[e + f*x] - 7
2*Cos[e + f*x]*Hypergeometric2F1[-1/4, 1, 3/4, -Tan[e + f*x]^2]*Sin[e + f*x]^2 + 8*Hypergeometric2F1[3/4, 1, 7
/4, -Tan[e + f*x]^2]*Sin[e + f*x]^3*Tan[e + f*x] - 9*Sqrt[2]*Cos[e + f*x]^3*(2*ArcTan[1 - Sqrt[2]*Sqrt[Tan[e +
f*x]]] - 2*ArcTan[1 + Sqrt[2]*Sqrt[Tan[e + f*x]]] + Log[1 - Sqrt[2]*Sqrt[Tan[e + f*x]] + Tan[e + f*x]] - Log[
1 + Sqrt[2]*Sqrt[Tan[e + f*x]] + Tan[e + f*x]])*Tan[e + f*x]^(5/2)))/(12*f*(Cos[e + f*x] + Sin[e + f*x])^3*(d*
Tan[e + f*x])^(5/2))

________________________________________________________________________________________

fricas [A]  time = 0.52, size = 228, normalized size = 1.95 \[ \left [\frac {3 \, \sqrt {2} a^{3} \sqrt {d} \log \left (\frac {\tan \left (f x + e\right )^{2} + \frac {2 \, \sqrt {2} \sqrt {d \tan \left (f x + e\right )} {\left (\tan \left (f x + e\right ) + 1\right )}}{\sqrt {d}} + 4 \, \tan \left (f x + e\right ) + 1}{\tan \left (f x + e\right )^{2} + 1}\right ) \tan \left (f x + e\right )^{2} - 2 \, {\left (9 \, a^{3} \tan \left (f x + e\right ) + a^{3}\right )} \sqrt {d \tan \left (f x + e\right )}}{3 \, d^{3} f \tan \left (f x + e\right )^{2}}, -\frac {2 \, {\left (3 \, \sqrt {2} a^{3} d \sqrt {-\frac {1}{d}} \arctan \left (\frac {\sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {-\frac {1}{d}} {\left (\tan \left (f x + e\right ) + 1\right )}}{2 \, \tan \left (f x + e\right )}\right ) \tan \left (f x + e\right )^{2} + {\left (9 \, a^{3} \tan \left (f x + e\right ) + a^{3}\right )} \sqrt {d \tan \left (f x + e\right )}\right )}}{3 \, d^{3} f \tan \left (f x + e\right )^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*tan(f*x+e))^3/(d*tan(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

[1/3*(3*sqrt(2)*a^3*sqrt(d)*log((tan(f*x + e)^2 + 2*sqrt(2)*sqrt(d*tan(f*x + e))*(tan(f*x + e) + 1)/sqrt(d) +
4*tan(f*x + e) + 1)/(tan(f*x + e)^2 + 1))*tan(f*x + e)^2 - 2*(9*a^3*tan(f*x + e) + a^3)*sqrt(d*tan(f*x + e)))/
(d^3*f*tan(f*x + e)^2), -2/3*(3*sqrt(2)*a^3*d*sqrt(-1/d)*arctan(1/2*sqrt(2)*sqrt(d*tan(f*x + e))*sqrt(-1/d)*(t
an(f*x + e) + 1)/tan(f*x + e))*tan(f*x + e)^2 + (9*a^3*tan(f*x + e) + a^3)*sqrt(d*tan(f*x + e)))/(d^3*f*tan(f*
x + e)^2)]

________________________________________________________________________________________

giac [B]  time = 1.59, size = 299, normalized size = 2.56 \[ \frac {\sqrt {2} {\left (a^{3} d \sqrt {{\left | d \right |}} + a^{3} {\left | d \right |}^{\frac {3}{2}}\right )} \log \left (d \tan \left (f x + e\right ) + \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {{\left | d \right |}} + {\left | d \right |}\right )}{2 \, d^{4} f} - \frac {\sqrt {2} {\left (a^{3} d \sqrt {{\left | d \right |}} + a^{3} {\left | d \right |}^{\frac {3}{2}}\right )} \log \left (d \tan \left (f x + e\right ) - \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {{\left | d \right |}} + {\left | d \right |}\right )}{2 \, d^{4} f} + \frac {{\left (\sqrt {2} a^{3} d \sqrt {{\left | d \right |}} - \sqrt {2} a^{3} {\left | d \right |}^{\frac {3}{2}}\right )} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {{\left | d \right |}} + 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {{\left | d \right |}}}\right )}{d^{4} f} + \frac {{\left (\sqrt {2} a^{3} d \sqrt {{\left | d \right |}} - \sqrt {2} a^{3} {\left | d \right |}^{\frac {3}{2}}\right )} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {{\left | d \right |}} - 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {{\left | d \right |}}}\right )}{d^{4} f} - \frac {2 \, {\left (9 \, a^{3} d \tan \left (f x + e\right ) + a^{3} d\right )}}{3 \, \sqrt {d \tan \left (f x + e\right )} d^{3} f \tan \left (f x + e\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*tan(f*x+e))^3/(d*tan(f*x+e))^(5/2),x, algorithm="giac")

[Out]

1/2*sqrt(2)*(a^3*d*sqrt(abs(d)) + a^3*abs(d)^(3/2))*log(d*tan(f*x + e) + sqrt(2)*sqrt(d*tan(f*x + e))*sqrt(abs
(d)) + abs(d))/(d^4*f) - 1/2*sqrt(2)*(a^3*d*sqrt(abs(d)) + a^3*abs(d)^(3/2))*log(d*tan(f*x + e) - sqrt(2)*sqrt
(d*tan(f*x + e))*sqrt(abs(d)) + abs(d))/(d^4*f) + (sqrt(2)*a^3*d*sqrt(abs(d)) - sqrt(2)*a^3*abs(d)^(3/2))*arct
an(1/2*sqrt(2)*(sqrt(2)*sqrt(abs(d)) + 2*sqrt(d*tan(f*x + e)))/sqrt(abs(d)))/(d^4*f) + (sqrt(2)*a^3*d*sqrt(abs
(d)) - sqrt(2)*a^3*abs(d)^(3/2))*arctan(-1/2*sqrt(2)*(sqrt(2)*sqrt(abs(d)) - 2*sqrt(d*tan(f*x + e)))/sqrt(abs(
d)))/(d^4*f) - 2/3*(9*a^3*d*tan(f*x + e) + a^3*d)/(sqrt(d*tan(f*x + e))*d^3*f*tan(f*x + e))

________________________________________________________________________________________

maple [B]  time = 0.20, size = 388, normalized size = 3.32 \[ \frac {a^{3} \left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )}{2 f \,d^{3}}+\frac {a^{3} \left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )}{f \,d^{3}}-\frac {a^{3} \left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )}{f \,d^{3}}-\frac {a^{3} \sqrt {2}\, \ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )}{2 f \,d^{2} \left (d^{2}\right )^{\frac {1}{4}}}-\frac {a^{3} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )}{f \,d^{2} \left (d^{2}\right )^{\frac {1}{4}}}+\frac {a^{3} \sqrt {2}\, \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )}{f \,d^{2} \left (d^{2}\right )^{\frac {1}{4}}}-\frac {2 a^{3}}{3 f d \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}-\frac {6 a^{3}}{d^{2} f \sqrt {d \tan \left (f x +e \right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*tan(f*x+e))^3/(d*tan(f*x+e))^(5/2),x)

[Out]

1/2/f*a^3/d^3*(d^2)^(1/4)*2^(1/2)*ln((d*tan(f*x+e)+(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)*2^(1/2)+(d^2)^(1/2))/(d*ta
n(f*x+e)-(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)*2^(1/2)+(d^2)^(1/2)))+1/f*a^3/d^3*(d^2)^(1/4)*2^(1/2)*arctan(2^(1/2)
/(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)+1)-1/f*a^3/d^3*(d^2)^(1/4)*2^(1/2)*arctan(-2^(1/2)/(d^2)^(1/4)*(d*tan(f*x+e)
)^(1/2)+1)-1/2/f*a^3/d^2*2^(1/2)/(d^2)^(1/4)*ln((d*tan(f*x+e)-(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)*2^(1/2)+(d^2)^(
1/2))/(d*tan(f*x+e)+(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)*2^(1/2)+(d^2)^(1/2)))-1/f*a^3/d^2*2^(1/2)/(d^2)^(1/4)*arc
tan(2^(1/2)/(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)+1)+1/f*a^3/d^2*2^(1/2)/(d^2)^(1/4)*arctan(-2^(1/2)/(d^2)^(1/4)*(d
*tan(f*x+e))^(1/2)+1)-2/3/f*a^3/d/(d*tan(f*x+e))^(3/2)-6*a^3/d^2/f/(d*tan(f*x+e))^(1/2)

________________________________________________________________________________________

maxima [A]  time = 0.89, size = 123, normalized size = 1.05 \[ \frac {\frac {3 \, a^{3} {\left (\frac {\sqrt {2} \log \left (d \tan \left (f x + e\right ) + \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right )}{\sqrt {d}} - \frac {\sqrt {2} \log \left (d \tan \left (f x + e\right ) - \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right )}{\sqrt {d}}\right )}}{d} - \frac {2 \, {\left (9 \, a^{3} d \tan \left (f x + e\right ) + a^{3} d\right )}}{\left (d \tan \left (f x + e\right )\right )^{\frac {3}{2}} d}}{3 \, d f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*tan(f*x+e))^3/(d*tan(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

1/3*(3*a^3*(sqrt(2)*log(d*tan(f*x + e) + sqrt(2)*sqrt(d*tan(f*x + e))*sqrt(d) + d)/sqrt(d) - sqrt(2)*log(d*tan
(f*x + e) - sqrt(2)*sqrt(d*tan(f*x + e))*sqrt(d) + d)/sqrt(d))/d - 2*(9*a^3*d*tan(f*x + e) + a^3*d)/((d*tan(f*
x + e))^(3/2)*d))/(d*f)

________________________________________________________________________________________

mupad [B]  time = 4.32, size = 102, normalized size = 0.87 \[ \frac {2\,\sqrt {2}\,a^3\,\mathrm {atanh}\left (\frac {32\,\sqrt {2}\,a^6\,d^{5/2}\,f\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{32\,a^6\,d^3\,f+32\,a^6\,d^3\,f\,\mathrm {tan}\left (e+f\,x\right )}\right )}{d^{5/2}\,f}-\frac {\frac {2\,a^3\,d}{3}+6\,a^3\,d\,\mathrm {tan}\left (e+f\,x\right )}{d^2\,f\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*tan(e + f*x))^3/(d*tan(e + f*x))^(5/2),x)

[Out]

(2*2^(1/2)*a^3*atanh((32*2^(1/2)*a^6*d^(5/2)*f*(d*tan(e + f*x))^(1/2))/(32*a^6*d^3*f + 32*a^6*d^3*f*tan(e + f*
x))))/(d^(5/2)*f) - ((2*a^3*d)/3 + 6*a^3*d*tan(e + f*x))/(d^2*f*(d*tan(e + f*x))^(3/2))

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ a^{3} \left (\int \frac {1}{\left (d \tan {\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx + \int \frac {3 \tan {\left (e + f x \right )}}{\left (d \tan {\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx + \int \frac {3 \tan ^{2}{\left (e + f x \right )}}{\left (d \tan {\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx + \int \frac {\tan ^{3}{\left (e + f x \right )}}{\left (d \tan {\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*tan(f*x+e))**3/(d*tan(f*x+e))**(5/2),x)

[Out]

a**3*(Integral((d*tan(e + f*x))**(-5/2), x) + Integral(3*tan(e + f*x)/(d*tan(e + f*x))**(5/2), x) + Integral(3
*tan(e + f*x)**2/(d*tan(e + f*x))**(5/2), x) + Integral(tan(e + f*x)**3/(d*tan(e + f*x))**(5/2), x))

________________________________________________________________________________________