3.12.10 \(\int \frac {(c+d \tan (e+f x))^{3/2}}{a+i a \tan (e+f x)} \, dx\) [1110]

Optimal. Leaf size=153 \[ -\frac {i (c-i d)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{2 a f}+\frac {\sqrt {c+i d} (i c+2 d) \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{2 a f}+\frac {(i c-d) \sqrt {c+d \tan (e+f x)}}{2 f (a+i a \tan (e+f x))} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.23, antiderivative size = 153, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3631, 3620, 3618, 65, 214} \begin {gather*} \frac {(-d+i c) \sqrt {c+d \tan (e+f x)}}{2 f (a+i a \tan (e+f x))}-\frac {i (c-i d)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{2 a f}+\frac {\sqrt {c+i d} (2 d+i c) \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{2 a f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 65

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 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]

Rule 3618

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c*(
d/f), Subst[Int[(a + (b/d)*x)^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 3620

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

Rule 3631

Int[((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)/((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*
c - a*d)*((c + d*Tan[e + f*x])^(n - 1)/(2*a*f*(a + b*Tan[e + f*x]))), x] + Dist[1/(2*a^2), Int[(c + d*Tan[e +
f*x])^(n - 2)*Simp[a*c^2 + a*d^2*(n - 1) - b*c*d*n - d*(a*c*(n - 2) + b*d*n)*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] && GtQ[n, 1]

Rubi steps

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

________________________________________________________________________________________

Mathematica [B] Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(376\) vs. \(2(153)=306\).
time = 4.52, size = 376, normalized size = 2.46 \begin {gather*} \frac {\sec (e+f x) (\cos (f x)+i \sin (f x)) \left (\left (-i (c-i d)^{3/2} \log \left (\frac {2 \left (-i d e^{2 i (e+f x)}+c \left (1+e^{2 i (e+f x)}\right )+\sqrt {c-i d} \left (1+e^{2 i (e+f x)}\right ) \sqrt {c-\frac {i d \left (-1+e^{2 i (e+f x)}\right )}{1+e^{2 i (e+f x)}}}\right )}{\sqrt {c-i d}}\right )+\frac {\left (i c^2+c d+2 i d^2\right ) \log \left (\frac {8 i e^{-2 i f x} \left (i d+c \left (1+e^{2 i (e+f x)}\right )+\sqrt {c+i d} \left (1+e^{2 i (e+f x)}\right ) \sqrt {c-\frac {i d \left (-1+e^{2 i (e+f x)}\right )}{1+e^{2 i (e+f x)}}}\right )}{\sqrt {c+i d} \left (c^2-i c d+2 d^2\right )}\right )}{\sqrt {c+i d}}\right ) (\cos (e)+i \sin (e))+2 (c+i d) \cos (e+f x) (i \cos (f x)+\sin (f x)) \sqrt {c+d \tan (e+f x)}\right )}{4 f (a+i a \tan (e+f x))} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(Sec[e + f*x]*(Cos[f*x] + I*Sin[f*x])*(((-I)*(c - I*d)^(3/2)*Log[(2*((-I)*d*E^((2*I)*(e + f*x)) + c*(1 + E^((2
*I)*(e + f*x))) + Sqrt[c - I*d]*(1 + E^((2*I)*(e + f*x)))*Sqrt[c - (I*d*(-1 + E^((2*I)*(e + f*x))))/(1 + E^((2
*I)*(e + f*x)))]))/Sqrt[c - I*d]] + ((I*c^2 + c*d + (2*I)*d^2)*Log[((8*I)*(I*d + c*(1 + E^((2*I)*(e + f*x))) +
Sqrt[c + I*d]*(1 + E^((2*I)*(e + f*x)))*Sqrt[c - (I*d*(-1 + E^((2*I)*(e + f*x))))/(1 + E^((2*I)*(e + f*x)))])
)/(Sqrt[c + I*d]*(c^2 - I*c*d + 2*d^2)*E^((2*I)*f*x))])/Sqrt[c + I*d])*(Cos[e] + I*Sin[e]) + 2*(c + I*d)*Cos[e
+ f*x]*(I*Cos[f*x] + Sin[f*x])*Sqrt[c + d*Tan[e + f*x]]))/(4*f*(a + I*a*Tan[e + f*x]))

________________________________________________________________________________________

Maple [A]
time = 0.37, size = 140, normalized size = 0.92

method	result	size
derivativedivides	\(\frac {2 d^{2} \left (\frac {i \left (i d -c \right )^{\frac {3}{2}} \arctan \left (\frac {\sqrt {c +d \tan \left (f x +e \right )}}{\sqrt {i d -c}}\right )}{4 d^{2}}+\frac {\left (i d +c \right ) \left (-\frac {d \sqrt {c +d \tan \left (f x +e \right )}}{-d \tan \left (f x +e \right )+i d}-\frac {\left (i c +2 d \right ) \arctan \left (\frac {\sqrt {c +d \tan \left (f x +e \right )}}{\sqrt {-i d -c}}\right )}{\sqrt {-i d -c}}\right )}{4 d^{2}}\right )}{f a}\)	\(140\)
default	\(\frac {2 d^{2} \left (\frac {i \left (i d -c \right )^{\frac {3}{2}} \arctan \left (\frac {\sqrt {c +d \tan \left (f x +e \right )}}{\sqrt {i d -c}}\right )}{4 d^{2}}+\frac {\left (i d +c \right ) \left (-\frac {d \sqrt {c +d \tan \left (f x +e \right )}}{-d \tan \left (f x +e \right )+i d}-\frac {\left (i c +2 d \right ) \arctan \left (\frac {\sqrt {c +d \tan \left (f x +e \right )}}{\sqrt {-i d -c}}\right )}{\sqrt {-i d -c}}\right )}{4 d^{2}}\right )}{f a}\)	\(140\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError >> ECL says: Error executing code in Maxima: expt: undefined: 0 to a negative e
xponent.

________________________________________________________________________________________

Fricas [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 808 vs. \(2 (119) = 238\).
time = 1.08, size = 808, normalized size = 5.28 \begin {gather*} \frac {{\left (a f \sqrt {-\frac {c^{3} - 3 i \, c^{2} d - 3 \, c d^{2} + i \, d^{3}}{a^{2} f^{2}}} e^{\left (2 i \, f x + 2 i \, e\right )} \log \left (-\frac {2 \, {\left (-i \, c^{2} - c d + {\left (a f e^{\left (2 i \, f x + 2 i \, e\right )} + a 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 {c^{3} - 3 i \, c^{2} d - 3 \, c d^{2} + i \, d^{3}}{a^{2} f^{2}}} + {\left (-i \, c^{2} - 2 \, c d + i \, d^{2}\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{i \, c + d}\right ) - a f \sqrt {-\frac {c^{3} - 3 i \, c^{2} d - 3 \, c d^{2} + i \, d^{3}}{a^{2} f^{2}}} e^{\left (2 i \, f x + 2 i \, e\right )} \log \left (-\frac {2 \, {\left (-i \, c^{2} - c d - {\left (a f e^{\left (2 i \, f x + 2 i \, e\right )} + a 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 {c^{3} - 3 i \, c^{2} d - 3 \, c d^{2} + i \, d^{3}}{a^{2} f^{2}}} + {\left (-i \, c^{2} - 2 \, c d + i \, d^{2}\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{i \, c + d}\right ) + a f \sqrt {-\frac {c^{3} - 3 i \, c^{2} d - 4 i \, d^{3}}{a^{2} f^{2}}} e^{\left (2 i \, f x + 2 i \, e\right )} \log \left (\frac {{\left (i \, c^{2} + c d + 2 i \, d^{2} + {\left (a f e^{\left (2 i \, f x + 2 i \, e\right )} + a 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 {c^{3} - 3 i \, c^{2} d - 4 i \, d^{3}}{a^{2} f^{2}}} + {\left (i \, c^{2} + 2 \, c d\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{2 \, a f}\right ) - a f \sqrt {-\frac {c^{3} - 3 i \, c^{2} d - 4 i \, d^{3}}{a^{2} f^{2}}} e^{\left (2 i \, f x + 2 i \, e\right )} \log \left (\frac {{\left (i \, c^{2} + c d + 2 i \, d^{2} - {\left (a f e^{\left (2 i \, f x + 2 i \, e\right )} + a 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 {c^{3} - 3 i \, c^{2} d - 4 i \, d^{3}}{a^{2} f^{2}}} + {\left (i \, c^{2} + 2 \, c d\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{2 \, a f}\right ) + 2 \, {\left ({\left (i \, c - d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + i \, c - d\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}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{8 \, a f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(a*f*sqrt(-(c^3 - 3*I*c^2*d - 3*c*d^2 + I*d^3)/(a^2*f^2))*e^(2*I*f*x + 2*I*e)*log(-2*(-I*c^2 - c*d + (a*f*
e^(2*I*f*x + 2*I*e) + a*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(-(c^
3 - 3*I*c^2*d - 3*c*d^2 + I*d^3)/(a^2*f^2)) + (-I*c^2 - 2*c*d + I*d^2)*e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I*
e)/(I*c + d)) - a*f*sqrt(-(c^3 - 3*I*c^2*d - 3*c*d^2 + I*d^3)/(a^2*f^2))*e^(2*I*f*x + 2*I*e)*log(-2*(-I*c^2 -
c*d - (a*f*e^(2*I*f*x + 2*I*e) + a*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(-(c^3 - 3*I*c^2*d - 3*c*d^2 + I*d^3)/(a^2*f^2)) + (-I*c^2 - 2*c*d + I*d^2)*e^(2*I*f*x + 2*I*e))*e^(-2*I
*f*x - 2*I*e)/(I*c + d)) + a*f*sqrt(-(c^3 - 3*I*c^2*d - 4*I*d^3)/(a^2*f^2))*e^(2*I*f*x + 2*I*e)*log(1/2*(I*c^2
+ c*d + 2*I*d^2 + (a*f*e^(2*I*f*x + 2*I*e) + a*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(-(c^3 - 3*I*c^2*d - 4*I*d^3)/(a^2*f^2)) + (I*c^2 + 2*c*d)*e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x
- 2*I*e)/(a*f)) - a*f*sqrt(-(c^3 - 3*I*c^2*d - 4*I*d^3)/(a^2*f^2))*e^(2*I*f*x + 2*I*e)*log(1/2*(I*c^2 + c*d +
2*I*d^2 - (a*f*e^(2*I*f*x + 2*I*e) + a*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(-(c^3 - 3*I*c^2*d - 4*I*d^3)/(a^2*f^2)) + (I*c^2 + 2*c*d)*e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I*e
)/(a*f)) + 2*((I*c - d)*e^(2*I*f*x + 2*I*e) + I*c - d)*sqrt(((c - I*d)*e^(2*I*f*x + 2*I*e) + c + I*d)/(e^(2*I*
f*x + 2*I*e) + 1)))*e^(-2*I*f*x - 2*I*e)/(a*f)

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \frac {i \left (\int \frac {c \sqrt {c + d \tan {\left (e + f x \right )}}}{\tan {\left (e + f x \right )} - i}\, dx + \int \frac {d \sqrt {c + d \tan {\left (e + f x \right )}} \tan {\left (e + f x \right )}}{\tan {\left (e + f x \right )} - i}\, dx\right )}{a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 407 vs. \(2 (119) = 238\).
time = 0.57, size = 407, normalized size = 2.66 \begin {gather*} \frac {{\left (-i \, c^{2} - c d - 2 i \, d^{2}\right )} \arctan \left (\frac {2 \, {\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 {-2 \, c + 2 \, \sqrt {c^{2} + d^{2}}} + i \, \sqrt {-2 \, c + 2 \, \sqrt {c^{2} + d^{2}}} d - \sqrt {c^{2} + d^{2}} \sqrt {-2 \, c + 2 \, \sqrt {c^{2} + d^{2}}}}\right )}{a \sqrt {-2 \, c + 2 \, \sqrt {c^{2} + d^{2}}} f {\left (\frac {i \, d}{c - \sqrt {c^{2} + d^{2}}} + 1\right )}} + \frac {{\left (i \, c^{2} + 2 \, c d - i \, d^{2}\right )} \arctan \left (\frac {2 \, {\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 {-2 \, c + 2 \, \sqrt {c^{2} + d^{2}}} - i \, \sqrt {-2 \, c + 2 \, \sqrt {c^{2} + d^{2}}} d - \sqrt {c^{2} + d^{2}} \sqrt {-2 \, c + 2 \, \sqrt {c^{2} + d^{2}}}}\right )}{a \sqrt {-2 \, c + 2 \, \sqrt {c^{2} + d^{2}}} f {\left (-\frac {i \, d}{c - \sqrt {c^{2} + d^{2}}} + 1\right )}} + \frac {\sqrt {d \tan \left (f x + e\right ) + c} c d + i \, \sqrt {d \tan \left (f x + e\right ) + c} d^{2}}{2 \, {\left (d \tan \left (f x + e\right ) - i \, d\right )} a f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-I*c^2 - c*d - 2*I*d^2)*arctan(2*(sqrt(d*tan(f*x + e) + c)*c - sqrt(c^2 + d^2)*sqrt(d*tan(f*x + e) + c))/(c*s
qrt(-2*c + 2*sqrt(c^2 + d^2)) + I*sqrt(-2*c + 2*sqrt(c^2 + d^2))*d - sqrt(c^2 + d^2)*sqrt(-2*c + 2*sqrt(c^2 +
d^2))))/(a*sqrt(-2*c + 2*sqrt(c^2 + d^2))*f*(I*d/(c - sqrt(c^2 + d^2)) + 1)) + (I*c^2 + 2*c*d - I*d^2)*arctan(
2*(sqrt(d*tan(f*x + e) + c)*c - sqrt(c^2 + d^2)*sqrt(d*tan(f*x + e) + c))/(c*sqrt(-2*c + 2*sqrt(c^2 + d^2)) -
I*sqrt(-2*c + 2*sqrt(c^2 + d^2))*d - sqrt(c^2 + d^2)*sqrt(-2*c + 2*sqrt(c^2 + d^2))))/(a*sqrt(-2*c + 2*sqrt(c^
2 + d^2))*f*(-I*d/(c - sqrt(c^2 + d^2)) + 1)) + 1/2*(sqrt(d*tan(f*x + e) + c)*c*d + I*sqrt(d*tan(f*x + e) + c)
*d^2)/((d*tan(f*x + e) - I*d)*a*f)

________________________________________________________________________________________

Mupad [B]
time = 7.34, size = 847, normalized size = 5.54 \begin {gather*} -2\,\mathrm {atanh}\left (\frac {20\,a^2\,d^6\,f^2\,\sqrt {c+d\,\mathrm {tan}\left (e+f\,x\right )}\,\sqrt {-\frac {c^3}{16\,a^2\,f^2}+\frac {d^3\,1{}\mathrm {i}}{4\,a^2\,f^2}+\frac {c^2\,d\,3{}\mathrm {i}}{16\,a^2\,f^2}}}{3{}\mathrm {i}\,a\,f\,c^4\,d^4+11\,a\,f\,c^3\,d^5-7{}\mathrm {i}\,a\,f\,c^2\,d^6+11\,a\,f\,c\,d^7-10{}\mathrm {i}\,a\,f\,d^8}+\frac {a^2\,c\,d^5\,f^2\,\sqrt {c+d\,\mathrm {tan}\left (e+f\,x\right )}\,\sqrt {-\frac {c^3}{16\,a^2\,f^2}+\frac {d^3\,1{}\mathrm {i}}{4\,a^2\,f^2}+\frac {c^2\,d\,3{}\mathrm {i}}{16\,a^2\,f^2}}\,32{}\mathrm {i}}{3{}\mathrm {i}\,a\,f\,c^4\,d^4+11\,a\,f\,c^3\,d^5-7{}\mathrm {i}\,a\,f\,c^2\,d^6+11\,a\,f\,c\,d^7-10{}\mathrm {i}\,a\,f\,d^8}-\frac {12\,a^2\,c^2\,d^4\,f^2\,\sqrt {c+d\,\mathrm {tan}\left (e+f\,x\right )}\,\sqrt {-\frac {c^3}{16\,a^2\,f^2}+\frac {d^3\,1{}\mathrm {i}}{4\,a^2\,f^2}+\frac {c^2\,d\,3{}\mathrm {i}}{16\,a^2\,f^2}}}{3{}\mathrm {i}\,a\,f\,c^4\,d^4+11\,a\,f\,c^3\,d^5-7{}\mathrm {i}\,a\,f\,c^2\,d^6+11\,a\,f\,c\,d^7-10{}\mathrm {i}\,a\,f\,d^8}\right )\,\sqrt {\frac {-2\,c^3+c^2\,d\,6{}\mathrm {i}+d^3\,8{}\mathrm {i}}{32\,a^2\,f^2}}-2\,\mathrm {atanh}\left (\frac {20\,a^2\,d^6\,f^2\,\sqrt {c+d\,\mathrm {tan}\left (e+f\,x\right )}\,\sqrt {\frac {3\,c\,d^2}{16\,a^2\,f^2}-\frac {d^3\,1{}\mathrm {i}}{16\,a^2\,f^2}-\frac {c^3}{16\,a^2\,f^2}+\frac {c^2\,d\,3{}\mathrm {i}}{16\,a^2\,f^2}}}{3{}\mathrm {i}\,a\,f\,c^4\,d^4+8\,a\,f\,c^3\,d^5-2{}\mathrm {i}\,a\,f\,c^2\,d^6+8\,a\,f\,c\,d^7-5{}\mathrm {i}\,a\,f\,d^8}-\frac {a^2\,c\,d^5\,f^2\,\sqrt {c+d\,\mathrm {tan}\left (e+f\,x\right )}\,\sqrt {\frac {3\,c\,d^2}{16\,a^2\,f^2}-\frac {d^3\,1{}\mathrm {i}}{16\,a^2\,f^2}-\frac {c^3}{16\,a^2\,f^2}+\frac {c^2\,d\,3{}\mathrm {i}}{16\,a^2\,f^2}}\,8{}\mathrm {i}}{3{}\mathrm {i}\,a\,f\,c^4\,d^4+8\,a\,f\,c^3\,d^5-2{}\mathrm {i}\,a\,f\,c^2\,d^6+8\,a\,f\,c\,d^7-5{}\mathrm {i}\,a\,f\,d^8}+\frac {12\,a^2\,c^2\,d^4\,f^2\,\sqrt {c+d\,\mathrm {tan}\left (e+f\,x\right )}\,\sqrt {\frac {3\,c\,d^2}{16\,a^2\,f^2}-\frac {d^3\,1{}\mathrm {i}}{16\,a^2\,f^2}-\frac {c^3}{16\,a^2\,f^2}+\frac {c^2\,d\,3{}\mathrm {i}}{16\,a^2\,f^2}}}{3{}\mathrm {i}\,a\,f\,c^4\,d^4+8\,a\,f\,c^3\,d^5-2{}\mathrm {i}\,a\,f\,c^2\,d^6+8\,a\,f\,c\,d^7-5{}\mathrm {i}\,a\,f\,d^8}\right )\,\sqrt {\frac {-2\,c^3+c^2\,d\,6{}\mathrm {i}+6\,c\,d^2-d^3\,2{}\mathrm {i}}{32\,a^2\,f^2}}-\frac {\left (d^2\,1{}\mathrm {i}+c\,d\right )\,\sqrt {c+d\,\mathrm {tan}\left (e+f\,x\right )}}{2\,a\,f\,\left (-d\,\mathrm {tan}\left (e+f\,x\right )+d\,1{}\mathrm {i}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- 2*atanh((20*a^2*d^6*f^2*(c + d*tan(e + f*x))^(1/2)*((d^3*1i)/(4*a^2*f^2) - c^3/(16*a^2*f^2) + (c^2*d*3i)/(16
*a^2*f^2))^(1/2))/(11*a*c*d^7*f - a*d^8*f*10i - a*c^2*d^6*f*7i + 11*a*c^3*d^5*f + a*c^4*d^4*f*3i) + (a^2*c*d^5
*f^2*(c + d*tan(e + f*x))^(1/2)*((d^3*1i)/(4*a^2*f^2) - c^3/(16*a^2*f^2) + (c^2*d*3i)/(16*a^2*f^2))^(1/2)*32i)
/(11*a*c*d^7*f - a*d^8*f*10i - a*c^2*d^6*f*7i + 11*a*c^3*d^5*f + a*c^4*d^4*f*3i) - (12*a^2*c^2*d^4*f^2*(c + d*
tan(e + f*x))^(1/2)*((d^3*1i)/(4*a^2*f^2) - c^3/(16*a^2*f^2) + (c^2*d*3i)/(16*a^2*f^2))^(1/2))/(11*a*c*d^7*f -
a*d^8*f*10i - a*c^2*d^6*f*7i + 11*a*c^3*d^5*f + a*c^4*d^4*f*3i))*((c^2*d*6i - 2*c^3 + d^3*8i)/(32*a^2*f^2))^(
1/2) - 2*atanh((20*a^2*d^6*f^2*(c + d*tan(e + f*x))^(1/2)*((3*c*d^2)/(16*a^2*f^2) - (d^3*1i)/(16*a^2*f^2) - c^
3/(16*a^2*f^2) + (c^2*d*3i)/(16*a^2*f^2))^(1/2))/(8*a*c*d^7*f - a*d^8*f*5i - a*c^2*d^6*f*2i + 8*a*c^3*d^5*f +
a*c^4*d^4*f*3i) - (a^2*c*d^5*f^2*(c + d*tan(e + f*x))^(1/2)*((3*c*d^2)/(16*a^2*f^2) - (d^3*1i)/(16*a^2*f^2) -
c^3/(16*a^2*f^2) + (c^2*d*3i)/(16*a^2*f^2))^(1/2)*8i)/(8*a*c*d^7*f - a*d^8*f*5i - a*c^2*d^6*f*2i + 8*a*c^3*d^5
*f + a*c^4*d^4*f*3i) + (12*a^2*c^2*d^4*f^2*(c + d*tan(e + f*x))^(1/2)*((3*c*d^2)/(16*a^2*f^2) - (d^3*1i)/(16*a
^2*f^2) - c^3/(16*a^2*f^2) + (c^2*d*3i)/(16*a^2*f^2))^(1/2))/(8*a*c*d^7*f - a*d^8*f*5i - a*c^2*d^6*f*2i + 8*a*
c^3*d^5*f + a*c^4*d^4*f*3i))*((6*c*d^2 + c^2*d*6i - 2*c^3 - d^3*2i)/(32*a^2*f^2))^(1/2) - ((c*d + d^2*1i)*(c +
d*tan(e + f*x))^(1/2))/(2*a*f*(d*1i - d*tan(e + f*x)))

________________________________________________________________________________________