3.231 \(\int \frac{1}{\tan ^{\frac{3}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=199 \[ -\frac{317 \sqrt{a+i a \tan (c+d x)}}{60 a^3 d \sqrt{\tan (c+d x)}}+\frac{151}{60 a^2 d \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}+\frac{\left (\frac{1}{8}+\frac{i}{8}\right ) \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{a^{5/2} d}+\frac{17}{30 a d \sqrt{\tan (c+d x)} (a+i a \tan (c+d x))^{3/2}}+\frac{1}{5 d \sqrt{\tan (c+d x)} (a+i a \tan (c+d x))^{5/2}} \]

[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) + 1/(5*d*Sq
rt[Tan[c + d*x]]*(a + I*a*Tan[c + d*x])^(5/2)) + 17/(30*a*d*Sqrt[Tan[c + d*x]]*(a + I*a*Tan[c + d*x])^(3/2)) +
151/(60*a^2*d*Sqrt[Tan[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]]) - (317*Sqrt[a + I*a*Tan[c + d*x]])/(60*a^3*d*Sqr
t[Tan[c + d*x]])

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Rubi [A]  time = 0.596893, antiderivative size = 199, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.214, Rules used = {3559, 3596, 3598, 12, 3544, 205} \[ -\frac{317 \sqrt{a+i a \tan (c+d x)}}{60 a^3 d \sqrt{\tan (c+d x)}}+\frac{151}{60 a^2 d \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}+\frac{\left (\frac{1}{8}+\frac{i}{8}\right ) \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{a^{5/2} d}+\frac{17}{30 a d \sqrt{\tan (c+d x)} (a+i a \tan (c+d x))^{3/2}}+\frac{1}{5 d \sqrt{\tan (c+d x)} (a+i a \tan (c+d x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/(Tan[c + d*x]^(3/2)*(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) + 1/(5*d*Sq
rt[Tan[c + d*x]]*(a + I*a*Tan[c + d*x])^(5/2)) + 17/(30*a*d*Sqrt[Tan[c + d*x]]*(a + I*a*Tan[c + d*x])^(3/2)) +
151/(60*a^2*d*Sqrt[Tan[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]]) - (317*Sqrt[a + I*a*Tan[c + d*x]])/(60*a^3*d*Sqr
t[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 3598

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*d - B*c)*(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n + 1))/(f
*(n + 1)*(c^2 + d^2)), x] - Dist[1/(a*(n + 1)*(c^2 + d^2)), Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n
+ 1)*Simp[A*(b*d*m - a*c*(n + 1)) - B*(b*c*m + a*d*(n + 1)) - a*(B*c - A*d)*(m + n + 1)*Tan[e + f*x], x], x],
x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && LtQ[n, -1]

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}{\tan ^{\frac{3}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}} \, dx &=\frac{1}{5 d \sqrt{\tan (c+d x)} (a+i a \tan (c+d x))^{5/2}}+\frac{\int \frac{\frac{11 a}{2}-3 i a \tan (c+d x)}{\tan ^{\frac{3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}} \, dx}{5 a^2}\\ &=\frac{1}{5 d \sqrt{\tan (c+d x)} (a+i a \tan (c+d x))^{5/2}}+\frac{17}{30 a d \sqrt{\tan (c+d x)} (a+i a \tan (c+d x))^{3/2}}+\frac{\int \frac{\frac{83 a^2}{4}-17 i a^2 \tan (c+d x)}{\tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}} \, dx}{15 a^4}\\ &=\frac{1}{5 d \sqrt{\tan (c+d x)} (a+i a \tan (c+d x))^{5/2}}+\frac{17}{30 a d \sqrt{\tan (c+d x)} (a+i a \tan (c+d x))^{3/2}}+\frac{151}{60 a^2 d \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}+\frac{\int \frac{\sqrt{a+i a \tan (c+d x)} \left (\frac{317 a^3}{8}-\frac{151}{4} i a^3 \tan (c+d x)\right )}{\tan ^{\frac{3}{2}}(c+d x)} \, dx}{15 a^6}\\ &=\frac{1}{5 d \sqrt{\tan (c+d x)} (a+i a \tan (c+d x))^{5/2}}+\frac{17}{30 a d \sqrt{\tan (c+d x)} (a+i a \tan (c+d x))^{3/2}}+\frac{151}{60 a^2 d \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}-\frac{317 \sqrt{a+i a \tan (c+d x)}}{60 a^3 d \sqrt{\tan (c+d x)}}+\frac{2 \int \frac{15 i a^4 \sqrt{a+i a \tan (c+d x)}}{16 \sqrt{\tan (c+d x)}} \, dx}{15 a^7}\\ &=\frac{1}{5 d \sqrt{\tan (c+d x)} (a+i a \tan (c+d x))^{5/2}}+\frac{17}{30 a d \sqrt{\tan (c+d x)} (a+i a \tan (c+d x))^{3/2}}+\frac{151}{60 a^2 d \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}-\frac{317 \sqrt{a+i a \tan (c+d x)}}{60 a^3 d \sqrt{\tan (c+d x)}}+\frac{i \int \frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{\tan (c+d x)}} \, dx}{8 a^3}\\ &=\frac{1}{5 d \sqrt{\tan (c+d x)} (a+i a \tan (c+d x))^{5/2}}+\frac{17}{30 a d \sqrt{\tan (c+d x)} (a+i a \tan (c+d x))^{3/2}}+\frac{151}{60 a^2 d \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}-\frac{317 \sqrt{a+i a \tan (c+d x)}}{60 a^3 d \sqrt{\tan (c+d x)}}+\frac{\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{1}{5 d \sqrt{\tan (c+d x)} (a+i a \tan (c+d x))^{5/2}}+\frac{17}{30 a d \sqrt{\tan (c+d x)} (a+i a \tan (c+d x))^{3/2}}+\frac{151}{60 a^2 d \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}-\frac{317 \sqrt{a+i a \tan (c+d x)}}{60 a^3 d \sqrt{\tan (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 1.28436, size = 203, normalized size = 1.02 \[ \frac{i e^{-4 i (c+d x)} \sqrt{\tan (c+d x)} \left (\sqrt{-1+e^{2 i (c+d x)}} \left (26 e^{2 i (c+d x)}+194 e^{4 i (c+d x)}-463 e^{6 i (c+d x)}+3\right )+15 e^{5 i (c+d x)} \left (-1+e^{2 i (c+d x)}\right ) \tanh ^{-1}\left (\frac{e^{i (c+d x)}}{\sqrt{-1+e^{2 i (c+d x)}}}\right )\right )}{60 \sqrt{2} a^2 d \left (-1+e^{2 i (c+d x)}\right )^{3/2} \sqrt{\frac{a e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(Tan[c + d*x]^(3/2)*(a + I*a*Tan[c + d*x])^(5/2)),x]

[Out]

((I/60)*(Sqrt[-1 + E^((2*I)*(c + d*x))]*(3 + 26*E^((2*I)*(c + d*x)) + 194*E^((4*I)*(c + d*x)) - 463*E^((6*I)*(
c + d*x))) + 15*E^((5*I)*(c + d*x))*(-1 + E^((2*I)*(c + d*x)))*ArcTanh[E^(I*(c + d*x))/Sqrt[-1 + E^((2*I)*(c +
d*x))]])*Sqrt[Tan[c + d*x]])/(Sqrt[2]*a^2*d*E^((4*I)*(c + d*x))*(-1 + E^((2*I)*(c + d*x)))^(3/2)*Sqrt[(a*E^((
2*I)*(c + d*x)))/(1 + E^((2*I)*(c + d*x)))])

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Maple [B]  time = 0.046, size = 615, normalized size = 3.1 \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)^(3/2)/(a+I*a*tan(d*x+c))^(5/2),x)

[Out]

-1/240/d*(a*(1+I*tan(d*x+c)))^(1/2)/a^3*(60*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-15*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)^5*a+1268*(a*tan(d*x+c)*(1+I*tan(d*x+c
)))^(1/2)*(-I*a)^(1/2)*tan(d*x+c)^4-60*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+90*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-5660*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(
1/2)*(-I*a)^(1/2)*tan(d*x+c)^2-4468*I*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*tan(d*x+c)^3-15*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+2940*I*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*tan(d*x+c)+480*(a*tan(d*x+c)*(1+I*tan(d*x+c))
)^(1/2)*(-I*a)^(1/2))/tan(d*x+c)^(1/2)/(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)^(3/2)/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

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Fricas [B]  time = 3.17248, size = 1246, normalized size = 6.26 \begin{align*} \frac{\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 (-463 i \, e^{\left (8 i \, d x + 8 i \, c\right )} - 269 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 220 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 29 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 3 i\right )} e^{\left (i \, d x + i \, c\right )} - 30 \,{\left (a^{3} d e^{\left (8 i \, d x + 8 i \, c\right )} - a^{3} d e^{\left (6 i \, d x + 6 i \, c\right )}\right )} \sqrt{\frac{i}{8 \, a^{5} d^{2}}} \log \left (\frac{1}{4} \,{\left (4 i \, 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 \,{\left (a^{3} d e^{\left (8 i \, d x + 8 i \, c\right )} - a^{3} d e^{\left (6 i \, d x + 6 i \, c\right )}\right )} \sqrt{\frac{i}{8 \, a^{5} d^{2}}} \log \left (\frac{1}{4} \,{\left (-4 i \, 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 )}{120 \,{\left (a^{3} d e^{\left (8 i \, d x + 8 i \, c\right )} - a^{3} d e^{\left (6 i \, d x + 6 i \, c\right )}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/tan(d*x+c)^(3/2)/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

1/120*(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))*
(-463*I*e^(8*I*d*x + 8*I*c) - 269*I*e^(6*I*d*x + 6*I*c) + 220*I*e^(4*I*d*x + 4*I*c) + 29*I*e^(2*I*d*x + 2*I*c)
+ 3*I)*e^(I*d*x + I*c) - 30*(a^3*d*e^(8*I*d*x + 8*I*c) - a^3*d*e^(6*I*d*x + 6*I*c))*sqrt(1/8*I/(a^5*d^2))*log
(1/4*(4*I*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*e^(8*I*d*x + 8*I*c) - a^3*d*e^(6*I*d*x + 6*I*c))*sqrt(1/8*I/(a^5*d^2))*log(1/4*(-4*I*a^3*d*sqr
t(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)))/(a^3*d*e^(8*I*
d*x + 8*I*c) - a^3*d*e^(6*I*d*x + 6*I*c))

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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)**(3/2)/(a+I*a*tan(d*x+c))**(5/2),x)

[Out]

Timed out

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Giac [A]  time = 1.33655, size = 150, normalized size = 0.75 \begin{align*} -\frac{2 \, \sqrt{-2 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )} a + 2 \, a^{2}} a^{2} \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 )}^{6} + \left (4 i + 4\right ) \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{5} a - \left (5 i + 5\right ) \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{4} a^{2} + \left (2 i + 2\right ) \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3} a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/tan(d*x+c)^(3/2)/(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^2*log(sqrt(I*a*tan(d*x + c) + a))/(-(I + 1)*(I*a*tan(d*x + c) +
a)^6 + (4*I + 4)*(I*a*tan(d*x + c) + a)^5*a - (5*I + 5)*(I*a*tan(d*x + c) + a)^4*a^2 + (2*I + 2)*(I*a*tan(d*x
+ c) + a)^3*a^3)