\(\int (a+i a \tan (e+f x)) (A+B \tan (e+f x)) (c-i c \tan (e+f x))^{7/2} \, dx\) [740]

Optimal result

Integrand size = 41, antiderivative size = 62 \[ \int (a+i a \tan (e+f x)) (A+B \tan (e+f x)) (c-i c \tan (e+f x))^{7/2} \, dx=\frac {2 a (i A+B) (c-i c \tan (e+f x))^{7/2}}{7 f}-\frac {2 a B (c-i c \tan (e+f x))^{9/2}}{9 c f} \]

[Out]

2/7*a*(I*A+B)*(c-I*c*tan(f*x+e))^(7/2)/f-2/9*a*B*(c-I*c*tan(f*x+e))^(9/2)/c/f

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.049, Rules used = {3669, 45} \[ \int (a+i a \tan (e+f x)) (A+B \tan (e+f x)) (c-i c \tan (e+f x))^{7/2} \, dx=\frac {2 a (B+i A) (c-i c \tan (e+f x))^{7/2}}{7 f}-\frac {2 a B (c-i c \tan (e+f x))^{9/2}}{9 c f} \]

[In]

Int[(a + I*a*Tan[e + f*x])*(A + B*Tan[e + f*x])*(c - I*c*Tan[e + f*x])^(7/2),x]

[Out]

(2*a*(I*A + B)*(c - I*c*Tan[e + f*x])^(7/2))/(7*f) - (2*a*B*(c - I*c*Tan[e + f*x])^(9/2))/(9*c*f)

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

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 (A+B x) (c-i c x)^{5/2} \, dx,x,\tan (e+f x)\right )}{f} \\ & = \frac {(a c) \text {Subst}\left (\int \left ((A-i B) (c-i c x)^{5/2}+\frac {i B (c-i c x)^{7/2}}{c}\right ) \, dx,x,\tan (e+f x)\right )}{f} \\ & = \frac {2 a (i A+B) (c-i c \tan (e+f x))^{7/2}}{7 f}-\frac {2 a B (c-i c \tan (e+f x))^{9/2}}{9 c f} \\ \end{align*}

Mathematica [A] (verified)

Time = 3.57 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.94 \[ \int (a+i a \tan (e+f x)) (A+B \tan (e+f x)) (c-i c \tan (e+f x))^{7/2} \, dx=-\frac {2 a c^3 (i+\tan (e+f x))^3 (9 A-2 i B+7 B \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}}{63 f} \]

[In]

Integrate[(a + I*a*Tan[e + f*x])*(A + B*Tan[e + f*x])*(c - I*c*Tan[e + f*x])^(7/2),x]

[Out]

(-2*a*c^3*(I + Tan[e + f*x])^3*(9*A - (2*I)*B + 7*B*Tan[e + f*x])*Sqrt[c - I*c*Tan[e + f*x]])/(63*f)

Maple [A] (verified)

Time = 0.39 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.89

method	result	size
derivativedivides	\(\frac {2 i a \left (\frac {i B \left (c -i c \tan \left (f x +e \right )\right )^{\frac {9}{2}}}{9}+\frac {\left (-i B c +c A \right ) \left (c -i c \tan \left (f x +e \right )\right )^{\frac {7}{2}}}{7}\right )}{f c}\)	\(55\)
default	\(\frac {2 i a \left (\frac {i B \left (c -i c \tan \left (f x +e \right )\right )^{\frac {9}{2}}}{9}+\frac {\left (-i B c +c A \right ) \left (c -i c \tan \left (f x +e \right )\right )^{\frac {7}{2}}}{7}\right )}{f c}\)	\(55\)
parts	\(\frac {2 i A a c \left (-\frac {\left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{5}-\frac {2 c \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}-4 \sqrt {c -i c \tan \left (f x +e \right )}\, c^{2}+4 c^{\frac {5}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )\right )}{f}+\frac {a \left (i A +B \right ) \left (\frac {2 \left (c -i c \tan \left (f x +e \right )\right )^{\frac {7}{2}}}{7}+\frac {2 c \left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{5}+\frac {4 \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}} c^{2}}{3}+8 \sqrt {c -i c \tan \left (f x +e \right )}\, c^{3}-8 c^{\frac {7}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )\right )}{f}-\frac {2 a B \left (\frac {\left (c -i c \tan \left (f x +e \right )\right )^{\frac {9}{2}}}{9}+\frac {c^{2} \left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{5}+\frac {2 \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}} c^{3}}{3}+4 \sqrt {c -i c \tan \left (f x +e \right )}\, c^{4}-4 c^{\frac {9}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )\right )}{f c}\)	\(324\)

[In]

int((a+I*a*tan(f*x+e))*(A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^(7/2),x,method=_RETURNVERBOSE)

[Out]

2*I/f*a/c*(1/9*I*B*(c-I*c*tan(f*x+e))^(9/2)+1/7*(-I*B*c+c*A)*(c-I*c*tan(f*x+e))^(7/2))

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 108 vs. \(2 (48) = 96\).

Time = 0.35 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.74 \[ \int (a+i a \tan (e+f x)) (A+B \tan (e+f x)) (c-i c \tan (e+f x))^{7/2} \, dx=-\frac {16 \, \sqrt {2} {\left (9 \, {\left (-i \, A - B\right )} a c^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (-9 i \, A + 5 \, B\right )} a c^{3}\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{63 \, {\left (f e^{\left (8 i \, f x + 8 i \, e\right )} + 4 \, f e^{\left (6 i \, f x + 6 i \, e\right )} + 6 \, f e^{\left (4 i \, f x + 4 i \, e\right )} + 4 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )}} \]

[In]

integrate((a+I*a*tan(f*x+e))*(A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^(7/2),x, algorithm="fricas")

[Out]

-16/63*sqrt(2)*(9*(-I*A - B)*a*c^3*e^(2*I*f*x + 2*I*e) + (-9*I*A + 5*B)*a*c^3)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1
))/(f*e^(8*I*f*x + 8*I*e) + 4*f*e^(6*I*f*x + 6*I*e) + 6*f*e^(4*I*f*x + 4*I*e) + 4*f*e^(2*I*f*x + 2*I*e) + f)

Sympy [F]

\[ \int (a+i a \tan (e+f x)) (A+B \tan (e+f x)) (c-i c \tan (e+f x))^{7/2} \, dx=i a \left (\int \left (- i A c^{3} \sqrt {- i c \tan {\left (e + f x \right )} + c}\right )\, dx + \int \left (- 2 A c^{3} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )}\right )\, dx + \int \left (- 2 A c^{3} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{3}{\left (e + f x \right )}\right )\, dx + \int \left (- 2 B c^{3} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{2}{\left (e + f x \right )}\right )\, dx + \int \left (- 2 B c^{3} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{4}{\left (e + f x \right )}\right )\, dx + \int i A c^{3} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{4}{\left (e + f x \right )}\, dx + \int \left (- i B c^{3} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )}\right )\, dx + \int i B c^{3} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{5}{\left (e + f x \right )}\, dx\right ) \]

[In]

integrate((a+I*a*tan(f*x+e))*(A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))**(7/2),x)

[Out]

I*a*(Integral(-I*A*c**3*sqrt(-I*c*tan(e + f*x) + c), x) + Integral(-2*A*c**3*sqrt(-I*c*tan(e + f*x) + c)*tan(e
+ f*x), x) + Integral(-2*A*c**3*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x)**3, x) + Integral(-2*B*c**3*sqrt(-I*
c*tan(e + f*x) + c)*tan(e + f*x)**2, x) + Integral(-2*B*c**3*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x)**4, x) +
Integral(I*A*c**3*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x)**4, x) + Integral(-I*B*c**3*sqrt(-I*c*tan(e + f*x)
+ c)*tan(e + f*x), x) + Integral(I*B*c**3*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x)**5, x))

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.77 \[ \int (a+i a \tan (e+f x)) (A+B \tan (e+f x)) (c-i c \tan (e+f x))^{7/2} \, dx=\frac {2 i \, {\left (7 i \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {9}{2}} B a + 9 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {7}{2}} {\left (A - i \, B\right )} a c\right )}}{63 \, c f} \]

[In]

integrate((a+I*a*tan(f*x+e))*(A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^(7/2),x, algorithm="maxima")

[Out]

2/63*I*(7*I*(-I*c*tan(f*x + e) + c)^(9/2)*B*a + 9*(-I*c*tan(f*x + e) + c)^(7/2)*(A - I*B)*a*c)/(c*f)

Giac [F]

\[ \int (a+i a \tan (e+f x)) (A+B \tan (e+f x)) (c-i c \tan (e+f x))^{7/2} \, dx=\int { {\left (B \tan \left (f x + e\right ) + A\right )} {\left (i \, a \tan \left (f x + e\right ) + a\right )} {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {7}{2}} \,d x } \]

[In]

integrate((a+I*a*tan(f*x+e))*(A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^(7/2),x, algorithm="giac")

[Out]

integrate((B*tan(f*x + e) + A)*(I*a*tan(f*x + e) + a)*(-I*c*tan(f*x + e) + c)^(7/2), x)

Mupad [B] (verification not implemented)

Time = 12.35 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.63 \[ \int (a+i a \tan (e+f x)) (A+B \tan (e+f x)) (c-i c \tan (e+f x))^{7/2} \, dx=\frac {16\,a\,c^3\,\sqrt {c+\frac {c\,\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1}}\,\left (A\,9{}\mathrm {i}-5\,B+A\,{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}\,9{}\mathrm {i}+9\,B\,{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}\right )}{63\,f\,{\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1\right )}^4} \]

[In]

int((A + B*tan(e + f*x))*(a + a*tan(e + f*x)*1i)*(c - c*tan(e + f*x)*1i)^(7/2),x)

[Out]

(16*a*c^3*(c + (c*(exp(e*2i + f*x*2i)*1i - 1i)*1i)/(exp(e*2i + f*x*2i) + 1))^(1/2)*(A*9i - 5*B + A*exp(e*2i +
f*x*2i)*9i + 9*B*exp(e*2i + f*x*2i)))/(63*f*(exp(e*2i + f*x*2i) + 1)^4)