3.2.60 \(\int \frac {(a+i a \tan (e+f x))^3}{(d \tan (e+f x))^{3/2}} \, dx\) [160]

Optimal. Leaf size=80 \[ -\frac {8 (-1)^{3/4} a^3 \text {ArcTan}\left (\frac {(-1)^{3/4} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{d^{3/2} f}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right )}{d f \sqrt {d \tan (e+f x)}} \]

[Out]

-8*(-1)^(3/4)*a^3*arctan((-1)^(3/4)*(d*tan(f*x+e))^(1/2)/d^(1/2))/d^(3/2)/f-2*(a^3+I*a^3*tan(f*x+e))/d/f/(d*ta
n(f*x+e))^(1/2)

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Rubi [A]
time = 0.07, antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3634, 12, 3614, 211} \begin {gather*} -\frac {8 (-1)^{3/4} a^3 \text {ArcTan}\left (\frac {(-1)^{3/4} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{d^{3/2} f}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right )}{d f \sqrt {d \tan (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-8*(-1)^(3/4)*a^3*ArcTan[((-1)^(3/4)*Sqrt[d*Tan[e + f*x]])/Sqrt[d]])/(d^(3/2)*f) - (2*(a^3 + I*a^3*Tan[e + f*
x]))/(d*f*Sqrt[d*Tan[e + f*x]])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 211

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

Rule 3614

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]

Rule 3634

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

Rubi steps

\begin {align*} \int \frac {(a+i a \tan (e+f x))^3}{(d \tan (e+f x))^{3/2}} \, dx &=-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right )}{d f \sqrt {d \tan (e+f x)}}-\frac {2 \int -\frac {2 i a^2 d (a+i a \tan (e+f x))}{\sqrt {d \tan (e+f x)}} \, dx}{d^2}\\ &=-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right )}{d f \sqrt {d \tan (e+f x)}}+\frac {\left (4 i a^2\right ) \int \frac {a+i a \tan (e+f x)}{\sqrt {d \tan (e+f x)}} \, dx}{d}\\ &=-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right )}{d f \sqrt {d \tan (e+f x)}}+\frac {\left (8 i a^4\right ) \text {Subst}\left (\int \frac {1}{a d-i a x^2} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{d f}\\ &=-\frac {8 (-1)^{3/4} a^3 \tan ^{-1}\left (\frac {(-1)^{3/4} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{d^{3/2} f}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right )}{d f \sqrt {d \tan (e+f x)}}\\ \end {align*}

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Mathematica [A]
time = 1.69, size = 156, normalized size = 1.95 \begin {gather*} \frac {2 a^3 e^{-3 i (e+f x)} (-i \cos (3 (e+f x))+\sin (3 (e+f x))) \left (-4 \tanh ^{-1}\left (\sqrt {\frac {-1+e^{2 i (e+f x)}}{1+e^{2 i (e+f x)}}}\right ) \tan (e+f x)+\sqrt {i \tan (e+f x)} (-i+\tan (e+f x))\right )}{d \sqrt {\frac {-1+e^{2 i (e+f x)}}{1+e^{2 i (e+f x)}}} f \sqrt {d \tan (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 308 vs. \(2 (67 ) = 134\).
time = 0.11, size = 309, normalized size = 3.86

method	result	size
derivativedivides	\(\frac {2 a^{3} \left (-i \sqrt {d \tan \left (f x +e \right )}+4 d \left (\frac {i \left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 d}-\frac {\sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 \left (d^{2}\right )^{\frac {1}{4}}}\right )-\frac {d}{\sqrt {d \tan \left (f x +e \right )}}\right )}{f \,d^{2}}\)	\(309\)
default	\(\frac {2 a^{3} \left (-i \sqrt {d \tan \left (f x +e \right )}+4 d \left (\frac {i \left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 d}-\frac {\sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 \left (d^{2}\right )^{\frac {1}{4}}}\right )-\frac {d}{\sqrt {d \tan \left (f x +e \right )}}\right )}{f \,d^{2}}\)	\(309\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/f*a^3/d^2*(-I*(d*tan(f*x+e))^(1/2)+4*d*(1/8*I/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)))+2*arctan(
2^(1/2)/(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)+1)-2*arctan(-2^(1/2)/(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)+1))-1/8/(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)))+2*arctan(2^(1/2)/(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)+1)-2*arctan(-2^(
1/2)/(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)+1)))-d/(d*tan(f*x+e))^(1/2))

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Maxima [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 202 vs. \(2 (71) = 142\).
time = 0.55, size = 202, normalized size = 2.52 \begin {gather*} -\frac {a^{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 )} + \frac {2 \, a^{3}}{\sqrt {d \tan \left (f x + e\right )}} + \frac {2 i \, \sqrt {d \tan \left (f x + e\right )} a^{3}}{d}}{d f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 357 vs. \(2 (71) = 142\).
time = 0.36, size = 357, normalized size = 4.46 \begin {gather*} \frac {-16 i \, a^{3} \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}} e^{\left (2 i \, f x + 2 i \, e\right )} - {\left (d^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} - d^{2} f\right )} \sqrt {\frac {64 i \, a^{6}}{d^{3} f^{2}}} \log \left (\frac {{\left (-8 i \, a^{3} d e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (i \, d^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d^{2} f\right )} \sqrt {\frac {64 i \, a^{6}}{d^{3} f^{2}}} \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 )}}{4 \, a^{3}}\right ) + {\left (d^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} - d^{2} f\right )} \sqrt {\frac {64 i \, a^{6}}{d^{3} f^{2}}} \log \left (\frac {{\left (-8 i \, a^{3} d e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (-i \, d^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} - i \, d^{2} f\right )} \sqrt {\frac {64 i \, a^{6}}{d^{3} f^{2}}} \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 )}}{4 \, a^{3}}\right )}{4 \, {\left (d^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} - d^{2} f\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(-16*I*a^3*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) - (d^2*f*e
^(2*I*f*x + 2*I*e) - d^2*f)*sqrt(64*I*a^6/(d^3*f^2))*log(1/4*(-8*I*a^3*d*e^(2*I*f*x + 2*I*e) + (I*d^2*f*e^(2*I
*f*x + 2*I*e) + I*d^2*f)*sqrt(64*I*a^6/(d^3*f^2))*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^3) + (d^2*f*e^(2*I*f*x + 2*I*e) - d^2*f)*sqrt(64*I*a^6/(d^3*f^2))*log(1/4*(-8*I*a
^3*d*e^(2*I*f*x + 2*I*e) + (-I*d^2*f*e^(2*I*f*x + 2*I*e) - I*d^2*f)*sqrt(64*I*a^6/(d^3*f^2))*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^3))/(d^2*f*e^(2*I*f*x + 2*I*e) - d^2*f
)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.72, size = 115, normalized size = 1.44 \begin {gather*} -\frac {2 \, {\left (-\frac {4 i \, \sqrt {2} a^{3} \arctan \left (-\frac {8 i \, \sqrt {d^{2}} \sqrt {d \tan \left (f x + e\right )}}{-4 i \, \sqrt {2} d^{\frac {3}{2}} + 4 \, \sqrt {2} \sqrt {d^{2}} \sqrt {d}}\right )}{\sqrt {d} f {\left (-\frac {i \, d}{\sqrt {d^{2}}} + 1\right )}} + \frac {a^{3}}{\sqrt {d \tan \left (f x + e\right )} f} + \frac {i \, \sqrt {d \tan \left (f x + e\right )} a^{3}}{d f}\right )}}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*(-4*I*sqrt(2)*a^3*arctan(-8*I*sqrt(d^2)*sqrt(d*tan(f*x + e))/(-4*I*sqrt(2)*d^(3/2) + 4*sqrt(2)*sqrt(d^2)*sq
rt(d)))/(sqrt(d)*f*(-I*d/sqrt(d^2) + 1)) + a^3/(sqrt(d*tan(f*x + e))*f) + I*sqrt(d*tan(f*x + e))*a^3/(d*f))/d

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Mupad [B]
time = 4.33, size = 77, normalized size = 0.96 \begin {gather*} -\frac {2\,a^3}{d\,f\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}-\frac {a^3\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,2{}\mathrm {i}}{d^2\,f}+\frac {2\,\sqrt {16{}\mathrm {i}}\,a^3\,\mathrm {atanh}\left (\frac {\sqrt {16{}\mathrm {i}}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{4\,\sqrt {d}}\right )}{d^{3/2}\,f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*16i^(1/2)*a^3*atanh((16i^(1/2)*(d*tan(e + f*x))^(1/2))/(4*d^(1/2))))/(d^(3/2)*f) - (a^3*(d*tan(e + f*x))^(1
/2)*2i)/(d^2*f) - (2*a^3)/(d*f*(d*tan(e + f*x))^(1/2))

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