3.1101 \(\int (a+i a \tan (e+f x))^3 \sqrt {c+d \tan (e+f x)} \, dx\)

Optimal. Leaf size=150 \[ \frac {4 a^3 (-6 d+i c) (c+d \tan (e+f x))^{3/2}}{15 d^2 f}+\frac {8 i a^3 \sqrt {c+d \tan (e+f x)}}{f}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right ) (c+d \tan (e+f x))^{3/2}}{5 d f}-\frac {8 i a^3 \sqrt {c-i d} \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{f} \]

[Out]

-8*I*a^3*arctanh((c+d*tan(f*x+e))^(1/2)/(c-I*d)^(1/2))*(c-I*d)^(1/2)/f+8*I*a^3*(c+d*tan(f*x+e))^(1/2)/f+4/15*a
^3*(I*c-6*d)*(c+d*tan(f*x+e))^(3/2)/d^2/f-2/5*(a^3+I*a^3*tan(f*x+e))*(c+d*tan(f*x+e))^(3/2)/d/f

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Rubi [A]  time = 0.42, antiderivative size = 150, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3556, 3592, 3528, 3537, 63, 208} \[ \frac {4 a^3 (-6 d+i c) (c+d \tan (e+f x))^{3/2}}{15 d^2 f}+\frac {8 i a^3 \sqrt {c+d \tan (e+f x)}}{f}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right ) (c+d \tan (e+f x))^{3/2}}{5 d f}-\frac {8 i a^3 \sqrt {c-i d} \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-8*I)*a^3*Sqrt[c - I*d]*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]])/f + ((8*I)*a^3*Sqrt[c + d*Tan[e + f
*x]])/f + (4*a^3*(I*c - 6*d)*(c + d*Tan[e + f*x])^(3/2))/(15*d^2*f) - (2*(a^3 + I*a^3*Tan[e + f*x])*(c + d*Tan
[e + f*x])^(3/2))/(5*d*f)

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 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 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 3556

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(b^2*(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^(n + 1))/(d*f*(m + n - 1)), x] + Dist[a/(d*(m + n - 1
)), Int[(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^n*Simp[b*c*(m - 2) + a*d*(m + 2*n) + (a*c*(m - 2) +
b*d*(3*m + 2*n - 4))*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a
^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && IntegerQ[2*m] && GtQ[m, 1] && NeQ[m + n - 1, 0] && (IntegerQ[m] || Intege
rsQ[2*m, 2*n])

Rule 3592

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

Rubi steps

\begin {align*} \int (a+i a \tan (e+f x))^3 \sqrt {c+d \tan (e+f x)} \, dx &=-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right ) (c+d \tan (e+f x))^{3/2}}{5 d f}+\frac {(2 a) \int (a+i a \tan (e+f x)) (a (i c+4 d)+a (c+6 i d) \tan (e+f x)) \sqrt {c+d \tan (e+f x)} \, dx}{5 d}\\ &=\frac {4 a^3 (i c-6 d) (c+d \tan (e+f x))^{3/2}}{15 d^2 f}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right ) (c+d \tan (e+f x))^{3/2}}{5 d f}+\frac {(2 a) \int \sqrt {c+d \tan (e+f x)} \left (10 a^2 d+10 i a^2 d \tan (e+f x)\right ) \, dx}{5 d}\\ &=\frac {8 i a^3 \sqrt {c+d \tan (e+f x)}}{f}+\frac {4 a^3 (i c-6 d) (c+d \tan (e+f x))^{3/2}}{15 d^2 f}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right ) (c+d \tan (e+f x))^{3/2}}{5 d f}+\frac {(2 a) \int \frac {10 a^2 (c-i d) d+10 a^2 d (i c+d) \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}} \, dx}{5 d}\\ &=\frac {8 i a^3 \sqrt {c+d \tan (e+f x)}}{f}+\frac {4 a^3 (i c-6 d) (c+d \tan (e+f x))^{3/2}}{15 d^2 f}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right ) (c+d \tan (e+f x))^{3/2}}{5 d f}+\frac {\left (40 i a^5 (c-i d)^2 d\right ) \operatorname {Subst}\left (\int \frac {1}{\left (100 a^4 d^2 (i c+d)^2+10 a^2 (c-i d) d x\right ) \sqrt {c+\frac {x}{10 a^2 (i c+d)}}} \, dx,x,10 a^2 d (i c+d) \tan (e+f x)\right )}{f}\\ &=\frac {8 i a^3 \sqrt {c+d \tan (e+f x)}}{f}+\frac {4 a^3 (i c-6 d) (c+d \tan (e+f x))^{3/2}}{15 d^2 f}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right ) (c+d \tan (e+f x))^{3/2}}{5 d f}-\frac {\left (800 a^7 (c-i d)^3 d\right ) \operatorname {Subst}\left (\int \frac {1}{-100 a^4 c (c-i d) d (i c+d)+100 a^4 d^2 (i c+d)^2+100 a^4 (c-i d) d (i c+d) x^2} \, dx,x,\sqrt {c+d \tan (e+f x)}\right )}{f}\\ &=-\frac {8 i a^3 \sqrt {c-i d} \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{f}+\frac {8 i a^3 \sqrt {c+d \tan (e+f x)}}{f}+\frac {4 a^3 (i c-6 d) (c+d \tan (e+f x))^{3/2}}{15 d^2 f}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right ) (c+d \tan (e+f x))^{3/2}}{5 d f}\\ \end {align*}

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Mathematica [A]  time = 5.62, size = 219, normalized size = 1.46 \[ \frac {a^3 (\cos (e+f x)+i \sin (e+f x))^3 \left (\frac {(\sin (3 e)+i \cos (3 e)) \sec ^2(e+f x) \sqrt {c+d \tan (e+f x)} \left (\left (2 c^2+15 i c d+63 d^2\right ) \cos (2 (e+f x))+2 c^2-d (c-15 i d) \sin (2 (e+f x))+15 i c d+57 d^2\right )}{15 d^2}-8 i e^{-3 i e} \sqrt {c-i d} \tanh ^{-1}\left (\frac {\sqrt {c-\frac {i d \left (-1+e^{2 i (e+f x)}\right )}{1+e^{2 i (e+f x)}}}}{\sqrt {c-i d}}\right )\right )}{f (\cos (f x)+i \sin (f x))^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*(Cos[e + f*x] + I*Sin[e + f*x])^3*(((-8*I)*Sqrt[c - I*d]*ArcTanh[Sqrt[c - (I*d*(-1 + E^((2*I)*(e + f*x)))
)/(1 + E^((2*I)*(e + f*x)))]/Sqrt[c - I*d]])/E^((3*I)*e) + (Sec[e + f*x]^2*(I*Cos[3*e] + Sin[3*e])*(2*c^2 + (1
5*I)*c*d + 57*d^2 + (2*c^2 + (15*I)*c*d + 63*d^2)*Cos[2*(e + f*x)] - (c - (15*I)*d)*d*Sin[2*(e + f*x)])*Sqrt[c
+ d*Tan[e + f*x]])/(15*d^2)))/(f*(Cos[f*x] + I*Sin[f*x])^3)

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fricas [B]  time = 0.67, size = 519, normalized size = 3.46 \[ \frac {15 \, {\left (d^{2} f e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, d^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + d^{2} f\right )} \sqrt {-\frac {64 \, a^{6} c - 64 i \, a^{6} d}{f^{2}}} \log \left (\frac {{\left (8 \, a^{3} c + {\left (i \, f e^{\left (2 i \, f x + 2 i \, e\right )} + i \, f\right )} \sqrt {\frac {{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {-\frac {64 \, a^{6} c - 64 i \, a^{6} d}{f^{2}}} + {\left (8 \, a^{3} c - 8 i \, a^{3} d\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{4 \, a^{3}}\right ) - 15 \, {\left (d^{2} f e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, d^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + d^{2} f\right )} \sqrt {-\frac {64 \, a^{6} c - 64 i \, a^{6} d}{f^{2}}} \log \left (\frac {{\left (8 \, a^{3} c + {\left (-i \, f e^{\left (2 i \, f x + 2 i \, e\right )} - i \, f\right )} \sqrt {\frac {{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {-\frac {64 \, a^{6} c - 64 i \, a^{6} d}{f^{2}}} + {\left (8 \, a^{3} c - 8 i \, a^{3} d\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{4 \, a^{3}}\right ) + {\left (16 i \, a^{3} c^{2} - 112 \, a^{3} c d + 384 i \, a^{3} d^{2} + {\left (16 i \, a^{3} c^{2} - 128 \, a^{3} c d + 624 i \, a^{3} d^{2}\right )} e^{\left (4 i \, f x + 4 i \, e\right )} + {\left (32 i \, a^{3} c^{2} - 240 \, a^{3} c d + 912 i \, a^{3} d^{2}\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} \sqrt {\frac {{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{60 \, {\left (d^{2} f e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, d^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + d^{2} f\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c+d*tan(f*x+e))^(1/2)*(a+I*a*tan(f*x+e))^3,x, algorithm="fricas")

[Out]

1/60*(15*(d^2*f*e^(4*I*f*x + 4*I*e) + 2*d^2*f*e^(2*I*f*x + 2*I*e) + d^2*f)*sqrt(-(64*a^6*c - 64*I*a^6*d)/f^2)*
log(1/4*(8*a^3*c + (I*f*e^(2*I*f*x + 2*I*e) + I*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))*sqrt(-(64*a^6*c - 64*I*a^6*d)/f^2) + (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) - 15*(d^2*f*e^(4*I*f*x + 4*I*e) + 2*d^2*f*e^(2*I*f*x + 2*I*e) + d^2*f)*sqrt(-(64*a^6*c - 64*I*a^6*d
)/f^2)*log(1/4*(8*a^3*c + (-I*f*e^(2*I*f*x + 2*I*e) - I*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))*sqrt(-(64*a^6*c - 64*I*a^6*d)/f^2) + (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^2 - 112*a^3*c*d + 384*I*a^3*d^2 + (16*I*a^3*c^2 - 128*a^3*c*d + 624*I*a^3*d^2
)*e^(4*I*f*x + 4*I*e) + (32*I*a^3*c^2 - 240*a^3*c*d + 912*I*a^3*d^2)*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)))/(d^2*f*e^(4*I*f*x + 4*I*e) + 2*d^2*f*e^(2*I*f*x + 2*I*e
) + d^2*f)

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giac [B]  time = 0.75, size = 277, normalized size = 1.85 \[ -\frac {32 \, {\left (-i \, a^{3} c - a^{3} d\right )} \arctan \left (\frac {4 \, {\left (\sqrt {d \tan \left (f x + e\right ) + c} c - \sqrt {c^{2} + d^{2}} \sqrt {d \tan \left (f x + e\right ) + c}\right )}}{c \sqrt {-8 \, c + 8 \, \sqrt {c^{2} + d^{2}}} - i \, \sqrt {-8 \, c + 8 \, \sqrt {c^{2} + d^{2}}} d - \sqrt {c^{2} + d^{2}} \sqrt {-8 \, c + 8 \, \sqrt {c^{2} + d^{2}}}}\right )}{\sqrt {-8 \, c + 8 \, \sqrt {c^{2} + d^{2}}} f {\left (-\frac {i \, d}{c - \sqrt {c^{2} + d^{2}}} + 1\right )}} - \frac {6 i \, {\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}} a^{3} d^{8} f^{4} - 10 i \, {\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} a^{3} c d^{8} f^{4} + 30 \, {\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} a^{3} d^{9} f^{4} - 120 i \, \sqrt {d \tan \left (f x + e\right ) + c} a^{3} d^{10} f^{4}}{15 \, d^{10} f^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c+d*tan(f*x+e))^(1/2)*(a+I*a*tan(f*x+e))^3,x, algorithm="giac")

[Out]

-32*(-I*a^3*c - a^3*d)*arctan(4*(sqrt(d*tan(f*x + e) + c)*c - sqrt(c^2 + d^2)*sqrt(d*tan(f*x + e) + c))/(c*sqr
t(-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))))/(sqrt(-8*c + 8*sqrt(c^2 + d^2))*f*(-I*d/(c - sqrt(c^2 + d^2)) + 1)) - 1/15*(6*I*(d*tan(f*x + e) + c)^(5/
2)*a^3*d^8*f^4 - 10*I*(d*tan(f*x + e) + c)^(3/2)*a^3*c*d^8*f^4 + 30*(d*tan(f*x + e) + c)^(3/2)*a^3*d^9*f^4 - 1
20*I*sqrt(d*tan(f*x + e) + c)*a^3*d^10*f^4)/(d^10*f^5)

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maple [B]  time = 0.44, size = 1272, normalized size = 8.48 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c+d*tan(f*x+e))^(1/2)*(a+I*a*tan(f*x+e))^3,x, algorithm="maxima")

[Out]

integrate((I*a*tan(f*x + e) + a)^3*sqrt(d*tan(f*x + e) + c), x)

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mupad [B]  time = 8.89, size = 200, normalized size = 1.33 \[ -\left (\left (c-d\,1{}\mathrm {i}\right )\,\left (\frac {a^3\,\left (c-d\,1{}\mathrm {i}\right )\,2{}\mathrm {i}}{d^2\,f}-\frac {a^3\,\left (c+d\,1{}\mathrm {i}\right )\,4{}\mathrm {i}}{d^2\,f}\right )+\frac {a^3\,{\left (c+d\,1{}\mathrm {i}\right )}^2\,2{}\mathrm {i}}{d^2\,f}\right )\,\sqrt {c+d\,\mathrm {tan}\left (e+f\,x\right )}-\left (\frac {a^3\,\left (c-d\,1{}\mathrm {i}\right )\,2{}\mathrm {i}}{3\,d^2\,f}-\frac {a^3\,\left (c+d\,1{}\mathrm {i}\right )\,4{}\mathrm {i}}{3\,d^2\,f}\right )\,{\left (c+d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}-\frac {a^3\,{\left (c+d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{5/2}\,2{}\mathrm {i}}{5\,d^2\,f}+\frac {\sqrt {16{}\mathrm {i}}\,a^3\,\mathrm {atan}\left (\frac {\sqrt {16{}\mathrm {i}}\,\sqrt {c+d\,\mathrm {tan}\left (e+f\,x\right )}\,1{}\mathrm {i}}{4\,\sqrt {d+c\,1{}\mathrm {i}}}\right )\,\sqrt {d+c\,1{}\mathrm {i}}\,2{}\mathrm {i}}{f} \]

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))^(1/2),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c+d*tan(f*x+e))**(1/2)*(a+I*a*tan(f*x+e))**3,x)

[Out]

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

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