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

Optimal. Leaf size=161 \[ -\frac{5 \sqrt{a+i a \tan (c+d x)}}{3 a d \tan ^{\frac{3}{2}}(c+d x)}+\frac{1}{d \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}+\frac{7 i \sqrt{a+i a \tan (c+d x)}}{3 a d \sqrt{\tan (c+d x)}}-\frac{\left (\frac{1}{2}-\frac{i}{2}\right ) \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{\sqrt{a} d} \]

[Out]

((-1/2 + I/2)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/(Sqrt[a]*d) + 1/(d*Tan
[c + d*x]^(3/2)*Sqrt[a + I*a*Tan[c + d*x]]) - (5*Sqrt[a + I*a*Tan[c + d*x]])/(3*a*d*Tan[c + d*x]^(3/2)) + (((7
*I)/3)*Sqrt[a + I*a*Tan[c + d*x]])/(a*d*Sqrt[Tan[c + d*x]])

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Rubi [A]  time = 0.409143, antiderivative size = 161, 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, 3598, 12, 3544, 205} \[ -\frac{5 \sqrt{a+i a \tan (c+d x)}}{3 a d \tan ^{\frac{3}{2}}(c+d x)}+\frac{1}{d \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}+\frac{7 i \sqrt{a+i a \tan (c+d x)}}{3 a d \sqrt{\tan (c+d x)}}-\frac{\left (\frac{1}{2}-\frac{i}{2}\right ) \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{\sqrt{a} d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-1/2 + I/2)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/(Sqrt[a]*d) + 1/(d*Tan
[c + d*x]^(3/2)*Sqrt[a + I*a*Tan[c + d*x]]) - (5*Sqrt[a + I*a*Tan[c + d*x]])/(3*a*d*Tan[c + d*x]^(3/2)) + (((7
*I)/3)*Sqrt[a + I*a*Tan[c + d*x]])/(a*d*Sqrt[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 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{5}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}} \, dx &=\frac{1}{d \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}+\frac{\int \frac{\sqrt{a+i a \tan (c+d x)} \left (\frac{5 a}{2}-2 i a \tan (c+d x)\right )}{\tan ^{\frac{5}{2}}(c+d x)} \, dx}{a^2}\\ &=\frac{1}{d \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}-\frac{5 \sqrt{a+i a \tan (c+d x)}}{3 a d \tan ^{\frac{3}{2}}(c+d x)}+\frac{2 \int \frac{\sqrt{a+i a \tan (c+d x)} \left (-\frac{7 i a^2}{4}-\frac{5}{2} a^2 \tan (c+d x)\right )}{\tan ^{\frac{3}{2}}(c+d x)} \, dx}{3 a^3}\\ &=\frac{1}{d \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}-\frac{5 \sqrt{a+i a \tan (c+d x)}}{3 a d \tan ^{\frac{3}{2}}(c+d x)}+\frac{7 i \sqrt{a+i a \tan (c+d x)}}{3 a d \sqrt{\tan (c+d x)}}+\frac{4 \int -\frac{3 a^3 \sqrt{a+i a \tan (c+d x)}}{8 \sqrt{\tan (c+d x)}} \, dx}{3 a^4}\\ &=\frac{1}{d \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}-\frac{5 \sqrt{a+i a \tan (c+d x)}}{3 a d \tan ^{\frac{3}{2}}(c+d x)}+\frac{7 i \sqrt{a+i a \tan (c+d x)}}{3 a d \sqrt{\tan (c+d x)}}-\frac{\int \frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{\tan (c+d x)}} \, dx}{2 a}\\ &=\frac{1}{d \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}-\frac{5 \sqrt{a+i a \tan (c+d x)}}{3 a d \tan ^{\frac{3}{2}}(c+d x)}+\frac{7 i \sqrt{a+i a \tan (c+d x)}}{3 a d \sqrt{\tan (c+d x)}}+\frac{(i a) \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 )}{d}\\ &=-\frac{\left (\frac{1}{2}-\frac{i}{2}\right ) \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{\sqrt{a} d}+\frac{1}{d \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}-\frac{5 \sqrt{a+i a \tan (c+d x)}}{3 a d \tan ^{\frac{3}{2}}(c+d x)}+\frac{7 i \sqrt{a+i a \tan (c+d x)}}{3 a d \sqrt{\tan (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 1.67475, size = 159, normalized size = 0.99 \[ \frac{i \left (-18 e^{2 i (c+d x)}+7 e^{4 i (c+d x)}+3 e^{i (c+d x)} \left (-1+e^{2 i (c+d x)}\right )^{3/2} \tanh ^{-1}\left (\frac{e^{i (c+d x)}}{\sqrt{-1+e^{2 i (c+d x)}}}\right )+3\right )}{3 \sqrt{2} d \left (-1+e^{4 i (c+d x)}\right ) \sqrt{\frac{a e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt{\tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((I/3)*(3 - 18*E^((2*I)*(c + d*x)) + 7*E^((4*I)*(c + d*x)) + 3*E^(I*(c + d*x))*(-1 + E^((2*I)*(c + d*x)))^(3/2
)*ArcTanh[E^(I*(c + d*x))/Sqrt[-1 + E^((2*I)*(c + d*x))]]))/(Sqrt[2]*d*Sqrt[(a*E^((2*I)*(c + d*x)))/(1 + E^((2
*I)*(c + d*x)))]*(-1 + E^((4*I)*(c + d*x)))*Sqrt[Tan[c + d*x]])

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Maple [B]  time = 0.072, size = 401, normalized size = 2.5 \begin{align*}{\frac{1}{12\,ad \left ( -\tan \left ( dx+c \right ) +i \right ) ^{2}}\sqrt{a \left ( 1+i\tan \left ( dx+c \right ) \right ) } \left ( 3\,i\sqrt{2}\ln \left ( -{\frac{1}{\tan \left ( dx+c \right ) +i} \left ( -2\,\sqrt{2}\sqrt{-ia}\sqrt{a\tan \left ( dx+c \right ) \left ( 1+i\tan \left ( dx+c \right ) \right ) }+ia-3\,a\tan \left ( dx+c \right ) \right ) } \right ) \left ( \tan \left ( dx+c \right ) \right ) ^{4}a+36\,\sqrt{a\tan \left ( dx+c \right ) \left ( 1+i\tan \left ( dx+c \right ) \right ) }\sqrt{-ia} \left ( \tan \left ( dx+c \right ) \right ) ^{2}-3\,i\sqrt{2}\ln \left ( -{\frac{1}{\tan \left ( dx+c \right ) +i} \left ( -2\,\sqrt{2}\sqrt{-ia}\sqrt{a\tan \left ( dx+c \right ) \left ( 1+i\tan \left ( dx+c \right ) \right ) }+ia-3\,a\tan \left ( dx+c \right ) \right ) } \right ) \left ( \tan \left ( dx+c \right ) \right ) ^{2}a+28\,i\sqrt{a\tan \left ( dx+c \right ) \left ( 1+i\tan \left ( dx+c \right ) \right ) }\sqrt{-ia} \left ( \tan \left ( dx+c \right ) \right ) ^{3}+6\,\sqrt{2}\ln \left ( -{\frac{-2\,\sqrt{2}\sqrt{-ia}\sqrt{a\tan \left ( dx+c \right ) \left ( 1+i\tan \left ( dx+c \right ) \right ) }+ia-3\,a\tan \left ( dx+c \right ) }{\tan \left ( dx+c \right ) +i}} \right ) \left ( \tan \left ( dx+c \right ) \right ) ^{3}a+8\,\sqrt{a\tan \left ( dx+c \right ) \left ( 1+i\tan \left ( dx+c \right ) \right ) }\sqrt{-ia} \right ) \left ( \tan \left ( dx+c \right ) \right ) ^{-{\frac{3}{2}}}{\frac{1}{\sqrt{a\tan \left ( dx+c \right ) \left ( 1+i\tan \left ( dx+c \right ) \right ) }}}{\frac{1}{\sqrt{-ia}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12/d*(a*(1+I*tan(d*x+c)))^(1/2)/a/tan(d*x+c)^(3/2)*(3*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+36*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(
1/2)*(-I*a)^(1/2)*tan(d*x+c)^2-3*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+28*I*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(-I*a)^(1/2)*tan
(d*x+c)^3+6*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))/(ta
n(d*x+c)+I))*tan(d*x+c)^3*a+8*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(-I*a)^(1/2))/(a*tan(d*x+c)*(1+I*tan(d*x+c
)))^(1/2)/(-tan(d*x+c)+I)^2/(-I*a)^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 2.8931, size = 1257, normalized size = 7.81 \begin{align*} -\frac{2 \, \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 (7 \, e^{\left (6 i \, d x + 6 i \, c\right )} - 11 \, e^{\left (4 i \, d x + 4 i \, c\right )} - 15 \, e^{\left (2 i \, d x + 2 i \, c\right )} + 3\right )} e^{\left (i \, d x + i \, c\right )} + 3 \,{\left (a d e^{\left (6 i \, d x + 6 i \, c\right )} - 2 \, a d e^{\left (4 i \, d x + 4 i \, c\right )} + a d e^{\left (2 i \, d x + 2 i \, c\right )}\right )} \sqrt{-\frac{2 i}{a d^{2}}} \log \left (\frac{1}{4} \,{\left (a d \sqrt{-\frac{2 i}{a 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 ) - 3 \,{\left (a d e^{\left (6 i \, d x + 6 i \, c\right )} - 2 \, a d e^{\left (4 i \, d x + 4 i \, c\right )} + a d e^{\left (2 i \, d x + 2 i \, c\right )}\right )} \sqrt{-\frac{2 i}{a d^{2}}} \log \left (-\frac{1}{4} \,{\left (a d \sqrt{-\frac{2 i}{a 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 )}{12 \,{\left (a d e^{\left (6 i \, d x + 6 i \, c\right )} - 2 \, a d e^{\left (4 i \, d x + 4 i \, c\right )} + a d e^{\left (2 i \, d x + 2 i \, c\right )}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(2*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)
)*(7*e^(6*I*d*x + 6*I*c) - 11*e^(4*I*d*x + 4*I*c) - 15*e^(2*I*d*x + 2*I*c) + 3)*e^(I*d*x + I*c) + 3*(a*d*e^(6*
I*d*x + 6*I*c) - 2*a*d*e^(4*I*d*x + 4*I*c) + a*d*e^(2*I*d*x + 2*I*c))*sqrt(-2*I/(a*d^2))*log(1/4*(a*d*sqrt(-2*
I/(a*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)) - 3*(a*d*e^(6*I*d*x + 6
*I*c) - 2*a*d*e^(4*I*d*x + 4*I*c) + a*d*e^(2*I*d*x + 2*I*c))*sqrt(-2*I/(a*d^2))*log(-1/4*(a*d*sqrt(-2*I/(a*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*d*e^(6*I*d*x + 6*I*c) - 2*a
*d*e^(4*I*d*x + 4*I*c) + a*d*e^(2*I*d*x + 2*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/(a+I*a*tan(d*x+c))**(1/2)/tan(d*x+c)**(5/2),x)

[Out]

Timed out

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Giac [A]  time = 1.27386, size = 171, normalized size = 1.06 \begin{align*} \frac{2 \, \sqrt{-2 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )} a + 2 \, a^{2}} a^{3} \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 (5 i - 5\right ) \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{4} a - \left (9 i - 9\right ) \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3} a^{2} + \left (7 i - 7\right ) \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} a^{3} - \left (2 i - 2\right ) \,{\left (i \, a \tan \left (d x + c\right ) + a\right )} a^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+I*a*tan(d*x+c))^(1/2)/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^3*log(sqrt(I*a*tan(d*x + c) + a))/(-(I - 1)*(I*a*tan(d*x + c) +
a)^5 + (5*I - 5)*(I*a*tan(d*x + c) + a)^4*a - (9*I - 9)*(I*a*tan(d*x + c) + a)^3*a^2 + (7*I - 7)*(I*a*tan(d*x
+ c) + a)^2*a^3 - (2*I - 2)*(I*a*tan(d*x + c) + a)*a^4)