3.759 \(\int (a+i a \tan (e+f x))^3 (A+B \tan (e+f x)) \sqrt {c-i c \tan (e+f x)} \, dx\)

Optimal. Leaf size=142 \[ \frac {2 a^3 (5 B+i A) (c-i c \tan (e+f x))^{5/2}}{5 c^2 f}-\frac {8 a^3 (2 B+i A) (c-i c \tan (e+f x))^{3/2}}{3 c f}+\frac {8 a^3 (B+i A) \sqrt {c-i c \tan (e+f x)}}{f}-\frac {2 a^3 B (c-i c \tan (e+f x))^{7/2}}{7 c^3 f} \]

[Out]

8*a^3*(I*A+B)*(c-I*c*tan(f*x+e))^(1/2)/f-8/3*a^3*(I*A+2*B)*(c-I*c*tan(f*x+e))^(3/2)/c/f+2/5*a^3*(I*A+5*B)*(c-I
*c*tan(f*x+e))^(5/2)/c^2/f-2/7*a^3*B*(c-I*c*tan(f*x+e))^(7/2)/c^3/f

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Rubi [A]  time = 0.18, antiderivative size = 142, 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} \[ \frac {2 a^3 (5 B+i A) (c-i c \tan (e+f x))^{5/2}}{5 c^2 f}-\frac {8 a^3 (2 B+i A) (c-i c \tan (e+f x))^{3/2}}{3 c f}+\frac {8 a^3 (B+i A) \sqrt {c-i c \tan (e+f x)}}{f}-\frac {2 a^3 B (c-i c \tan (e+f x))^{7/2}}{7 c^3 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 7.47, size = 124, normalized size = 0.87 \[ \frac {a^3 \sec ^2(e+f x) (\cos (3 f x)+i \sin (3 f x)) \sqrt {c-i c \tan (e+f x)} ((-98 A+100 i B) \tan (e+f x)+\cos (2 (e+f x)) ((-98 A+130 i B) \tan (e+f x)+322 i A+290 B)+280 i A+170 B)}{105 f (\cos (f x)+i \sin (f x))^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*Sec[e + f*x]^2*(Cos[3*f*x] + I*Sin[3*f*x])*Sqrt[c - I*c*Tan[e + f*x]]*((280*I)*A + 170*B + (-98*A + (100*
I)*B)*Tan[e + f*x] + Cos[2*(e + f*x)]*((322*I)*A + 290*B + (-98*A + (130*I)*B)*Tan[e + f*x])))/(105*f*(Cos[f*x
] + I*Sin[f*x])^3)

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fricas [A]  time = 0.99, size = 133, normalized size = 0.94 \[ \frac {\sqrt {2} {\left ({\left (840 i \, A + 840 \, B\right )} a^{3} e^{\left (6 i \, f x + 6 i \, e\right )} + {\left (1960 i \, A + 1400 \, B\right )} a^{3} e^{\left (4 i \, f x + 4 i \, e\right )} + {\left (1568 i \, A + 1120 \, B\right )} a^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (448 i \, A + 320 \, B\right )} a^{3}\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{105 \, {\left (f e^{\left (6 i \, f x + 6 i \, e\right )} + 3 \, f e^{\left (4 i \, f x + 4 i \, e\right )} + 3 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )}} \]

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))^3*(A+B*tan(f*x+e)),x, algorithm="fricas")

[Out]

1/105*sqrt(2)*((840*I*A + 840*B)*a^3*e^(6*I*f*x + 6*I*e) + (1960*I*A + 1400*B)*a^3*e^(4*I*f*x + 4*I*e) + (1568
*I*A + 1120*B)*a^3*e^(2*I*f*x + 2*I*e) + (448*I*A + 320*B)*a^3)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))/(f*e^(6*I*f*
x + 6*I*e) + 3*f*e^(4*I*f*x + 4*I*e) + 3*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))^3*(A+B*tan(f*x+e)),x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 0.41, size = 121, normalized size = 0.85 \[ \frac {2 i a^{3} \left (\frac {i B \left (c -i c \tan \left (f x +e \right )\right )^{\frac {7}{2}}}{7}+\frac {\left (-5 i B c +c A \right ) \left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{5}+\frac {\left (-4 \left (-i B c +c A \right ) c +4 i B \,c^{2}\right ) \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}+4 \left (-i B c +c A \right ) c^{2} \sqrt {c -i c \tan \left (f x +e \right )}\right )}{f \,c^{3}} \]

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))^3*(A+B*tan(f*x+e)),x)

[Out]

2*I/f*a^3/c^3*(1/7*I*B*(c-I*c*tan(f*x+e))^(7/2)+1/5*(-5*I*B*c+c*A)*(c-I*c*tan(f*x+e))^(5/2)+1/3*(-4*(-I*B*c+c*
A)*c+4*I*B*c^2)*(c-I*c*tan(f*x+e))^(3/2)+4*(-I*B*c+c*A)*c^2*(c-I*c*tan(f*x+e))^(1/2))

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maxima [A]  time = 0.76, size = 104, normalized size = 0.73 \[ \frac {2 i \, {\left (15 i \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {7}{2}} B a^{3} + 21 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}} {\left (A - 5 i \, B\right )} a^{3} c - 140 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} {\left (A - 2 i \, B\right )} a^{3} c^{2} + 420 \, \sqrt {-i \, c \tan \left (f x + e\right ) + c} {\left (A - i \, B\right )} a^{3} c^{3}\right )}}{105 \, c^{3} f} \]

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))^3*(A+B*tan(f*x+e)),x, algorithm="maxima")

[Out]

2/105*I*(15*I*(-I*c*tan(f*x + e) + c)^(7/2)*B*a^3 + 21*(-I*c*tan(f*x + e) + c)^(5/2)*(A - 5*I*B)*a^3*c - 140*(
-I*c*tan(f*x + e) + c)^(3/2)*(A - 2*I*B)*a^3*c^2 + 420*sqrt(-I*c*tan(f*x + e) + c)*(A - I*B)*a^3*c^3)/(c^3*f)

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mupad [B]  time = 13.58, size = 313, normalized size = 2.20 \[ -\frac {\sqrt {c+\frac {c\,\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1}}\,\left (\frac {a^3\,\left (A-B\,1{}\mathrm {i}\right )\,8{}\mathrm {i}}{3\,f}+\frac {a^3\,\left (A-B\,3{}\mathrm {i}\right )\,8{}\mathrm {i}}{3\,f}\right )}{{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1}-\frac {\sqrt {c+\frac {c\,\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1}}\,\left (\frac {a^3\,\left (A-B\,1{}\mathrm {i}\right )\,8{}\mathrm {i}}{7\,f}-\frac {a^3\,\left (A+B\,1{}\mathrm {i}\right )\,8{}\mathrm {i}}{7\,f}\right )}{{\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1\right )}^3}+\frac {\sqrt {c+\frac {c\,\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1}}\,\left (\frac {32\,B\,a^3}{5\,f}+\frac {a^3\,\left (A-B\,1{}\mathrm {i}\right )\,8{}\mathrm {i}}{5\,f}\right )}{{\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1\right )}^2}+\frac {a^3\,\left (A-B\,1{}\mathrm {i}\right )\,\sqrt {c+\frac {c\,\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1}}\,8{}\mathrm {i}}{f} \]

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)^3*(c - c*tan(e + f*x)*1i)^(1/2),x)

[Out]

((c + (c*(exp(e*2i + f*x*2i)*1i - 1i)*1i)/(exp(e*2i + f*x*2i) + 1))^(1/2)*((a^3*(A - B*1i)*8i)/(5*f) + (32*B*a
^3)/(5*f)))/(exp(e*2i + f*x*2i) + 1)^2 - ((c + (c*(exp(e*2i + f*x*2i)*1i - 1i)*1i)/(exp(e*2i + f*x*2i) + 1))^(
1/2)*((a^3*(A - B*1i)*8i)/(7*f) - (a^3*(A + B*1i)*8i)/(7*f)))/(exp(e*2i + f*x*2i) + 1)^3 - ((c + (c*(exp(e*2i
+ f*x*2i)*1i - 1i)*1i)/(exp(e*2i + f*x*2i) + 1))^(1/2)*((a^3*(A - B*1i)*8i)/(3*f) + (a^3*(A - B*3i)*8i)/(3*f))
)/(exp(e*2i + f*x*2i) + 1) + (a^3*(A - B*1i)*(c + (c*(exp(e*2i + f*x*2i)*1i - 1i)*1i)/(exp(e*2i + f*x*2i) + 1)
)^(1/2)*8i)/f

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - i a^{3} \left (\int i A \sqrt {- i c \tan {\left (e + f x \right )} + c}\, dx + \int \left (- 3 A \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )}\right )\, dx + \int A \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{3}{\left (e + f x \right )}\, dx + \int \left (- 3 B \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{2}{\left (e + f x \right )}\right )\, dx + \int B \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{4}{\left (e + f x \right )}\, dx + \int \left (- 3 i A \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{2}{\left (e + f x \right )}\right )\, dx + \int i B \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )}\, dx + \int \left (- 3 i B \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{3}{\left (e + f x \right )}\right )\, dx\right ) \]

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))**3*(A+B*tan(f*x+e)),x)

[Out]

-I*a**3*(Integral(I*A*sqrt(-I*c*tan(e + f*x) + c), x) + Integral(-3*A*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x)
, x) + Integral(A*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x)**3, x) + Integral(-3*B*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)**4, x) + Integral(-3*I*A*sqrt(-I*c*t
an(e + f*x) + c)*tan(e + f*x)**2, x) + Integral(I*B*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x), x) + Integral(-3
*I*B*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x)**3, x))

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