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

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

[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]))

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Rubi [A]  time = 0.372151, antiderivative size = 157, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {3558, 3596, 12, 16, 3549, 3533, 205} \[ \frac{\sqrt [4]{-1} d^{3/2} \tan ^{-1}\left (\frac{(-1)^{3/4} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{8 a^3 f}+\frac{d \sqrt{d \tan (e+f x)}}{8 f \left (a^3+i a^3 \tan (e+f x)\right )}+\frac{d \sqrt{d \tan (e+f x)}}{6 a f (a+i a \tan (e+f x))^2}-\frac{d \sqrt{d \tan (e+f x)}}{6 f (a+i a \tan (e+f x))^3} \]

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 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 3596

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]

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 3549

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)), Int[
(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 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]

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]

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 ) \operatorname{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*}

Mathematica [A]  time = 2.18219, size = 158, normalized size = 1.01 \[ \frac{d^2 (\sin (3 (e+f x))+i \cos (3 (e+f x))) \left (3 i \sin (e+f x)-3 i \sin (3 (e+f x))+5 \cos (e+f x)-5 \cos (3 (e+f x))+6 \sqrt{i \tan (e+f x)} \tanh ^{-1}\left (\sqrt{\frac{-1+e^{2 i (e+f x)}}{1+e^{2 i (e+f x)}}}\right ) (\cos (3 (e+f x))+i \sin (3 (e+f x)))\right )}{48 a^3 f \sqrt{d \tan (e+f x)}} \]

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 = 0.054, size = 148, normalized size = 0.9 \begin{align*}{\frac{-{\frac{i}{8}}{d}^{2}}{f{a}^{3} \left ( -id+d\tan \left ( fx+e \right ) \right ) ^{3}} \left ( d\tan \left ( fx+e \right ) \right ) ^{{\frac{5}{2}}}}-{\frac{5\,{d}^{3}}{12\,f{a}^{3} \left ( -id+d\tan \left ( fx+e \right ) \right ) ^{3}} \left ( d\tan \left ( fx+e \right ) \right ) ^{{\frac{3}{2}}}}+{\frac{{\frac{i}{8}}{d}^{4}}{f{a}^{3} \left ( -id+d\tan \left ( fx+e \right ) \right ) ^{3}}\sqrt{d\tan \left ( fx+e \right ) }}-{\frac{{\frac{i}{8}}{d}^{2}}{f{a}^{3}}\arctan \left ({\sqrt{d\tan \left ( fx+e \right ) }{\frac{1}{\sqrt{id}}}} \right ){\frac{1}{\sqrt{id}}}} \end{align*}

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)

[Out]

-1/8*I/f/a^3*d^2/(-I*d+d*tan(f*x+e))^3*(d*tan(f*x+e))^(5/2)-5/12/f/a^3*d^3/(-I*d+d*tan(f*x+e))^3*(d*tan(f*x+e)
)^(3/2)+1/8*I/f/a^3*d^4/(-I*d+d*tan(f*x+e))^3*(d*tan(f*x+e))^(1/2)-1/8*I/f/a^3*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., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: RuntimeError} \end{align*}

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

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Fricas [B]  time = 2.40073, size = 927, normalized size = 5.9 \begin{align*} -\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{{\left (-2 i \, d^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + 16 \,{\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{{\left (-2 i \, d^{2} e^{\left (2 i \, f x + 2 i \, e\right )} - 16 \,{\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{align*}

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) + 16*(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) - 16*(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(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: AttributeError} \end{align*}

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]

Exception raised: AttributeError

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Giac [A]  time = 1.17094, size = 223, normalized size = 1.42 \begin{align*} -\frac{1}{24} \, d^{4}{\left (\frac{3 i \, \sqrt{2} \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 )}{a^{3} d^{\frac{5}{2}} f{\left (\frac{i \, d}{\sqrt{d^{2}}} + 1\right )}} + \frac{3 i \, \sqrt{d \tan \left (f x + e\right )} d^{2} \tan \left (f x + e\right )^{2} + 10 \, \sqrt{d \tan \left (f x + e\right )} d^{2} \tan \left (f x + e\right ) - 3 i \, \sqrt{d \tan \left (f x + e\right )} d^{2}}{{\left (d \tan \left (f x + e\right ) - i \, d\right )}^{3} a^{3} d^{2} f}\right )} \end{align*}

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^4*(3*I*sqrt(2)*arctan(16*sqrt(d^2)*sqrt(d*tan(f*x + e))/(8*I*sqrt(2)*d^(3/2) + 8*sqrt(2)*sqrt(d^2)*sqr
t(d)))/(a^3*d^(5/2)*f*(I*d/sqrt(d^2) + 1)) + (3*I*sqrt(d*tan(f*x + e))*d^2*tan(f*x + e)^2 + 10*sqrt(d*tan(f*x
+ e))*d^2*tan(f*x + e) - 3*I*sqrt(d*tan(f*x + e))*d^2)/((d*tan(f*x + e) - I*d)^3*a^3*d^2*f))