3.140 \(\int \frac {a+i a \tan (e+f x)}{(d \tan (e+f x))^{5/2}} \, dx\)

Optimal. Leaf size=87 \[ \frac {2 \sqrt [4]{-1} a \tan ^{-1}\left (\frac {(-1)^{3/4} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{d^{5/2} f}-\frac {2 i a}{d^2 f \sqrt {d \tan (e+f x)}}-\frac {2 a}{3 d f (d \tan (e+f x))^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.12, antiderivative size = 87, 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 = {3529, 3533, 205} \[ \frac {2 \sqrt [4]{-1} a \tan ^{-1}\left (\frac {(-1)^{3/4} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{d^{5/2} f}-\frac {2 i a}{d^2 f \sqrt {d \tan (e+f x)}}-\frac {2 a}{3 d f (d \tan (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + I*a*Tan[e + f*x])/(d*Tan[e + f*x])^(5/2),x]

[Out]

(2*(-1)^(1/4)*a*ArcTan[((-1)^(3/4)*Sqrt[d*Tan[e + f*x]])/Sqrt[d]])/(d^(5/2)*f) - (2*a)/(3*d*f*(d*Tan[e + f*x])
^(3/2)) - ((2*I)*a)/(d^2*f*Sqrt[d*Tan[e + f*x]])

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 3529

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

Rubi steps

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

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Mathematica [A]  time = 1.33, size = 140, normalized size = 1.61 \[ \frac {2 a e^{-i (e+f x)} \left (1+e^{2 i (e+f x)}\right ) \cos (e+f x) (\tan (e+f x)-i) \left (3 i \tan (e+f x)-3 (i \tan (e+f x))^{3/2} \tanh ^{-1}\left (\sqrt {\frac {-1+e^{2 i (e+f x)}}{1+e^{2 i (e+f x)}}}\right )+1\right )}{3 d^2 f \left (-1+e^{2 i (e+f x)}\right ) \sqrt {d \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + I*a*Tan[e + f*x])/(d*Tan[e + f*x])^(5/2),x]

[Out]

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

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fricas [B]  time = 0.45, size = 390, normalized size = 4.48 \[ -\frac {3 \, {\left (d^{3} f e^{\left (4 i \, f x + 4 i \, e\right )} - 2 \, d^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} + d^{3} f\right )} \sqrt {-\frac {4 i \, a^{2}}{d^{5} f^{2}}} \log \left (\frac {{\left (-2 i \, a d e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (d^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} + d^{3} f\right )} \sqrt {\frac {-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {-\frac {4 i \, a^{2}}{d^{5} f^{2}}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{a}\right ) - 3 \, {\left (d^{3} f e^{\left (4 i \, f x + 4 i \, e\right )} - 2 \, d^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} + d^{3} f\right )} \sqrt {-\frac {4 i \, a^{2}}{d^{5} f^{2}}} \log \left (\frac {{\left (-2 i \, a d e^{\left (2 i \, f x + 2 i \, e\right )} - {\left (d^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} + d^{3} f\right )} \sqrt {\frac {-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {-\frac {4 i \, a^{2}}{d^{5} f^{2}}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{a}\right ) - 16 \, {\left (2 \, a e^{\left (4 i \, f x + 4 i \, e\right )} + a e^{\left (2 i \, f x + 2 i \, e\right )} - a\right )} \sqrt {\frac {-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{12 \, {\left (d^{3} f e^{\left (4 i \, f x + 4 i \, e\right )} - 2 \, d^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} + d^{3} f\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))/(d*tan(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

-1/12*(3*(d^3*f*e^(4*I*f*x + 4*I*e) - 2*d^3*f*e^(2*I*f*x + 2*I*e) + d^3*f)*sqrt(-4*I*a^2/(d^5*f^2))*log((-2*I*
a*d*e^(2*I*f*x + 2*I*e) + (d^3*f*e^(2*I*f*x + 2*I*e) + d^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(-4*I*a^2/(d^5*f^2)))*e^(-2*I*f*x - 2*I*e)/a) - 3*(d^3*f*e^(4*I*f*x + 4*I*e) - 2*d^3*f*e^
(2*I*f*x + 2*I*e) + d^3*f)*sqrt(-4*I*a^2/(d^5*f^2))*log((-2*I*a*d*e^(2*I*f*x + 2*I*e) - (d^3*f*e^(2*I*f*x + 2*
I*e) + d^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(-4*I*a^2/(d^5*f^2)))*e^(-2
*I*f*x - 2*I*e)/a) - 16*(2*a*e^(4*I*f*x + 4*I*e) + a*e^(2*I*f*x + 2*I*e) - a)*sqrt((-I*d*e^(2*I*f*x + 2*I*e) +
I*d)/(e^(2*I*f*x + 2*I*e) + 1)))/(d^3*f*e^(4*I*f*x + 4*I*e) - 2*d^3*f*e^(2*I*f*x + 2*I*e) + d^3*f)

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giac [A]  time = 3.38, size = 109, normalized size = 1.25 \[ -\frac {2}{3} \, a {\left (\frac {3 i \, \sqrt {2} \arctan \left (\frac {16 \, \sqrt {d^{2}} \sqrt {d \tan \left (f x + e\right )}}{8 i \, \sqrt {2} d^{\frac {3}{2}} + 8 \, \sqrt {2} \sqrt {d^{2}} \sqrt {d}}\right )}{d^{\frac {5}{2}} f {\left (\frac {i \, d}{\sqrt {d^{2}}} + 1\right )}} + \frac {3 i \, d \tan \left (f x + e\right ) + d}{\sqrt {d \tan \left (f x + e\right )} d^{3} f \tan \left (f x + e\right )}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))/(d*tan(f*x+e))^(5/2),x, algorithm="giac")

[Out]

-2/3*a*(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)*sqrt(d
)))/(d^(5/2)*f*(I*d/sqrt(d^2) + 1)) + (3*I*d*tan(f*x + e) + d)/(sqrt(d*tan(f*x + e))*d^3*f*tan(f*x + e)))

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maple [B]  time = 0.17, size = 378, normalized size = 4.34 \[ -\frac {a \left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )}{4 f \,d^{3}}-\frac {a \left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )}{2 f \,d^{3}}+\frac {a \left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )}{2 f \,d^{3}}-\frac {i a \sqrt {2}\, \ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )}{4 f \,d^{2} \left (d^{2}\right )^{\frac {1}{4}}}-\frac {i a \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )}{2 f \,d^{2} \left (d^{2}\right )^{\frac {1}{4}}}+\frac {i a \sqrt {2}\, \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )}{2 f \,d^{2} \left (d^{2}\right )^{\frac {1}{4}}}-\frac {2 a}{3 d f \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}-\frac {2 i a}{d^{2} f \sqrt {d \tan \left (f x +e \right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+I*a*tan(f*x+e))/(d*tan(f*x+e))^(5/2),x)

[Out]

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

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maxima [B]  time = 0.85, size = 192, normalized size = 2.21 \[ \frac {\frac {3 \, a {\left (-\frac {\left (2 i + 2\right ) \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} + 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {d}}\right )}{\sqrt {d}} - \frac {\left (2 i + 2\right ) \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} - 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {d}}\right )}{\sqrt {d}} + \frac {\left (i - 1\right ) \, \sqrt {2} \log \left (d \tan \left (f x + e\right ) + \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right )}{\sqrt {d}} - \frac {\left (i - 1\right ) \, \sqrt {2} \log \left (d \tan \left (f x + e\right ) - \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right )}{\sqrt {d}}\right )}}{d} - \frac {8 \, {\left (3 i \, a d \tan \left (f x + e\right ) + a d\right )}}{\left (d \tan \left (f x + e\right )\right )^{\frac {3}{2}} d}}{12 \, d f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))/(d*tan(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

1/12*(3*a*(-(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)*s
qrt(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))/d - 8*(3*I*a*d*tan(f*x + e) + a*d)/((d*tan(f*x + e))
^(3/2)*d))/(d*f)

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mupad [B]  time = 4.94, size = 70, normalized size = 0.80 \[ -\frac {a\,2{}\mathrm {i}}{d^2\,f\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}-\frac {2\,a}{3\,d\,f\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}}+\frac {{\left (-1\right )}^{1/4}\,a\,\mathrm {atanh}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{\sqrt {d}}\right )\,2{}\mathrm {i}}{d^{5/2}\,f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*tan(e + f*x)*1i)/(d*tan(e + f*x))^(5/2),x)

[Out]

((-1)^(1/4)*a*atanh(((-1)^(1/4)*(d*tan(e + f*x))^(1/2))/d^(1/2))*2i)/(d^(5/2)*f) - (2*a)/(3*d*f*(d*tan(e + f*x
))^(3/2)) - (a*2i)/(d^2*f*(d*tan(e + f*x))^(1/2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ i a \left (\int \left (- \frac {i}{\left (d \tan {\left (e + f x \right )}\right )^{\frac {5}{2}}}\right )\, dx + \int \frac {\tan {\left (e + f x \right )}}{\left (d \tan {\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))/(d*tan(f*x+e))**(5/2),x)

[Out]

I*a*(Integral(-I/(d*tan(e + f*x))**(5/2), x) + Integral(tan(e + f*x)/(d*tan(e + f*x))**(5/2), x))

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