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

Optimal. Leaf size=226 \[ \frac{5 (B+7 i A) \tanh ^{-1}\left (\frac{\sqrt{c-i c \tan (e+f x)}}{\sqrt{2} \sqrt{c}}\right )}{64 \sqrt{2} a^2 c^{3/2} f}+\frac{-B+i A}{4 a^2 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{3/2}}-\frac{5 (B+7 i A)}{64 a^2 c f \sqrt{c-i c \tan (e+f x)}}-\frac{5 (B+7 i A)}{96 a^2 f (c-i c \tan (e+f x))^{3/2}}+\frac{B+7 i A}{16 a^2 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}} \]

[Out]

(5*((7*I)*A + B)*ArcTanh[Sqrt[c - I*c*Tan[e + f*x]]/(Sqrt[2]*Sqrt[c])])/(64*Sqrt[2]*a^2*c^(3/2)*f) - (5*((7*I)
*A + B))/(96*a^2*f*(c - I*c*Tan[e + f*x])^(3/2)) + (I*A - B)/(4*a^2*f*(1 + I*Tan[e + f*x])^2*(c - I*c*Tan[e +
f*x])^(3/2)) + ((7*I)*A + B)/(16*a^2*f*(1 + I*Tan[e + f*x])*(c - I*c*Tan[e + f*x])^(3/2)) - (5*((7*I)*A + B))/
(64*a^2*c*f*Sqrt[c - I*c*Tan[e + f*x]])

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Rubi [A]  time = 0.290363, antiderivative size = 226, normalized size of antiderivative = 1., number of steps used = 7, 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} \[ \frac{5 (B+7 i A) \tanh ^{-1}\left (\frac{\sqrt{c-i c \tan (e+f x)}}{\sqrt{2} \sqrt{c}}\right )}{64 \sqrt{2} a^2 c^{3/2} f}+\frac{-B+i A}{4 a^2 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{3/2}}-\frac{5 (B+7 i A)}{64 a^2 c f \sqrt{c-i c \tan (e+f x)}}-\frac{5 (B+7 i A)}{96 a^2 f (c-i c \tan (e+f x))^{3/2}}+\frac{B+7 i A}{16 a^2 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(5*((7*I)*A + B)*ArcTanh[Sqrt[c - I*c*Tan[e + f*x]]/(Sqrt[2]*Sqrt[c])])/(64*Sqrt[2]*a^2*c^(3/2)*f) - (5*((7*I)
*A + B))/(96*a^2*f*(c - I*c*Tan[e + f*x])^(3/2)) + (I*A - B)/(4*a^2*f*(1 + I*Tan[e + f*x])^2*(c - I*c*Tan[e +
f*x])^(3/2)) + ((7*I)*A + B)/(16*a^2*f*(1 + I*Tan[e + f*x])*(c - I*c*Tan[e + f*x])^(3/2)) - (5*((7*I)*A + B))/
(64*a^2*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))^2 (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)^3 (c-i c x)^{5/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{i A-B}{4 a^2 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{3/2}}+\frac{((7 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 )}{8 f}\\ &=\frac{i A-B}{4 a^2 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{3/2}}+\frac{7 i A+B}{16 a^2 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}+\frac{(5 (7 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 )}{32 a f}\\ &=-\frac{5 (7 i A+B)}{96 a^2 f (c-i c \tan (e+f x))^{3/2}}+\frac{i A-B}{4 a^2 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{3/2}}+\frac{7 i A+B}{16 a^2 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}+\frac{(5 (7 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 )}{64 a f}\\ &=-\frac{5 (7 i A+B)}{96 a^2 f (c-i c \tan (e+f x))^{3/2}}+\frac{i A-B}{4 a^2 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{3/2}}+\frac{7 i A+B}{16 a^2 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}-\frac{5 (7 i A+B)}{64 a^2 c f \sqrt{c-i c \tan (e+f x)}}+\frac{(5 (7 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 )}{128 a c f}\\ &=-\frac{5 (7 i A+B)}{96 a^2 f (c-i c \tan (e+f x))^{3/2}}+\frac{i A-B}{4 a^2 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{3/2}}+\frac{7 i A+B}{16 a^2 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}-\frac{5 (7 i A+B)}{64 a^2 c f \sqrt{c-i c \tan (e+f x)}}+\frac{(5 (7 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 )}{64 a c^2 f}\\ &=\frac{5 (7 i A+B) \tanh ^{-1}\left (\frac{\sqrt{c-i c \tan (e+f x)}}{\sqrt{2} \sqrt{c}}\right )}{64 \sqrt{2} a^2 c^{3/2} f}-\frac{5 (7 i A+B)}{96 a^2 f (c-i c \tan (e+f x))^{3/2}}+\frac{i A-B}{4 a^2 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{3/2}}+\frac{7 i A+B}{16 a^2 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}-\frac{5 (7 i A+B)}{64 a^2 c f \sqrt{c-i c \tan (e+f x)}}\\ \end{align*}

Mathematica [A]  time = 5.91634, size = 204, normalized size = 0.9 \[ \frac{e^{-4 i (e+f x)} \sqrt{c-i c \tan (e+f x)} \left (15 (B+7 i A) e^{4 i (e+f x)} \sqrt{1+e^{2 i (e+f x)}} \tanh ^{-1}\left (\sqrt{1+e^{2 i (e+f x)}}\right )-i \left (1+e^{2 i (e+f x)}\right ) \left (A \left (-39 e^{2 i (e+f x)}+80 e^{4 i (e+f x)}+8 e^{6 i (e+f x)}-6\right )-i B \left (15 e^{2 i (e+f x)}+32 e^{4 i (e+f x)}+8 e^{6 i (e+f x)}+6\right )\right )\right )}{384 a^2 c^2 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((-I)*(1 + E^((2*I)*(e + f*x)))*((-I)*B*(6 + 15*E^((2*I)*(e + f*x)) + 32*E^((4*I)*(e + f*x)) + 8*E^((6*I)*(e
+ f*x))) + A*(-6 - 39*E^((2*I)*(e + f*x)) + 80*E^((4*I)*(e + f*x)) + 8*E^((6*I)*(e + f*x)))) + 15*((7*I)*A + B
)*E^((4*I)*(e + f*x))*Sqrt[1 + E^((2*I)*(e + f*x))]*ArcTanh[Sqrt[1 + E^((2*I)*(e + f*x))]])*Sqrt[c - I*c*Tan[e
+ f*x]])/(384*a^2*c^2*E^((4*I)*(e + f*x))*f)

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Maple [A]  time = 0.109, size = 179, normalized size = 0.8 \begin{align*}{\frac{-2\,i{c}^{2}}{f{a}^{2}} \left ({\frac{1}{16\,{c}^{3}} \left ({\frac{1}{ \left ( -c-ic\tan \left ( fx+e \right ) \right ) ^{2}} \left ( \left ({\frac{3\,i}{8}}B+{\frac{11\,A}{8}} \right ) \left ( c-ic\tan \left ( fx+e \right ) \right ) ^{{\frac{3}{2}}}+ \left ( -{\frac{13\,Ac}{4}}-{\frac{5\,i}{4}}Bc \right ) \sqrt{c-ic\tan \left ( fx+e \right ) } \right ) }-{\frac{ \left ( 35\,A-5\,iB \right ) \sqrt{2}}{16}{\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{-3\,A+iB}{16\,{c}^{3}}{\frac{1}{\sqrt{c-ic\tan \left ( fx+e \right ) }}}}-{\frac{-A+iB}{24\,{c}^{2}} \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))^2/(c-I*c*tan(f*x+e))^(3/2),x)

[Out]

-2*I/f/a^2*c^2*(1/16/c^3*(((3/8*I*B+11/8*A)*(c-I*c*tan(f*x+e))^(3/2)+(-13/4*A*c-5/4*I*B*c)*(c-I*c*tan(f*x+e))^
(1/2))/(-c-I*c*tan(f*x+e))^2-5/16*(7*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^3*(-3*A+I*B)/(c-I*c*tan(f*x+e))^(1/2)-1/24/c^2*(-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))^2/(c-I*c*tan(f*x+e))^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.53654, size = 1139, normalized size = 5.04 \begin{align*} \frac{{\left (3 \, \sqrt{\frac{1}{2}} a^{2} c^{2} f \sqrt{-\frac{1225 \, A^{2} - 350 i \, A B - 25 \, B^{2}}{a^{4} c^{3} f^{2}}} e^{\left (4 i \, f x + 4 i \, e\right )} \log \left (\frac{{\left (\sqrt{2} \sqrt{\frac{1}{2}}{\left (a^{2} c f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{2} c f\right )} \sqrt{\frac{c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt{-\frac{1225 \, A^{2} - 350 i \, A B - 25 \, B^{2}}{a^{4} c^{3} f^{2}}} + 35 i \, A + 5 \, B\right )} e^{\left (-i \, f x - i \, e\right )}}{32 \, a^{2} c f}\right ) - 3 \, \sqrt{\frac{1}{2}} a^{2} c^{2} f \sqrt{-\frac{1225 \, A^{2} - 350 i \, A B - 25 \, B^{2}}{a^{4} c^{3} f^{2}}} e^{\left (4 i \, f x + 4 i \, e\right )} \log \left (-\frac{{\left (\sqrt{2} \sqrt{\frac{1}{2}}{\left (a^{2} c f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{2} c f\right )} \sqrt{\frac{c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt{-\frac{1225 \, A^{2} - 350 i \, A B - 25 \, B^{2}}{a^{4} c^{3} f^{2}}} - 35 i \, A - 5 \, B\right )} e^{\left (-i \, f x - i \, e\right )}}{32 \, a^{2} c f}\right ) + \sqrt{2}{\left ({\left (-8 i \, A - 8 \, B\right )} e^{\left (8 i \, f x + 8 i \, e\right )} +{\left (-88 i \, A - 40 \, B\right )} e^{\left (6 i \, f x + 6 i \, e\right )} +{\left (-41 i \, A - 47 \, B\right )} e^{\left (4 i \, f x + 4 i \, e\right )} +{\left (45 i \, A - 21 \, B\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + 6 i \, A - 6 \, B\right )} \sqrt{\frac{c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-4 i \, f x - 4 i \, e\right )}}{384 \, a^{2} 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))^2/(c-I*c*tan(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

1/384*(3*sqrt(1/2)*a^2*c^2*f*sqrt(-(1225*A^2 - 350*I*A*B - 25*B^2)/(a^4*c^3*f^2))*e^(4*I*f*x + 4*I*e)*log(1/32
*(sqrt(2)*sqrt(1/2)*(a^2*c*f*e^(2*I*f*x + 2*I*e) + a^2*c*f)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(-(1225*A^2
- 350*I*A*B - 25*B^2)/(a^4*c^3*f^2)) + 35*I*A + 5*B)*e^(-I*f*x - I*e)/(a^2*c*f)) - 3*sqrt(1/2)*a^2*c^2*f*sqrt(
-(1225*A^2 - 350*I*A*B - 25*B^2)/(a^4*c^3*f^2))*e^(4*I*f*x + 4*I*e)*log(-1/32*(sqrt(2)*sqrt(1/2)*(a^2*c*f*e^(2
*I*f*x + 2*I*e) + a^2*c*f)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(-(1225*A^2 - 350*I*A*B - 25*B^2)/(a^4*c^3*f^
2)) - 35*I*A - 5*B)*e^(-I*f*x - I*e)/(a^2*c*f)) + sqrt(2)*((-8*I*A - 8*B)*e^(8*I*f*x + 8*I*e) + (-88*I*A - 40*
B)*e^(6*I*f*x + 6*I*e) + (-41*I*A - 47*B)*e^(4*I*f*x + 4*I*e) + (45*I*A - 21*B)*e^(2*I*f*x + 2*I*e) + 6*I*A -
6*B)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1)))*e^(-4*I*f*x - 4*I*e)/(a^2*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))**2/(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 )}^{2}{\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))^2/(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)^2*(-I*c*tan(f*x + e) + c)^(3/2)), x)