3.136 \(\int (d \tan (e+f x))^{3/2} (a+i a \tan (e+f x)) \, dx\)

Optimal. Leaf size=82 \[ \frac {2 \sqrt [4]{-1} a d^{3/2} \tan ^{-1}\left (\frac {(-1)^{3/4} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{f}+\frac {2 a d \sqrt {d \tan (e+f x)}}{f}+\frac {2 i a (d \tan (e+f x))^{3/2}}{3 f} \]

[Out]

2*(-1)^(1/4)*a*d^(3/2)*arctan((-1)^(3/4)*(d*tan(f*x+e))^(1/2)/d^(1/2))/f+2*a*d*(d*tan(f*x+e))^(1/2)/f+2/3*I*a*
(d*tan(f*x+e))^(3/2)/f

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Rubi [A]  time = 0.12, antiderivative size = 82, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {3528, 3533, 205} \[ \frac {2 \sqrt [4]{-1} a d^{3/2} \tan ^{-1}\left (\frac {(-1)^{3/4} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{f}+\frac {2 a d \sqrt {d \tan (e+f x)}}{f}+\frac {2 i a (d \tan (e+f x))^{3/2}}{3 f} \]

Antiderivative was successfully verified.

[In]

Int[(d*Tan[e + f*x])^(3/2)*(a + I*a*Tan[e + f*x]),x]

[Out]

(2*(-1)^(1/4)*a*d^(3/2)*ArcTan[((-1)^(3/4)*Sqrt[d*Tan[e + f*x]])/Sqrt[d]])/f + (2*a*d*Sqrt[d*Tan[e + f*x]])/f
+ (((2*I)/3)*a*(d*Tan[e + f*x])^(3/2))/f

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 3528

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] + Int[(a + b*Tan[e + f*x])^(m - 1)*Simp[a*c - b*d + (b*c + a*d)*Tan[e + f*x
], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && GtQ[m, 0]

Rule 3533

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[(2*c^2)/f, S
ubst[Int[1/(b*c - d*x^2), x], x, Sqrt[b*Tan[e + f*x]]], x] /; FreeQ[{b, c, d, e, f}, x] && EqQ[c^2 + d^2, 0]

Rubi steps

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

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Mathematica [A]  time = 1.04, size = 148, normalized size = 1.80 \[ -\frac {2 a d \left (2 e^{2 i (e+f x)}-4 e^{4 i (e+f x)}+3 \sqrt {\frac {-1+e^{2 i (e+f x)}}{1+e^{2 i (e+f x)}}} \left (1+e^{2 i (e+f x)}\right )^2 \tanh ^{-1}\left (\sqrt {\frac {-1+e^{2 i (e+f x)}}{1+e^{2 i (e+f x)}}}\right )+2\right ) \sqrt {d \tan (e+f x)}}{3 f \left (-1+e^{4 i (e+f x)}\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[(d*Tan[e + f*x])^(3/2)*(a + I*a*Tan[e + f*x]),x]

[Out]

(-2*a*d*(2 + 2*E^((2*I)*(e + f*x)) - 4*E^((4*I)*(e + f*x)) + 3*Sqrt[(-1 + E^((2*I)*(e + f*x)))/(1 + E^((2*I)*(
e + f*x)))]*(1 + E^((2*I)*(e + f*x)))^2*ArcTanh[Sqrt[(-1 + E^((2*I)*(e + f*x)))/(1 + E^((2*I)*(e + f*x)))]])*S
qrt[d*Tan[e + f*x]])/(3*(-1 + E^((4*I)*(e + f*x)))*f)

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fricas [B]  time = 0.44, size = 310, normalized size = 3.78 \[ -\frac {3 \, \sqrt {-\frac {4 i \, a^{2} d^{3}}{f^{2}}} {\left (f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )} \log \left (\frac {{\left (-2 i \, a d^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + \sqrt {-\frac {4 i \, a^{2} d^{3}}{f^{2}}} {\left (f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )} \sqrt {\frac {-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{a d}\right ) - 3 \, \sqrt {-\frac {4 i \, a^{2} d^{3}}{f^{2}}} {\left (f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )} \log \left (\frac {{\left (-2 i \, a d^{2} e^{\left (2 i \, f x + 2 i \, e\right )} - \sqrt {-\frac {4 i \, a^{2} d^{3}}{f^{2}}} {\left (f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )} \sqrt {\frac {-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{a d}\right ) - 16 \, {\left (2 \, a d e^{\left (2 i \, f x + 2 i \, e\right )} + a d\right )} \sqrt {\frac {-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{12 \, {\left (f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(3*sqrt(-4*I*a^2*d^3/f^2)*(f*e^(2*I*f*x + 2*I*e) + f)*log((-2*I*a*d^2*e^(2*I*f*x + 2*I*e) + sqrt(-4*I*a^
2*d^3/f^2)*(f*e^(2*I*f*x + 2*I*e) + f)*sqrt((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x + 2*I*e) + 1)))*e^(-2
*I*f*x - 2*I*e)/(a*d)) - 3*sqrt(-4*I*a^2*d^3/f^2)*(f*e^(2*I*f*x + 2*I*e) + f)*log((-2*I*a*d^2*e^(2*I*f*x + 2*I
*e) - sqrt(-4*I*a^2*d^3/f^2)*(f*e^(2*I*f*x + 2*I*e) + f)*sqrt((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x + 2
*I*e) + 1)))*e^(-2*I*f*x - 2*I*e)/(a*d)) - 16*(2*a*d*e^(2*I*f*x + 2*I*e) + a*d)*sqrt((-I*d*e^(2*I*f*x + 2*I*e)
+ I*d)/(e^(2*I*f*x + 2*I*e) + 1)))/(f*e^(2*I*f*x + 2*I*e) + f)

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giac [A]  time = 1.92, size = 124, normalized size = 1.51 \[ -\frac {2}{3} \, a d {\left (\frac {3 i \, \sqrt {2} \sqrt {d} \arctan \left (\frac {16 \, \sqrt {d^{2}} \sqrt {d \tan \left (f x + e\right )}}{8 i \, \sqrt {2} d^{\frac {3}{2}} + 8 \, \sqrt {2} \sqrt {d^{2}} \sqrt {d}}\right )}{f {\left (\frac {i \, d}{\sqrt {d^{2}}} + 1\right )}} + \frac {-i \, \sqrt {d \tan \left (f x + e\right )} d^{3} f^{2} \tan \left (f x + e\right ) - 3 \, \sqrt {d \tan \left (f x + e\right )} d^{3} f^{2}}{d^{3} f^{3}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*a*d*(3*I*sqrt(2)*sqrt(d)*arctan(16*sqrt(d^2)*sqrt(d*tan(f*x + e))/(8*I*sqrt(2)*d^(3/2) + 8*sqrt(2)*sqrt(d
^2)*sqrt(d)))/(f*(I*d/sqrt(d^2) + 1)) + (-I*sqrt(d*tan(f*x + e))*d^3*f^2*tan(f*x + e) - 3*sqrt(d*tan(f*x + e))
*d^3*f^2)/(d^3*f^3))

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maple [B]  time = 0.18, size = 367, normalized size = 4.48 \[ \frac {2 i a \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3 f}+\frac {2 a d \sqrt {d \tan \left (f x +e \right )}}{f}-\frac {a d \left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )}{4 f}-\frac {a d \left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )}{2 f}+\frac {a d \left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )}{2 f}-\frac {i a \,d^{2} \sqrt {2}\, \ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )}{4 f \left (d^{2}\right )^{\frac {1}{4}}}-\frac {i a \,d^{2} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )}{2 f \left (d^{2}\right )^{\frac {1}{4}}}+\frac {i a \,d^{2} \sqrt {2}\, \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )}{2 f \left (d^{2}\right )^{\frac {1}{4}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*tan(f*x+e))^(3/2)*(a+I*a*tan(f*x+e)),x)

[Out]

2/3*I*a*(d*tan(f*x+e))^(3/2)/f+2*a*d*(d*tan(f*x+e))^(1/2)/f-1/4*a/f*d*(d^2)^(1/4)*2^(1/2)*ln((d*tan(f*x+e)+(d^
2)^(1/4)*(d*tan(f*x+e))^(1/2)*2^(1/2)+(d^2)^(1/2))/(d*tan(f*x+e)-(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)*2^(1/2)+(d^2
)^(1/2)))-1/2*a/f*d*(d^2)^(1/4)*2^(1/2)*arctan(2^(1/2)/(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)+1)+1/2*a/f*d*(d^2)^(1/
4)*2^(1/2)*arctan(-2^(1/2)/(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)+1)-1/4*I*a/f*d^2/(d^2)^(1/4)*2^(1/2)*ln((d*tan(f*x
+e)-(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)*2^(1/2)+(d^2)^(1/2))/(d*tan(f*x+e)+(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)*2^(1/
2)+(d^2)^(1/2)))-1/2*I*a/f*d^2/(d^2)^(1/4)*2^(1/2)*arctan(2^(1/2)/(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)+1)+1/2*I*a/
f*d^2/(d^2)^(1/4)*2^(1/2)*arctan(-2^(1/2)/(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)+1)

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maxima [B]  time = 0.72, size = 193, normalized size = 2.35 \[ \frac {3 \, a d^{3} {\left (-\frac {\left (2 i + 2\right ) \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} + 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {d}}\right )}{\sqrt {d}} - \frac {\left (2 i + 2\right ) \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} - 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {d}}\right )}{\sqrt {d}} + \frac {\left (i - 1\right ) \, \sqrt {2} \log \left (d \tan \left (f x + e\right ) + \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right )}{\sqrt {d}} - \frac {\left (i - 1\right ) \, \sqrt {2} \log \left (d \tan \left (f x + e\right ) - \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right )}{\sqrt {d}}\right )} + 8 i \, \left (d \tan \left (f x + e\right )\right )^{\frac {3}{2}} a d + 24 \, \sqrt {d \tan \left (f x + e\right )} a d^{2}}{12 \, d f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(3*a*d^3*(-(2*I + 2)*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2)*sqrt(d) + 2*sqrt(d*tan(f*x + e)))/sqrt(d))/sqrt(
d) - (2*I + 2)*sqrt(2)*arctan(-1/2*sqrt(2)*(sqrt(2)*sqrt(d) - 2*sqrt(d*tan(f*x + e)))/sqrt(d))/sqrt(d) + (I -
1)*sqrt(2)*log(d*tan(f*x + e) + sqrt(2)*sqrt(d*tan(f*x + e))*sqrt(d) + d)/sqrt(d) - (I - 1)*sqrt(2)*log(d*tan(
f*x + e) - sqrt(2)*sqrt(d*tan(f*x + e))*sqrt(d) + d)/sqrt(d)) + 8*I*(d*tan(f*x + e))^(3/2)*a*d + 24*sqrt(d*tan
(f*x + e))*a*d^2)/(d*f)

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mupad [B]  time = 4.60, size = 65, normalized size = 0.79 \[ \frac {a\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}\,2{}\mathrm {i}}{3\,f}+\frac {2\,a\,d\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{f}+\frac {{\left (-1\right )}^{1/4}\,a\,d^{3/2}\,\mathrm {atanh}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{\sqrt {d}}\right )\,2{}\mathrm {i}}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a*(d*tan(e + f*x))^(3/2)*2i)/(3*f) + (2*a*d*(d*tan(e + f*x))^(1/2))/f + ((-1)^(1/4)*a*d^(3/2)*atanh(((-1)^(1/
4)*(d*tan(e + f*x))^(1/2))/d^(1/2))*2i)/f

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ i a \left (\int \left (- i \left (d \tan {\left (e + f x \right )}\right )^{\frac {3}{2}}\right )\, dx + \int \left (d \tan {\left (e + f x \right )}\right )^{\frac {3}{2}} \tan {\left (e + f x \right )}\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*tan(f*x+e))**(3/2)*(a+I*a*tan(f*x+e)),x)

[Out]

I*a*(Integral(-I*(d*tan(e + f*x))**(3/2), x) + Integral((d*tan(e + f*x))**(3/2)*tan(e + f*x), x))

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