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

Optimal. Leaf size=126 \[ -\frac {5 (a+i a \tan (c+d x))^{3/2}}{3 a^2 d}-\frac {\tan ^2(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}+\frac {4 \sqrt {a+i a \tan (c+d x)}}{a d}+\frac {\tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{\sqrt {2} \sqrt {a} d} \]

[Out]

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

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Rubi [A]  time = 0.20, antiderivative size = 126, normalized size of antiderivative = 1.00, 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 = {3558, 3592, 3527, 3480, 206} \[ -\frac {5 (a+i a \tan (c+d x))^{3/2}}{3 a^2 d}-\frac {\tan ^2(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}+\frac {4 \sqrt {a+i a \tan (c+d x)}}{a d}+\frac {\tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{\sqrt {2} \sqrt {a} d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3558

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

Rubi steps

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

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Mathematica [A]  time = 1.11, size = 129, normalized size = 1.02 \[ \frac {18 e^{2 i (c+d x)}+7 e^{4 i (c+d x)}+3 e^{i (c+d x)} \left (1+e^{2 i (c+d x)}\right )^{3/2} \sinh ^{-1}\left (e^{i (c+d x)}\right )+3}{3 \sqrt {2} d \left (1+e^{2 i (c+d x)}\right )^2 \sqrt {\frac {a e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3 + 18*E^((2*I)*(c + d*x)) + 7*E^((4*I)*(c + d*x)) + 3*E^(I*(c + d*x))*(1 + E^((2*I)*(c + d*x)))^(3/2)*ArcSin
h[E^(I*(c + d*x))])/(3*Sqrt[2]*d*Sqrt[(a*E^((2*I)*(c + d*x)))/(1 + E^((2*I)*(c + d*x)))]*(1 + E^((2*I)*(c + d*
x)))^2)

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fricas [B]  time = 0.44, size = 289, normalized size = 2.29 \[ \frac {3 \, \sqrt {2} {\left (a d e^{\left (3 i \, d x + 3 i \, c\right )} + a d e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {1}{a d^{2}}} \log \left (4 \, {\left ({\left (a d e^{\left (2 i \, d x + 2 i \, c\right )} + a d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a d^{2}}} + a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) - 3 \, \sqrt {2} {\left (a d e^{\left (3 i \, d x + 3 i \, c\right )} + a d e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {1}{a d^{2}}} \log \left (-4 \, {\left ({\left (a d e^{\left (2 i \, d x + 2 i \, c\right )} + a d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a d^{2}}} - a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) + 2 \, \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (7 \, e^{\left (4 i \, d x + 4 i \, c\right )} + 18 \, e^{\left (2 i \, d x + 2 i \, c\right )} + 3\right )}}{12 \, {\left (a d e^{\left (3 i \, d x + 3 i \, c\right )} + a d e^{\left (i \, d x + i \, c\right )}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(3*sqrt(2)*(a*d*e^(3*I*d*x + 3*I*c) + a*d*e^(I*d*x + I*c))*sqrt(1/(a*d^2))*log(4*((a*d*e^(2*I*d*x + 2*I*c
) + a*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(1/(a*d^2)) + a*e^(I*d*x + I*c))*e^(-I*d*x - I*c)) - 3*sqrt(2)*
(a*d*e^(3*I*d*x + 3*I*c) + a*d*e^(I*d*x + I*c))*sqrt(1/(a*d^2))*log(-4*((a*d*e^(2*I*d*x + 2*I*c) + a*d)*sqrt(a
/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(1/(a*d^2)) - a*e^(I*d*x + I*c))*e^(-I*d*x - I*c)) + 2*sqrt(2)*sqrt(a/(e^(2*I*
d*x + 2*I*c) + 1))*(7*e^(4*I*d*x + 4*I*c) + 18*e^(2*I*d*x + 2*I*c) + 3))/(a*d*e^(3*I*d*x + 3*I*c) + a*d*e^(I*d
*x + I*c))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\tan \left (d x + c\right )^{3}}{\sqrt {i \, a \tan \left (d x + c\right ) + a}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(tan(d*x + c)^3/sqrt(I*a*tan(d*x + c) + a), x)

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maple [A]  time = 0.18, size = 93, normalized size = 0.74 \[ -\frac {2 \left (\frac {\left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3}-a \sqrt {a +i a \tan \left (d x +c \right )}-\frac {a^{\frac {3}{2}} \sqrt {2}\, \arctanh \left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{4}-\frac {a^{2}}{2 \sqrt {a +i a \tan \left (d x +c \right )}}\right )}{d \,a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.52, size = 120, normalized size = 0.95 \[ -\frac {3 \, \sqrt {2} a^{\frac {7}{2}} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right ) + 8 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} a^{2} - 24 \, \sqrt {i \, a \tan \left (d x + c\right ) + a} a^{3} - \frac {12 \, a^{4}}{\sqrt {i \, a \tan \left (d x + c\right ) + a}}}{12 \, a^{4} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(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))) + 8*(I*a*tan(d*x + c) + a)^(3/2)*a^2 - 24*sqrt(I*a*tan(d*x + c) + a)*a^3 - 12*a^4/sqrt(I*a*tan(
d*x + c) + a))/(a^4*d)

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mupad [B]  time = 4.17, size = 99, normalized size = 0.79 \[ \frac {1}{d\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}+\frac {2\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{a\,d}-\frac {2\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}}{3\,a^2\,d}-\frac {\sqrt {2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2\,\sqrt {a}}\right )\,1{}\mathrm {i}}{2\,\sqrt {a}\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(c + d*x)^3/(a + a*tan(c + d*x)*1i)^(1/2),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\tan ^{3}{\left (c + d x \right )}}{\sqrt {i a \left (\tan {\left (c + d x \right )} - i\right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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