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

Optimal. Leaf size=125 \[ -\frac {\tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{4 \sqrt {2} a^{5/2} d}+\frac {1}{4 a^2 d \sqrt {a+i a \tan (c+d x)}}+\frac {1}{6 a d (a+i a \tan (c+d x))^{3/2}}-\frac {1}{5 d (a+i a \tan (c+d x))^{5/2}} \]

[Out]

-1/8*arctanh(1/2*(a+I*a*tan(d*x+c))^(1/2)*2^(1/2)/a^(1/2))/a^(5/2)/d*2^(1/2)+1/4/a^2/d/(a+I*a*tan(d*x+c))^(1/2
)-1/5/d/(a+I*a*tan(d*x+c))^(5/2)+1/6/a/d/(a+I*a*tan(d*x+c))^(3/2)

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Rubi [A]  time = 0.10, antiderivative size = 125, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3526, 3479, 3480, 206} \[ \frac {1}{4 a^2 d \sqrt {a+i a \tan (c+d x)}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{4 \sqrt {2} a^{5/2} d}+\frac {1}{6 a d (a+i a \tan (c+d x))^{3/2}}-\frac {1}{5 d (a+i a \tan (c+d x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])]/(4*Sqrt[2]*a^(5/2)*d) - 1/(5*d*(a + I*a*Tan[c + d*x])^(
5/2)) + 1/(6*a*d*(a + I*a*Tan[c + d*x])^(3/2)) + 1/(4*a^2*d*Sqrt[a + I*a*Tan[c + d*x]])

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 3479

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(a*(a + b*Tan[c + d*x])^n)/(2*b*d*n), x] +
Dist[1/(2*a), Int[(a + b*Tan[c + d*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 + b^2, 0] && LtQ[n
, 0]

Rule 3480

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

Rule 3526

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

Rubi steps

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

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Mathematica [A]  time = 0.74, size = 135, normalized size = 1.08 \[ \frac {e^{-6 i (c+d x)} \left (1+e^{2 i (c+d x)}\right )^{3/2} \sec ^2(c+d x) \left (\sqrt {1+e^{2 i (c+d x)}} \left (-e^{2 i (c+d x)}+17 e^{4 i (c+d x)}-3\right )-15 e^{5 i (c+d x)} \sinh ^{-1}\left (e^{i (c+d x)}\right )\right )}{240 a^2 d \sqrt {a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((1 + E^((2*I)*(c + d*x)))^(3/2)*(Sqrt[1 + E^((2*I)*(c + d*x))]*(-3 - E^((2*I)*(c + d*x)) + 17*E^((4*I)*(c + d
*x))) - 15*E^((5*I)*(c + d*x))*ArcSinh[E^(I*(c + d*x))])*Sec[c + d*x]^2)/(240*a^2*d*E^((6*I)*(c + d*x))*Sqrt[a
+ I*a*Tan[c + d*x]])

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fricas [B]  time = 0.44, size = 284, normalized size = 2.27 \[ -\frac {{\left (15 \, \sqrt {\frac {1}{2}} a^{3} d \sqrt {\frac {1}{a^{5} d^{2}}} e^{\left (5 i \, d x + 5 i \, c\right )} \log \left (4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{3} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a^{5} d^{2}}} + a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) - 15 \, \sqrt {\frac {1}{2}} a^{3} d \sqrt {\frac {1}{a^{5} d^{2}}} e^{\left (5 i \, d x + 5 i \, c\right )} \log \left (-4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{3} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a^{5} d^{2}}} - a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) - \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (17 \, e^{\left (6 i \, d x + 6 i \, c\right )} + 16 \, e^{\left (4 i \, d x + 4 i \, c\right )} - 4 \, e^{\left (2 i \, d x + 2 i \, c\right )} - 3\right )}\right )} e^{\left (-5 i \, d x - 5 i \, c\right )}}{120 \, a^{3} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/120*(15*sqrt(1/2)*a^3*d*sqrt(1/(a^5*d^2))*e^(5*I*d*x + 5*I*c)*log(4*(sqrt(2)*sqrt(1/2)*(a^3*d*e^(2*I*d*x +
2*I*c) + a^3*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(1/(a^5*d^2)) + a*e^(I*d*x + I*c))*e^(-I*d*x - I*c)) - 1
5*sqrt(1/2)*a^3*d*sqrt(1/(a^5*d^2))*e^(5*I*d*x + 5*I*c)*log(-4*(sqrt(2)*sqrt(1/2)*(a^3*d*e^(2*I*d*x + 2*I*c) +
a^3*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(1/(a^5*d^2)) - a*e^(I*d*x + I*c))*e^(-I*d*x - I*c)) - sqrt(2)*s
qrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(17*e^(6*I*d*x + 6*I*c) + 16*e^(4*I*d*x + 4*I*c) - 4*e^(2*I*d*x + 2*I*c) - 3)
)*e^(-5*I*d*x - 5*I*c)/(a^3*d)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\tan \left (d x + c\right )}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.16, size = 91, normalized size = 0.73 \[ \frac {-\frac {1}{5 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}+\frac {1}{4 a^{2} \sqrt {a +i a \tan \left (d x +c \right )}}+\frac {1}{6 a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}-\frac {\sqrt {2}\, \arctanh \left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{8 a^{\frac {5}{2}}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(-1/5/(a+I*a*tan(d*x+c))^(5/2)+1/4/a^2/(a+I*a*tan(d*x+c))^(1/2)+1/6/a/(a+I*a*tan(d*x+c))^(3/2)-1/8/a^(5/2)
*2^(1/2)*arctanh(1/2*(a+I*a*tan(d*x+c))^(1/2)*2^(1/2)/a^(1/2)))

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maxima [A]  time = 0.58, size = 116, normalized size = 0.93 \[ \frac {\frac {15 \, \sqrt {2} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right )}{\sqrt {a}} + \frac {4 \, {\left (15 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} + 10 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )} a - 12 \, a^{2}\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}}{240 \, a^{2} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/240*(15*sqrt(2)*log(-(sqrt(2)*sqrt(a) - sqrt(I*a*tan(d*x + c) + a))/(sqrt(2)*sqrt(a) + sqrt(I*a*tan(d*x + c)
+ a)))/sqrt(a) + 4*(15*(I*a*tan(d*x + c) + a)^2 + 10*(I*a*tan(d*x + c) + a)*a - 12*a^2)/(I*a*tan(d*x + c) + a
)^(5/2))/(a^2*d)

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mupad [B]  time = 0.20, size = 91, normalized size = 0.73 \[ \frac {\frac {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}{6\,a}+\frac {{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^2}{4\,a^2}-\frac {1}{5}}{d\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}}-\frac {\sqrt {2}\,\mathrm {atanh}\left (\frac {\sqrt {2}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {a}}\right )}{8\,a^{5/2}\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((a + a*tan(c + d*x)*1i)/(6*a) + (a + a*tan(c + d*x)*1i)^2/(4*a^2) - 1/5)/(d*(a + a*tan(c + d*x)*1i)^(5/2)) -
(2^(1/2)*atanh((2^(1/2)*(a + a*tan(c + d*x)*1i)^(1/2))/(2*a^(1/2))))/(8*a^(5/2)*d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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