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

Optimal. Leaf size=204 \[ -\frac{2 a^2 \tan ^4(c+d x) \sqrt{a+i a \tan (c+d x)}}{9 d}+\frac{38 i a^2 \tan ^3(c+d x) \sqrt{a+i a \tan (c+d x)}}{63 d}+\frac{92 a^2 \tan ^2(c+d x) \sqrt{a+i a \tan (c+d x)}}{105 d}-\frac{368 a^2 \sqrt{a+i a \tan (c+d x)}}{105 d}+\frac{4 \sqrt{2} a^{5/2} \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{d}-\frac{472 a (a+i a \tan (c+d x))^{3/2}}{315 d} \]

[Out]

(4*Sqrt[2]*a^(5/2)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/d - (368*a^2*Sqrt[a + I*a*Tan[c + d*
x]])/(105*d) + (92*a^2*Tan[c + d*x]^2*Sqrt[a + I*a*Tan[c + d*x]])/(105*d) + (((38*I)/63)*a^2*Tan[c + d*x]^3*Sq
rt[a + I*a*Tan[c + d*x]])/d - (2*a^2*Tan[c + d*x]^4*Sqrt[a + I*a*Tan[c + d*x]])/(9*d) - (472*a*(a + I*a*Tan[c
+ d*x])^(3/2))/(315*d)

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Rubi [A]  time = 0.495793, antiderivative size = 204, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.231, Rules used = {3556, 3597, 3592, 3527, 3480, 206} \[ -\frac{2 a^2 \tan ^4(c+d x) \sqrt{a+i a \tan (c+d x)}}{9 d}+\frac{38 i a^2 \tan ^3(c+d x) \sqrt{a+i a \tan (c+d x)}}{63 d}+\frac{92 a^2 \tan ^2(c+d x) \sqrt{a+i a \tan (c+d x)}}{105 d}-\frac{368 a^2 \sqrt{a+i a \tan (c+d x)}}{105 d}+\frac{4 \sqrt{2} a^{5/2} \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{d}-\frac{472 a (a+i a \tan (c+d x))^{3/2}}{315 d} \]

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^3*(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 - (368*a^2*Sqrt[a + I*a*Tan[c + d*
x]])/(105*d) + (92*a^2*Tan[c + d*x]^2*Sqrt[a + I*a*Tan[c + d*x]])/(105*d) + (((38*I)/63)*a^2*Tan[c + d*x]^3*Sq
rt[a + I*a*Tan[c + d*x]])/d - (2*a^2*Tan[c + d*x]^4*Sqrt[a + I*a*Tan[c + d*x]])/(9*d) - (472*a*(a + I*a*Tan[c
+ d*x])^(3/2))/(315*d)

Rule 3556

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(b^2*(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^(n + 1))/(d*f*(m + n - 1)), x] + Dist[a/(d*(m + n - 1
)), Int[(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^n*Simp[b*c*(m - 2) + a*d*(m + 2*n) + (a*c*(m - 2) +
b*d*(3*m + 2*n - 4))*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] && IntegerQ[2*m] && GtQ[m, 1] && NeQ[m + n - 1, 0] && (IntegerQ[m] || Intege
rsQ[2*m, 2*n])

Rule 3597

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(B*(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n)/(f*(m + n)), x] +
Dist[1/(a*(m + n)), Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n - 1)*Simp[a*A*c*(m + n) - B*(b*c*m + a*
d*n) + (a*A*d*(m + n) - B*(b*d*m - a*c*n))*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] && GtQ[n, 0]

Rule 3592

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

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]

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 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])

Rubi steps

\begin{align*} \int \tan ^3(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx &=-\frac{2 a^2 \tan ^4(c+d x) \sqrt{a+i a \tan (c+d x)}}{9 d}+\frac{1}{9} (2 a) \int \tan ^3(c+d x) \sqrt{a+i a \tan (c+d x)} \left (\frac{17 a}{2}+\frac{19}{2} i a \tan (c+d x)\right ) \, dx\\ &=\frac{38 i a^2 \tan ^3(c+d x) \sqrt{a+i a \tan (c+d x)}}{63 d}-\frac{2 a^2 \tan ^4(c+d x) \sqrt{a+i a \tan (c+d x)}}{9 d}+\frac{4}{63} \int \tan ^2(c+d x) \sqrt{a+i a \tan (c+d x)} \left (-\frac{57 i a^2}{2}+\frac{69}{2} a^2 \tan (c+d x)\right ) \, dx\\ &=\frac{92 a^2 \tan ^2(c+d x) \sqrt{a+i a \tan (c+d x)}}{105 d}+\frac{38 i a^2 \tan ^3(c+d x) \sqrt{a+i a \tan (c+d x)}}{63 d}-\frac{2 a^2 \tan ^4(c+d x) \sqrt{a+i a \tan (c+d x)}}{9 d}+\frac{8 \int \tan (c+d x) \sqrt{a+i a \tan (c+d x)} \left (-69 a^3-\frac{177}{2} i a^3 \tan (c+d x)\right ) \, dx}{315 a}\\ &=\frac{92 a^2 \tan ^2(c+d x) \sqrt{a+i a \tan (c+d x)}}{105 d}+\frac{38 i a^2 \tan ^3(c+d x) \sqrt{a+i a \tan (c+d x)}}{63 d}-\frac{2 a^2 \tan ^4(c+d x) \sqrt{a+i a \tan (c+d x)}}{9 d}-\frac{472 a (a+i a \tan (c+d x))^{3/2}}{315 d}+\frac{8 \int \sqrt{a+i a \tan (c+d x)} \left (\frac{177 i a^3}{2}-69 a^3 \tan (c+d x)\right ) \, dx}{315 a}\\ &=-\frac{368 a^2 \sqrt{a+i a \tan (c+d x)}}{105 d}+\frac{92 a^2 \tan ^2(c+d x) \sqrt{a+i a \tan (c+d x)}}{105 d}+\frac{38 i a^2 \tan ^3(c+d x) \sqrt{a+i a \tan (c+d x)}}{63 d}-\frac{2 a^2 \tan ^4(c+d x) \sqrt{a+i a \tan (c+d x)}}{9 d}-\frac{472 a (a+i a \tan (c+d x))^{3/2}}{315 d}+\left (4 i a^2\right ) \int \sqrt{a+i a \tan (c+d x)} \, dx\\ &=-\frac{368 a^2 \sqrt{a+i a \tan (c+d x)}}{105 d}+\frac{92 a^2 \tan ^2(c+d x) \sqrt{a+i a \tan (c+d x)}}{105 d}+\frac{38 i a^2 \tan ^3(c+d x) \sqrt{a+i a \tan (c+d x)}}{63 d}-\frac{2 a^2 \tan ^4(c+d x) \sqrt{a+i a \tan (c+d x)}}{9 d}-\frac{472 a (a+i a \tan (c+d x))^{3/2}}{315 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{368 a^2 \sqrt{a+i a \tan (c+d x)}}{105 d}+\frac{92 a^2 \tan ^2(c+d x) \sqrt{a+i a \tan (c+d x)}}{105 d}+\frac{38 i a^2 \tan ^3(c+d x) \sqrt{a+i a \tan (c+d x)}}{63 d}-\frac{2 a^2 \tan ^4(c+d x) \sqrt{a+i a \tan (c+d x)}}{9 d}-\frac{472 a (a+i a \tan (c+d x))^{3/2}}{315 d}\\ \end{align*}

Mathematica [A]  time = 1.99794, size = 176, normalized size = 0.86 \[ -\frac{a^2 e^{-i (c+2 d x)} \sqrt{1+e^{2 i (c+d x)}} \sqrt{\frac{a e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}} (\cos (d x)+i \sin (d x)) \left (\sqrt{1+e^{2 i (c+d x)}} \sec ^5(c+d x) (282 i \sin (2 (c+d x))+331 i \sin (4 (c+d x))+3012 \cos (2 (c+d x))+961 \cos (4 (c+d x))[phone] \sinh ^{-1}\left (e^{i (c+d x)}\right )\right )}{1260 \sqrt{2} d} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[c + d*x]^3*(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])*(-10080*ArcSinh[E^(I*(c + d*x))] + Sqrt[1 + E^((2*I)*(c + d*x))]*Sec[c + d*x]^5*(2331 + 3012*Cos[2*(c +
d*x)] + 961*Cos[4*(c + d*x)] + (282*I)*Sin[2*(c + d*x)] + (331*I)*Sin[4*(c + d*x)])))/(1260*Sqrt[2]*d*E^(I*(c
+ 2*d*x)))

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Maple [A]  time = 0.019, size = 131, normalized size = 0.6 \begin{align*} -2\,{\frac{1}{{a}^{2}d} \left ( 1/9\, \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{9/2}-1/7\,a \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{7/2}+1/5\,{a}^{2} \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{5/2}+1/3\,{a}^{3} \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{3/2}+2\,{a}^{4}\sqrt{a+ia\tan \left ( dx+c \right ) }-2\,{a}^{9/2}\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{a+ia\tan \left ( dx+c \right ) }\sqrt{2}}{\sqrt{a}}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/d/a^2*(1/9*(a+I*a*tan(d*x+c))^(9/2)-1/7*a*(a+I*a*tan(d*x+c))^(7/2)+1/5*a^2*(a+I*a*tan(d*x+c))^(5/2)+1/3*a^3
*(a+I*a*tan(d*x+c))^(3/2)+2*a^4*(a+I*a*tan(d*x+c))^(1/2)-2*a^(9/2)*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 [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.4192, size = 1234, normalized size = 6.05 \begin{align*} -\frac{2 \,{\left (2 \, \sqrt{2}{\left (646 \, a^{2} e^{\left (8 i \, d x + 8 i \, c\right )} + 1647 \, a^{2} e^{\left (6 i \, d x + 6 i \, c\right )} + 2331 \, a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} + 1365 \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + 315 \, a^{2}\right )} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (i \, d x + i \, c\right )} - 315 \, \sqrt{2} \sqrt{\frac{a^{5}}{d^{2}}}{\left (d e^{\left (8 i \, d x + 8 i \, c\right )} + 4 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 4 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (\frac{{\left (\sqrt{2} \sqrt{\frac{a^{5}}{d^{2}}} d e^{\left (2 i \, d x + 2 i \, c\right )} + \sqrt{2}{\left (a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2}\right )} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{a^{2}}\right ) + 315 \, \sqrt{2} \sqrt{\frac{a^{5}}{d^{2}}}{\left (d e^{\left (8 i \, d x + 8 i \, c\right )} + 4 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 4 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (-\frac{{\left (\sqrt{2} \sqrt{\frac{a^{5}}{d^{2}}} d e^{\left (2 i \, d x + 2 i \, c\right )} - \sqrt{2}{\left (a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2}\right )} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{a^{2}}\right )\right )}}{315 \,{\left (d e^{\left (8 i \, d x + 8 i \, c\right )} + 4 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 4 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/315*(2*sqrt(2)*(646*a^2*e^(8*I*d*x + 8*I*c) + 1647*a^2*e^(6*I*d*x + 6*I*c) + 2331*a^2*e^(4*I*d*x + 4*I*c) +
1365*a^2*e^(2*I*d*x + 2*I*c) + 315*a^2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*e^(I*d*x + I*c) - 315*sqrt(2)*sqrt(
a^5/d^2)*(d*e^(8*I*d*x + 8*I*c) + 4*d*e^(6*I*d*x + 6*I*c) + 6*d*e^(4*I*d*x + 4*I*c) + 4*d*e^(2*I*d*x + 2*I*c)
+ d)*log((sqrt(2)*sqrt(a^5/d^2)*d*e^(2*I*d*x + 2*I*c) + sqrt(2)*(a^2*e^(2*I*d*x + 2*I*c) + a^2)*sqrt(a/(e^(2*I
*d*x + 2*I*c) + 1))*e^(I*d*x + I*c))*e^(-2*I*d*x - 2*I*c)/a^2) + 315*sqrt(2)*sqrt(a^5/d^2)*(d*e^(8*I*d*x + 8*I
*c) + 4*d*e^(6*I*d*x + 6*I*c) + 6*d*e^(4*I*d*x + 4*I*c) + 4*d*e^(2*I*d*x + 2*I*c) + d)*log(-(sqrt(2)*sqrt(a^5/
d^2)*d*e^(2*I*d*x + 2*I*c) - sqrt(2)*(a^2*e^(2*I*d*x + 2*I*c) + a^2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*e^(I*d*
x + I*c))*e^(-2*I*d*x - 2*I*c)/a^2))/(d*e^(8*I*d*x + 8*I*c) + 4*d*e^(6*I*d*x + 6*I*c) + 6*d*e^(4*I*d*x + 4*I*c
) + 4*d*e^(2*I*d*x + 2*I*c) + d)

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

[Out]

Timed out

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Giac [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(tan(d*x+c)^3*(a+I*a*tan(d*x+c))^(5/2),x, algorithm="giac")

[Out]

Timed out