\(\int (a+i a \tan (e+f x))^3 (A+B \tan (e+f x)) \sqrt {c-i c \tan (e+f x)} \, dx\) [759]

Optimal result

Integrand size = 43, antiderivative size = 142 \[ \int (a+i a \tan (e+f x))^3 (A+B \tan (e+f x)) \sqrt {c-i c \tan (e+f x)} \, dx=\frac {8 a^3 (i A+B) \sqrt {c-i c \tan (e+f x)}}{f}-\frac {8 a^3 (i A+2 B) (c-i c \tan (e+f x))^{3/2}}{3 c f}+\frac {2 a^3 (i A+5 B) (c-i c \tan (e+f x))^{5/2}}{5 c^2 f}-\frac {2 a^3 B (c-i c \tan (e+f x))^{7/2}}{7 c^3 f} \]

[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

Rubi [A] (verified)

Time = 0.22 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.047, Rules used = {3669, 78} \[ \int (a+i a \tan (e+f x))^3 (A+B \tan (e+f x)) \sqrt {c-i c \tan (e+f x)} \, dx=\frac {2 a^3 (5 B+i A) (c-i c \tan (e+f x))^{5/2}}{5 c^2 f}-\frac {8 a^3 (2 B+i A) (c-i c \tan (e+f x))^{3/2}}{3 c f}+\frac {8 a^3 (B+i A) \sqrt {c-i c \tan (e+f x)}}{f}-\frac {2 a^3 B (c-i c \tan (e+f x))^{7/2}}{7 c^3 f} \]

[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 78

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 3669

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*} \text {integral}& = \frac {(a c) \text {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) \text {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*}

Mathematica [A] (verified)

Time = 3.96 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.65 \[ \int (a+i a \tan (e+f x))^3 (A+B \tan (e+f x)) \sqrt {c-i c \tan (e+f x)} \, dx=-\frac {2 a^3 c (i+\tan (e+f x)) \left (-301 A+230 i B+(-98 i A-115 B) \tan (e+f x)+3 (7 A-20 i B) \tan ^2(e+f x)+15 B \tan ^3(e+f x)\right )}{105 f \sqrt {c-i c \tan (e+f x)}} \]

[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]

(-2*a^3*c*(I + Tan[e + f*x])*(-301*A + (230*I)*B + ((-98*I)*A - 115*B)*Tan[e + f*x] + 3*(7*A - (20*I)*B)*Tan[e
+ f*x]^2 + 15*B*Tan[e + f*x]^3))/(105*f*Sqrt[c - I*c*Tan[e + f*x]])

Maple [A] (verified)

Time = 0.37 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.85

method	result	size
derivativedivides	\(\frac {2 i a^{3} \left (\frac {i B \left (c -i c \tan \left (f x +e \right )\right )^{\frac {7}{2}}}{7}+\frac {\left (-5 i B c +c A \right ) \left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{5}+\frac {\left (-4 \left (-i B c +c A \right ) c +4 i B \,c^{2}\right ) \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}+4 \sqrt {c -i c \tan \left (f x +e \right )}\, \left (-i B c +c A \right ) c^{2}\right )}{f \,c^{3}}\)	\(121\)
default	\(\frac {2 i a^{3} \left (\frac {i B \left (c -i c \tan \left (f x +e \right )\right )^{\frac {7}{2}}}{7}+\frac {\left (-5 i B c +c A \right ) \left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{5}+\frac {\left (-4 \left (-i B c +c A \right ) c +4 i B \,c^{2}\right ) \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}+4 \sqrt {c -i c \tan \left (f x +e \right )}\, \left (-i B c +c A \right ) c^{2}\right )}{f \,c^{3}}\)	\(121\)
parts	\(-\frac {6 i a^{3} \left (-i B +A \right ) \left (\frac {\left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}-\frac {c^{\frac {3}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{2}\right )}{f c}-\frac {2 a^{3} \left (i A +3 B \right ) \left (-\frac {\left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{5}+\frac {c \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}-\sqrt {c -i c \tan \left (f x +e \right )}\, c^{2}+\frac {c^{\frac {5}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{2}\right )}{f \,c^{2}}+\frac {a^{3} \left (3 i A +B \right ) \left (2 \sqrt {c -i c \tan \left (f x +e \right )}-\sqrt {c}\, \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )\right )}{f}+\frac {i a^{3} A \sqrt {c}\, \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{f}+\frac {2 B \,a^{3} \left (-\frac {\left (c -i c \tan \left (f x +e \right )\right )^{\frac {7}{2}}}{7}+\frac {2 c \left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{5}-\frac {2 \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}} c^{2}}{3}+\frac {c^{\frac {7}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{2}\right )}{f \,c^{3}}\)	\(367\)

[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,method=_RETURNVERBOSE)

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.96 \[ \int (a+i a \tan (e+f x))^3 (A+B \tan (e+f x)) \sqrt {c-i c \tan (e+f x)} \, dx=-\frac {8 \, \sqrt {2} {\left (105 \, {\left (-i \, A - B\right )} a^{3} e^{\left (6 i \, f x + 6 i \, e\right )} + 35 \, {\left (-7 i \, A - 5 \, B\right )} a^{3} e^{\left (4 i \, f x + 4 i \, e\right )} + 28 \, {\left (-7 i \, A - 5 \, B\right )} a^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + 8 \, {\left (-7 i \, A - 5 \, B\right )} a^{3}\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{105 \, {\left (f e^{\left (6 i \, f x + 6 i \, e\right )} + 3 \, f e^{\left (4 i \, f x + 4 i \, e\right )} + 3 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )}} \]

[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]

-8/105*sqrt(2)*(105*(-I*A - B)*a^3*e^(6*I*f*x + 6*I*e) + 35*(-7*I*A - 5*B)*a^3*e^(4*I*f*x + 4*I*e) + 28*(-7*I*
A - 5*B)*a^3*e^(2*I*f*x + 2*I*e) + 8*(-7*I*A - 5*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)

Sympy [F]

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

[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))

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.73 \[ \int (a+i a \tan (e+f x))^3 (A+B \tan (e+f x)) \sqrt {c-i c \tan (e+f x)} \, dx=\frac {2 i \, {\left (15 i \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {7}{2}} B a^{3} + 21 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}} {\left (A - 5 i \, B\right )} a^{3} c - 140 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} {\left (A - 2 i \, B\right )} a^{3} c^{2} + 420 \, \sqrt {-i \, c \tan \left (f x + e\right ) + c} {\left (A - i \, B\right )} a^{3} c^{3}\right )}}{105 \, c^{3} f} \]

[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)

Giac [F]

\[ \int (a+i a \tan (e+f x))^3 (A+B \tan (e+f x)) \sqrt {c-i c \tan (e+f x)} \, dx=\int { {\left (B \tan \left (f x + e\right ) + A\right )} {\left (i \, a \tan \left (f x + e\right ) + a\right )}^{3} \sqrt {-i \, c \tan \left (f x + e\right ) + c} \,d x } \]

[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]

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

Mupad [B] (verification not implemented)

Time = 12.55 (sec) , antiderivative size = 313, normalized size of antiderivative = 2.20 \[ \int (a+i a \tan (e+f x))^3 (A+B \tan (e+f x)) \sqrt {c-i c \tan (e+f x)} \, dx=-\frac {\sqrt {c+\frac {c\,\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1}}\,\left (\frac {a^3\,\left (A-B\,1{}\mathrm {i}\right )\,8{}\mathrm {i}}{3\,f}+\frac {a^3\,\left (A-B\,3{}\mathrm {i}\right )\,8{}\mathrm {i}}{3\,f}\right )}{{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1}-\frac {\sqrt {c+\frac {c\,\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1}}\,\left (\frac {a^3\,\left (A-B\,1{}\mathrm {i}\right )\,8{}\mathrm {i}}{7\,f}-\frac {a^3\,\left (A+B\,1{}\mathrm {i}\right )\,8{}\mathrm {i}}{7\,f}\right )}{{\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1\right )}^3}+\frac {\sqrt {c+\frac {c\,\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1}}\,\left (\frac {32\,B\,a^3}{5\,f}+\frac {a^3\,\left (A-B\,1{}\mathrm {i}\right )\,8{}\mathrm {i}}{5\,f}\right )}{{\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1\right )}^2}+\frac {a^3\,\left (A-B\,1{}\mathrm {i}\right )\,\sqrt {c+\frac {c\,\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1}}\,8{}\mathrm {i}}{f} \]

[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