3.706 \(\int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^4}{a+i a \tan (e+f x)} \, dx\)

Optimal. Leaf size=157 \[ -\frac {c^4 (-5 B+i A) \tan ^2(e+f x)}{2 a f}+\frac {c^4 (5 A+12 i B) \tan (e+f x)}{a f}-\frac {8 c^4 (A+i B)}{a f (-\tan (e+f x)+i)}-\frac {4 c^4 (-5 B+3 i A) \log (\cos (e+f x))}{a f}-\frac {4 c^4 x (3 A+5 i B)}{a}-\frac {i B c^4 \tan ^3(e+f x)}{3 a f} \]

[Out]

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

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Rubi [A]  time = 0.22, antiderivative size = 157, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.049, Rules used = {3588, 77} \[ -\frac {c^4 (-5 B+i A) \tan ^2(e+f x)}{2 a f}+\frac {c^4 (5 A+12 i B) \tan (e+f x)}{a f}-\frac {8 c^4 (A+i B)}{a f (-\tan (e+f x)+i)}-\frac {4 c^4 (-5 B+3 i A) \log (\cos (e+f x))}{a f}-\frac {4 c^4 x (3 A+5 i B)}{a}-\frac {i B c^4 \tan ^3(e+f x)}{3 a f} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*Tan[e + f*x])*(c - I*c*Tan[e + f*x])^4)/(a + I*a*Tan[e + f*x]),x]

[Out]

(-4*(3*A + (5*I)*B)*c^4*x)/a - (4*((3*I)*A - 5*B)*c^4*Log[Cos[e + f*x]])/(a*f) - (8*(A + I*B)*c^4)/(a*f*(I - T
an[e + f*x])) + ((5*A + (12*I)*B)*c^4*Tan[e + f*x])/(a*f) - ((I*A - 5*B)*c^4*Tan[e + f*x]^2)/(2*a*f) - ((I/3)*
B*c^4*Tan[e + f*x]^3)/(a*f)

Rule 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

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]

Rubi steps

\begin {align*} \int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^4}{a+i a \tan (e+f x)} \, dx &=\frac {(a c) \operatorname {Subst}\left (\int \frac {(A+B x) (c-i c x)^3}{(a+i a x)^2} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {(a c) \operatorname {Subst}\left (\int \left (\frac {(5 A+12 i B) c^3}{a^2}+\frac {(-i A+5 B) c^3 x}{a^2}-\frac {i B c^3 x^2}{a^2}-\frac {8 (A+i B) c^3}{a^2 (-i+x)^2}+\frac {4 i (3 A+5 i B) c^3}{a^2 (-i+x)}\right ) \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac {4 (3 A+5 i B) c^4 x}{a}-\frac {4 (3 i A-5 B) c^4 \log (\cos (e+f x))}{a f}-\frac {8 (A+i B) c^4}{a f (i-\tan (e+f x))}+\frac {(5 A+12 i B) c^4 \tan (e+f x)}{a f}-\frac {(i A-5 B) c^4 \tan ^2(e+f x)}{2 a f}-\frac {i B c^4 \tan ^3(e+f x)}{3 a f}\\ \end {align*}

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Mathematica [A]  time = 3.89, size = 260, normalized size = 1.66 \[ \frac {c^4 (\cos (f x)+i \sin (f x)) (A+B \tan (e+f x)) \left (24 (A+i B) (\sin (e)+i \cos (e)) \cos (2 f x)+24 (A+i B) (\cos (e)-i \sin (e)) \sin (2 f x)+12 (5 B-3 i A) \left (\cos \left (\frac {e}{2}\right )+i \sin \left (\frac {e}{2}\right )\right )^2 \log \left (\cos ^2(e+f x)\right )-24 (3 A+5 i B) (\cos (e)+i \sin (e)) \tan ^{-1}(\tan (f x))+\cos (e) (\tan (e)-i) (2 B \tan (e)+3 (A+5 i B)) \sec ^2(e+f x)+2 (15 A+37 i B) (1+i \tan (e)) \sin (f x) \sec (e+f x)+2 B (\tan (e)-i) \sin (f x) \sec ^3(e+f x)\right )}{6 f (a+i a \tan (e+f x)) (A \cos (e+f x)+B \sin (e+f x))} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*Tan[e + f*x])*(c - I*c*Tan[e + f*x])^4)/(a + I*a*Tan[e + f*x]),x]

[Out]

(c^4*(Cos[f*x] + I*Sin[f*x])*(12*((-3*I)*A + 5*B)*Log[Cos[e + f*x]^2]*(Cos[e/2] + I*Sin[e/2])^2 - 24*(3*A + (5
*I)*B)*ArcTan[Tan[f*x]]*(Cos[e] + I*Sin[e]) + 24*(A + I*B)*Cos[2*f*x]*(I*Cos[e] + Sin[e]) + 24*(A + I*B)*(Cos[
e] - I*Sin[e])*Sin[2*f*x] + 2*(15*A + (37*I)*B)*Sec[e + f*x]*Sin[f*x]*(1 + I*Tan[e]) + 2*B*Sec[e + f*x]^3*Sin[
f*x]*(-I + Tan[e]) + Cos[e]*Sec[e + f*x]^2*(-I + Tan[e])*(3*(A + (5*I)*B) + 2*B*Tan[e]))*(A + B*Tan[e + f*x]))
/(6*f*(A*Cos[e + f*x] + B*Sin[e + f*x])*(a + I*a*Tan[e + f*x]))

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fricas [B]  time = 0.89, size = 297, normalized size = 1.89 \[ -\frac {24 \, {\left (3 \, A + 5 i \, B\right )} c^{4} f x e^{\left (8 i \, f x + 8 i \, e\right )} - {\left (12 i \, A - 12 \, B\right )} c^{4} + {\left (72 \, {\left (3 \, A + 5 i \, B\right )} c^{4} f x - {\left (36 i \, A - 60 \, B\right )} c^{4}\right )} e^{\left (6 i \, f x + 6 i \, e\right )} + {\left (72 \, {\left (3 \, A + 5 i \, B\right )} c^{4} f x - {\left (90 i \, A - 150 \, B\right )} c^{4}\right )} e^{\left (4 i \, f x + 4 i \, e\right )} + {\left (24 \, {\left (3 \, A + 5 i \, B\right )} c^{4} f x - {\left (66 i \, A - 110 \, B\right )} c^{4}\right )} e^{\left (2 i \, f x + 2 i \, e\right )} - {\left ({\left (-36 i \, A + 60 \, B\right )} c^{4} e^{\left (8 i \, f x + 8 i \, e\right )} + {\left (-108 i \, A + 180 \, B\right )} c^{4} e^{\left (6 i \, f x + 6 i \, e\right )} + {\left (-108 i \, A + 180 \, B\right )} c^{4} e^{\left (4 i \, f x + 4 i \, e\right )} + {\left (-36 i \, A + 60 \, B\right )} c^{4} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} \log \left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right )}{3 \, {\left (a f e^{\left (8 i \, f x + 8 i \, e\right )} + 3 \, a f e^{\left (6 i \, f x + 6 i \, e\right )} + 3 \, a f e^{\left (4 i \, f x + 4 i \, e\right )} + a f e^{\left (2 i \, f x + 2 i \, e\right )}\right )}} \]

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))^4/(a+I*a*tan(f*x+e)),x, algorithm="fricas")

[Out]

-1/3*(24*(3*A + 5*I*B)*c^4*f*x*e^(8*I*f*x + 8*I*e) - (12*I*A - 12*B)*c^4 + (72*(3*A + 5*I*B)*c^4*f*x - (36*I*A
- 60*B)*c^4)*e^(6*I*f*x + 6*I*e) + (72*(3*A + 5*I*B)*c^4*f*x - (90*I*A - 150*B)*c^4)*e^(4*I*f*x + 4*I*e) + (2
4*(3*A + 5*I*B)*c^4*f*x - (66*I*A - 110*B)*c^4)*e^(2*I*f*x + 2*I*e) - ((-36*I*A + 60*B)*c^4*e^(8*I*f*x + 8*I*e
) + (-108*I*A + 180*B)*c^4*e^(6*I*f*x + 6*I*e) + (-108*I*A + 180*B)*c^4*e^(4*I*f*x + 4*I*e) + (-36*I*A + 60*B)
*c^4*e^(2*I*f*x + 2*I*e))*log(e^(2*I*f*x + 2*I*e) + 1))/(a*f*e^(8*I*f*x + 8*I*e) + 3*a*f*e^(6*I*f*x + 6*I*e) +
3*a*f*e^(4*I*f*x + 4*I*e) + a*f*e^(2*I*f*x + 2*I*e))

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giac [B]  time = 3.28, size = 444, normalized size = 2.83 \[ -\frac {2 \, {\left (\frac {3 \, {\left (6 i \, A c^{4} - 10 \, B c^{4}\right )} \log \left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}{a} - \frac {3 \, {\left (12 i \, A c^{4} - 20 \, B c^{4}\right )} \log \left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - i\right )}{a} - \frac {3 \, {\left (-6 i \, A c^{4} + 10 \, B c^{4}\right )} \log \left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}{a} - \frac {3 \, {\left (-18 i \, A c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 30 \, B c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 44 \, A c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 68 i \, B c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 18 i \, A c^{4} - 30 \, B c^{4}\right )}}{a {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - i\right )}^{2}} + \frac {-33 i \, A c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{6} + 55 \, B c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{6} + 15 \, A c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} + 36 i \, B c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} + 102 i \, A c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 180 \, B c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 30 \, A c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 76 i \, B c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 102 i \, A c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 180 \, B c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 15 \, A c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 36 i \, B c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 33 i \, A c^{4} - 55 \, B c^{4}}{{\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 1\right )}^{3} a}\right )}}{3 \, f} \]

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))^4/(a+I*a*tan(f*x+e)),x, algorithm="giac")

[Out]

-2/3*(3*(6*I*A*c^4 - 10*B*c^4)*log(tan(1/2*f*x + 1/2*e) + 1)/a - 3*(12*I*A*c^4 - 20*B*c^4)*log(tan(1/2*f*x + 1
/2*e) - I)/a - 3*(-6*I*A*c^4 + 10*B*c^4)*log(tan(1/2*f*x + 1/2*e) - 1)/a - 3*(-18*I*A*c^4*tan(1/2*f*x + 1/2*e)
^2 + 30*B*c^4*tan(1/2*f*x + 1/2*e)^2 - 44*A*c^4*tan(1/2*f*x + 1/2*e) - 68*I*B*c^4*tan(1/2*f*x + 1/2*e) + 18*I*
A*c^4 - 30*B*c^4)/(a*(tan(1/2*f*x + 1/2*e) - I)^2) + (-33*I*A*c^4*tan(1/2*f*x + 1/2*e)^6 + 55*B*c^4*tan(1/2*f*
x + 1/2*e)^6 + 15*A*c^4*tan(1/2*f*x + 1/2*e)^5 + 36*I*B*c^4*tan(1/2*f*x + 1/2*e)^5 + 102*I*A*c^4*tan(1/2*f*x +
1/2*e)^4 - 180*B*c^4*tan(1/2*f*x + 1/2*e)^4 - 30*A*c^4*tan(1/2*f*x + 1/2*e)^3 - 76*I*B*c^4*tan(1/2*f*x + 1/2*
e)^3 - 102*I*A*c^4*tan(1/2*f*x + 1/2*e)^2 + 180*B*c^4*tan(1/2*f*x + 1/2*e)^2 + 15*A*c^4*tan(1/2*f*x + 1/2*e) +
36*I*B*c^4*tan(1/2*f*x + 1/2*e) + 33*I*A*c^4 - 55*B*c^4)/((tan(1/2*f*x + 1/2*e)^2 - 1)^3*a))/f

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maple [A]  time = 0.19, size = 193, normalized size = 1.23 \[ \frac {5 c^{4} B \left (\tan ^{2}\left (f x +e \right )\right )}{2 f a}-\frac {i B \,c^{4} \left (\tan ^{3}\left (f x +e \right )\right )}{3 a f}+\frac {5 c^{4} A \tan \left (f x +e \right )}{f a}-\frac {i c^{4} A \left (\tan ^{2}\left (f x +e \right )\right )}{2 f a}+\frac {12 i c^{4} B \tan \left (f x +e \right )}{f a}+\frac {8 i c^{4} B}{f a \left (\tan \left (f x +e \right )-i\right )}+\frac {8 c^{4} A}{f a \left (\tan \left (f x +e \right )-i\right )}+\frac {12 i c^{4} A \ln \left (\tan \left (f x +e \right )-i\right )}{f a}-\frac {20 c^{4} B \ln \left (\tan \left (f x +e \right )-i\right )}{f a} \]

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))^4/(a+I*a*tan(f*x+e)),x)

[Out]

5/2/f*c^4/a*B*tan(f*x+e)^2-1/3*I*B*c^4*tan(f*x+e)^3/a/f+5/f*c^4/a*A*tan(f*x+e)-1/2*I/f*c^4/a*A*tan(f*x+e)^2+12
*I/f*c^4/a*B*tan(f*x+e)+8*I/f*c^4/a/(tan(f*x+e)-I)*B+8/f*c^4/a/(tan(f*x+e)-I)*A+12*I/f*c^4/a*A*ln(tan(f*x+e)-I
)-20/f*c^4/a*B*ln(tan(f*x+e)-I)

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]

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))^4/(a+I*a*tan(f*x+e)),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: Error executing code in Maxima: expt: undefined: 0 to a negative e
xponent.

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mupad [B]  time = 8.80, size = 205, normalized size = 1.31 \[ \frac {\ln \left (\mathrm {tan}\left (e+f\,x\right )-\mathrm {i}\right )\,\left (-\frac {20\,B\,c^4}{a}+\frac {A\,c^4\,12{}\mathrm {i}}{a}\right )}{f}-\frac {{\mathrm {tan}\left (e+f\,x\right )}^2\,\left (-\frac {B\,c^4}{a}+\frac {c^4\,\left (A+B\,3{}\mathrm {i}\right )\,1{}\mathrm {i}}{2\,a}\right )}{f}-\frac {\frac {\left (4\,A\,c^4+B\,c^4\,12{}\mathrm {i}\right )\,1{}\mathrm {i}}{a}-\frac {\left (12\,A\,c^4+B\,c^4\,20{}\mathrm {i}\right )\,1{}\mathrm {i}}{a}}{f\,\left (1+\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}+\frac {\mathrm {tan}\left (e+f\,x\right )\,\left (\frac {2\,c^4\,\left (A+B\,3{}\mathrm {i}\right )}{a}+\frac {B\,c^4\,3{}\mathrm {i}}{a}-\frac {c^4\,\left (-B+A\,1{}\mathrm {i}\right )\,3{}\mathrm {i}}{a}\right )}{f}-\frac {B\,c^4\,{\mathrm {tan}\left (e+f\,x\right )}^3\,1{}\mathrm {i}}{3\,a\,f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*tan(e + f*x))*(c - c*tan(e + f*x)*1i)^4)/(a + a*tan(e + f*x)*1i),x)

[Out]

(log(tan(e + f*x) - 1i)*((A*c^4*12i)/a - (20*B*c^4)/a))/f - (tan(e + f*x)^2*((c^4*(A + B*3i)*1i)/(2*a) - (B*c^
4)/a))/f - (((4*A*c^4 + B*c^4*12i)*1i)/a - ((12*A*c^4 + B*c^4*20i)*1i)/a)/(f*(tan(e + f*x)*1i + 1)) + (tan(e +
f*x)*((2*c^4*(A + B*3i))/a + (B*c^4*3i)/a - (c^4*(A*1i - B)*3i)/a))/f - (B*c^4*tan(e + f*x)^3*1i)/(3*a*f)

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sympy [A]  time = 1.09, size = 340, normalized size = 2.17 \[ \frac {- 30 A c^{4} - 74 i B c^{4} + \left (- 54 A c^{4} e^{2 i e} - 114 i B c^{4} e^{2 i e}\right ) e^{2 i f x} + \left (- 24 A c^{4} e^{4 i e} - 48 i B c^{4} e^{4 i e}\right ) e^{4 i f x}}{3 i a f e^{6 i e} e^{6 i f x} + 9 i a f e^{4 i e} e^{4 i f x} + 9 i a f e^{2 i e} e^{2 i f x} + 3 i a f} + \begin {cases} - \frac {\left (- 4 i A c^{4} + 4 B c^{4}\right ) e^{- 2 i e} e^{- 2 i f x}}{a f} & \text {for}\: a f e^{2 i e} \neq 0 \\x \left (- \frac {- 24 A c^{4} - 40 i B c^{4}}{a} + \frac {i \left (24 i A c^{4} e^{2 i e} - 8 i A c^{4} - 40 B c^{4} e^{2 i e} + 8 B c^{4}\right ) e^{- 2 i e}}{a}\right ) & \text {otherwise} \end {cases} - \frac {4 i c^{4} \left (3 A + 5 i B\right ) \log {\left (e^{2 i f x} + e^{- 2 i e} \right )}}{a f} - \frac {x \left (24 A c^{4} + 40 i B c^{4}\right )}{a} \]

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))**4/(a+I*a*tan(f*x+e)),x)

[Out]

(-30*A*c**4 - 74*I*B*c**4 + (-54*A*c**4*exp(2*I*e) - 114*I*B*c**4*exp(2*I*e))*exp(2*I*f*x) + (-24*A*c**4*exp(4
*I*e) - 48*I*B*c**4*exp(4*I*e))*exp(4*I*f*x))/(3*I*a*f*exp(6*I*e)*exp(6*I*f*x) + 9*I*a*f*exp(4*I*e)*exp(4*I*f*
x) + 9*I*a*f*exp(2*I*e)*exp(2*I*f*x) + 3*I*a*f) + Piecewise((-(-4*I*A*c**4 + 4*B*c**4)*exp(-2*I*e)*exp(-2*I*f*
x)/(a*f), Ne(a*f*exp(2*I*e), 0)), (x*(-(-24*A*c**4 - 40*I*B*c**4)/a + I*(24*I*A*c**4*exp(2*I*e) - 8*I*A*c**4 -
40*B*c**4*exp(2*I*e) + 8*B*c**4)*exp(-2*I*e)/a), True)) - 4*I*c**4*(3*A + 5*I*B)*log(exp(2*I*f*x) + exp(-2*I*
e))/(a*f) - x*(24*A*c**4 + 40*I*B*c**4)/a

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