3.1111 \(\int \frac {(c+d \tan (e+f x))^{3/2}}{(a+i a \tan (e+f x))^2} \, dx\)

Optimal. Leaf size=209 \[ \frac {\left (2 c d+i \left (2 c^2+d^2\right )\right ) \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{8 a^2 f \sqrt {c+i d}}+\frac {(3 d+2 i c) \sqrt {c+d \tan (e+f x)}}{8 a^2 f (1+i \tan (e+f x))}-\frac {i (c-i d)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{4 a^2 f}+\frac {(-d+i c) \sqrt {c+d \tan (e+f x)}}{4 f (a+i a \tan (e+f x))^2} \]

[Out]

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

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Rubi [A]  time = 0.65, antiderivative size = 209, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3558, 3596, 3539, 3537, 63, 208} \[ \frac {\left (2 c d+i \left (2 c^2+d^2\right )\right ) \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{8 a^2 f \sqrt {c+i d}}+\frac {(3 d+2 i c) \sqrt {c+d \tan (e+f x)}}{8 a^2 f (1+i \tan (e+f x))}-\frac {i (c-i d)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{4 a^2 f}+\frac {(-d+i c) \sqrt {c+d \tan (e+f x)}}{4 f (a+i a \tan (e+f x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-I/4)*(c - I*d)^(3/2)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]])/(a^2*f) + ((2*c*d + I*(2*c^2 + d^2))*
ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c + I*d]])/(8*a^2*Sqrt[c + I*d]*f) + (((2*I)*c + 3*d)*Sqrt[c + d*Tan[e +
f*x]])/(8*a^2*f*(1 + I*Tan[e + f*x])) + ((I*c - d)*Sqrt[c + d*Tan[e + f*x]])/(4*f*(a + I*a*Tan[e + f*x])^2)

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 3539

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 3558

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Si
mp[((b*c - a*d)*(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n - 1))/(2*a*f*m), x] + Dist[1/(2*a^2*m), Int[(a
+ b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n - 2)*Simp[c*(a*c*m + b*d*(n - 1)) - d*(b*c*m + a*d*(n - 1))
- d*(b*d*(m - n + 1) - a*c*(m + n - 1))*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] && LtQ[m, 0] && GtQ[n, 1] && (IntegerQ[m] || IntegersQ[2*m,
2*n])

Rule 3596

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

Rubi steps

\begin {align*} \int \frac {(c+d \tan (e+f x))^{3/2}}{(a+i a \tan (e+f x))^2} \, dx &=\frac {(i c-d) \sqrt {c+d \tan (e+f x)}}{4 f (a+i a \tan (e+f x))^2}-\frac {\int \frac {-\frac {1}{2} a \left (4 c^2-5 i c d+d^2\right )-\frac {1}{2} a (3 c-5 i d) d \tan (e+f x)}{(a+i a \tan (e+f x)) \sqrt {c+d \tan (e+f x)}} \, dx}{4 a^2}\\ &=\frac {(2 i c+3 d) \sqrt {c+d \tan (e+f x)}}{8 a^2 f (1+i \tan (e+f x))}+\frac {(i c-d) \sqrt {c+d \tan (e+f x)}}{4 f (a+i a \tan (e+f x))^2}+\frac {\int \frac {\frac {1}{2} a^2 \left (4 i c^3+2 c^2 d+5 i c d^2+d^3\right )+\frac {1}{2} a^2 (i c-d) (2 c-3 i d) d \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}} \, dx}{8 a^4 (i c-d)}\\ &=\frac {(2 i c+3 d) \sqrt {c+d \tan (e+f x)}}{8 a^2 f (1+i \tan (e+f x))}+\frac {(i c-d) \sqrt {c+d \tan (e+f x)}}{4 f (a+i a \tan (e+f x))^2}+\frac {(c-i d)^2 \int \frac {1+i \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}} \, dx}{8 a^2}+\frac {\left (2 c^2-2 i c d+d^2\right ) \int \frac {1-i \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}} \, dx}{16 a^2}\\ &=\frac {(2 i c+3 d) \sqrt {c+d \tan (e+f x)}}{8 a^2 f (1+i \tan (e+f x))}+\frac {(i c-d) \sqrt {c+d \tan (e+f x)}}{4 f (a+i a \tan (e+f x))^2}+\frac {\left (i (c-i d)^2\right ) \operatorname {Subst}\left (\int \frac {1}{(-1+x) \sqrt {c-i d x}} \, dx,x,i \tan (e+f x)\right )}{8 a^2 f}-\frac {\left (i \left (2 c^2-2 i c d+d^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{(-1+x) \sqrt {c+i d x}} \, dx,x,-i \tan (e+f x)\right )}{16 a^2 f}\\ &=\frac {(2 i c+3 d) \sqrt {c+d \tan (e+f x)}}{8 a^2 f (1+i \tan (e+f x))}+\frac {(i c-d) \sqrt {c+d \tan (e+f x)}}{4 f (a+i a \tan (e+f x))^2}-\frac {(c-i d)^2 \operatorname {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 )}{4 a^2 d f}-\frac {\left (2 c^2-2 i c d+d^2\right ) \operatorname {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 )}{8 a^2 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 )}{4 a^2 f}+\frac {\left (2 i c^2+2 c d+i d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{8 a^2 \sqrt {c+i d} f}+\frac {(2 i c+3 d) \sqrt {c+d \tan (e+f x)}}{8 a^2 f (1+i \tan (e+f x))}+\frac {(i c-d) \sqrt {c+d \tan (e+f x)}}{4 f (a+i a \tan (e+f x))^2}\\ \end {align*}

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Mathematica [A]  time = 1.83, size = 272, normalized size = 1.30 \[ \frac {\sec ^2(e+f x) (\cos (f x)+i \sin (f x))^2 \left (\frac {2 (\cos (2 e)+i \sin (2 e)) \left (2 i \sqrt {-c-i d} (c-i d)^2 \tan ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {-c+i d}}\right )-i \sqrt {-c+i d} \left (2 c^2-2 i c d+d^2\right ) \tan ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {-c-i d}}\right )\right )}{\sqrt {-c-i d} \sqrt {-c+i d}}+2 \cos (e+f x) (\cos (2 f x)-i \sin (2 f x)) \sqrt {c+d \tan (e+f x)} ((-2 c+3 i d) \sin (e+f x)+(d+4 i c) \cos (e+f x))\right )}{16 f (a+i a \tan (e+f x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[e + f*x]^2*(Cos[f*x] + I*Sin[f*x])^2*((2*((-I)*Sqrt[-c + I*d]*(2*c^2 - (2*I)*c*d + d^2)*ArcTan[Sqrt[c + d
*Tan[e + f*x]]/Sqrt[-c - I*d]] + (2*I)*Sqrt[-c - I*d]*(c - I*d)^2*ArcTan[Sqrt[c + d*Tan[e + f*x]]/Sqrt[-c + I*
d]])*(Cos[2*e] + I*Sin[2*e]))/(Sqrt[-c - I*d]*Sqrt[-c + I*d]) + 2*Cos[e + f*x]*(Cos[2*f*x] - I*Sin[2*f*x])*(((
4*I)*c + d)*Cos[e + f*x] + (-2*c + (3*I)*d)*Sin[e + f*x])*Sqrt[c + d*Tan[e + f*x]]))/(16*f*(a + I*a*Tan[e + f*
x])^2)

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fricas [B]  time = 0.67, size = 982, normalized size = 4.70 \[ \text {result too large to display} \]

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))^2,x, algorithm="fricas")

[Out]

-1/32*(2*a^2*f*sqrt(-(c^3 - 3*I*c^2*d - 3*c*d^2 + I*d^3)/(a^4*f^2))*e^(4*I*f*x + 4*I*e)*log((2*I*c^2 + 2*c*d +
2*(a^2*f*e^(2*I*f*x + 2*I*e) + a^2*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^4*f^2)) + (2*I*c^2 + 4*c*d - 2*I*d^2)*e^(2*I*f*x + 2*I*e))*e^(
-2*I*f*x - 2*I*e)/(I*c + d)) - 2*a^2*f*sqrt(-(c^3 - 3*I*c^2*d - 3*c*d^2 + I*d^3)/(a^4*f^2))*e^(4*I*f*x + 4*I*e
)*log((2*I*c^2 + 2*c*d - 2*(a^2*f*e^(2*I*f*x + 2*I*e) + a^2*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^4*f^2)) + (2*I*c^2 + 4*c*d - 2*I*d^2)*
e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I*e)/(I*c + d)) - a^2*f*sqrt(-(4*I*c^4 + 8*c^3*d + 4*c*d^3 + I*d^4)/((I*a
^4*c - a^4*d)*f^2))*e^(4*I*f*x + 4*I*e)*log(-1/8*(2*c^3 + 3*c*d^2 + I*d^3 - ((I*a^2*c - a^2*d)*f*e^(2*I*f*x +
2*I*e) + (I*a^2*c - a^2*d)*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(-
(4*I*c^4 + 8*c^3*d + 4*c*d^3 + I*d^4)/((I*a^4*c - a^4*d)*f^2)) + (2*c^3 - 2*I*c^2*d + c*d^2)*e^(2*I*f*x + 2*I*
e))*e^(-2*I*f*x - 2*I*e)/((I*a^2*c - a^2*d)*f)) + a^2*f*sqrt(-(4*I*c^4 + 8*c^3*d + 4*c*d^3 + I*d^4)/((I*a^4*c
- a^4*d)*f^2))*e^(4*I*f*x + 4*I*e)*log(-1/8*(2*c^3 + 3*c*d^2 + I*d^3 - ((-I*a^2*c + a^2*d)*f*e^(2*I*f*x + 2*I*
e) + (-I*a^2*c + a^2*d)*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(-(4*
I*c^4 + 8*c^3*d + 4*c*d^3 + I*d^4)/((I*a^4*c - a^4*d)*f^2)) + (2*c^3 - 2*I*c^2*d + c*d^2)*e^(2*I*f*x + 2*I*e))
*e^(-2*I*f*x - 2*I*e)/((I*a^2*c - a^2*d)*f)) - 2*((3*I*c + 2*d)*e^(4*I*f*x + 4*I*e) + (4*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^(-4*I*f*x - 4*
I*e)/(a^2*f)

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giac [B]  time = 0.87, size = 458, normalized size = 2.19 \[ -\frac {\sqrt {2} {\left (c^{2} - 2 i \, c d - d^{2}\right )} \arctan \left (\frac {16 i \, \sqrt {d \tan \left (f x + e\right ) + c} c + 16 i \, \sqrt {c^{2} + d^{2}} \sqrt {d \tan \left (f x + e\right ) + c}}{8 \, {\left (\sqrt {2} \sqrt {c + \sqrt {c^{2} + d^{2}}} c - i \, \sqrt {2} \sqrt {c + \sqrt {c^{2} + d^{2}}} d + \sqrt {2} \sqrt {c^{2} + d^{2}} \sqrt {c + \sqrt {c^{2} + d^{2}}}\right )}}\right )}{4 \, a^{2} \sqrt {c + \sqrt {c^{2} + d^{2}}} f {\left (-\frac {i \, d}{c + \sqrt {c^{2} + d^{2}}} + 1\right )}} - \frac {{\left (2 i \, c^{2} + 2 \, c d + i \, d^{2}\right )} \arctan \left (\frac {4 \, {\left (\sqrt {d \tan \left (f x + e\right ) + c} c - \sqrt {c^{2} + d^{2}} \sqrt {d \tan \left (f x + e\right ) + c}\right )}}{c \sqrt {-8 \, c + 8 \, \sqrt {c^{2} + d^{2}}} + i \, \sqrt {-8 \, c + 8 \, \sqrt {c^{2} + d^{2}}} d - \sqrt {c^{2} + d^{2}} \sqrt {-8 \, c + 8 \, \sqrt {c^{2} + d^{2}}}}\right )}{2 \, a^{2} \sqrt {-8 \, c + 8 \, \sqrt {c^{2} + d^{2}}} f {\left (\frac {i \, d}{c - \sqrt {c^{2} + d^{2}}} + 1\right )}} + \frac {2 \, {\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} c d - 2 \, \sqrt {d \tan \left (f x + e\right ) + c} c^{2} d - 3 i \, {\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} d^{2} - i \, \sqrt {d \tan \left (f x + e\right ) + c} c d^{2} - \sqrt {d \tan \left (f x + e\right ) + c} d^{3}}{8 \, {\left (d \tan \left (f x + e\right ) - i \, d\right )}^{2} a^{2} f} \]

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))^2,x, algorithm="giac")

[Out]

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

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maple [B]  time = 0.39, size = 861, normalized size = 4.12 \[ \frac {d \left (c +d \tan \left (f x +e \right )\right )^{\frac {3}{2}} c^{3}}{4 f \,a^{2} \left (d \tan \left (f x +e \right )-i d \right )^{2} \left (2 i c d +c^{2}-d^{2}\right )}+\frac {d^{3} \left (c +d \tan \left (f x +e \right )\right )^{\frac {3}{2}} c}{2 f \,a^{2} \left (d \tan \left (f x +e \right )-i d \right )^{2} \left (2 i c d +c^{2}-d^{2}\right )}+\frac {i d^{2} \left (c +d \tan \left (f x +e \right )\right )^{\frac {3}{2}} c^{2}}{8 f \,a^{2} \left (d \tan \left (f x +e \right )-i d \right )^{2} \left (2 i c d +c^{2}-d^{2}\right )}+\frac {3 i d^{4} \left (c +d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{8 f \,a^{2} \left (d \tan \left (f x +e \right )-i d \right )^{2} \left (2 i c d +c^{2}-d^{2}\right )}-\frac {d \sqrt {c +d \tan \left (f x +e \right )}\, c^{4}}{4 f \,a^{2} \left (d \tan \left (f x +e \right )-i d \right )^{2} \left (2 i c d +c^{2}-d^{2}\right )}+\frac {3 d^{3} \sqrt {c +d \tan \left (f x +e \right )}\, c^{2}}{8 f \,a^{2} \left (d \tan \left (f x +e \right )-i d \right )^{2} \left (2 i c d +c^{2}-d^{2}\right )}+\frac {d^{5} \sqrt {c +d \tan \left (f x +e \right )}}{8 f \,a^{2} \left (d \tan \left (f x +e \right )-i d \right )^{2} \left (2 i c d +c^{2}-d^{2}\right )}-\frac {5 i d^{2} \sqrt {c +d \tan \left (f x +e \right )}\, c^{3}}{8 f \,a^{2} \left (d \tan \left (f x +e \right )-i d \right )^{2} \left (2 i c d +c^{2}-d^{2}\right )}-\frac {i d^{4} \sqrt {c +d \tan \left (f x +e \right )}\, c}{8 f \,a^{2} \left (d \tan \left (f x +e \right )-i d \right )^{2} \left (2 i c d +c^{2}-d^{2}\right )}+\frac {d \arctan \left (\frac {\sqrt {c +d \tan \left (f x +e \right )}}{\sqrt {-i d -c}}\right ) c^{3}}{4 f \,a^{2} \left (2 i c d +c^{2}-d^{2}\right ) \sqrt {-i d -c}}+\frac {d^{3} \arctan \left (\frac {\sqrt {c +d \tan \left (f x +e \right )}}{\sqrt {-i d -c}}\right ) c}{2 f \,a^{2} \left (2 i c d +c^{2}-d^{2}\right ) \sqrt {-i d -c}}-\frac {i \arctan \left (\frac {\sqrt {c +d \tan \left (f x +e \right )}}{\sqrt {-i d -c}}\right ) c^{4}}{4 f \,a^{2} \left (2 i c d +c^{2}-d^{2}\right ) \sqrt {-i d -c}}-\frac {3 i d^{2} \arctan \left (\frac {\sqrt {c +d \tan \left (f x +e \right )}}{\sqrt {-i d -c}}\right ) c^{2}}{8 f \,a^{2} \left (2 i c d +c^{2}-d^{2}\right ) \sqrt {-i d -c}}+\frac {i d^{4} \arctan \left (\frac {\sqrt {c +d \tan \left (f x +e \right )}}{\sqrt {-i d -c}}\right )}{8 f \,a^{2} \left (2 i c d +c^{2}-d^{2}\right ) \sqrt {-i d -c}}+\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 f \,a^{2}} \]

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))^2,x)

[Out]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]

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))^2,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 7.76, size = 1580, normalized size = 7.56 \[ \text {result too large to display} \]

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)^2,x)

[Out]

- atan((a^4*d^6*f^2*(c + d*tan(e + f*x))^(1/2)*((3*c*d^2)/(64*a^4*f^2) - (d^3*1i)/(64*a^4*f^2) - c^3/(64*a^4*f
^2) + (c^2*d*3i)/(64*a^4*f^2))^(1/2)*80i)/(a^2*d^8*f*10i - a^2*c^2*d^6*f*26i + 8*a^2*c^3*d^5*f - 28*a^2*c*d^7*
f) - (64*a^4*c*d^5*f^2*(c + d*tan(e + f*x))^(1/2)*((3*c*d^2)/(64*a^4*f^2) - (d^3*1i)/(64*a^4*f^2) - c^3/(64*a^
4*f^2) + (c^2*d*3i)/(64*a^4*f^2))^(1/2))/(a^2*d^8*f*10i - a^2*c^2*d^6*f*26i + 8*a^2*c^3*d^5*f - 28*a^2*c*d^7*f
))*((6*c*d^2 + c^2*d*6i - 2*c^3 - d^3*2i)/(128*a^4*f^2))^(1/2)*2i - (((c + d*tan(e + f*x))^(1/2)*(c*d^2*3i + 6
*c^2*d + 3*d^3))/(24*a^2*f) + (d*(c*2i + 3*d)*(c + d*tan(e + f*x))^(3/2)*1i)/(8*a^2*f))/(c*d*2i - (2*c + d*2i)
*(c + d*tan(e + f*x)) + (c + d*tan(e + f*x))^2 + c^2 - d^2) - atan((((a^2*f*(128*a^4*d^5*f^2 + a^4*c*d^4*f^2*3
84i - 256*a^4*c^2*d^3*f^2) - 4096*a^8*c*d^2*f^4*(c + d*tan(e + f*x))^(1/2)*(-(4*c*d^3 + 8*c^3*d + c^4*4i + d^4
*1i)/(256*a^4*f^2*(c*1i - d)))^(1/2))*(-(4*c*d^3 + 8*c^3*d + c^4*4i + d^4*1i)/(256*a^4*f^2*(c*1i - d)))^(1/2)
- 8*a^4*f^2*(c + d*tan(e + f*x))^(1/2)*(c*d^5*12i + 5*d^6 - 24*c^2*d^4 - c^3*d^3*24i + 8*c^4*d^2))*(-(4*c*d^3
+ 8*c^3*d + c^4*4i + d^4*1i)/(256*a^4*f^2*(c*1i - d)))^(1/2)*1i - ((a^2*f*(128*a^4*d^5*f^2 + a^4*c*d^4*f^2*384
i - 256*a^4*c^2*d^3*f^2) + 4096*a^8*c*d^2*f^4*(c + d*tan(e + f*x))^(1/2)*(-(4*c*d^3 + 8*c^3*d + c^4*4i + d^4*1
i)/(256*a^4*f^2*(c*1i - d)))^(1/2))*(-(4*c*d^3 + 8*c^3*d + c^4*4i + d^4*1i)/(256*a^4*f^2*(c*1i - d)))^(1/2) +
8*a^4*f^2*(c + d*tan(e + f*x))^(1/2)*(c*d^5*12i + 5*d^6 - 24*c^2*d^4 - c^3*d^3*24i + 8*c^4*d^2))*(-(4*c*d^3 +
8*c^3*d + c^4*4i + d^4*1i)/(256*a^4*f^2*(c*1i - d)))^(1/2)*1i)/(((a^2*f*(128*a^4*d^5*f^2 + a^4*c*d^4*f^2*384i
- 256*a^4*c^2*d^3*f^2) - 4096*a^8*c*d^2*f^4*(c + d*tan(e + f*x))^(1/2)*(-(4*c*d^3 + 8*c^3*d + c^4*4i + d^4*1i)
/(256*a^4*f^2*(c*1i - d)))^(1/2))*(-(4*c*d^3 + 8*c^3*d + c^4*4i + d^4*1i)/(256*a^4*f^2*(c*1i - d)))^(1/2) - 8*
a^4*f^2*(c + d*tan(e + f*x))^(1/2)*(c*d^5*12i + 5*d^6 - 24*c^2*d^4 - c^3*d^3*24i + 8*c^4*d^2))*(-(4*c*d^3 + 8*
c^3*d + c^4*4i + d^4*1i)/(256*a^4*f^2*(c*1i - d)))^(1/2) + ((a^2*f*(128*a^4*d^5*f^2 + a^4*c*d^4*f^2*384i - 256
*a^4*c^2*d^3*f^2) + 4096*a^8*c*d^2*f^4*(c + d*tan(e + f*x))^(1/2)*(-(4*c*d^3 + 8*c^3*d + c^4*4i + d^4*1i)/(256
*a^4*f^2*(c*1i - d)))^(1/2))*(-(4*c*d^3 + 8*c^3*d + c^4*4i + d^4*1i)/(256*a^4*f^2*(c*1i - d)))^(1/2) + 8*a^4*f
^2*(c + d*tan(e + f*x))^(1/2)*(c*d^5*12i + 5*d^6 - 24*c^2*d^4 - c^3*d^3*24i + 8*c^4*d^2))*(-(4*c*d^3 + 8*c^3*d
+ c^4*4i + d^4*1i)/(256*a^4*f^2*(c*1i - d)))^(1/2) + 2*a^2*f*(2*c*d^7 - d^8*3i - c^2*d^6*15i + 28*c^3*d^5 + c
^4*d^4*18i - 4*c^5*d^3)))*(-(4*c*d^3 + 8*c^3*d + c^4*4i + d^4*1i)/(256*a^4*f^2*(c*1i - d)))^(1/2)*2i

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \frac {\int \frac {c \sqrt {c + d \tan {\left (e + f x \right )}}}{\tan ^{2}{\left (e + f x \right )} - 2 i \tan {\left (e + f x \right )} - 1}\, dx + \int \frac {d \sqrt {c + d \tan {\left (e + f x \right )}} \tan {\left (e + f x \right )}}{\tan ^{2}{\left (e + f x \right )} - 2 i \tan {\left (e + f x \right )} - 1}\, dx}{a^{2}} \]

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))**2,x)

[Out]

-(Integral(c*sqrt(c + d*tan(e + f*x))/(tan(e + f*x)**2 - 2*I*tan(e + f*x) - 1), x) + Integral(d*sqrt(c + d*tan
(e + f*x))*tan(e + f*x)/(tan(e + f*x)**2 - 2*I*tan(e + f*x) - 1), x))/a**2

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