3.230 \(\int \frac{1}{\sqrt{\tan (c+d x)} (a+i a \tan (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=162 \[ \frac{67 \sqrt{\tan (c+d x)}}{60 a^2 d \sqrt{a+i a \tan (c+d x)}}+\frac{\left (\frac{1}{8}-\frac{i}{8}\right ) \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{a^{5/2} d}+\frac{13 \sqrt{\tan (c+d x)}}{30 a d (a+i a \tan (c+d x))^{3/2}}+\frac{\sqrt{\tan (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}} \]

[Out]

((1/8 - I/8)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/(a^(5/2)*d) + Sqrt[Tan[
c + d*x]]/(5*d*(a + I*a*Tan[c + d*x])^(5/2)) + (13*Sqrt[Tan[c + d*x]])/(30*a*d*(a + I*a*Tan[c + d*x])^(3/2)) +
(67*Sqrt[Tan[c + d*x]])/(60*a^2*d*Sqrt[a + I*a*Tan[c + d*x]])

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Rubi [A]  time = 0.439259, antiderivative size = 162, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.179, Rules used = {3559, 3596, 12, 3544, 205} \[ \frac{67 \sqrt{\tan (c+d x)}}{60 a^2 d \sqrt{a+i a \tan (c+d x)}}+\frac{\left (\frac{1}{8}-\frac{i}{8}\right ) \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{a^{5/2} d}+\frac{13 \sqrt{\tan (c+d x)}}{30 a d (a+i a \tan (c+d x))^{3/2}}+\frac{\sqrt{\tan (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[Tan[c + d*x]]*(a + I*a*Tan[c + d*x])^(5/2)),x]

[Out]

((1/8 - I/8)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/(a^(5/2)*d) + Sqrt[Tan[
c + d*x]]/(5*d*(a + I*a*Tan[c + d*x])^(5/2)) + (13*Sqrt[Tan[c + d*x]])/(30*a*d*(a + I*a*Tan[c + d*x])^(3/2)) +
(67*Sqrt[Tan[c + d*x]])/(60*a^2*d*Sqrt[a + I*a*Tan[c + d*x]])

Rule 3559

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

Rule 3596

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[((a*A + b*B)*(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n + 1))/(2
*f*m*(b*c - a*d)), x] + Dist[1/(2*a*m*(b*c - a*d)), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Si
mp[A*(b*c*m - a*d*(2*m + n + 1)) + B*(a*c*m - b*d*(n + 1)) + d*(A*b - a*B)*(m + n + 1)*Tan[e + f*x], x], x], x
] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && LtQ[m, 0] &&  !GtQ[n,
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 3544

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

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}{\sqrt{\tan (c+d x)} (a+i a \tan (c+d x))^{5/2}} \, dx &=\frac{\sqrt{\tan (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac{\int \frac{\frac{9 a}{2}-2 i a \tan (c+d x)}{\sqrt{\tan (c+d x)} (a+i a \tan (c+d x))^{3/2}} \, dx}{5 a^2}\\ &=\frac{\sqrt{\tan (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac{13 \sqrt{\tan (c+d x)}}{30 a d (a+i a \tan (c+d x))^{3/2}}+\frac{\int \frac{\frac{41 a^2}{4}-\frac{13}{2} i a^2 \tan (c+d x)}{\sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}} \, dx}{15 a^4}\\ &=\frac{\sqrt{\tan (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac{13 \sqrt{\tan (c+d x)}}{30 a d (a+i a \tan (c+d x))^{3/2}}+\frac{67 \sqrt{\tan (c+d x)}}{60 a^2 d \sqrt{a+i a \tan (c+d x)}}+\frac{\int \frac{15 a^3 \sqrt{a+i a \tan (c+d x)}}{8 \sqrt{\tan (c+d x)}} \, dx}{15 a^6}\\ &=\frac{\sqrt{\tan (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac{13 \sqrt{\tan (c+d x)}}{30 a d (a+i a \tan (c+d x))^{3/2}}+\frac{67 \sqrt{\tan (c+d x)}}{60 a^2 d \sqrt{a+i a \tan (c+d x)}}+\frac{\int \frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{\tan (c+d x)}} \, dx}{8 a^3}\\ &=\frac{\sqrt{\tan (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac{13 \sqrt{\tan (c+d x)}}{30 a d (a+i a \tan (c+d x))^{3/2}}+\frac{67 \sqrt{\tan (c+d x)}}{60 a^2 d \sqrt{a+i a \tan (c+d x)}}-\frac{i \operatorname{Subst}\left (\int \frac{1}{-i a-2 a^2 x^2} \, dx,x,\frac{\sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{4 a d}\\ &=\frac{\left (\frac{1}{8}-\frac{i}{8}\right ) \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{a^{5/2} d}+\frac{\sqrt{\tan (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac{13 \sqrt{\tan (c+d x)}}{30 a d (a+i a \tan (c+d x))^{3/2}}+\frac{67 \sqrt{\tan (c+d x)}}{60 a^2 d \sqrt{a+i a \tan (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 2.02125, size = 171, normalized size = 1.06 \[ -\frac{i e^{-6 i (c+d x)} \sqrt{\frac{a e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}} \left (-16 e^{2 i (c+d x)}-64 e^{4 i (c+d x)}+83 e^{6 i (c+d x)}+15 e^{5 i (c+d x)} \sqrt{-1+e^{2 i (c+d x)}} \tanh ^{-1}\left (\frac{e^{i (c+d x)}}{\sqrt{-1+e^{2 i (c+d x)}}}\right )-3\right )}{60 \sqrt{2} a^3 d \sqrt{\tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(Sqrt[Tan[c + d*x]]*(a + I*a*Tan[c + d*x])^(5/2)),x]

[Out]

((-I/60)*Sqrt[(a*E^((2*I)*(c + d*x)))/(1 + E^((2*I)*(c + d*x)))]*(-3 - 16*E^((2*I)*(c + d*x)) - 64*E^((4*I)*(c
+ d*x)) + 83*E^((6*I)*(c + d*x)) + 15*E^((5*I)*(c + d*x))*Sqrt[-1 + E^((2*I)*(c + d*x))]*ArcTanh[E^(I*(c + d*
x))/Sqrt[-1 + E^((2*I)*(c + d*x))]]))/(Sqrt[2]*a^3*d*E^((6*I)*(c + d*x))*Sqrt[Tan[c + d*x]])

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Maple [B]  time = 0.057, size = 576, normalized size = 3.6 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/240/d*tan(d*x+c)^(1/2)*(a*(1+I*tan(d*x+c)))^(1/2)*(15*I*2^(1/2)*ln(-(-2*2^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*
(1+I*tan(d*x+c)))^(1/2)+I*a-3*a*tan(d*x+c))/(tan(d*x+c)+I))*tan(d*x+c)^4*a-90*I*2^(1/2)*ln(-(-2*2^(1/2)*(-I*a)
^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)+I*a-3*a*tan(d*x+c))/(tan(d*x+c)+I))*tan(d*x+c)^2*a+268*I*(a*tan(d
*x+c)*(1+I*tan(d*x+c)))^(1/2)*(-I*a)^(1/2)*tan(d*x+c)^3+60*2^(1/2)*ln(-(-2*2^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*
(1+I*tan(d*x+c)))^(1/2)+I*a-3*a*tan(d*x+c))/(tan(d*x+c)+I))*tan(d*x+c)^3*a+15*I*2^(1/2)*ln(-(-2*2^(1/2)*(-I*a)
^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)+I*a-3*a*tan(d*x+c))/(tan(d*x+c)+I))*a-1060*I*(a*tan(d*x+c)*(1+I*t
an(d*x+c)))^(1/2)*(-I*a)^(1/2)*tan(d*x+c)-60*2^(1/2)*ln(-(-2*2^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c
)))^(1/2)+I*a-3*a*tan(d*x+c))/(tan(d*x+c)+I))*tan(d*x+c)*a+908*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(-I*a)^(1
/2)*tan(d*x+c)^2-420*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(-I*a)^(1/2))/a^3/(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(
1/2)/(-tan(d*x+c)+I)^4/(-I*a)^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError

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Fricas [B]  time = 3.003, size = 1077, normalized size = 6.65 \begin{align*} \frac{{\left (30 \, a^{3} d \sqrt{-\frac{i}{8 \, a^{5} d^{2}}} e^{\left (6 i \, d x + 6 i \, c\right )} \log \left (\frac{1}{4} \,{\left (4 \, a^{3} d \sqrt{-\frac{i}{8 \, a^{5} d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} + \sqrt{2} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt{\frac{-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}{\left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) - 30 \, a^{3} d \sqrt{-\frac{i}{8 \, a^{5} d^{2}}} e^{\left (6 i \, d x + 6 i \, c\right )} \log \left (-\frac{1}{4} \,{\left (4 \, a^{3} d \sqrt{-\frac{i}{8 \, a^{5} d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} - \sqrt{2} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt{\frac{-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}{\left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) + \sqrt{2} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt{\frac{-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}{\left (83 \, e^{\left (6 i \, d x + 6 i \, c\right )} + 102 \, e^{\left (4 i \, d x + 4 i \, c\right )} + 22 \, e^{\left (2 i \, d x + 2 i \, c\right )} + 3\right )} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-6 i \, d x - 6 i \, c\right )}}{120 \, a^{3} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/120*(30*a^3*d*sqrt(-1/8*I/(a^5*d^2))*e^(6*I*d*x + 6*I*c)*log(1/4*(4*a^3*d*sqrt(-1/8*I/(a^5*d^2))*e^(2*I*d*x
+ 2*I*c) + sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) +
1))*(e^(2*I*d*x + 2*I*c) + 1)*e^(I*d*x + I*c))*e^(-I*d*x - I*c)) - 30*a^3*d*sqrt(-1/8*I/(a^5*d^2))*e^(6*I*d*x
+ 6*I*c)*log(-1/4*(4*a^3*d*sqrt(-1/8*I/(a^5*d^2))*e^(2*I*d*x + 2*I*c) - sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) +
1))*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1))*(e^(2*I*d*x + 2*I*c) + 1)*e^(I*d*x + I*c))*e^
(-I*d*x - I*c)) + sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*
I*c) + 1))*(83*e^(6*I*d*x + 6*I*c) + 102*e^(4*I*d*x + 4*I*c) + 22*e^(2*I*d*x + 2*I*c) + 3)*e^(I*d*x + I*c))*e^
(-6*I*d*x - 6*I*c)/(a^3*d)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/tan(d*x+c)**(1/2)/(a+I*a*tan(d*x+c))**(5/2),x)

[Out]

Timed out

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Giac [A]  time = 1.29846, size = 123, normalized size = 0.76 \begin{align*} -\frac{2 \, \sqrt{-2 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )} a + 2 \, a^{2}} a \log \left (\sqrt{i \, a \tan \left (d x + c\right ) + a}\right )}{-\left (i - 1\right ) \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{5} + \left (3 i - 3\right ) \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{4} a - \left (2 i - 2\right ) \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3} a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*sqrt(-2*(I*a*tan(d*x + c) + a)*a + 2*a^2)*a*log(sqrt(I*a*tan(d*x + c) + a))/(-(I - 1)*(I*a*tan(d*x + c) + a
)^5 + (3*I - 3)*(I*a*tan(d*x + c) + a)^4*a - (2*I - 2)*(I*a*tan(d*x + c) + a)^3*a^2)