3.1110 \(\int \frac{(c+d \tan (e+f x))^{3/2}}{a+i a \tan (e+f x)} \, dx\)

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

[Out]

((-I/2)*(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)*Ar
cTanh[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*Ta
n[e + f*x]))

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Rubi [A]  time = 0.334361, antiderivative size = 153, normalized size of antiderivative = 1., 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 = {3550, 3539, 3537, 63, 208} \[ \frac{(-d+i c) \sqrt{c+d \tan (e+f x)}}{2 f (a+i a \tan (e+f x))}-\frac{i (c-i d)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{2 a f}+\frac{\sqrt{c+i d} (2 d+i c) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{2 a f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-I/2)*(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)*Ar
cTanh[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*Ta
n[e + f*x]))

Rule 3550

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]

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 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 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]

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 ) \operatorname{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)) \operatorname{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 \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 )}{2 a d f}-\frac{((c+i d) (c-2 i d)) \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 )}{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]  time = 3.93634, size = 376, normalized size = 2.46 \[ \frac{\sec (e+f x) (\cos (f x)+i \sin (f x)) \left ((\cos (e)+i \sin (e)) \left (\frac{\left (i c^2+c d+2 i d^2\right ) \log \left (\frac{8 i e^{-2 i f x} \left (\sqrt{c+i d} \left (1+e^{2 i (e+f x)}\right ) \sqrt{c-\frac{i d \left (-1+e^{2 i (e+f x)}\right )}{1+e^{2 i (e+f x)}}}+c \left (1+e^{2 i (e+f x)}\right )+i d\right )}{\sqrt{c+i d} \left (c^2-i c d+2 d^2\right )}\right )}{\sqrt{c+i d}}-i (c-i d)^{3/2} \log \left (\frac{2 \left (\sqrt{c-i d} \left (1+e^{2 i (e+f x)}\right ) \sqrt{c-\frac{i d \left (-1+e^{2 i (e+f x)}\right )}{1+e^{2 i (e+f x)}}}+c \left (1+e^{2 i (e+f x)}\right )-i d e^{2 i (e+f x)}\right )}{\sqrt{c-i d}}\right )\right )+2 (c+i d) \cos (e+f x) (\sin (f x)+i \cos (f x)) \sqrt{c+d \tan (e+f x)}\right )}{4 f (a+i a \tan (e+f x))} \]

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]))

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Maple [B]  time = 0.056, size = 257, normalized size = 1.7 \begin{align*}{\frac{{\frac{i}{2}}}{af} \left ( id-c \right ) ^{{\frac{3}{2}}}\arctan \left ({\sqrt{c+d\tan \left ( fx+e \right ) }{\frac{1}{\sqrt{id-c}}}} \right ) }+{\frac{{\frac{i}{2}}{d}^{2}}{af \left ( -id+d\tan \left ( fx+e \right ) \right ) }\sqrt{c+d\tan \left ( fx+e \right ) }}+{\frac{cd}{2\,af \left ( -id+d\tan \left ( fx+e \right ) \right ) }\sqrt{c+d\tan \left ( fx+e \right ) }}-{\frac{cd}{2\,af}\arctan \left ({\sqrt{c+d\tan \left ( fx+e \right ) }{\frac{1}{\sqrt{-id-c}}}} \right ){\frac{1}{\sqrt{-id-c}}}}-{\frac{{\frac{i}{2}}{c}^{2}}{af}\arctan \left ({\sqrt{c+d\tan \left ( fx+e \right ) }{\frac{1}{\sqrt{-id-c}}}} \right ){\frac{1}{\sqrt{-id-c}}}}-{\frac{i{d}^{2}}{af}\arctan \left ({\sqrt{c+d\tan \left ( fx+e \right ) }{\frac{1}{\sqrt{-id-c}}}} \right ){\frac{1}{\sqrt{-id-c}}}} \end{align*}

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)

[Out]

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

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

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

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Fricas [B]  time = 2.74397, size = 2010, normalized size = 13.14 \begin{align*} \text{result too large to display} \end{align*}

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 + 2*c*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 - 3*c*d^2 + I*d^3)/(a^2*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*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 + 2*c*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 - 3*c*d^2 + I*d^3)/(a^2*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*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)

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Sympy [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: AttributeError} \end{align*}

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]

Exception raised: AttributeError

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Giac [B]  time = 1.51572, size = 564, normalized size = 3.69 \begin{align*} \frac{1}{2} \, d^{2}{\left (\frac{\sqrt{d \tan \left (f x + e\right ) + c} c + i \, \sqrt{d \tan \left (f x + e\right ) + c} d}{{\left (d \tan \left (f x + e\right ) - i \, d\right )} a d f} - \frac{4 \,{\left (i \, c^{2} + c d + 2 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 )}{a \sqrt{-8 \, c + 8 \, \sqrt{c^{2} + d^{2}}} d^{2} f{\left (\frac{i \, d}{c - \sqrt{c^{2} + d^{2}}} + 1\right )}} - \frac{4 \,{\left (-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 )}{a \sqrt{-8 \, c + 8 \, \sqrt{c^{2} + d^{2}}} d^{2} f{\left (-\frac{i \, d}{c - \sqrt{c^{2} + d^{2}}} + 1\right )}}\right )} \end{align*}

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]

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