3.85 \(\int \tan ^3(c+d x) \sqrt{a+i a \tan (c+d x)} \, dx\)

Optimal. Leaf size=127 \[ \frac{2 \tan ^2(c+d x) \sqrt{a+i a \tan (c+d x)}}{5 d}-\frac{2 (a+i a \tan (c+d x))^{3/2}}{15 a d}-\frac{8 \sqrt{a+i a \tan (c+d x)}}{5 d}+\frac{\sqrt{2} \sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{d} \]

[Out]

(Sqrt[2]*Sqrt[a]*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/d - (8*Sqrt[a + I*a*Tan[c + d*x]])/(5*
d) + (2*Tan[c + d*x]^2*Sqrt[a + I*a*Tan[c + d*x]])/(5*d) - (2*(a + I*a*Tan[c + d*x])^(3/2))/(15*a*d)

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Rubi [A]  time = 0.212139, antiderivative size = 127, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.192, Rules used = {3560, 3592, 3527, 3480, 206} \[ \frac{2 \tan ^2(c+d x) \sqrt{a+i a \tan (c+d x)}}{5 d}-\frac{2 (a+i a \tan (c+d x))^{3/2}}{15 a d}-\frac{8 \sqrt{a+i a \tan (c+d x)}}{5 d}+\frac{\sqrt{2} \sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{d} \]

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^3*Sqrt[a + I*a*Tan[c + d*x]],x]

[Out]

(Sqrt[2]*Sqrt[a]*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/d - (8*Sqrt[a + I*a*Tan[c + d*x]])/(5*
d) + (2*Tan[c + d*x]^2*Sqrt[a + I*a*Tan[c + d*x]])/(5*d) - (2*(a + I*a*Tan[c + d*x])^(3/2))/(15*a*d)

Rule 3560

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

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) \sqrt{a+i a \tan (c+d x)} \, dx &=\frac{2 \tan ^2(c+d x) \sqrt{a+i a \tan (c+d x)}}{5 d}-\frac{2 \int \tan (c+d x) \left (2 a+\frac{1}{2} i a \tan (c+d x)\right ) \sqrt{a+i a \tan (c+d x)} \, dx}{5 a}\\ &=\frac{2 \tan ^2(c+d x) \sqrt{a+i a \tan (c+d x)}}{5 d}-\frac{2 (a+i a \tan (c+d x))^{3/2}}{15 a d}-\frac{2 \int \sqrt{a+i a \tan (c+d x)} \left (-\frac{i a}{2}+2 a \tan (c+d x)\right ) \, dx}{5 a}\\ &=-\frac{8 \sqrt{a+i a \tan (c+d x)}}{5 d}+\frac{2 \tan ^2(c+d x) \sqrt{a+i a \tan (c+d x)}}{5 d}-\frac{2 (a+i a \tan (c+d x))^{3/2}}{15 a d}+i \int \sqrt{a+i a \tan (c+d x)} \, dx\\ &=-\frac{8 \sqrt{a+i a \tan (c+d x)}}{5 d}+\frac{2 \tan ^2(c+d x) \sqrt{a+i a \tan (c+d x)}}{5 d}-\frac{2 (a+i a \tan (c+d x))^{3/2}}{15 a d}+\frac{(2 a) \operatorname{Subst}\left (\int \frac{1}{2 a-x^2} \, dx,x,\sqrt{a+i a \tan (c+d x)}\right )}{d}\\ &=\frac{\sqrt{2} \sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{d}-\frac{8 \sqrt{a+i a \tan (c+d x)}}{5 d}+\frac{2 \tan ^2(c+d x) \sqrt{a+i a \tan (c+d x)}}{5 d}-\frac{2 (a+i a \tan (c+d x))^{3/2}}{15 a d}\\ \end{align*}

Mathematica [A]  time = 1.53763, size = 95, normalized size = 0.75 \[ \frac{\sec ^2(c+d x) \sqrt{a+i a \tan (c+d x)} \left (-i \sin (2 (c+d x))-16 \cos (2 (c+d x))+\frac{30 \cos ^3(c+d x) \sinh ^{-1}\left (e^{i (c+d x)}\right )}{\sqrt{1+e^{2 i (c+d x)}}}-10\right )}{15 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[c + d*x]^3*Sqrt[a + I*a*Tan[c + d*x]],x]

[Out]

(Sec[c + d*x]^2*(-10 + (30*ArcSinh[E^(I*(c + d*x))]*Cos[c + d*x]^3)/Sqrt[1 + E^((2*I)*(c + d*x))] - 16*Cos[2*(
c + d*x)] - I*Sin[2*(c + d*x)])*Sqrt[a + I*a*Tan[c + d*x]])/(15*d)

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Maple [A]  time = 0.041, size = 92, normalized size = 0.7 \begin{align*} -2\,{\frac{1}{{a}^{2}d} \left ( 1/5\, \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{5/2}-1/3\, \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{3/2}a+{a}^{2}\sqrt{a+ia\tan \left ( dx+c \right ) }-1/2\,{a}^{5/2}\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((a+I*a*tan(d*x+c))^(1/2)*tan(d*x+c)^3,x)

[Out]

-2/d/a^2*(1/5*(a+I*a*tan(d*x+c))^(5/2)-1/3*(a+I*a*tan(d*x+c))^(3/2)*a+a^2*(a+I*a*tan(d*x+c))^(1/2)-1/2*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 [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((a+I*a*tan(d*x+c))^(1/2)*tan(d*x+c)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.32852, size = 876, normalized size = 6.9 \begin{align*} -\frac{4 \, \sqrt{2} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}{\left (17 \, e^{\left (4 i \, d x + 4 i \, c\right )} + 20 \, e^{\left (2 i \, d x + 2 i \, c\right )} + 15\right )} e^{\left (i \, d x + i \, c\right )} - 15 \, \sqrt{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 )} \sqrt{\frac{a}{d^{2}}} \log \left ({\left (\sqrt{2} d \sqrt{\frac{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}}{\left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}\right ) + 15 \, \sqrt{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 )} \sqrt{\frac{a}{d^{2}}} \log \left (-{\left (\sqrt{2} d \sqrt{\frac{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}}{\left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}\right )}{30 \,{\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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))^(1/2)*tan(d*x+c)^3,x, algorithm="fricas")

[Out]

-1/30*(4*sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(17*e^(4*I*d*x + 4*I*c) + 20*e^(2*I*d*x + 2*I*c) + 15)*e^(I
*d*x + I*c) - 15*sqrt(2)*(d*e^(4*I*d*x + 4*I*c) + 2*d*e^(2*I*d*x + 2*I*c) + d)*sqrt(a/d^2)*log((sqrt(2)*d*sqrt
(a/d^2)*e^(2*I*d*x + 2*I*c) + sqrt(2)*sqrt(a/(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^(-2*I*d*x - 2*I*c)) + 15*sqrt(2)*(d*e^(4*I*d*x + 4*I*c) + 2*d*e^(2*I*d*x + 2*I*c) + d)*sqrt(a/d^2)*log(
-(sqrt(2)*d*sqrt(a/d^2)*e^(2*I*d*x + 2*I*c) - sqrt(2)*sqrt(a/(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^(-2*I*d*x - 2*I*c)))/(d*e^(4*I*d*x + 4*I*c) + 2*d*e^(2*I*d*x + 2*I*c) + d)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{a \left (i \tan{\left (c + d x \right )} + 1\right )} \tan ^{3}{\left (c + d x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))**(1/2)*tan(d*x+c)**3,x)

[Out]

Integral(sqrt(a*(I*tan(c + d*x) + 1))*tan(c + d*x)**3, x)

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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((a+I*a*tan(d*x+c))^(1/2)*tan(d*x+c)^3,x, algorithm="giac")

[Out]

Timed out