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

Optimal. Leaf size=119 \[ -\frac {4 \sqrt {2} a^{5/2} \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}+\frac {4 a^2 \sqrt {a+i a \tan (c+d x)}}{d}+\frac {2 a (a+i a \tan (c+d x))^{3/2}}{3 d}+\frac {2 (a+i a \tan (c+d x))^{5/2}}{5 d} \]

[Out]

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

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Rubi [A]  time = 0.10, antiderivative size = 119, 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 = {3527, 3478, 3480, 206} \[ \frac {4 a^2 \sqrt {a+i a \tan (c+d x)}}{d}-\frac {4 \sqrt {2} a^{5/2} \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}+\frac {2 a (a+i a \tan (c+d x))^{3/2}}{3 d}+\frac {2 (a+i a \tan (c+d x))^{5/2}}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3527

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

Rubi steps

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

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Mathematica [A]  time = 1.28, size = 154, normalized size = 1.29 \[ \frac {a^2 e^{-i (c+2 d x)} \sqrt {1+e^{2 i (c+d x)}} \sqrt {\frac {a e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}} (\cos (d x)+i \sin (d x)) \left (\sqrt {1+e^{2 i (c+d x)}} \sec ^3(c+d x) (11 i \sin (2 (c+d x))+41 \cos (2 (c+d x))+35)-120 \sinh ^{-1}\left (e^{i (c+d x)}\right )\right )}{15 \sqrt {2} d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(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])*(-120*ArcSinh[E^(I*(c + d*x))] + Sqrt[1 + E^((2*I)*(c + d*x))]*Sec[c + d*x]^3*(35 + 41*Cos[2*(c + d*x)] +
(11*I)*Sin[2*(c + d*x)])))/(15*Sqrt[2]*d*E^(I*(c + 2*d*x)))

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fricas [B]  time = 0.44, size = 312, normalized size = 2.62 \[ -\frac {2 \, {\left (15 \, \sqrt {2} \sqrt {\frac {a^{5}}{d^{2}}} {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (\frac {4 \, {\left (a^{3} e^{\left (i \, d x + i \, c\right )} + \sqrt {\frac {a^{5}}{d^{2}}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{a^{2}}\right ) - 15 \, \sqrt {2} \sqrt {\frac {a^{5}}{d^{2}}} {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (\frac {4 \, {\left (a^{3} e^{\left (i \, d x + i \, c\right )} - \sqrt {\frac {a^{5}}{d^{2}}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{a^{2}}\right ) - 2 \, \sqrt {2} {\left (26 \, a^{2} e^{\left (5 i \, d x + 5 i \, c\right )} + 35 \, a^{2} e^{\left (3 i \, d x + 3 i \, c\right )} + 15 \, a^{2} e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )}}{15 \, {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/15*(15*sqrt(2)*sqrt(a^5/d^2)*(d*e^(4*I*d*x + 4*I*c) + 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)*(d*e^(2*I*d*x + 2*I*c) + d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)))*e^(-I*d*x - I*c)/a^2) - 15*s
qrt(2)*sqrt(a^5/d^2)*(d*e^(4*I*d*x + 4*I*c) + 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)*(d*e^(2*I*d*x + 2*I*c) + d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)))*e^(-I*d*x - I*c)/a^2) - 2*sqrt(2)*(26*a
^2*e^(5*I*d*x + 5*I*c) + 35*a^2*e^(3*I*d*x + 3*I*c) + 15*a^2*e^(I*d*x + I*c))*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)
))/(d*e^(4*I*d*x + 4*I*c) + 2*d*e^(2*I*d*x + 2*I*c) + d)

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

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

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maple [A]  time = 0.14, size = 89, normalized size = 0.75 \[ \frac {\frac {2 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}{5}+\frac {2 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}} a}{3}+4 a^{2} \sqrt {a +i a \tan \left (d x +c \right )}-4 a^{\frac {5}{2}} \sqrt {2}\, \arctanh \left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{d} \]

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*(2/5*(a+I*a*tan(d*x+c))^(5/2)+2/3*(a+I*a*tan(d*x+c))^(3/2)*a+4*a^2*(a+I*a*tan(d*x+c))^(1/2)-4*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 = 0.59, size = 120, normalized size = 1.01 \[ \frac {2 \, {\left (15 \, \sqrt {2} a^{\frac {9}{2}} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right ) + 3 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} a^{2} + 5 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} a^{3} + 30 \, \sqrt {i \, a \tan \left (d x + c\right ) + a} a^{4}\right )}}{15 \, a^{2} d} \]

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]

2/15*(15*sqrt(2)*a^(9/2)*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))) + 3*(I*a*tan(d*x + c) + a)^(5/2)*a^2 + 5*(I*a*tan(d*x + c) + a)^(3/2)*a^3 + 30*sqrt(I*a*tan(d*x
+ c) + a)*a^4)/(a^2*d)

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mupad [B]  time = 0.32, size = 98, normalized size = 0.82 \[ \frac {2\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}}{5\,d}+\frac {4\,a^2\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{d}+\frac {2\,a\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}}{3\,d}+\frac {\sqrt {2}\,a^{5/2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2\,\sqrt {a}}\right )\,4{}\mathrm {i}}{d} \]

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]

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

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

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

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