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

Optimal. Leaf size=101 \[ -\frac {4 i \sqrt {2} a^{5/2} \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}+\frac {4 i a^2 \sqrt {a+i a \tan (c+d x)}}{d}+\frac {2 i a (a+i a \tan (c+d x))^{3/2}}{3 d} \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 101, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {3478, 3480, 206} \[ \frac {4 i a^2 \sqrt {a+i a \tan (c+d x)}}{d}-\frac {4 i \sqrt {2} a^{5/2} \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}+\frac {2 i a (a+i a \tan (c+d x))^{3/2}}{3 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + I*a*Tan[c + d*x])^(5/2),x]

[Out]

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

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 3478

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(a + b*Tan[c + d*x])^(n - 1))/(d*(n - 1)
), x] + Dist[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] && G
tQ[n, 1]

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]

Rubi steps

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

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Mathematica [A]  time = 0.95, size = 140, normalized size = 1.39 \[ -\frac {\sqrt {2} a^2 e^{-i (c+2 d x)} \sqrt {1+e^{2 i (c+d x)}} \sqrt {\frac {a e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}} (\cos (d x)+i \sin (d x)) \left (12 i \sinh ^{-1}\left (e^{i (c+d x)}\right )+\sqrt {1+e^{2 i (c+d x)}} (\tan (c+d x)-7 i) \sec (c+d x)\right )}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/3*(Sqrt[2]*a^2*Sqrt[(a*E^((2*I)*(c + d*x)))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x))]*(Cos[d
*x] + I*Sin[d*x])*((12*I)*ArcSinh[E^(I*(c + d*x))] + Sqrt[1 + E^((2*I)*(c + d*x))]*Sec[c + d*x]*(-7*I + Tan[c
+ d*x])))/(d*E^(I*(c + 2*d*x)))

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fricas [B]  time = 0.44, size = 270, normalized size = 2.67 \[ \frac {24 \, \sqrt {2} \sqrt {-\frac {a^{5}}{d^{2}}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (\frac {4 \, {\left (a^{3} e^{\left (i \, d x + i \, c\right )} + \sqrt {-\frac {a^{5}}{d^{2}}} {\left (i \, d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{a^{2}}\right ) - 24 \, \sqrt {2} \sqrt {-\frac {a^{5}}{d^{2}}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (\frac {4 \, {\left (a^{3} e^{\left (i \, d x + i \, c\right )} + \sqrt {-\frac {a^{5}}{d^{2}}} {\left (-i \, d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{a^{2}}\right ) + \sqrt {2} {\left (64 i \, a^{2} e^{\left (3 i \, d x + 3 i \, c\right )} + 48 i \, a^{2} e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{12 \, {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(24*sqrt(2)*sqrt(-a^5/d^2)*(d*e^(2*I*d*x + 2*I*c) + d)*log(4*(a^3*e^(I*d*x + I*c) + sqrt(-a^5/d^2)*(I*d*e
^(2*I*d*x + 2*I*c) + I*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)))*e^(-I*d*x - I*c)/a^2) - 24*sqrt(2)*sqrt(-a^5/d^2)
*(d*e^(2*I*d*x + 2*I*c) + d)*log(4*(a^3*e^(I*d*x + I*c) + sqrt(-a^5/d^2)*(-I*d*e^(2*I*d*x + 2*I*c) - I*d)*sqrt
(a/(e^(2*I*d*x + 2*I*c) + 1)))*e^(-I*d*x - I*c)/a^2) + sqrt(2)*(64*I*a^2*e^(3*I*d*x + 3*I*c) + 48*I*a^2*e^(I*d
*x + I*c))*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)))/(d*e^(2*I*d*x + 2*I*c) + d)

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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maple [A]  time = 0.11, size = 73, normalized size = 0.72 \[ \frac {2 i a \left (\frac {\left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3}+2 a \sqrt {a +i a \tan \left (d x +c \right )}-2 a^{\frac {3}{2}} \sqrt {2}\, \arctanh \left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*I/d*a*(1/3*(a+I*a*tan(d*x+c))^(3/2)+2*a*(a+I*a*tan(d*x+c))^(1/2)-2*a^(3/2)*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 = 0.72, size = 101, normalized size = 1.00 \[ \frac {2 i \, {\left (3 \, \sqrt {2} a^{\frac {7}{2}} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right ) + {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} a^{2} + 6 \, \sqrt {i \, a \tan \left (d x + c\right ) + a} a^{3}\right )}}{3 \, a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*I*(3*sqrt(2)*a^(7/2)*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))) + (I*a*tan(d*x + c) + a)^(3/2)*a^2 + 6*sqrt(I*a*tan(d*x + c) + a)*a^3)/(a*d)

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mupad [B]  time = 4.08, size = 84, normalized size = 0.83 \[ \frac {a^2\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}\,4{}\mathrm {i}}{d}+\frac {a\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}\,2{}\mathrm {i}}{3\,d}-\frac {\sqrt {2}\,{\left (-a\right )}^{5/2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {-a}}\right )\,4{}\mathrm {i}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*tan(c + d*x)*1i)^(5/2),x)

[Out]

(a^2*(a + a*tan(c + d*x)*1i)^(1/2)*4i)/d + (a*(a + a*tan(c + d*x)*1i)^(3/2)*2i)/(3*d) - (2^(1/2)*(-a)^(5/2)*at
an((2^(1/2)*(a + a*tan(c + d*x)*1i)^(1/2))/(2*(-a)^(1/2)))*4i)/d

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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