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

Optimal. Leaf size=274 \[ \frac{35 (B+3 i A) \tanh ^{-1}\left (\frac{\sqrt{c-i c \tan (e+f x)}}{\sqrt{2} \sqrt{c}}\right )}{256 \sqrt{2} a^3 c^{3/2} f}+\frac{-B+i A}{6 a^3 f (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{3/2}}-\frac{35 (B+3 i A)}{256 a^3 c f \sqrt{c-i c \tan (e+f x)}}-\frac{35 (B+3 i A)}{384 a^3 f (c-i c \tan (e+f x))^{3/2}}+\frac{7 (B+3 i A)}{64 a^3 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}+\frac{B+3 i A}{16 a^3 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{3/2}} \]

[Out]

(35*((3*I)*A + B)*ArcTanh[Sqrt[c - I*c*Tan[e + f*x]]/(Sqrt[2]*Sqrt[c])])/(256*Sqrt[2]*a^3*c^(3/2)*f) - (35*((3
*I)*A + B))/(384*a^3*f*(c - I*c*Tan[e + f*x])^(3/2)) + (I*A - B)/(6*a^3*f*(1 + I*Tan[e + f*x])^3*(c - I*c*Tan[
e + f*x])^(3/2)) + ((3*I)*A + B)/(16*a^3*f*(1 + I*Tan[e + f*x])^2*(c - I*c*Tan[e + f*x])^(3/2)) + (7*((3*I)*A
+ B))/(64*a^3*f*(1 + I*Tan[e + f*x])*(c - I*c*Tan[e + f*x])^(3/2)) - (35*((3*I)*A + B))/(256*a^3*c*f*Sqrt[c -
I*c*Tan[e + f*x]])

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Rubi [A]  time = 0.309907, antiderivative size = 274, normalized size of antiderivative = 1., number of steps used = 8, 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} \[ \frac{35 (B+3 i A) \tanh ^{-1}\left (\frac{\sqrt{c-i c \tan (e+f x)}}{\sqrt{2} \sqrt{c}}\right )}{256 \sqrt{2} a^3 c^{3/2} f}+\frac{-B+i A}{6 a^3 f (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{3/2}}-\frac{35 (B+3 i A)}{256 a^3 c f \sqrt{c-i c \tan (e+f x)}}-\frac{35 (B+3 i A)}{384 a^3 f (c-i c \tan (e+f x))^{3/2}}+\frac{7 (B+3 i A)}{64 a^3 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}+\frac{B+3 i A}{16 a^3 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(35*((3*I)*A + B)*ArcTanh[Sqrt[c - I*c*Tan[e + f*x]]/(Sqrt[2]*Sqrt[c])])/(256*Sqrt[2]*a^3*c^(3/2)*f) - (35*((3
*I)*A + B))/(384*a^3*f*(c - I*c*Tan[e + f*x])^(3/2)) + (I*A - B)/(6*a^3*f*(1 + I*Tan[e + f*x])^3*(c - I*c*Tan[
e + f*x])^(3/2)) + ((3*I)*A + B)/(16*a^3*f*(1 + I*Tan[e + f*x])^2*(c - I*c*Tan[e + f*x])^(3/2)) + (7*((3*I)*A
+ B))/(64*a^3*f*(1 + I*Tan[e + f*x])*(c - I*c*Tan[e + f*x])^(3/2)) - (35*((3*I)*A + B))/(256*a^3*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))^3 (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)^4 (c-i c x)^{5/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{i A-B}{6 a^3 f (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{3/2}}+\frac{((3 A-i B) c) \operatorname{Subst}\left (\int \frac{1}{(a+i a x)^3 (c-i c x)^{5/2}} \, dx,x,\tan (e+f x)\right )}{4 f}\\ &=\frac{i A-B}{6 a^3 f (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{3/2}}+\frac{3 i A+B}{16 a^3 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{3/2}}+\frac{(7 (3 A-i B) c) \operatorname{Subst}\left (\int \frac{1}{(a+i a x)^2 (c-i c x)^{5/2}} \, dx,x,\tan (e+f x)\right )}{32 a f}\\ &=\frac{i A-B}{6 a^3 f (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{3/2}}+\frac{3 i A+B}{16 a^3 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{3/2}}+\frac{7 (3 i A+B)}{64 a^3 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}+\frac{(35 (3 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 )}{128 a^2 f}\\ &=-\frac{35 (3 i A+B)}{384 a^3 f (c-i c \tan (e+f x))^{3/2}}+\frac{i A-B}{6 a^3 f (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{3/2}}+\frac{3 i A+B}{16 a^3 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{3/2}}+\frac{7 (3 i A+B)}{64 a^3 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}+\frac{(35 (3 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 )}{256 a^2 f}\\ &=-\frac{35 (3 i A+B)}{384 a^3 f (c-i c \tan (e+f x))^{3/2}}+\frac{i A-B}{6 a^3 f (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{3/2}}+\frac{3 i A+B}{16 a^3 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{3/2}}+\frac{7 (3 i A+B)}{64 a^3 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}-\frac{35 (3 i A+B)}{256 a^3 c f \sqrt{c-i c \tan (e+f x)}}+\frac{(35 (3 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 )}{512 a^2 c f}\\ &=-\frac{35 (3 i A+B)}{384 a^3 f (c-i c \tan (e+f x))^{3/2}}+\frac{i A-B}{6 a^3 f (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{3/2}}+\frac{3 i A+B}{16 a^3 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{3/2}}+\frac{7 (3 i A+B)}{64 a^3 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}-\frac{35 (3 i A+B)}{256 a^3 c f \sqrt{c-i c \tan (e+f x)}}+\frac{(35 (3 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 )}{256 a^2 c^2 f}\\ &=\frac{35 (3 i A+B) \tanh ^{-1}\left (\frac{\sqrt{c-i c \tan (e+f x)}}{\sqrt{2} \sqrt{c}}\right )}{256 \sqrt{2} a^3 c^{3/2} f}-\frac{35 (3 i A+B)}{384 a^3 f (c-i c \tan (e+f x))^{3/2}}+\frac{i A-B}{6 a^3 f (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{3/2}}+\frac{3 i A+B}{16 a^3 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{3/2}}+\frac{7 (3 i A+B)}{64 a^3 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}-\frac{35 (3 i A+B)}{256 a^3 c f \sqrt{c-i c \tan (e+f x)}}\\ \end{align*}

Mathematica [A]  time = 7.99622, size = 206, normalized size = 0.75 \[ \frac{\sqrt{c-i c \tan (e+f x)} (\sin (e+f x)+i \cos (e+f x)) \left (105 (3 A-i B) e^{i (e+f x)} \sqrt{1+e^{2 i (e+f x)}} \tanh ^{-1}\left (\sqrt{1+e^{2 i (e+f x)}}\right )-2 \cos (e+f x) (2 (79 A-69 i B) \cos (2 (e+f x))+8 (A-3 i B) \cos (4 (e+f x))+258 i A \sin (2 (e+f x))+24 i A \sin (4 (e+f x))-165 A+86 B \sin (2 (e+f x))+8 B \sin (4 (e+f x))-9 i B)\right )}{1536 a^3 c^2 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((I*Cos[e + f*x] + Sin[e + f*x])*(105*(3*A - I*B)*E^(I*(e + f*x))*Sqrt[1 + E^((2*I)*(e + f*x))]*ArcTanh[Sqrt[1
+ E^((2*I)*(e + f*x))]] - 2*Cos[e + f*x]*(-165*A - (9*I)*B + 2*(79*A - (69*I)*B)*Cos[2*(e + f*x)] + 8*(A - (3
*I)*B)*Cos[4*(e + f*x)] + (258*I)*A*Sin[2*(e + f*x)] + 86*B*Sin[2*(e + f*x)] + (24*I)*A*Sin[4*(e + f*x)] + 8*B
*Sin[4*(e + f*x)]))*Sqrt[c - I*c*Tan[e + f*x]])/(1536*a^3*c^2*f)

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Maple [A]  time = 0.115, size = 206, normalized size = 0.8 \begin{align*}{\frac{2\,i{c}^{3}}{f{a}^{3}} \left ( -{\frac{1}{16\,{c}^{4}} \left ({\frac{1}{ \left ( -c-ic\tan \left ( fx+e \right ) \right ) ^{3}} \left ( \left ( -{\frac{3\,i}{32}}B+{\frac{41\,A}{32}} \right ) \left ( c-ic\tan \left ( fx+e \right ) \right ) ^{{\frac{5}{2}}}+ \left ({\frac{i}{6}}Bc-{\frac{35\,Ac}{6}} \right ) \left ( c-ic\tan \left ( fx+e \right ) \right ) ^{{\frac{3}{2}}}+ \left ({\frac{55\,A{c}^{2}}{8}}+{\frac{3\,i}{8}}B{c}^{2} \right ) \sqrt{c-ic\tan \left ( fx+e \right ) } \right ) }-{\frac{ \left ( 105\,A-35\,iB \right ) \sqrt{2}}{64}{\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{2\,A-iB}{16\,{c}^{4}}{\frac{1}{\sqrt{c-ic\tan \left ( fx+e \right ) }}}}-{\frac{A-iB}{48\,{c}^{3}} \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))^3/(c-I*c*tan(f*x+e))^(3/2),x)

[Out]

2*I/f/a^3*c^3*(-1/16/c^4*(((-3/32*I*B+41/32*A)*(c-I*c*tan(f*x+e))^(5/2)+(1/6*I*B*c-35/6*A*c)*(c-I*c*tan(f*x+e)
)^(3/2)+(55/8*A*c^2+3/8*I*B*c^2)*(c-I*c*tan(f*x+e))^(1/2))/(-c-I*c*tan(f*x+e))^3-35/64*(3*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)))-1/16/c^4*(2*A-I*B)/(c-I*c*tan(f*x+e))^(1/2)-1/48/c^3
*(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))^3/(c-I*c*tan(f*x+e))^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.51609, size = 1233, normalized size = 4.5 \begin{align*} \frac{{\left (3 \, \sqrt{\frac{1}{2}} a^{3} c^{2} f \sqrt{-\frac{11025 \, A^{2} - 7350 i \, A B - 1225 \, B^{2}}{a^{6} c^{3} f^{2}}} e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (\frac{{\left (\sqrt{2} \sqrt{\frac{1}{2}}{\left (a^{3} c f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{3} c f\right )} \sqrt{\frac{c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt{-\frac{11025 \, A^{2} - 7350 i \, A B - 1225 \, B^{2}}{a^{6} c^{3} f^{2}}} + 105 i \, A + 35 \, B\right )} e^{\left (-i \, f x - i \, e\right )}}{128 \, a^{3} c f}\right ) - 3 \, \sqrt{\frac{1}{2}} a^{3} c^{2} f \sqrt{-\frac{11025 \, A^{2} - 7350 i \, A B - 1225 \, B^{2}}{a^{6} c^{3} f^{2}}} e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (-\frac{{\left (\sqrt{2} \sqrt{\frac{1}{2}}{\left (a^{3} c f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{3} c f\right )} \sqrt{\frac{c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt{-\frac{11025 \, A^{2} - 7350 i \, A B - 1225 \, B^{2}}{a^{6} c^{3} f^{2}}} - 105 i \, A - 35 \, B\right )} e^{\left (-i \, f x - i \, e\right )}}{128 \, a^{3} c f}\right ) + \sqrt{2}{\left ({\left (-16 i \, A - 16 \, B\right )} e^{\left (10 i \, f x + 10 i \, e\right )} +{\left (-224 i \, A - 128 \, B\right )} e^{\left (8 i \, f x + 8 i \, e\right )} +{\left (-43 i \, A - 121 \, B\right )} e^{\left (6 i \, f x + 6 i \, e\right )} +{\left (215 i \, A - 35 \, B\right )} e^{\left (4 i \, f x + 4 i \, e\right )} +{\left (58 i \, A - 34 \, B\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + 8 i \, A - 8 \, B\right )} \sqrt{\frac{c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-6 i \, f x - 6 i \, e\right )}}{1536 \, a^{3} 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))^3/(c-I*c*tan(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

1/1536*(3*sqrt(1/2)*a^3*c^2*f*sqrt(-(11025*A^2 - 7350*I*A*B - 1225*B^2)/(a^6*c^3*f^2))*e^(6*I*f*x + 6*I*e)*log
(1/128*(sqrt(2)*sqrt(1/2)*(a^3*c*f*e^(2*I*f*x + 2*I*e) + a^3*c*f)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(-(110
25*A^2 - 7350*I*A*B - 1225*B^2)/(a^6*c^3*f^2)) + 105*I*A + 35*B)*e^(-I*f*x - I*e)/(a^3*c*f)) - 3*sqrt(1/2)*a^3
*c^2*f*sqrt(-(11025*A^2 - 7350*I*A*B - 1225*B^2)/(a^6*c^3*f^2))*e^(6*I*f*x + 6*I*e)*log(-1/128*(sqrt(2)*sqrt(1
/2)*(a^3*c*f*e^(2*I*f*x + 2*I*e) + a^3*c*f)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(-(11025*A^2 - 7350*I*A*B -
1225*B^2)/(a^6*c^3*f^2)) - 105*I*A - 35*B)*e^(-I*f*x - I*e)/(a^3*c*f)) + sqrt(2)*((-16*I*A - 16*B)*e^(10*I*f*x
+ 10*I*e) + (-224*I*A - 128*B)*e^(8*I*f*x + 8*I*e) + (-43*I*A - 121*B)*e^(6*I*f*x + 6*I*e) + (215*I*A - 35*B)
*e^(4*I*f*x + 4*I*e) + (58*I*A - 34*B)*e^(2*I*f*x + 2*I*e) + 8*I*A - 8*B)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1)))*e
^(-6*I*f*x - 6*I*e)/(a^3*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))**3/(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 )}^{3}{\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))^3/(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)^3*(-I*c*tan(f*x + e) + c)^(3/2)), x)