3.751 \(\int (a+i a \tan (e+f x))^2 (A+B \tan (e+f x)) \sqrt {c-i c \tan (e+f x)} \, dx\)

Optimal. Leaf size=103 \[ -\frac {2 a^2 (3 B+i A) (c-i c \tan (e+f x))^{3/2}}{3 c f}+\frac {4 a^2 (B+i A) \sqrt {c-i c \tan (e+f x)}}{f}+\frac {2 a^2 B (c-i c \tan (e+f x))^{5/2}}{5 c^2 f} \]

[Out]

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

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Rubi [A]  time = 0.16, antiderivative size = 103, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.047, Rules used = {3588, 77} \[ -\frac {2 a^2 (3 B+i A) (c-i c \tan (e+f x))^{3/2}}{3 c f}+\frac {4 a^2 (B+i A) \sqrt {c-i c \tan (e+f x)}}{f}+\frac {2 a^2 B (c-i c \tan (e+f x))^{5/2}}{5 c^2 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*a^2*(I*A + B)*Sqrt[c - I*c*Tan[e + f*x]])/f - (2*a^2*(I*A + 3*B)*(c - I*c*Tan[e + f*x])^(3/2))/(3*c*f) + (2
*a^2*B*(c - I*c*Tan[e + f*x])^(5/2))/(5*c^2*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 (a+i a \tan (e+f x))^2 (A+B \tan (e+f x)) \sqrt {c-i c \tan (e+f x)} \, dx &=\frac {(a c) \operatorname {Subst}\left (\int \frac {(a+i a x) (A+B x)}{\sqrt {c-i c x}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {(a c) \operatorname {Subst}\left (\int \left (\frac {2 a (A-i B)}{\sqrt {c-i c x}}-\frac {a (A-3 i B) \sqrt {c-i c x}}{c}-\frac {i a B (c-i c x)^{3/2}}{c^2}\right ) \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {4 a^2 (i A+B) \sqrt {c-i c \tan (e+f x)}}{f}-\frac {2 a^2 (i A+3 B) (c-i c \tan (e+f x))^{3/2}}{3 c f}+\frac {2 a^2 B (c-i c \tan (e+f x))^{5/2}}{5 c^2 f}\\ \end {align*}

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Mathematica [A]  time = 4.96, size = 83, normalized size = 0.81 \[ \frac {a^2 \sec ^2(e+f x) \sqrt {c-i c \tan (e+f x)} ((-5 A+9 i B) \sin (2 (e+f x))+(21 B+25 i A) \cos (2 (e+f x))+5 (3 B+5 i A))}{15 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*Sec[e + f*x]^2*(5*((5*I)*A + 3*B) + ((25*I)*A + 21*B)*Cos[2*(e + f*x)] + (-5*A + (9*I)*B)*Sin[2*(e + f*x)
])*Sqrt[c - I*c*Tan[e + f*x]])/(15*f)

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fricas [A]  time = 1.49, size = 101, normalized size = 0.98 \[ \frac {\sqrt {2} {\left ({\left (60 i \, A + 60 \, B\right )} a^{2} e^{\left (4 i \, f x + 4 i \, e\right )} + {\left (100 i \, A + 60 \, B\right )} a^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (40 i \, A + 24 \, B\right )} a^{2}\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{15 \, {\left (f e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-I*c*tan(f*x+e))^(1/2)*(a+I*a*tan(f*x+e))^2*(A+B*tan(f*x+e)),x, algorithm="fricas")

[Out]

1/15*sqrt(2)*((60*I*A + 60*B)*a^2*e^(4*I*f*x + 4*I*e) + (100*I*A + 60*B)*a^2*e^(2*I*f*x + 2*I*e) + (40*I*A + 2
4*B)*a^2)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))/(f*e^(4*I*f*x + 4*I*e) + 2*f*e^(2*I*f*x + 2*I*e) + f)

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-I*c*tan(f*x+e))^(1/2)*(a+I*a*tan(f*x+e))^2*(A+B*tan(f*x+e)),x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 0.27, size = 83, normalized size = 0.81 \[ -\frac {2 i a^{2} \left (\frac {i B \left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{5}+\frac {\left (-3 i B c +c A \right ) \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}-2 \left (-i B c +c A \right ) c \sqrt {c -i c \tan \left (f x +e \right )}\right )}{f \,c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c-I*c*tan(f*x+e))^(1/2)*(a+I*a*tan(f*x+e))^2*(A+B*tan(f*x+e)),x)

[Out]

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

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maxima [A]  time = 0.40, size = 78, normalized size = 0.76 \[ -\frac {2 i \, {\left (3 i \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}} B a^{2} + 5 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} {\left (A - 3 i \, B\right )} a^{2} c - 30 \, \sqrt {-i \, c \tan \left (f x + e\right ) + c} {\left (A - i \, B\right )} a^{2} c^{2}\right )}}{15 \, c^{2} f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-I*c*tan(f*x+e))^(1/2)*(a+I*a*tan(f*x+e))^2*(A+B*tan(f*x+e)),x, algorithm="maxima")

[Out]

-2/15*I*(3*I*(-I*c*tan(f*x + e) + c)^(5/2)*B*a^2 + 5*(-I*c*tan(f*x + e) + c)^(3/2)*(A - 3*I*B)*a^2*c - 30*sqrt
(-I*c*tan(f*x + e) + c)*(A - I*B)*a^2*c^2)/(c^2*f)

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mupad [B]  time = 11.99, size = 241, normalized size = 2.34 \[ \frac {2\,a^2\,\sqrt {\frac {c\,\left (\cos \left (2\,e+2\,f\,x\right )+1-\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}\,\left (A\,250{}\mathrm {i}+174\,B+A\,\cos \left (2\,e+2\,f\,x\right )\,375{}\mathrm {i}+A\,\cos \left (4\,e+4\,f\,x\right )\,150{}\mathrm {i}+A\,\cos \left (6\,e+6\,f\,x\right )\,25{}\mathrm {i}+267\,B\,\cos \left (2\,e+2\,f\,x\right )+114\,B\,\cos \left (4\,e+4\,f\,x\right )+21\,B\,\cos \left (6\,e+6\,f\,x\right )-25\,A\,\sin \left (2\,e+2\,f\,x\right )-20\,A\,\sin \left (4\,e+4\,f\,x\right )-5\,A\,\sin \left (6\,e+6\,f\,x\right )+B\,\sin \left (2\,e+2\,f\,x\right )\,45{}\mathrm {i}+B\,\sin \left (4\,e+4\,f\,x\right )\,36{}\mathrm {i}+B\,\sin \left (6\,e+6\,f\,x\right )\,9{}\mathrm {i}\right )}{15\,f\,\left (15\,\cos \left (2\,e+2\,f\,x\right )+6\,\cos \left (4\,e+4\,f\,x\right )+\cos \left (6\,e+6\,f\,x\right )+10\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*tan(e + f*x))*(a + a*tan(e + f*x)*1i)^2*(c - c*tan(e + f*x)*1i)^(1/2),x)

[Out]

(2*a^2*((c*(cos(2*e + 2*f*x) - sin(2*e + 2*f*x)*1i + 1))/(cos(2*e + 2*f*x) + 1))^(1/2)*(A*250i + 174*B + A*cos
(2*e + 2*f*x)*375i + A*cos(4*e + 4*f*x)*150i + A*cos(6*e + 6*f*x)*25i + 267*B*cos(2*e + 2*f*x) + 114*B*cos(4*e
+ 4*f*x) + 21*B*cos(6*e + 6*f*x) - 25*A*sin(2*e + 2*f*x) - 20*A*sin(4*e + 4*f*x) - 5*A*sin(6*e + 6*f*x) + B*s
in(2*e + 2*f*x)*45i + B*sin(4*e + 4*f*x)*36i + B*sin(6*e + 6*f*x)*9i))/(15*f*(15*cos(2*e + 2*f*x) + 6*cos(4*e
+ 4*f*x) + cos(6*e + 6*f*x) + 10))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - a^{2} \left (\int \left (- A \sqrt {- i c \tan {\left (e + f x \right )} + c}\right )\, dx + \int A \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{2}{\left (e + f x \right )}\, dx + \int \left (- B \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )}\right )\, dx + \int B \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{3}{\left (e + f x \right )}\, dx + \int \left (- 2 i A \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )}\right )\, dx + \int \left (- 2 i B \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{2}{\left (e + f x \right )}\right )\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-I*c*tan(f*x+e))**(1/2)*(a+I*a*tan(f*x+e))**2*(A+B*tan(f*x+e)),x)

[Out]

-a**2*(Integral(-A*sqrt(-I*c*tan(e + f*x) + c), x) + Integral(A*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x)**2, x
) + Integral(-B*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x), x) + Integral(B*sqrt(-I*c*tan(e + f*x) + c)*tan(e +
f*x)**3, x) + Integral(-2*I*A*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x), x) + Integral(-2*I*B*sqrt(-I*c*tan(e +
f*x) + c)*tan(e + f*x)**2, x))

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