3.1131 \(\int \frac {(a+i a \tan (e+f x))^3}{(c+d \tan (e+f x))^{5/2}} \, dx\)

Optimal. Leaf size=158 \[ \frac {4 a^3 (-d+i c) (c-4 i d)}{3 d^2 f (c-i d)^2 \sqrt {c+d \tan (e+f x)}}+\frac {2 (c+i d) \left (a^3+i a^3 \tan (e+f x)\right )}{3 d f (c-i d) (c+d \tan (e+f x))^{3/2}}-\frac {8 i a^3 \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{f (c-i d)^{5/2}} \]

[Out]

-8*I*a^3*arctanh((c+d*tan(f*x+e))^(1/2)/(c-I*d)^(1/2))/(c-I*d)^(5/2)/f+4/3*a^3*(I*c-d)*(c-4*I*d)/(c-I*d)^2/d^2
/f/(c+d*tan(f*x+e))^(1/2)+2/3*(c+I*d)*(a^3+I*a^3*tan(f*x+e))/(c-I*d)/d/f/(c+d*tan(f*x+e))^(3/2)

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Rubi [A]  time = 0.42, antiderivative size = 158, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3553, 3591, 3537, 63, 208} \[ \frac {4 a^3 (-d+i c) (c-4 i d)}{3 d^2 f (c-i d)^2 \sqrt {c+d \tan (e+f x)}}+\frac {2 (c+i d) \left (a^3+i a^3 \tan (e+f x)\right )}{3 d f (c-i d) (c+d \tan (e+f x))^{3/2}}-\frac {8 i a^3 \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{f (c-i d)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + I*a*Tan[e + f*x])^3/(c + d*Tan[e + f*x])^(5/2),x]

[Out]

((-8*I)*a^3*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]])/((c - I*d)^(5/2)*f) + (2*(c + I*d)*(a^3 + I*a^3*T
an[e + f*x]))/(3*(c - I*d)*d*f*(c + d*Tan[e + f*x])^(3/2)) + (4*a^3*(I*c - d)*(c - (4*I)*d))/(3*(c - I*d)^2*d^
2*f*Sqrt[c + d*Tan[e + f*x]])

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

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 3537

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

Rule 3553

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Si
mp[(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] /; Fr
eeQ[{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] && LtQ[
n, -1] && (IntegerQ[m] || IntegersQ[2*m, 2*n])

Rule 3591

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

Rubi steps

\begin {align*} \int \frac {(a+i a \tan (e+f x))^3}{(c+d \tan (e+f x))^{5/2}} \, dx &=\frac {2 (c+i d) \left (a^3+i a^3 \tan (e+f x)\right )}{3 (c-i d) d f (c+d \tan (e+f x))^{3/2}}-\frac {2 \int \frac {(a+i a \tan (e+f x)) \left (-a^2 (c+4 i d)+a^2 (i c+2 d) \tan (e+f x)\right )}{(c+d \tan (e+f x))^{3/2}} \, dx}{3 d (i c+d)}\\ &=\frac {2 (c+i d) \left (a^3+i a^3 \tan (e+f x)\right )}{3 (c-i d) d f (c+d \tan (e+f x))^{3/2}}+\frac {4 a^3 (i c-d) (c-4 i d)}{3 (c-i d)^2 d^2 f \sqrt {c+d \tan (e+f x)}}-\frac {2 \int \frac {-6 a^3 (i c-d) d+6 a^3 (c+i d) d \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}} \, dx}{3 d (i c+d) \left (c^2+d^2\right )}\\ &=\frac {2 (c+i d) \left (a^3+i a^3 \tan (e+f x)\right )}{3 (c-i d) d f (c+d \tan (e+f x))^{3/2}}+\frac {4 a^3 (i c-d) (c-4 i d)}{3 (c-i d)^2 d^2 f \sqrt {c+d \tan (e+f x)}}+\frac {\left (24 a^6 (c+i d) d\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {c+\frac {x}{6 a^3 (c+i d)}} \left (36 a^6 (c+i d)^2 d^2-6 a^3 (i c-d) d x\right )} \, dx,x,6 a^3 (c+i d) d \tan (e+f x)\right )}{(c-i d)^2 f}\\ &=\frac {2 (c+i d) \left (a^3+i a^3 \tan (e+f x)\right )}{3 (c-i d) d f (c+d \tan (e+f x))^{3/2}}+\frac {4 a^3 (i c-d) (c-4 i d)}{3 (c-i d)^2 d^2 f \sqrt {c+d \tan (e+f x)}}+\frac {\left (288 a^9 (c+i d)^2 d\right ) \operatorname {Subst}\left (\int \frac {1}{36 a^6 c (i c-d) (c+i d) d+36 a^6 (c+i d)^2 d^2-36 a^6 (i c-d) (c+i d) d x^2} \, dx,x,\sqrt {c+d \tan (e+f x)}\right )}{(c-i d)^2 f}\\ &=-\frac {8 i a^3 \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{(c-i d)^{5/2} f}+\frac {2 (c+i d) \left (a^3+i a^3 \tan (e+f x)\right )}{3 (c-i d) d f (c+d \tan (e+f x))^{3/2}}+\frac {4 a^3 (i c-d) (c-4 i d)}{3 (c-i d)^2 d^2 f \sqrt {c+d \tan (e+f x)}}\\ \end {align*}

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Mathematica [A]  time = 9.24, size = 232, normalized size = 1.47 \[ \frac {a^3 (\cos (e+f x)+i \sin (e+f x))^3 \left (\frac {2 (c+i d) (\sin (3 e)+i \cos (3 e)) \cos (e+f x) \sqrt {c+d \tan (e+f x)} \left (\left (2 c^2-9 i c d-d^2\right ) \cos (e+f x)+3 d (c-3 i d) \sin (e+f x)\right )}{3 d^2 (c-i d)^2 (c \cos (e+f x)+d \sin (e+f x))^2}-\frac {8 i e^{-3 i e} \tanh ^{-1}\left (\frac {\sqrt {c-\frac {i d \left (-1+e^{2 i (e+f x)}\right )}{1+e^{2 i (e+f x)}}}}{\sqrt {c-i d}}\right )}{(c-i d)^{5/2}}\right )}{f (\cos (f x)+i \sin (f x))^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + I*a*Tan[e + f*x])^3/(c + d*Tan[e + f*x])^(5/2),x]

[Out]

(a^3*(Cos[e + f*x] + I*Sin[e + f*x])^3*(((-8*I)*ArcTanh[Sqrt[c - (I*d*(-1 + E^((2*I)*(e + f*x))))/(1 + E^((2*I
)*(e + f*x)))]/Sqrt[c - I*d]])/((c - I*d)^(5/2)*E^((3*I)*e)) + (2*(c + I*d)*Cos[e + f*x]*(I*Cos[3*e] + Sin[3*e
])*((2*c^2 - (9*I)*c*d - d^2)*Cos[e + f*x] + 3*(c - (3*I)*d)*d*Sin[e + f*x])*Sqrt[c + d*Tan[e + f*x]])/(3*(c -
I*d)^2*d^2*(c*Cos[e + f*x] + d*Sin[e + f*x])^2)))/(f*(Cos[f*x] + I*Sin[f*x])^3)

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fricas [B]  time = 0.62, size = 972, normalized size = 6.15 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^3/(c+d*tan(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

(sqrt(-64*I*a^6/((I*c^5 + 5*c^4*d - 10*I*c^3*d^2 - 10*c^2*d^3 + 5*I*c*d^4 + d^5)*f^2))*((3*c^4*d^2 - 12*I*c^3*
d^3 - 18*c^2*d^4 + 12*I*c*d^5 + 3*d^6)*f*e^(4*I*f*x + 4*I*e) + (6*c^4*d^2 - 12*I*c^3*d^3 - 12*I*c*d^5 - 6*d^6)
*f*e^(2*I*f*x + 2*I*e) + 3*(c^4*d^2 + 2*c^2*d^4 + d^6)*f)*log(1/4*(8*a^3*c + sqrt(-64*I*a^6/((I*c^5 + 5*c^4*d
- 10*I*c^3*d^2 - 10*c^2*d^3 + 5*I*c*d^4 + d^5)*f^2))*((I*c^3 + 3*c^2*d - 3*I*c*d^2 - d^3)*f*e^(2*I*f*x + 2*I*e
) + (I*c^3 + 3*c^2*d - 3*I*c*d^2 - d^3)*f)*sqrt(((c - I*d)*e^(2*I*f*x + 2*I*e) + c + I*d)/(e^(2*I*f*x + 2*I*e)
+ 1)) + (8*a^3*c - 8*I*a^3*d)*e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I*e)/a^3) - sqrt(-64*I*a^6/((I*c^5 + 5*c^4
*d - 10*I*c^3*d^2 - 10*c^2*d^3 + 5*I*c*d^4 + d^5)*f^2))*((3*c^4*d^2 - 12*I*c^3*d^3 - 18*c^2*d^4 + 12*I*c*d^5 +
3*d^6)*f*e^(4*I*f*x + 4*I*e) + (6*c^4*d^2 - 12*I*c^3*d^3 - 12*I*c*d^5 - 6*d^6)*f*e^(2*I*f*x + 2*I*e) + 3*(c^4
*d^2 + 2*c^2*d^4 + d^6)*f)*log(1/4*(8*a^3*c + sqrt(-64*I*a^6/((I*c^5 + 5*c^4*d - 10*I*c^3*d^2 - 10*c^2*d^3 + 5
*I*c*d^4 + d^5)*f^2))*((-I*c^3 - 3*c^2*d + 3*I*c*d^2 + d^3)*f*e^(2*I*f*x + 2*I*e) + (-I*c^3 - 3*c^2*d + 3*I*c*
d^2 + d^3)*f)*sqrt(((c - I*d)*e^(2*I*f*x + 2*I*e) + c + I*d)/(e^(2*I*f*x + 2*I*e) + 1)) + (8*a^3*c - 8*I*a^3*d
)*e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I*e)/a^3) + (16*I*a^3*c^3 + 32*a^3*c^2*d + 112*I*a^3*c*d^2 - 64*a^3*d^3
+ (16*I*a^3*c^3 + 80*a^3*c^2*d + 16*I*a^3*c*d^2 + 80*a^3*d^3)*e^(4*I*f*x + 4*I*e) + (32*I*a^3*c^3 + 112*a^3*c
^2*d + 128*I*a^3*c*d^2 + 16*a^3*d^3)*e^(2*I*f*x + 2*I*e))*sqrt(((c - I*d)*e^(2*I*f*x + 2*I*e) + c + I*d)/(e^(2
*I*f*x + 2*I*e) + 1)))/((12*c^4*d^2 - 48*I*c^3*d^3 - 72*c^2*d^4 + 48*I*c*d^5 + 12*d^6)*f*e^(4*I*f*x + 4*I*e) +
(24*c^4*d^2 - 48*I*c^3*d^3 - 48*I*c*d^5 - 24*d^6)*f*e^(2*I*f*x + 2*I*e) + 12*(c^4*d^2 + 2*c^2*d^4 + d^6)*f)

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giac [B]  time = 1.87, size = 312, normalized size = 1.97 \[ \frac {32 \, a^{3} \arctan \left (\frac {4 \, {\left (\sqrt {d \tan \left (f x + e\right ) + c} c - \sqrt {c^{2} + d^{2}} \sqrt {d \tan \left (f x + e\right ) + c}\right )}}{c \sqrt {-8 \, c + 8 \, \sqrt {c^{2} + d^{2}}} - i \, \sqrt {-8 \, c + 8 \, \sqrt {c^{2} + d^{2}}} d - \sqrt {c^{2} + d^{2}} \sqrt {-8 \, c + 8 \, \sqrt {c^{2} + d^{2}}}}\right )}{{\left (-i \, c^{2} f - 2 \, c d f + i \, d^{2} f\right )} \sqrt {-8 \, c + 8 \, \sqrt {c^{2} + d^{2}}} {\left (-\frac {i \, d}{c - \sqrt {c^{2} + d^{2}}} + 1\right )}} + \frac {{\left (6 i \, d \tan \left (f x + e\right ) + 6 i \, c\right )} a^{3} c^{2} - 2 i \, a^{3} c^{3} + 12 \, {\left (d \tan \left (f x + e\right ) + c\right )} a^{3} c d + 2 \, a^{3} c^{2} d + {\left (18 i \, d \tan \left (f x + e\right ) + 18 i \, c\right )} a^{3} d^{2} - 2 i \, a^{3} c d^{2} + 2 \, a^{3} d^{3}}{{\left (3 \, c^{2} d^{2} f - 6 i \, c d^{3} f - 3 \, d^{4} f\right )} {\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^3/(c+d*tan(f*x+e))^(5/2),x, algorithm="giac")

[Out]

32*a^3*arctan(4*(sqrt(d*tan(f*x + e) + c)*c - sqrt(c^2 + d^2)*sqrt(d*tan(f*x + e) + c))/(c*sqrt(-8*c + 8*sqrt(
c^2 + d^2)) - I*sqrt(-8*c + 8*sqrt(c^2 + d^2))*d - sqrt(c^2 + d^2)*sqrt(-8*c + 8*sqrt(c^2 + d^2))))/((-I*c^2*f
- 2*c*d*f + I*d^2*f)*sqrt(-8*c + 8*sqrt(c^2 + d^2))*(-I*d/(c - sqrt(c^2 + d^2)) + 1)) + ((6*I*d*tan(f*x + e)
+ 6*I*c)*a^3*c^2 - 2*I*a^3*c^3 + 12*(d*tan(f*x + e) + c)*a^3*c*d + 2*a^3*c^2*d + (18*I*d*tan(f*x + e) + 18*I*c
)*a^3*d^2 - 2*I*a^3*c*d^2 + 2*a^3*d^3)/((3*c^2*d^2*f - 6*I*c*d^3*f - 3*d^4*f)*(d*tan(f*x + e) + c)^(3/2))

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maple [B]  time = 0.30, size = 2772, normalized size = 17.54 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+I*a*tan(f*x+e))^3/(c+d*tan(f*x+e))^(5/2),x)

[Out]

-2/3/f*a^3*d/(c^2+d^2)/(c+d*tan(f*x+e))^(3/2)-6*I/f*a^3*d^2/(c^2+d^2)^(5/2)/(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*ln((
c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)-d*tan(f*x+e)-c-(c^2+d^2)^(1/2))*c+6*I/f*a^3*d^2/(c^2+d^2)^
(5/2)/(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*ln(d*tan(f*x+e)+c+(c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)+(c^
2+d^2)^(1/2))*c+12*I/f*a^3*d^2/(c^2+d^2)^(5/2)/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan((2*(c+d*tan(f*x+e))^(1/2)+
(2*(c^2+d^2)^(1/2)+2*c)^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*c-12*I/f*a^3*d^2/(c^2+d^2)^(5/2)/(2*(c^2+d^2)^(1
/2)-2*c)^(1/2)*arctan(((2*(c^2+d^2)^(1/2)+2*c)^(1/2)-2*(c+d*tan(f*x+e))^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*
c-4/f*a^3*d/(c^2+d^2)^2/(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*ln((c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)-
d*tan(f*x+e)-c-(c^2+d^2)^(1/2))*c+6/f*a^3*d/(c^2+d^2)^(5/2)/(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*ln(d*tan(f*x+e)+c+(c
+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)+(c^2+d^2)^(1/2))*c^2-12/f*a^3*d/(c^2+d^2)^(5/2)/(2*(c^2+d^2
)^(1/2)-2*c)^(1/2)*arctan(((2*(c^2+d^2)^(1/2)+2*c)^(1/2)-2*(c+d*tan(f*x+e))^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/
2))*c^2+8/f*a^3*d/(c^2+d^2)^2/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan(((2*(c^2+d^2)^(1/2)+2*c)^(1/2)-2*(c+d*tan(f
*x+e))^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*c-8/f*a^3*d/(c^2+d^2)^2/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan((2*(
c+d*tan(f*x+e))^(1/2)+(2*(c^2+d^2)^(1/2)+2*c)^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*c+4*I/f*a^3/(c^2+d^2)^(5/2
)/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan(((2*(c^2+d^2)^(1/2)+2*c)^(1/2)-2*(c+d*tan(f*x+e))^(1/2))/(2*(c^2+d^2)^(
1/2)-2*c)^(1/2))*c^3-4*I/f*a^3/(c^2+d^2)^(5/2)/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan((2*(c+d*tan(f*x+e))^(1/2)+
(2*(c^2+d^2)^(1/2)+2*c)^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*c^3-2*I/f*a^3/(c^2+d^2)^(5/2)/(2*(c^2+d^2)^(1/2)
+2*c)^(1/2)*ln(d*tan(f*x+e)+c+(c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)+(c^2+d^2)^(1/2))*c^3-2/3*I/
f*a^3/d^2/(c^2+d^2)/(c+d*tan(f*x+e))^(3/2)*c^3-2*I/f*a^3*d^2/(c^2+d^2)^2/(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*ln((c+d
*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)-d*tan(f*x+e)-c-(c^2+d^2)^(1/2))+2*I/f*a^3/d^2/(c^2+d^2)^2/(c+
d*tan(f*x+e))^(1/2)*c^4+2*I/f*a^3*d^2/(c^2+d^2)^2/(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*ln(d*tan(f*x+e)+c+(c+d*tan(f*x
+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)+(c^2+d^2)^(1/2))-4*I/f*a^3*d^2/(c^2+d^2)^2/(2*(c^2+d^2)^(1/2)-2*c)^(1
/2)*arctan((2*(c+d*tan(f*x+e))^(1/2)+(2*(c^2+d^2)^(1/2)+2*c)^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))+4*I/f*a^3*d
^2/(c^2+d^2)^2/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan(((2*(c^2+d^2)^(1/2)+2*c)^(1/2)-2*(c+d*tan(f*x+e))^(1/2))/(
2*(c^2+d^2)^(1/2)-2*c)^(1/2))+2*I/f*a^3/(c^2+d^2)^(5/2)/(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*ln((c+d*tan(f*x+e))^(1/2
)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)-d*tan(f*x+e)-c-(c^2+d^2)^(1/2))*c^3+12/f*a^3*d/(c^2+d^2)^(5/2)/(2*(c^2+d^2)^(1
/2)-2*c)^(1/2)*arctan((2*(c+d*tan(f*x+e))^(1/2)+(2*(c^2+d^2)^(1/2)+2*c)^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*
c^2-6/f*a^3*d/(c^2+d^2)^(5/2)/(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*ln((c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^
(1/2)-d*tan(f*x+e)-c-(c^2+d^2)^(1/2))*c^2-4*I/f*a^3/(c^2+d^2)^2/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan(((2*(c^2+
d^2)^(1/2)+2*c)^(1/2)-2*(c+d*tan(f*x+e))^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*c^2+4*I/f*a^3/(c^2+d^2)^2/(2*(c
^2+d^2)^(1/2)-2*c)^(1/2)*arctan((2*(c+d*tan(f*x+e))^(1/2)+(2*(c^2+d^2)^(1/2)+2*c)^(1/2))/(2*(c^2+d^2)^(1/2)-2*
c)^(1/2))*c^2+4/f*a^3*d/(c^2+d^2)^2/(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*ln(d*tan(f*x+e)+c+(c+d*tan(f*x+e))^(1/2)*(2*
(c^2+d^2)^(1/2)+2*c)^(1/2)+(c^2+d^2)^(1/2))*c-2*I/f*a^3/(c^2+d^2)^2/(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*ln(d*tan(f*x
+e)+c+(c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)+(c^2+d^2)^(1/2))*c^2+2*I/f*a^3/(c^2+d^2)^2/(2*(c^2+
d^2)^(1/2)+2*c)^(1/2)*ln((c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)-d*tan(f*x+e)-c-(c^2+d^2)^(1/2))*
c^2+2*I/f*a^3/(c^2+d^2)/(c+d*tan(f*x+e))^(3/2)*c-6*I/f*a^3*d^2/(c^2+d^2)^2/(c+d*tan(f*x+e))^(1/2)+2/f*a^3*d^3/
(c^2+d^2)^(5/2)/(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*ln((c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)-d*tan(f*
x+e)-c-(c^2+d^2)^(1/2))+4/f*a^3*d^3/(c^2+d^2)^(5/2)/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan(((2*(c^2+d^2)^(1/2)+2
*c)^(1/2)-2*(c+d*tan(f*x+e))^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))+2/f*a^3/d/(c^2+d^2)/(c+d*tan(f*x+e))^(3/2)*
c^2-2/f*a^3*d^3/(c^2+d^2)^(5/2)/(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*ln(d*tan(f*x+e)+c+(c+d*tan(f*x+e))^(1/2)*(2*(c^2
+d^2)^(1/2)+2*c)^(1/2)+(c^2+d^2)^(1/2))-4/f*a^3*d^3/(c^2+d^2)^(5/2)/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan((2*(c
+d*tan(f*x+e))^(1/2)+(2*(c^2+d^2)^(1/2)+2*c)^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))-16/f*a^3*d/(c^2+d^2)^2/(c+d
*tan(f*x+e))^(1/2)*c+12*I/f*a^3/(c^2+d^2)^2/(c+d*tan(f*x+e))^(1/2)*c^2

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maxima [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^3/(c+d*tan(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

Timed out

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mupad [B]  time = 9.15, size = 255, normalized size = 1.61 \[ -\frac {\frac {\left (a^3\,c^2+a^3\,c\,d\,2{}\mathrm {i}-a^3\,d^2\right )\,2{}\mathrm {i}}{3\,d^2\,f\,\left (c-d\,1{}\mathrm {i}\right )}-\frac {a^3\,\left (c+d\,\mathrm {tan}\left (e+f\,x\right )\right )\,\left (c^2-c\,d\,2{}\mathrm {i}+3\,d^2\right )\,2{}\mathrm {i}}{d^2\,f\,{\left (c-d\,1{}\mathrm {i}\right )}^2}}{{\left (c+d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}}+\frac {a^3\,\mathrm {atan}\left (\frac {\sqrt {c+d\,\mathrm {tan}\left (e+f\,x\right )}\,\left (2\,c^8\,f^2+8\,c^6\,d^2\,f^2+12\,c^4\,d^4\,f^2+8\,c^2\,d^6\,f^2+2\,d^8\,f^2\right )}{2\,f\,{\left (-c+d\,1{}\mathrm {i}\right )}^{5/2}\,\left (f\,c^6+2{}\mathrm {i}\,f\,c^5\,d+f\,c^4\,d^2+4{}\mathrm {i}\,f\,c^3\,d^3-f\,c^2\,d^4+2{}\mathrm {i}\,f\,c\,d^5-f\,d^6\right )}\right )\,8{}\mathrm {i}}{f\,{\left (-c+d\,1{}\mathrm {i}\right )}^{5/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*tan(e + f*x)*1i)^3/(c + d*tan(e + f*x))^(5/2),x)

[Out]

(a^3*atan(((c + d*tan(e + f*x))^(1/2)*(2*c^8*f^2 + 2*d^8*f^2 + 8*c^2*d^6*f^2 + 12*c^4*d^4*f^2 + 8*c^6*d^2*f^2)
)/(2*f*(d*1i - c)^(5/2)*(c^6*f - d^6*f - c^2*d^4*f + c^3*d^3*f*4i + c^4*d^2*f + c*d^5*f*2i + c^5*d*f*2i)))*8i)
/(f*(d*1i - c)^(5/2)) - (((a^3*c^2 - a^3*d^2 + a^3*c*d*2i)*2i)/(3*d^2*f*(c - d*1i)) - (a^3*(c + d*tan(e + f*x)
)*(c^2 - c*d*2i + 3*d^2)*2i)/(d^2*f*(c - d*1i)^2))/(c + d*tan(e + f*x))^(3/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))**3/(c+d*tan(f*x+e))**(5/2),x)

[Out]

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

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