3.778 \(\int \frac{A+B \tan (e+f x)}{(a+i a \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2}} \, dx\)

Optimal. Leaf size=273 \[ -\frac{7 (-B+9 i A)}{128 a^2 c^2 f \sqrt{c-i c \tan (e+f x)}}+\frac{7 (-B+9 i A) \tanh ^{-1}\left (\frac{\sqrt{c-i c \tan (e+f x)}}{\sqrt{2} \sqrt{c}}\right )}{128 \sqrt{2} a^2 c^{5/2} f}+\frac{-B+i A}{4 a^2 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2}}-\frac{7 (-B+9 i A)}{192 a^2 c f (c-i c \tan (e+f x))^{3/2}}-\frac{7 (-B+9 i A)}{160 a^2 f (c-i c \tan (e+f x))^{5/2}}+\frac{-B+9 i A}{16 a^2 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}} \]

[Out]

(7*((9*I)*A - B)*ArcTanh[Sqrt[c - I*c*Tan[e + f*x]]/(Sqrt[2]*Sqrt[c])])/(128*Sqrt[2]*a^2*c^(5/2)*f) - (7*((9*I
)*A - B))/(160*a^2*f*(c - I*c*Tan[e + f*x])^(5/2)) + (I*A - B)/(4*a^2*f*(1 + I*Tan[e + f*x])^2*(c - I*c*Tan[e
+ f*x])^(5/2)) + ((9*I)*A - B)/(16*a^2*f*(1 + I*Tan[e + f*x])*(c - I*c*Tan[e + f*x])^(5/2)) - (7*((9*I)*A - B)
)/(192*a^2*c*f*(c - I*c*Tan[e + f*x])^(3/2)) - (7*((9*I)*A - B))/(128*a^2*c^2*f*Sqrt[c - I*c*Tan[e + f*x]])

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Rubi [A]  time = 0.315856, antiderivative size = 273, 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} \[ -\frac{7 (-B+9 i A)}{128 a^2 c^2 f \sqrt{c-i c \tan (e+f x)}}+\frac{7 (-B+9 i A) \tanh ^{-1}\left (\frac{\sqrt{c-i c \tan (e+f x)}}{\sqrt{2} \sqrt{c}}\right )}{128 \sqrt{2} a^2 c^{5/2} f}+\frac{-B+i A}{4 a^2 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2}}-\frac{7 (-B+9 i A)}{192 a^2 c f (c-i c \tan (e+f x))^{3/2}}-\frac{7 (-B+9 i A)}{160 a^2 f (c-i c \tan (e+f x))^{5/2}}+\frac{-B+9 i A}{16 a^2 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}} \]

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])^(5/2)),x]

[Out]

(7*((9*I)*A - B)*ArcTanh[Sqrt[c - I*c*Tan[e + f*x]]/(Sqrt[2]*Sqrt[c])])/(128*Sqrt[2]*a^2*c^(5/2)*f) - (7*((9*I
)*A - B))/(160*a^2*f*(c - I*c*Tan[e + f*x])^(5/2)) + (I*A - B)/(4*a^2*f*(1 + I*Tan[e + f*x])^2*(c - I*c*Tan[e
+ f*x])^(5/2)) + ((9*I)*A - B)/(16*a^2*f*(1 + I*Tan[e + f*x])*(c - I*c*Tan[e + f*x])^(5/2)) - (7*((9*I)*A - B)
)/(192*a^2*c*f*(c - I*c*Tan[e + f*x])^(3/2)) - (7*((9*I)*A - B))/(128*a^2*c^2*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))^{5/2}} \, dx &=\frac{(a c) \operatorname{Subst}\left (\int \frac{A+B x}{(a+i a x)^3 (c-i c x)^{7/2}} \, 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))^{5/2}}+\frac{((9 A+i B) c) \operatorname{Subst}\left (\int \frac{1}{(a+i a x)^2 (c-i c x)^{7/2}} \, 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))^{5/2}}+\frac{9 i A-B}{16 a^2 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}+\frac{(7 (9 A+i B) c) \operatorname{Subst}\left (\int \frac{1}{(a+i a x) (c-i c x)^{7/2}} \, dx,x,\tan (e+f x)\right )}{32 a f}\\ &=-\frac{7 (9 i A-B)}{160 a^2 f (c-i c \tan (e+f x))^{5/2}}+\frac{i A-B}{4 a^2 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2}}+\frac{9 i A-B}{16 a^2 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}+\frac{(7 (9 A+i B)) \operatorname{Subst}\left (\int \frac{1}{(a+i a x) (c-i c x)^{5/2}} \, dx,x,\tan (e+f x)\right )}{64 a f}\\ &=-\frac{7 (9 i A-B)}{160 a^2 f (c-i c \tan (e+f x))^{5/2}}+\frac{i A-B}{4 a^2 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2}}+\frac{9 i A-B}{16 a^2 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}-\frac{7 (9 i A-B)}{192 a^2 c f (c-i c \tan (e+f x))^{3/2}}+\frac{(7 (9 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 )}{128 a c f}\\ &=-\frac{7 (9 i A-B)}{160 a^2 f (c-i c \tan (e+f x))^{5/2}}+\frac{i A-B}{4 a^2 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2}}+\frac{9 i A-B}{16 a^2 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}-\frac{7 (9 i A-B)}{192 a^2 c f (c-i c \tan (e+f x))^{3/2}}-\frac{7 (9 i A-B)}{128 a^2 c^2 f \sqrt{c-i c \tan (e+f x)}}+\frac{(7 (9 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 )}{256 a c^2 f}\\ &=-\frac{7 (9 i A-B)}{160 a^2 f (c-i c \tan (e+f x))^{5/2}}+\frac{i A-B}{4 a^2 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2}}+\frac{9 i A-B}{16 a^2 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}-\frac{7 (9 i A-B)}{192 a^2 c f (c-i c \tan (e+f x))^{3/2}}-\frac{7 (9 i A-B)}{128 a^2 c^2 f \sqrt{c-i c \tan (e+f x)}}+\frac{(7 (9 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 )}{128 a c^3 f}\\ &=\frac{7 (9 i A-B) \tanh ^{-1}\left (\frac{\sqrt{c-i c \tan (e+f x)}}{\sqrt{2} \sqrt{c}}\right )}{128 \sqrt{2} a^2 c^{5/2} f}-\frac{7 (9 i A-B)}{160 a^2 f (c-i c \tan (e+f x))^{5/2}}+\frac{i A-B}{4 a^2 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2}}+\frac{9 i A-B}{16 a^2 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}-\frac{7 (9 i A-B)}{192 a^2 c f (c-i c \tan (e+f x))^{3/2}}-\frac{7 (9 i A-B)}{128 a^2 c^2 f \sqrt{c-i c \tan (e+f x)}}\\ \end{align*}

Mathematica [A]  time = 9.1252, size = 209, normalized size = 0.77 \[ \frac{\sqrt{c-i c \tan (e+f x)} (\cos (e+f x)+i \sin (e+f x)) \left (105 i (9 A+i B) e^{-i (e+f x)} \sqrt{1+e^{2 i (e+f x)}} \tanh ^{-1}\left (\sqrt{1+e^{2 i (e+f x)}}\right )+2 \cos (e+f x) ((-223 B+87 i A) \cos (2 (e+f x))+6 i (A+9 i B) \cos (4 (e+f x))+423 A \sin (2 (e+f x))+54 A \sin (4 (e+f x))-864 i A+47 i B \sin (2 (e+f x))+6 i B \sin (4 (e+f x))-64 B)\right )}{3840 a^2 c^3 f} \]

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])^(5/2)),x]

[Out]

((Cos[e + f*x] + I*Sin[e + f*x])*(((105*I)*(9*A + I*B)*Sqrt[1 + E^((2*I)*(e + f*x))]*ArcTanh[Sqrt[1 + E^((2*I)
*(e + f*x))]])/E^(I*(e + f*x)) + 2*Cos[e + f*x]*((-864*I)*A - 64*B + ((87*I)*A - 223*B)*Cos[2*(e + f*x)] + (6*
I)*(A + (9*I)*B)*Cos[4*(e + f*x)] + 423*A*Sin[2*(e + f*x)] + (47*I)*B*Sin[2*(e + f*x)] + 54*A*Sin[4*(e + f*x)]
+ (6*I)*B*Sin[4*(e + f*x)]))*Sqrt[c - I*c*Tan[e + f*x]])/(3840*a^2*c^3*f)

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Maple [A]  time = 0.118, size = 199, normalized size = 0.7 \begin{align*}{\frac{-2\,i{c}^{2}}{f{a}^{2}} \left ({\frac{1}{16\,{c}^{4}} \left ({\frac{1}{ \left ( -c-ic\tan \left ( fx+e \right ) \right ) ^{2}} \left ( \left ({\frac{7\,i}{16}}B+{\frac{15\,A}{16}} \right ) \left ( c-ic\tan \left ( fx+e \right ) \right ) ^{{\frac{3}{2}}}+ \left ( -{\frac{9\,i}{8}}Bc-{\frac{17\,Ac}{8}} \right ) \sqrt{c-ic\tan \left ( fx+e \right ) } \right ) }-{\frac{ \left ( 7\,iB+63\,A \right ) \sqrt{2}}{32}{\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}{16\,{c}^{4}}{\frac{1}{\sqrt{c-ic\tan \left ( fx+e \right ) }}}}-{\frac{-3\,A+iB}{48\,{c}^{3}} \left ( c-ic\tan \left ( fx+e \right ) \right ) ^{-{\frac{3}{2}}}}-{\frac{-A+iB}{40\,{c}^{2}} \left ( c-ic\tan \left ( fx+e \right ) \right ) ^{-{\frac{5}{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))^(5/2),x)

[Out]

-2*I/f/a^2*c^2*(1/16/c^4*(((7/16*I*B+15/16*A)*(c-I*c*tan(f*x+e))^(3/2)+(-9/8*I*B*c-17/8*A*c)*(c-I*c*tan(f*x+e)
)^(1/2))/(-c-I*c*tan(f*x+e))^2-7/32*(I*B+9*A)*2^(1/2)/c^(1/2)*arctanh(1/2*(c-I*c*tan(f*x+e))^(1/2)*2^(1/2)/c^(
1/2)))+3/16*A/c^4/(c-I*c*tan(f*x+e))^(1/2)-1/48/c^3*(-3*A+I*B)/(c-I*c*tan(f*x+e))^(3/2)-1/40/c^2*(-A+I*B)/(c-I
*c*tan(f*x+e))^(5/2))

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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))^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.82368, size = 1233, normalized size = 4.52 \begin{align*} \frac{{\left (15 \, \sqrt{\frac{1}{2}} a^{2} c^{3} f \sqrt{-\frac{3969 \, A^{2} + 882 i \, A B - 49 \, B^{2}}{a^{4} c^{5} 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^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{2} c^{2} f\right )} \sqrt{\frac{c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt{-\frac{3969 \, A^{2} + 882 i \, A B - 49 \, B^{2}}{a^{4} c^{5} f^{2}}} + 63 i \, A - 7 \, B\right )} e^{\left (-i \, f x - i \, e\right )}}{64 \, a^{2} c^{2} f}\right ) - 15 \, \sqrt{\frac{1}{2}} a^{2} c^{3} f \sqrt{-\frac{3969 \, A^{2} + 882 i \, A B - 49 \, B^{2}}{a^{4} c^{5} 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^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{2} c^{2} f\right )} \sqrt{\frac{c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt{-\frac{3969 \, A^{2} + 882 i \, A B - 49 \, B^{2}}{a^{4} c^{5} f^{2}}} - 63 i \, A + 7 \, B\right )} e^{\left (-i \, f x - i \, e\right )}}{64 \, a^{2} c^{2} f}\right ) + \sqrt{2}{\left ({\left (-24 i \, A - 24 \, B\right )} e^{\left (10 i \, f x + 10 i \, e\right )} +{\left (-192 i \, A - 112 \, B\right )} e^{\left (8 i \, f x + 8 i \, e\right )} +{\left (-1032 i \, A - 152 \, B\right )} e^{\left (6 i \, f x + 6 i \, e\right )} +{\left (-609 i \, A - 199 \, B\right )} e^{\left (4 i \, f x + 4 i \, e\right )} +{\left (285 i \, A - 165 \, B\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + 30 i \, A - 30 \, 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 )}}{3840 \, a^{2} c^{3} 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))^(5/2),x, algorithm="fricas")

[Out]

1/3840*(15*sqrt(1/2)*a^2*c^3*f*sqrt(-(3969*A^2 + 882*I*A*B - 49*B^2)/(a^4*c^5*f^2))*e^(4*I*f*x + 4*I*e)*log(1/
64*(sqrt(2)*sqrt(1/2)*(a^2*c^2*f*e^(2*I*f*x + 2*I*e) + a^2*c^2*f)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(-(396
9*A^2 + 882*I*A*B - 49*B^2)/(a^4*c^5*f^2)) + 63*I*A - 7*B)*e^(-I*f*x - I*e)/(a^2*c^2*f)) - 15*sqrt(1/2)*a^2*c^
3*f*sqrt(-(3969*A^2 + 882*I*A*B - 49*B^2)/(a^4*c^5*f^2))*e^(4*I*f*x + 4*I*e)*log(-1/64*(sqrt(2)*sqrt(1/2)*(a^2
*c^2*f*e^(2*I*f*x + 2*I*e) + a^2*c^2*f)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(-(3969*A^2 + 882*I*A*B - 49*B^2
)/(a^4*c^5*f^2)) - 63*I*A + 7*B)*e^(-I*f*x - I*e)/(a^2*c^2*f)) + sqrt(2)*((-24*I*A - 24*B)*e^(10*I*f*x + 10*I*
e) + (-192*I*A - 112*B)*e^(8*I*f*x + 8*I*e) + (-1032*I*A - 152*B)*e^(6*I*f*x + 6*I*e) + (-609*I*A - 199*B)*e^(
4*I*f*x + 4*I*e) + (285*I*A - 165*B)*e^(2*I*f*x + 2*I*e) + 30*I*A - 30*B)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1)))*e
^(-4*I*f*x - 4*I*e)/(a^2*c^3*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))**(5/2),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{5}{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))^(5/2),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)^(5/2)), x)