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

Optimal. Leaf size=107 \[ \frac {2 (-1)^{3/4} a d^{5/2} \tanh ^{-1}\left (\frac {(-1)^{3/4} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{f}+\frac {2 i a d^2 \sqrt {d \tan (e+f x)}}{f}+\frac {2 a d (d \tan (e+f x))^{3/2}}{3 f}-\frac {2 i a (d \tan (e+f x))^{5/2}}{5 f} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.15, antiderivative size = 107, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {3528, 3533, 208} \[ \frac {2 i a d^2 \sqrt {d \tan (e+f x)}}{f}+\frac {2 (-1)^{3/4} a d^{5/2} \tanh ^{-1}\left (\frac {(-1)^{3/4} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{f}+\frac {2 a d (d \tan (e+f x))^{3/2}}{3 f}-\frac {2 i a (d \tan (e+f x))^{5/2}}{5 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[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))^{5/2} (a-i a \tan (e+f x)) \, dx &=-\frac {2 i a (d \tan (e+f x))^{5/2}}{5 f}+\int (d \tan (e+f x))^{3/2} (i a d+a d \tan (e+f x)) \, dx\\ &=\frac {2 a d (d \tan (e+f x))^{3/2}}{3 f}-\frac {2 i a (d \tan (e+f x))^{5/2}}{5 f}+\int \sqrt {d \tan (e+f x)} \left (-a d^2+i a d^2 \tan (e+f x)\right ) \, dx\\ &=\frac {2 i a d^2 \sqrt {d \tan (e+f x)}}{f}+\frac {2 a d (d \tan (e+f x))^{3/2}}{3 f}-\frac {2 i a (d \tan (e+f x))^{5/2}}{5 f}+\int \frac {-i a d^3-a d^3 \tan (e+f x)}{\sqrt {d \tan (e+f x)}} \, dx\\ &=\frac {2 i a d^2 \sqrt {d \tan (e+f x)}}{f}+\frac {2 a d (d \tan (e+f x))^{3/2}}{3 f}-\frac {2 i a (d \tan (e+f x))^{5/2}}{5 f}-\frac {\left (2 a^2 d^6\right ) \operatorname {Subst}\left (\int \frac {1}{-i a d^4+a d^3 x^2} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{f}\\ &=\frac {2 (-1)^{3/4} a d^{5/2} \tanh ^{-1}\left (\frac {(-1)^{3/4} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{f}+\frac {2 i a d^2 \sqrt {d \tan (e+f x)}}{f}+\frac {2 a d (d \tan (e+f x))^{3/2}}{3 f}-\frac {2 i a (d \tan (e+f x))^{5/2}}{5 f}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.51, size = 90, normalized size = 0.84 \[ \frac {2 a (d \tan (e+f x))^{5/2} \left (15 (-1)^{3/4} \tanh ^{-1}\left ((-1)^{3/4} \sqrt {\tan (e+f x)}\right )+\sqrt {\tan (e+f x)} \left (-3 i \tan ^2(e+f x)+5 \tan (e+f x)+15 i\right )\right )}{15 f \tan ^{\frac {5}{2}}(e+f x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a*(d*Tan[e + f*x])^(5/2)*(15*(-1)^(3/4)*ArcTanh[(-1)^(3/4)*Sqrt[Tan[e + f*x]]] + Sqrt[Tan[e + f*x]]*(15*I +
5*Tan[e + f*x] - (3*I)*Tan[e + f*x]^2)))/(15*f*Tan[e + f*x]^(5/2))

________________________________________________________________________________________

fricas [B]  time = 0.47, size = 341, normalized size = 3.19 \[ \frac {15 \, \sqrt {-\frac {4 i \, a^{2} d^{5}}{f^{2}}} {\left (f e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )} \log \left (\frac {{\left (2 \, a d^{3} + \sqrt {-\frac {4 i \, a^{2} d^{5}}{f^{2}}} {\left (f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )} \sqrt {\frac {-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{f}\right ) - 15 \, \sqrt {-\frac {4 i \, a^{2} d^{5}}{f^{2}}} {\left (f e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )} \log \left (\frac {{\left (2 \, a d^{3} - \sqrt {-\frac {4 i \, a^{2} d^{5}}{f^{2}}} {\left (f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )} \sqrt {\frac {-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{f}\right ) + {\left (104 i \, a d^{2} e^{\left (4 i \, f x + 4 i \, e\right )} + 192 i \, a d^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + 184 i \, a d^{2}\right )} \sqrt {\frac {-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{60 \, {\left (f e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/60*(15*sqrt(-4*I*a^2*d^5/f^2)*(f*e^(4*I*f*x + 4*I*e) + 2*f*e^(2*I*f*x + 2*I*e) + f)*log((2*a*d^3 + sqrt(-4*I
*a^2*d^5/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)/f) - 15*sqrt(-4*I*a^2*d^5/f^2)*(f*e^(4*I*f*x + 4*I*e) + 2*f*e^(2*I*f*x + 2*I*e) + f)*log((2
*a*d^3 - sqrt(-4*I*a^2*d^5/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)/f) + (104*I*a*d^2*e^(4*I*f*x + 4*I*e) + 192*I*a*d^2*e^(2*I*f*x + 2*I*e) +
184*I*a*d^2)*sqrt((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x + 2*I*e) + 1)))/(f*e^(4*I*f*x + 4*I*e) + 2*f*e
^(2*I*f*x + 2*I*e) + f)

________________________________________________________________________________________

giac [A]  time = 0.93, size = 154, normalized size = 1.44 \[ -\frac {2}{15} \, a d^{2} {\left (\frac {15 \, \sqrt {2} \sqrt {d} \arctan \left (\frac {16 \, \sqrt {d^{2}} \sqrt {d \tan \left (f x + e\right )}}{-8 i \, \sqrt {2} d^{\frac {3}{2}} + 8 \, \sqrt {2} \sqrt {d^{2}} \sqrt {d}}\right )}{f {\left (-\frac {i \, d}{\sqrt {d^{2}}} + 1\right )}} + \frac {3 i \, \sqrt {d \tan \left (f x + e\right )} d^{10} f^{4} \tan \left (f x + e\right )^{2} - 5 \, \sqrt {d \tan \left (f x + e\right )} d^{10} f^{4} \tan \left (f x + e\right ) - 15 i \, \sqrt {d \tan \left (f x + e\right )} d^{10} f^{4}}{d^{10} f^{5}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/15*a*d^2*(15*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)*sqr
t(d^2)*sqrt(d)))/(f*(-I*d/sqrt(d^2) + 1)) + (3*I*sqrt(d*tan(f*x + e))*d^10*f^4*tan(f*x + e)^2 - 5*sqrt(d*tan(f
*x + e))*d^10*f^4*tan(f*x + e) - 15*I*sqrt(d*tan(f*x + e))*d^10*f^4)/(d^10*f^5))

________________________________________________________________________________________

maple [B]  time = 0.17, size = 393, normalized size = 3.67 \[ -\frac {2 i a \left (d \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{5 f}+\frac {2 a d \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3 f}+\frac {2 i a \,d^{2} \sqrt {d \tan \left (f x +e \right )}}{f}-\frac {i a \,d^{2} \left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )}{4 f}-\frac {i a \,d^{2} \left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )}{2 f}+\frac {i a \,d^{2} \left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )}{2 f}-\frac {a \,d^{3} \sqrt {2}\, \ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )}{4 f \left (d^{2}\right )^{\frac {1}{4}}}-\frac {a \,d^{3} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )}{2 f \left (d^{2}\right )^{\frac {1}{4}}}+\frac {a \,d^{3} \sqrt {2}\, \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )}{2 f \left (d^{2}\right )^{\frac {1}{4}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [B]  time = 0.59, size = 209, normalized size = 1.95 \[ \frac {15 \, a d^{4} {\left (-\frac {\left (2 i + 2\right ) \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} + 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {d}}\right )}{\sqrt {d}} - \frac {\left (2 i + 2\right ) \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} - 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {d}}\right )}{\sqrt {d}} - \frac {\left (i - 1\right ) \, \sqrt {2} \log \left (d \tan \left (f x + e\right ) + \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right )}{\sqrt {d}} + \frac {\left (i - 1\right ) \, \sqrt {2} \log \left (d \tan \left (f x + e\right ) - \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right )}{\sqrt {d}}\right )} - 24 i \, \left (d \tan \left (f x + e\right )\right )^{\frac {5}{2}} a d + 40 \, \left (d \tan \left (f x + e\right )\right )^{\frac {3}{2}} a d^{2} + 120 i \, \sqrt {d \tan \left (f x + e\right )} a d^{3}}{60 \, d f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/60*(15*a*d^4*(-(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)) - 24*I*(d*tan(f*x + e))^(5/2)*a*d + 40*(d*tan(f
*x + e))^(3/2)*a*d^2 + 120*I*sqrt(d*tan(f*x + e))*a*d^3)/(d*f)

________________________________________________________________________________________

mupad [B]  time = 4.58, size = 143, normalized size = 1.34 \[ \frac {2\,a\,d\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}}{3\,f}-\frac {2\,{\left (-1\right )}^{1/4}\,a\,d^{5/2}\,\mathrm {atan}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{\sqrt {d}}\right )}{f}+\frac {{\left (-1\right )}^{1/4}\,a\,d^{5/2}\,\mathrm {atanh}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{\sqrt {d}}\right )}{f}-\frac {a\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{5/2}\,2{}\mathrm {i}}{5\,f}+\frac {a\,d^2\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,2{}\mathrm {i}}{f}+\frac {{\left (-1\right )}^{1/4}\,a\,d^{5/2}\,\mathrm {atan}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,1{}\mathrm {i}}{\sqrt {d}}\right )\,1{}\mathrm {i}}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________