3.766 \(\int \frac{(A+B \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}{a+i a \tan (e+f x)} \, dx\)

Optimal. Leaf size=144 \[ -\frac{c^{3/2} (-5 B+i A) \tanh ^{-1}\left (\frac{\sqrt{c-i c \tan (e+f x)}}{\sqrt{2} \sqrt{c}}\right )}{\sqrt{2} a f}+\frac{c (-5 B+i A) \sqrt{c-i c \tan (e+f x)}}{2 a f}+\frac{(-B+i A) (c-i c \tan (e+f x))^{3/2}}{2 a f (1+i \tan (e+f x))} \]

[Out]

-(((I*A - 5*B)*c^(3/2)*ArcTanh[Sqrt[c - I*c*Tan[e + f*x]]/(Sqrt[2]*Sqrt[c])])/(Sqrt[2]*a*f)) + ((I*A - 5*B)*c*
Sqrt[c - I*c*Tan[e + f*x]])/(2*a*f) + ((I*A - B)*(c - I*c*Tan[e + f*x])^(3/2))/(2*a*f*(1 + I*Tan[e + f*x]))

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Rubi [A]  time = 0.220348, antiderivative size = 144, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 43, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.116, Rules used = {3588, 78, 50, 63, 208} \[ -\frac{c^{3/2} (-5 B+i A) \tanh ^{-1}\left (\frac{\sqrt{c-i c \tan (e+f x)}}{\sqrt{2} \sqrt{c}}\right )}{\sqrt{2} a f}+\frac{c (-5 B+i A) \sqrt{c-i c \tan (e+f x)}}{2 a f}+\frac{(-B+i A) (c-i c \tan (e+f x))^{3/2}}{2 a f (1+i \tan (e+f x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((I*A - 5*B)*c^(3/2)*ArcTanh[Sqrt[c - I*c*Tan[e + f*x]]/(Sqrt[2]*Sqrt[c])])/(Sqrt[2]*a*f)) + ((I*A - 5*B)*c*
Sqrt[c - I*c*Tan[e + f*x]])/(2*a*f) + ((I*A - B)*(c - I*c*Tan[e + f*x])^(3/2))/(2*a*f*(1 + I*Tan[e + f*x]))

Rule 3588

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

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ
[p, n]))))

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, 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]

Rubi steps

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

Mathematica [F]  time = 180.002, size = 0, normalized size = 0. \[ \text{\$Aborted} \]

Verification is Not applicable to the result.

[In]

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

[Out]

$Aborted

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Maple [A]  time = 0.101, size = 109, normalized size = 0.8 \begin{align*}{\frac{2\,ic}{af} \left ( iB\sqrt{c-ic\tan \left ( fx+e \right ) }+c \left ({\frac{1}{-c-ic\tan \left ( fx+e \right ) } \left ( -{\frac{A}{2}}-{\frac{i}{2}}B \right ) \sqrt{c-ic\tan \left ( fx+e \right ) }}-{\frac{ \left ( A+5\,iB \right ) \sqrt{2}}{4}{\it Artanh} \left ({\frac{\sqrt{2}}{2}\sqrt{c-ic\tan \left ( fx+e \right ) }{\frac{1}{\sqrt{c}}}} \right ){\frac{1}{\sqrt{c}}}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^(3/2)/(a+I*a*tan(f*x+e)),x)

[Out]

2*I/f/a*c*(I*B*(c-I*c*tan(f*x+e))^(1/2)+c*((-1/2*A-1/2*I*B)*(c-I*c*tan(f*x+e))^(1/2)/(-c-I*c*tan(f*x+e))-1/4*(
A+5*I*B)*2^(1/2)/c^(1/2)*arctanh(1/2*(c-I*c*tan(f*x+e))^(1/2)*2^(1/2)/c^(1/2))))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.12786, size = 855, normalized size = 5.94 \begin{align*} \frac{{\left (a f \sqrt{-\frac{{\left (2 \, A^{2} + 20 i \, A B - 50 \, B^{2}\right )} c^{3}}{a^{2} f^{2}}} e^{\left (2 i \, f x + 2 i \, e\right )} \log \left (\frac{{\left ({\left (-2 i \, A + 10 \, B\right )} c^{2} + \sqrt{2}{\left (a f e^{\left (2 i \, f x + 2 i \, e\right )} + a f\right )} \sqrt{-\frac{{\left (2 \, A^{2} + 20 i \, A B - 50 \, B^{2}\right )} c^{3}}{a^{2} f^{2}}} \sqrt{\frac{c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-i \, f x - i \, e\right )}}{a f}\right ) - a f \sqrt{-\frac{{\left (2 \, A^{2} + 20 i \, A B - 50 \, B^{2}\right )} c^{3}}{a^{2} f^{2}}} e^{\left (2 i \, f x + 2 i \, e\right )} \log \left (\frac{{\left ({\left (-2 i \, A + 10 \, B\right )} c^{2} - \sqrt{2}{\left (a f e^{\left (2 i \, f x + 2 i \, e\right )} + a f\right )} \sqrt{-\frac{{\left (2 \, A^{2} + 20 i \, A B - 50 \, B^{2}\right )} c^{3}}{a^{2} f^{2}}} \sqrt{\frac{c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-i \, f x - i \, e\right )}}{a f}\right ) + \sqrt{2}{\left ({\left (2 i \, A - 10 \, B\right )} c e^{\left (2 i \, f x + 2 i \, e\right )} +{\left (2 i \, A - 2 \, B\right )} c\right )} \sqrt{\frac{c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{4 \, a f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(a*f*sqrt(-(2*A^2 + 20*I*A*B - 50*B^2)*c^3/(a^2*f^2))*e^(2*I*f*x + 2*I*e)*log(((-2*I*A + 10*B)*c^2 + sqrt(
2)*(a*f*e^(2*I*f*x + 2*I*e) + a*f)*sqrt(-(2*A^2 + 20*I*A*B - 50*B^2)*c^3/(a^2*f^2))*sqrt(c/(e^(2*I*f*x + 2*I*e
) + 1)))*e^(-I*f*x - I*e)/(a*f)) - a*f*sqrt(-(2*A^2 + 20*I*A*B - 50*B^2)*c^3/(a^2*f^2))*e^(2*I*f*x + 2*I*e)*lo
g(((-2*I*A + 10*B)*c^2 - sqrt(2)*(a*f*e^(2*I*f*x + 2*I*e) + a*f)*sqrt(-(2*A^2 + 20*I*A*B - 50*B^2)*c^3/(a^2*f^
2))*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1)))*e^(-I*f*x - I*e)/(a*f)) + sqrt(2)*((2*I*A - 10*B)*c*e^(2*I*f*x + 2*I*e)
+ (2*I*A - 2*B)*c)*sqrt(c/(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((A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))**(3/2)/(a+I*a*tan(f*x+e)),x)

[Out]

Exception raised: AttributeError

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B \tan \left (f x + e\right ) + A\right )}{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac{3}{2}}}{i \, a \tan \left (f x + e\right ) + a}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((B*tan(f*x + e) + A)*(-I*c*tan(f*x + e) + c)^(3/2)/(I*a*tan(f*x + e) + a), x)