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

Optimal. Leaf size=184 \[ \frac{(-B+5 i A) \tanh ^{-1}\left (\frac{\sqrt{c-i c \tan (e+f x)}}{\sqrt{2} \sqrt{c}}\right )}{8 \sqrt{2} a c^{3/2} f}+\frac{-B+i A}{2 a f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}-\frac{-B+5 i A}{8 a c f \sqrt{c-i c \tan (e+f x)}}-\frac{-B+5 i A}{12 a f (c-i c \tan (e+f x))^{3/2}} \]

[Out]

(((5*I)*A - B)*ArcTanh[Sqrt[c - I*c*Tan[e + f*x]]/(Sqrt[2]*Sqrt[c])])/(8*Sqrt[2]*a*c^(3/2)*f) - ((5*I)*A - B)/
(12*a*f*(c - I*c*Tan[e + f*x])^(3/2)) + (I*A - B)/(2*a*f*(1 + I*Tan[e + f*x])*(c - I*c*Tan[e + f*x])^(3/2)) -
((5*I)*A - B)/(8*a*c*f*Sqrt[c - I*c*Tan[e + f*x]])

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Rubi [A]  time = 0.26758, antiderivative size = 184, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 43, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.116, Rules used = {3588, 78, 51, 63, 208} \[ \frac{(-B+5 i A) \tanh ^{-1}\left (\frac{\sqrt{c-i c \tan (e+f x)}}{\sqrt{2} \sqrt{c}}\right )}{8 \sqrt{2} a c^{3/2} f}+\frac{-B+i A}{2 a f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}-\frac{-B+5 i A}{8 a c f \sqrt{c-i c \tan (e+f x)}}-\frac{-B+5 i A}{12 a f (c-i c \tan (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((5*I)*A - B)*ArcTanh[Sqrt[c - I*c*Tan[e + f*x]]/(Sqrt[2]*Sqrt[c])])/(8*Sqrt[2]*a*c^(3/2)*f) - ((5*I)*A - B)/
(12*a*f*(c - I*c*Tan[e + f*x])^(3/2)) + (I*A - B)/(2*a*f*(1 + I*Tan[e + f*x])*(c - I*c*Tan[e + f*x])^(3/2)) -
((5*I)*A - B)/(8*a*c*f*Sqrt[c - I*c*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 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && 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)}{(a+i a \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}} \, dx &=\frac{(a c) \operatorname{Subst}\left (\int \frac{A+B x}{(a+i a x)^2 (c-i c x)^{5/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{i A-B}{2 a f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}+\frac{((5 A+i B) c) \operatorname{Subst}\left (\int \frac{1}{(a+i a x) (c-i c x)^{5/2}} \, dx,x,\tan (e+f x)\right )}{4 f}\\ &=-\frac{5 i A-B}{12 a f (c-i c \tan (e+f x))^{3/2}}+\frac{i A-B}{2 a f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}+\frac{(5 A+i B) \operatorname{Subst}\left (\int \frac{1}{(a+i a x) (c-i c x)^{3/2}} \, dx,x,\tan (e+f x)\right )}{8 f}\\ &=-\frac{5 i A-B}{12 a f (c-i c \tan (e+f x))^{3/2}}+\frac{i A-B}{2 a f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}-\frac{5 i A-B}{8 a c f \sqrt{c-i c \tan (e+f x)}}+\frac{(5 A+i B) \operatorname{Subst}\left (\int \frac{1}{(a+i a x) \sqrt{c-i c x}} \, dx,x,\tan (e+f x)\right )}{16 c f}\\ &=-\frac{5 i A-B}{12 a f (c-i c \tan (e+f x))^{3/2}}+\frac{i A-B}{2 a f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}-\frac{5 i A-B}{8 a c f \sqrt{c-i c \tan (e+f x)}}+\frac{(5 i A-B) \operatorname{Subst}\left (\int \frac{1}{2 a-\frac{a x^2}{c}} \, dx,x,\sqrt{c-i c \tan (e+f x)}\right )}{8 c^2 f}\\ &=\frac{(5 i A-B) \tanh ^{-1}\left (\frac{\sqrt{c-i c \tan (e+f x)}}{\sqrt{2} \sqrt{c}}\right )}{8 \sqrt{2} a c^{3/2} f}-\frac{5 i A-B}{12 a f (c-i c \tan (e+f x))^{3/2}}+\frac{i A-B}{2 a f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}-\frac{5 i A-B}{8 a c f \sqrt{c-i c \tan (e+f x)}}\\ \end{align*}

Mathematica [A]  time = 6.11132, size = 239, normalized size = 1.3 \[ -\frac{e^{-i (e+2 f x)} \sqrt{\frac{c}{1+e^{2 i (e+f x)}}} (\cos (f x)+i \sin (f x)) (A+B \tan (e+f x)) \left (\left (1+e^{2 i (e+f x)}\right ) \left (i A \left (14 e^{2 i (e+f x)}+2 e^{4 i (e+f x)}-3\right )+B \left (2 e^{2 i (e+f x)}+2 e^{4 i (e+f x)}+3\right )\right )+3 (B-5 i A) e^{2 i (e+f x)} \sqrt{1+e^{2 i (e+f x)}} \tanh ^{-1}\left (\sqrt{1+e^{2 i (e+f x)}}\right )\right )}{24 \sqrt{2} c^2 f (a+i a \tan (e+f x)) (A \cos (e+f x)+B \sin (e+f x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sqrt[c/(1 + E^((2*I)*(e + f*x)))]*((1 + E^((2*I)*(e + f*x)))*(B*(3 + 2*E^((2*I)*(e + f*x)) + 2*E^((4*I)*(e +
f*x))) + I*A*(-3 + 14*E^((2*I)*(e + f*x)) + 2*E^((4*I)*(e + f*x)))) + 3*((-5*I)*A + B)*E^((2*I)*(e + f*x))*Sq
rt[1 + E^((2*I)*(e + f*x))]*ArcTanh[Sqrt[1 + E^((2*I)*(e + f*x))]])*(Cos[f*x] + I*Sin[f*x])*(A + B*Tan[e + f*x
]))/(24*Sqrt[2]*c^2*E^(I*(e + 2*f*x))*f*(A*Cos[e + f*x] + B*Sin[e + f*x])*(a + I*a*Tan[e + f*x]))

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Maple [A]  time = 0.1, size = 141, normalized size = 0.8 \begin{align*}{\frac{2\,ic}{af} \left ( -{\frac{1}{4\,{c}^{2}} \left ({\frac{1}{-c-ic\tan \left ( fx+e \right ) } \left ({\frac{A}{4}}+{\frac{i}{4}}B \right ) \sqrt{c-ic\tan \left ( fx+e \right ) }}-{\frac{ \left ( 5\,A+iB \right ) \sqrt{2}}{8}{\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 ) }-{\frac{A}{4\,{c}^{2}}{\frac{1}{\sqrt{c-ic\tan \left ( fx+e \right ) }}}}-{\frac{A-iB}{12\,c} \left ( c-ic\tan \left ( fx+e \right ) \right ) ^{-{\frac{3}{2}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*I/f/a*c*(-1/4/c^2*((1/4*A+1/4*I*B)*(c-I*c*tan(f*x+e))^(1/2)/(-c-I*c*tan(f*x+e))-1/8*(5*A+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)))-1/4*A/c^2/(c-I*c*tan(f*x+e))^(1/2)-1/12/c*(A-I*B)/(c-
I*c*tan(f*x+e))^(3/2))

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.21914, size = 1017, normalized size = 5.53 \begin{align*} \frac{{\left (3 \, \sqrt{\frac{1}{2}} a c^{2} f \sqrt{-\frac{25 \, A^{2} + 10 i \, A B - B^{2}}{a^{2} c^{3} f^{2}}} e^{\left (2 i \, f x + 2 i \, e\right )} \log \left (\frac{{\left (\sqrt{2} \sqrt{\frac{1}{2}}{\left (a c f e^{\left (2 i \, f x + 2 i \, e\right )} + a c f\right )} \sqrt{\frac{c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt{-\frac{25 \, A^{2} + 10 i \, A B - B^{2}}{a^{2} c^{3} f^{2}}} + 5 i \, A - B\right )} e^{\left (-i \, f x - i \, e\right )}}{4 \, a c f}\right ) - 3 \, \sqrt{\frac{1}{2}} a c^{2} f \sqrt{-\frac{25 \, A^{2} + 10 i \, A B - B^{2}}{a^{2} c^{3} f^{2}}} e^{\left (2 i \, f x + 2 i \, e\right )} \log \left (-\frac{{\left (\sqrt{2} \sqrt{\frac{1}{2}}{\left (a c f e^{\left (2 i \, f x + 2 i \, e\right )} + a c f\right )} \sqrt{\frac{c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt{-\frac{25 \, A^{2} + 10 i \, A B - B^{2}}{a^{2} c^{3} f^{2}}} - 5 i \, A + B\right )} e^{\left (-i \, f x - i \, e\right )}}{4 \, a c f}\right ) + \sqrt{2}{\left ({\left (-2 i \, A - 2 \, B\right )} e^{\left (6 i \, f x + 6 i \, e\right )} +{\left (-16 i \, A - 4 \, B\right )} e^{\left (4 i \, f x + 4 i \, e\right )} +{\left (-11 i \, A - 5 \, B\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + 3 i \, A - 3 \, B\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 )}}{48 \, a c^{2} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(3*sqrt(1/2)*a*c^2*f*sqrt(-(25*A^2 + 10*I*A*B - B^2)/(a^2*c^3*f^2))*e^(2*I*f*x + 2*I*e)*log(1/4*(sqrt(2)*
sqrt(1/2)*(a*c*f*e^(2*I*f*x + 2*I*e) + a*c*f)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(-(25*A^2 + 10*I*A*B - B^2
)/(a^2*c^3*f^2)) + 5*I*A - B)*e^(-I*f*x - I*e)/(a*c*f)) - 3*sqrt(1/2)*a*c^2*f*sqrt(-(25*A^2 + 10*I*A*B - B^2)/
(a^2*c^3*f^2))*e^(2*I*f*x + 2*I*e)*log(-1/4*(sqrt(2)*sqrt(1/2)*(a*c*f*e^(2*I*f*x + 2*I*e) + a*c*f)*sqrt(c/(e^(
2*I*f*x + 2*I*e) + 1))*sqrt(-(25*A^2 + 10*I*A*B - B^2)/(a^2*c^3*f^2)) - 5*I*A + B)*e^(-I*f*x - I*e)/(a*c*f)) +
sqrt(2)*((-2*I*A - 2*B)*e^(6*I*f*x + 6*I*e) + (-16*I*A - 4*B)*e^(4*I*f*x + 4*I*e) + (-11*I*A - 5*B)*e^(2*I*f*
x + 2*I*e) + 3*I*A - 3*B)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1)))*e^(-2*I*f*x - 2*I*e)/(a*c^2*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))/(a+I*a*tan(f*x+e))/(c-I*c*tan(f*x+e))**(3/2),x)

[Out]

Exception raised: AttributeError

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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