3.363 \(\int \frac{1}{(d \tan (e+f x))^{5/2} (a+a \tan (e+f x))} \, dx\)

Optimal. Leaf size=135 \[ \frac{\tan ^{-1}\left (\frac{\sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{a d^{5/2} f}+\frac{2}{a d^2 f \sqrt{d \tan (e+f x)}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{d} \tan (e+f x)+\sqrt{d}}{\sqrt{2} \sqrt{d \tan (e+f x)}}\right )}{\sqrt{2} a d^{5/2} f}-\frac{2}{3 a d f (d \tan (e+f x))^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.489828, antiderivative size = 135, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 10, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {3569, 3649, 12, 16, 3573, 3532, 208, 3634, 63, 205} \[ \frac{\tan ^{-1}\left (\frac{\sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{a d^{5/2} f}+\frac{2}{a d^2 f \sqrt{d \tan (e+f x)}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{d} \tan (e+f x)+\sqrt{d}}{\sqrt{2} \sqrt{d \tan (e+f x)}}\right )}{\sqrt{2} a d^{5/2} f}-\frac{2}{3 a d f (d \tan (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/((d*Tan[e + f*x])^(5/2)*(a + a*Tan[e + f*x])),x]

[Out]

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

Rule 3569

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

Rule 3649

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*t
an[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[((A*b^2 - a*(b*B - a*C))*(a + b*T
an[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n + 1))/(f*(m + 1)*(b*c - a*d)*(a^2 + b^2)), x] + Dist[1/((m + 1)*(
b*c - a*d)*(a^2 + b^2)), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[A*(a*(b*c - a*d)*(m + 1)
- b^2*d*(m + n + 2)) + (b*B - a*C)*(b*c*(m + 1) + a*d*(n + 1)) - (m + 1)*(b*c - a*d)*(A*b - a*B - b*C)*Tan[e
+ f*x] - d*(A*b^2 - a*(b*B - a*C))*(m + n + 2)*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C,
n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && LtQ[m, -1] &&  !(ILtQ[n, -1] && ( !I
ntegerQ[m] || (EqQ[c, 0] && NeQ[a, 0])))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 16

Int[(u_.)*(v_)^(m_.)*((b_)*(v_))^(n_), x_Symbol] :> Dist[1/b^m, Int[u*(b*v)^(m + n), x], x] /; FreeQ[{b, n}, x
] && IntegerQ[m]

Rule 3573

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(3/2)/((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[1
/(c^2 + d^2), Int[Simp[a^2*c - b^2*c + 2*a*b*d + (2*a*b*c - a^2*d + b^2*d)*Tan[e + f*x], x]/Sqrt[a + b*Tan[e +
f*x]], x], x] + Dist[(b*c - a*d)^2/(c^2 + d^2), Int[(1 + Tan[e + f*x]^2)/(Sqrt[a + b*Tan[e + f*x]]*(c + d*Tan
[e + f*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]

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 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 3634

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_.)*((A_) + (C_.)*
tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Dist[A/f, Subst[Int[(a + b*x)^m*(c + d*x)^n, x], x, Tan[e + f*x]], x]
/; FreeQ[{a, b, c, d, e, f, A, C, m, n}, x] && EqQ[A, C]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 205

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

Rubi steps

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

Mathematica [A]  time = 1.29216, size = 130, normalized size = 0.96 \[ \frac{12 \tan ^{-1}\left (\sqrt{\tan (e+f x)}\right ) \tan ^{\frac{3}{2}}(e+f x)+24 \tan (e+f x)+3 \sqrt{2} \tan ^{\frac{3}{2}}(e+f x) \left (\log \left (-\tan (e+f x)+\sqrt{2} \sqrt{\tan (e+f x)}-1\right )-\log \left (\tan (e+f x)+\sqrt{2} \sqrt{\tan (e+f x)}+1\right )\right )-8}{12 a d f (d \tan (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((d*Tan[e + f*x])^(5/2)*(a + a*Tan[e + f*x])),x]

[Out]

(-8 + 24*Tan[e + f*x] + 12*ArcTan[Sqrt[Tan[e + f*x]]]*Tan[e + f*x]^(3/2) + 3*Sqrt[2]*(Log[-1 + Sqrt[2]*Sqrt[Ta
n[e + f*x]] - Tan[e + f*x]] - Log[1 + Sqrt[2]*Sqrt[Tan[e + f*x]] + Tan[e + f*x]])*Tan[e + f*x]^(3/2))/(12*a*d*
f*(d*Tan[e + f*x])^(3/2))

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Maple [B]  time = 0.03, size = 415, normalized size = 3.1 \begin{align*} -{\frac{\sqrt{2}}{8\,fa{d}^{3}}\sqrt [4]{{d}^{2}}\ln \left ({ \left ( d\tan \left ( fx+e \right ) +\sqrt [4]{{d}^{2}}\sqrt{d\tan \left ( fx+e \right ) }\sqrt{2}+\sqrt{{d}^{2}} \right ) \left ( d\tan \left ( fx+e \right ) -\sqrt [4]{{d}^{2}}\sqrt{d\tan \left ( fx+e \right ) }\sqrt{2}+\sqrt{{d}^{2}} \right ) ^{-1}} \right ) }-{\frac{\sqrt{2}}{4\,fa{d}^{3}}\sqrt [4]{{d}^{2}}\arctan \left ({\sqrt{2}\sqrt{d\tan \left ( fx+e \right ) }{\frac{1}{\sqrt [4]{{d}^{2}}}}}+1 \right ) }+{\frac{\sqrt{2}}{4\,fa{d}^{3}}\sqrt [4]{{d}^{2}}\arctan \left ( -{\sqrt{2}\sqrt{d\tan \left ( fx+e \right ) }{\frac{1}{\sqrt [4]{{d}^{2}}}}}+1 \right ) }+{\frac{\sqrt{2}}{8\,fa{d}^{2}}\ln \left ({ \left ( d\tan \left ( fx+e \right ) -\sqrt [4]{{d}^{2}}\sqrt{d\tan \left ( fx+e \right ) }\sqrt{2}+\sqrt{{d}^{2}} \right ) \left ( d\tan \left ( fx+e \right ) +\sqrt [4]{{d}^{2}}\sqrt{d\tan \left ( fx+e \right ) }\sqrt{2}+\sqrt{{d}^{2}} \right ) ^{-1}} \right ){\frac{1}{\sqrt [4]{{d}^{2}}}}}+{\frac{\sqrt{2}}{4\,fa{d}^{2}}\arctan \left ({\sqrt{2}\sqrt{d\tan \left ( fx+e \right ) }{\frac{1}{\sqrt [4]{{d}^{2}}}}}+1 \right ){\frac{1}{\sqrt [4]{{d}^{2}}}}}-{\frac{\sqrt{2}}{4\,fa{d}^{2}}\arctan \left ( -{\sqrt{2}\sqrt{d\tan \left ( fx+e \right ) }{\frac{1}{\sqrt [4]{{d}^{2}}}}}+1 \right ){\frac{1}{\sqrt [4]{{d}^{2}}}}}-{\frac{2}{3\,adf} \left ( d\tan \left ( fx+e \right ) \right ) ^{-{\frac{3}{2}}}}+2\,{\frac{1}{fa{d}^{2}\sqrt{d\tan \left ( fx+e \right ) }}}+{\frac{1}{fa}\arctan \left ({\sqrt{d\tan \left ( fx+e \right ) }{\frac{1}{\sqrt{d}}}} \right ){d}^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8/f/a/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*tan
(f*x+e)-(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)*2^(1/2)+(d^2)^(1/2)))-1/4/f/a/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)+1/4/f/a/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)+1/8/f/a/d^2/(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*tan(f*x+e)+(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)*2^(1/2)+(d^2)^(1/2)))+1/4/f/a/d^2/(d^2)^(1/4)*2^(1/2)*arctan
(2^(1/2)/(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)+1)-1/4/f/a/d^2/(d^2)^(1/4)*2^(1/2)*arctan(-2^(1/2)/(d^2)^(1/4)*(d*ta
n(f*x+e))^(1/2)+1)-2/3/a/d/f/(d*tan(f*x+e))^(3/2)+2/a/d^2/f/(d*tan(f*x+e))^(1/2)+arctan((d*tan(f*x+e))^(1/2)/d
^(1/2))/a/d^(5/2)/f

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d*tan(f*x+e))^(5/2)/(a+a*tan(f*x+e)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.05209, size = 829, normalized size = 6.14 \begin{align*} \left [\frac{3 \, \sqrt{2} \sqrt{-d} \arctan \left (\frac{\sqrt{2} \sqrt{d \tan \left (f x + e\right )} \sqrt{-d}{\left (\tan \left (f x + e\right ) + 1\right )}}{2 \, d \tan \left (f x + e\right )}\right ) \tan \left (f x + e\right )^{2} - 3 \, \sqrt{-d} \log \left (\frac{d \tan \left (f x + e\right ) - 2 \, \sqrt{d \tan \left (f x + e\right )} \sqrt{-d} - d}{\tan \left (f x + e\right ) + 1}\right ) \tan \left (f x + e\right )^{2} + 4 \, \sqrt{d \tan \left (f x + e\right )}{\left (3 \, \tan \left (f x + e\right ) - 1\right )}}{6 \, a d^{3} f \tan \left (f x + e\right )^{2}}, \frac{3 \, \sqrt{2} \sqrt{d} \log \left (\frac{d \tan \left (f x + e\right )^{2} - 2 \, \sqrt{2} \sqrt{d \tan \left (f x + e\right )} \sqrt{d}{\left (\tan \left (f x + e\right ) + 1\right )} + 4 \, d \tan \left (f x + e\right ) + d}{\tan \left (f x + e\right )^{2} + 1}\right ) \tan \left (f x + e\right )^{2} + 12 \, \sqrt{d} \arctan \left (\frac{\sqrt{d \tan \left (f x + e\right )}}{\sqrt{d}}\right ) \tan \left (f x + e\right )^{2} + 8 \, \sqrt{d \tan \left (f x + e\right )}{\left (3 \, \tan \left (f x + e\right ) - 1\right )}}{12 \, a d^{3} f \tan \left (f x + e\right )^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d*tan(f*x+e))^(5/2)/(a+a*tan(f*x+e)),x, algorithm="fricas")

[Out]

[1/6*(3*sqrt(2)*sqrt(-d)*arctan(1/2*sqrt(2)*sqrt(d*tan(f*x + e))*sqrt(-d)*(tan(f*x + e) + 1)/(d*tan(f*x + e)))
*tan(f*x + e)^2 - 3*sqrt(-d)*log((d*tan(f*x + e) - 2*sqrt(d*tan(f*x + e))*sqrt(-d) - d)/(tan(f*x + e) + 1))*ta
n(f*x + e)^2 + 4*sqrt(d*tan(f*x + e))*(3*tan(f*x + e) - 1))/(a*d^3*f*tan(f*x + e)^2), 1/12*(3*sqrt(2)*sqrt(d)*
log((d*tan(f*x + e)^2 - 2*sqrt(2)*sqrt(d*tan(f*x + e))*sqrt(d)*(tan(f*x + e) + 1) + 4*d*tan(f*x + e) + d)/(tan
(f*x + e)^2 + 1))*tan(f*x + e)^2 + 12*sqrt(d)*arctan(sqrt(d*tan(f*x + e))/sqrt(d))*tan(f*x + e)^2 + 8*sqrt(d*t
an(f*x + e))*(3*tan(f*x + e) - 1))/(a*d^3*f*tan(f*x + e)^2)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{1}{\left (d \tan{\left (e + f x \right )}\right )^{\frac{5}{2}} \tan{\left (e + f x \right )} + \left (d \tan{\left (e + f x \right )}\right )^{\frac{5}{2}}}\, dx}{a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d*tan(f*x+e))**(5/2)/(a+a*tan(f*x+e)),x)

[Out]

Integral(1/((d*tan(e + f*x))**(5/2)*tan(e + f*x) + (d*tan(e + f*x))**(5/2)), x)/a

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Giac [B]  time = 1.3756, size = 420, normalized size = 3.11 \begin{align*} -\frac{1}{24} \, d^{2}{\left (\frac{6 \, \sqrt{2}{\left (d \sqrt{{\left | d \right |}} -{\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 )}{a d^{6} f} + \frac{6 \, \sqrt{2}{\left (d \sqrt{{\left | d \right |}} -{\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 )}{a d^{6} f} - \frac{24 \, \arctan \left (\frac{\sqrt{d \tan \left (f x + e\right )}}{\sqrt{d}}\right )}{a d^{\frac{9}{2}} f} + \frac{3 \, \sqrt{2}{\left (d \sqrt{{\left | d \right |}} +{\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 )}{a d^{6} f} - \frac{3 \, \sqrt{2}{\left (d \sqrt{{\left | d \right |}} +{\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 )}{a d^{6} f} - \frac{16 \,{\left (3 \, d \tan \left (f x + e\right ) - d\right )}}{\sqrt{d \tan \left (f x + e\right )} a d^{5} f \tan \left (f x + e\right )}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d*tan(f*x+e))^(5/2)/(a+a*tan(f*x+e)),x, algorithm="giac")

[Out]

-1/24*d^2*(6*sqrt(2)*(d*sqrt(abs(d)) - abs(d)^(3/2))*arctan(1/2*sqrt(2)*(sqrt(2)*sqrt(abs(d)) + 2*sqrt(d*tan(f
*x + e)))/sqrt(abs(d)))/(a*d^6*f) + 6*sqrt(2)*(d*sqrt(abs(d)) - abs(d)^(3/2))*arctan(-1/2*sqrt(2)*(sqrt(2)*sqr
t(abs(d)) - 2*sqrt(d*tan(f*x + e)))/sqrt(abs(d)))/(a*d^6*f) - 24*arctan(sqrt(d*tan(f*x + e))/sqrt(d))/(a*d^(9/
2)*f) + 3*sqrt(2)*(d*sqrt(abs(d)) + abs(d)^(3/2))*log(d*tan(f*x + e) + sqrt(2)*sqrt(d*tan(f*x + e))*sqrt(abs(d
)) + abs(d))/(a*d^6*f) - 3*sqrt(2)*(d*sqrt(abs(d)) + abs(d)^(3/2))*log(d*tan(f*x + e) - sqrt(2)*sqrt(d*tan(f*x
+ e))*sqrt(abs(d)) + abs(d))/(a*d^6*f) - 16*(3*d*tan(f*x + e) - d)/(sqrt(d*tan(f*x + e))*a*d^5*f*tan(f*x + e)
))