3.2.82 \(\int \frac {(d \tan (e+f x))^{3/2}}{(a+i a \tan (e+f x))^3} \, dx\) [182]

Optimal. Leaf size=157 \[ \frac {\sqrt [4]{-1} d^{3/2} \text {ArcTan}\left (\frac {(-1)^{3/4} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{8 a^3 f}-\frac {d \sqrt {d \tan (e+f x)}}{6 f (a+i a \tan (e+f x))^3}+\frac {d \sqrt {d \tan (e+f x)}}{6 a f (a+i a \tan (e+f x))^2}+\frac {d \sqrt {d \tan (e+f x)}}{8 f \left (a^3+i a^3 \tan (e+f x)\right )} \]

[Out]

1/8*(-1)^(1/4)*d^(3/2)*arctan((-1)^(3/4)*(d*tan(f*x+e))^(1/2)/d^(1/2))/a^3/f-1/6*d*(d*tan(f*x+e))^(1/2)/f/(a+I
*a*tan(f*x+e))^3+1/6*d*(d*tan(f*x+e))^(1/2)/a/f/(a+I*a*tan(f*x+e))^2+1/8*d*(d*tan(f*x+e))^(1/2)/f/(a^3+I*a^3*t
an(f*x+e))

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Rubi [A]
time = 0.24, antiderivative size = 157, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3639, 3677, 12, 16, 3630, 3614, 211} \begin {gather*} \frac {\sqrt [4]{-1} d^{3/2} \text {ArcTan}\left (\frac {(-1)^{3/4} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{8 a^3 f}+\frac {d \sqrt {d \tan (e+f x)}}{8 f \left (a^3+i a^3 \tan (e+f x)\right )}+\frac {d \sqrt {d \tan (e+f x)}}{6 a f (a+i a \tan (e+f x))^2}-\frac {d \sqrt {d \tan (e+f x)}}{6 f (a+i a \tan (e+f x))^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-1)^(1/4)*d^(3/2)*ArcTan[((-1)^(3/4)*Sqrt[d*Tan[e + f*x]])/Sqrt[d]])/(8*a^3*f) - (d*Sqrt[d*Tan[e + f*x]])/(6
*f*(a + I*a*Tan[e + f*x])^3) + (d*Sqrt[d*Tan[e + f*x]])/(6*a*f*(a + I*a*Tan[e + f*x])^2) + (d*Sqrt[d*Tan[e + f
*x]])/(8*f*(a^3 + I*a^3*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 16

Int[(u_.)*(v_)^(m_.)*((b_)*(v_))^(n_), x_Symbol] :> Dist[1/b^m, Int[u*(b*v)^(m + n), x], x] /; FreeQ[{b, n}, x
] && IntegerQ[m]

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 3630

Int[((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)/((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-(
a*c + b*d))*((c + d*Tan[e + f*x])^n/(2*(b*c - a*d)*f*(a + b*Tan[e + f*x]))), x] + Dist[1/(2*a*(b*c - a*d)), In
t[(c + d*Tan[e + f*x])^(n - 1)*Simp[a*c*d*(n - 1) + b*c^2 + b*d^2*n - d*(b*c - a*d)*(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[0
, n, 1]

Rule 3639

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(-(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 3677

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

Rubi steps

\begin {align*} \int \frac {(d \tan (e+f x))^{3/2}}{(a+i a \tan (e+f x))^3} \, dx &=-\frac {d \sqrt {d \tan (e+f x)}}{6 f (a+i a \tan (e+f x))^3}-\frac {\int \frac {-\frac {a d^2}{2}+\frac {7}{2} i a d^2 \tan (e+f x)}{\sqrt {d \tan (e+f x)} (a+i a \tan (e+f x))^2} \, dx}{6 a^2}\\ &=-\frac {d \sqrt {d \tan (e+f x)}}{6 f (a+i a \tan (e+f x))^3}+\frac {d \sqrt {d \tan (e+f x)}}{6 a f (a+i a \tan (e+f x))^2}-\frac {\int \frac {6 i a^2 d^3 \tan (e+f x)}{\sqrt {d \tan (e+f x)} (a+i a \tan (e+f x))} \, dx}{24 a^4 d}\\ &=-\frac {d \sqrt {d \tan (e+f x)}}{6 f (a+i a \tan (e+f x))^3}+\frac {d \sqrt {d \tan (e+f x)}}{6 a f (a+i a \tan (e+f x))^2}-\frac {\left (i d^2\right ) \int \frac {\tan (e+f x)}{\sqrt {d \tan (e+f x)} (a+i a \tan (e+f x))} \, dx}{4 a^2}\\ &=-\frac {d \sqrt {d \tan (e+f x)}}{6 f (a+i a \tan (e+f x))^3}+\frac {d \sqrt {d \tan (e+f x)}}{6 a f (a+i a \tan (e+f x))^2}-\frac {(i d) \int \frac {\sqrt {d \tan (e+f x)}}{a+i a \tan (e+f x)} \, dx}{4 a^2}\\ &=-\frac {d \sqrt {d \tan (e+f x)}}{6 f (a+i a \tan (e+f x))^3}+\frac {d \sqrt {d \tan (e+f x)}}{6 a f (a+i a \tan (e+f x))^2}+\frac {d \sqrt {d \tan (e+f x)}}{8 f \left (a^3+i a^3 \tan (e+f x)\right )}+\frac {i \int \frac {\frac {1}{2} i a d^2-\frac {1}{2} a d^2 \tan (e+f x)}{\sqrt {d \tan (e+f x)}} \, dx}{8 a^4}\\ &=-\frac {d \sqrt {d \tan (e+f x)}}{6 f (a+i a \tan (e+f x))^3}+\frac {d \sqrt {d \tan (e+f x)}}{6 a f (a+i a \tan (e+f x))^2}+\frac {d \sqrt {d \tan (e+f x)}}{8 f \left (a^3+i a^3 \tan (e+f x)\right )}-\frac {\left (i d^4\right ) \text {Subst}\left (\int \frac {1}{\frac {1}{2} i a d^3+\frac {1}{2} a d^2 x^2} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{16 a^2 f}\\ &=\frac {\sqrt [4]{-1} d^{3/2} \tan ^{-1}\left (\frac {(-1)^{3/4} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{8 a^3 f}-\frac {d \sqrt {d \tan (e+f x)}}{6 f (a+i a \tan (e+f x))^3}+\frac {d \sqrt {d \tan (e+f x)}}{6 a f (a+i a \tan (e+f x))^2}+\frac {d \sqrt {d \tan (e+f x)}}{8 f \left (a^3+i a^3 \tan (e+f x)\right )}\\ \end {align*}

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Mathematica [A]
time = 2.38, size = 158, normalized size = 1.01 \begin {gather*} \frac {d^2 (i \cos (3 (e+f x))+\sin (3 (e+f x))) \left (5 \cos (e+f x)-5 \cos (3 (e+f x))+3 i \sin (e+f x)-3 i \sin (3 (e+f x))+6 \tanh ^{-1}\left (\sqrt {\frac {-1+e^{2 i (e+f x)}}{1+e^{2 i (e+f x)}}}\right ) (\cos (3 (e+f x))+i \sin (3 (e+f x))) \sqrt {i \tan (e+f x)}\right )}{48 a^3 f \sqrt {d \tan (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(d^2*(I*Cos[3*(e + f*x)] + Sin[3*(e + f*x)])*(5*Cos[e + f*x] - 5*Cos[3*(e + f*x)] + (3*I)*Sin[e + f*x] - (3*I)
*Sin[3*(e + f*x)] + 6*ArcTanh[Sqrt[(-1 + E^((2*I)*(e + f*x)))/(1 + E^((2*I)*(e + f*x)))]]*(Cos[3*(e + f*x)] +
I*Sin[3*(e + f*x)])*Sqrt[I*Tan[e + f*x]]))/(48*a^3*f*Sqrt[d*Tan[e + f*x]])

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Maple [A]
time = 0.20, size = 103, normalized size = 0.66

method	result	size
derivativedivides	\(\frac {2 d^{4} \left (\frac {-\left (d \tan \left (f x +e \right )\right )^{\frac {5}{2}}+\frac {10 i d \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}+d^{2} \sqrt {d \tan \left (f x +e \right )}}{16 d^{2} \left (i d \tan \left (f x +e \right )+d \right )^{3}}-\frac {i \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {i d}}\right )}{16 d^{2} \sqrt {i d}}\right )}{f \,a^{3}}\)	\(103\)
default	\(\frac {2 d^{4} \left (\frac {-\left (d \tan \left (f x +e \right )\right )^{\frac {5}{2}}+\frac {10 i d \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}+d^{2} \sqrt {d \tan \left (f x +e \right )}}{16 d^{2} \left (i d \tan \left (f x +e \right )+d \right )^{3}}-\frac {i \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {i d}}\right )}{16 d^{2} \sqrt {i d}}\right )}{f \,a^{3}}\)	\(103\)

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))^3,x,method=_RETURNVERBOSE)

[Out]

2/f/a^3*d^4*(1/16/d^2*(-(d*tan(f*x+e))^(5/2)+10/3*I*d*(d*tan(f*x+e))^(3/2)+d^2*(d*tan(f*x+e))^(1/2))/(I*d*tan(
f*x+e)+d)^3-1/16*I/d^2/(I*d)^(1/2)*arctan((d*tan(f*x+e))^(1/2)/(I*d)^(1/2)))

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

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))^3,x, algorithm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: Error executing code in Maxima: expt: undefined: 0 to a negative e
xponent.

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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. 359 vs. \(2 (134) = 268\).
time = 0.38, size = 359, normalized size = 2.29 \begin {gather*} \frac {{\left (12 \, a^{3} f \sqrt {-\frac {i \, d^{3}}{64 \, a^{6} f^{2}}} e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (-\frac {2 \, {\left (i \, d^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + 8 \, {\left (a^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{3} 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}} \sqrt {-\frac {i \, d^{3}}{64 \, a^{6} f^{2}}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{d}\right ) - 12 \, a^{3} f \sqrt {-\frac {i \, d^{3}}{64 \, a^{6} f^{2}}} e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (-\frac {2 \, {\left (i \, d^{2} e^{\left (2 i \, f x + 2 i \, e\right )} - 8 \, {\left (a^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{3} 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}} \sqrt {-\frac {i \, d^{3}}{64 \, a^{6} f^{2}}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{d}\right ) + {\left (4 \, d e^{\left (6 i \, f x + 6 i \, e\right )} + 4 \, d e^{\left (4 i \, f x + 4 i \, e\right )} - d e^{\left (2 i \, f x + 2 i \, e\right )} - 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}}\right )} e^{\left (-6 i \, f x - 6 i \, e\right )}}{48 \, a^{3} f} \end {gather*}

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))^3,x, algorithm="fricas")

[Out]

1/48*(12*a^3*f*sqrt(-1/64*I*d^3/(a^6*f^2))*e^(6*I*f*x + 6*I*e)*log(-2*(I*d^2*e^(2*I*f*x + 2*I*e) + 8*(a^3*f*e^
(2*I*f*x + 2*I*e) + a^3*f)*sqrt((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(-1/64*I*d^3/(
a^6*f^2)))*e^(-2*I*f*x - 2*I*e)/d) - 12*a^3*f*sqrt(-1/64*I*d^3/(a^6*f^2))*e^(6*I*f*x + 6*I*e)*log(-2*(I*d^2*e^
(2*I*f*x + 2*I*e) - 8*(a^3*f*e^(2*I*f*x + 2*I*e) + a^3*f)*sqrt((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x +
2*I*e) + 1))*sqrt(-1/64*I*d^3/(a^6*f^2)))*e^(-2*I*f*x - 2*I*e)/d) + (4*d*e^(6*I*f*x + 6*I*e) + 4*d*e^(4*I*f*x
+ 4*I*e) - d*e^(2*I*f*x + 2*I*e) - d)*sqrt((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x + 2*I*e) + 1)))*e^(-6*
I*f*x - 6*I*e)/(a^3*f)

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

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))**3,x)

[Out]

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

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Giac [A]
time = 0.73, size = 160, normalized size = 1.02 \begin {gather*} -\frac {1}{24} \, d {\left (\frac {3 i \, \sqrt {2} \sqrt {d} \arctan \left (\frac {8 \, \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 )}{a^{3} f {\left (\frac {i \, d}{\sqrt {d^{2}}} + 1\right )}} + \frac {3 i \, \sqrt {d \tan \left (f x + e\right )} d^{3} \tan \left (f x + e\right )^{2} + 10 \, \sqrt {d \tan \left (f x + e\right )} d^{3} \tan \left (f x + e\right ) - 3 i \, \sqrt {d \tan \left (f x + e\right )} d^{3}}{{\left (d \tan \left (f x + e\right ) - i \, d\right )}^{3} a^{3} f}\right )} \end {gather*}

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))^3,x, algorithm="giac")

[Out]

-1/24*d*(3*I*sqrt(2)*sqrt(d)*arctan(8*sqrt(d^2)*sqrt(d*tan(f*x + e))/(4*I*sqrt(2)*d^(3/2) + 4*sqrt(2)*sqrt(d^2
)*sqrt(d)))/(a^3*f*(I*d/sqrt(d^2) + 1)) + (3*I*sqrt(d*tan(f*x + e))*d^3*tan(f*x + e)^2 + 10*sqrt(d*tan(f*x + e
))*d^3*tan(f*x + e) - 3*I*sqrt(d*tan(f*x + e))*d^3)/((d*tan(f*x + e) - I*d)^3*a^3*f))

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Mupad [B]
time = 4.22, size = 155, normalized size = 0.99 \begin {gather*} \frac {\frac {5\,d^3\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}}{12\,a^3\,f}-\frac {d^4\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,1{}\mathrm {i}}{8\,a^3\,f}+\frac {d^2\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{5/2}\,1{}\mathrm {i}}{8\,a^3\,f}}{-d^3\,{\mathrm {tan}\left (e+f\,x\right )}^3+d^3\,{\mathrm {tan}\left (e+f\,x\right )}^2\,3{}\mathrm {i}+3\,d^3\,\mathrm {tan}\left (e+f\,x\right )-d^3\,1{}\mathrm {i}}-\frac {\sqrt {\frac {1}{256}{}\mathrm {i}}\,{\left (-d\right )}^{3/2}\,\mathrm {atan}\left (\frac {16\,\sqrt {\frac {1}{256}{}\mathrm {i}}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{\sqrt {-d}}\right )\,2{}\mathrm {i}}{a^3\,f} \end {gather*}

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)^3,x)

[Out]

((5*d^3*(d*tan(e + f*x))^(3/2))/(12*a^3*f) - (d^4*(d*tan(e + f*x))^(1/2)*1i)/(8*a^3*f) + (d^2*(d*tan(e + f*x))
^(5/2)*1i)/(8*a^3*f))/(3*d^3*tan(e + f*x) - d^3*1i + d^3*tan(e + f*x)^2*3i - d^3*tan(e + f*x)^3) - ((1i/256)^(
1/2)*(-d)^(3/2)*atan((16*(1i/256)^(1/2)*(d*tan(e + f*x))^(1/2))/(-d)^(1/2))*2i)/(a^3*f)

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