\(\int (a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))^{3/2} \, dx\) [962]

Optimal result

Integrand size = 33, antiderivative size = 94 \[ \int (a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))^{3/2} \, dx=\frac {8 i a^3 (c-i c \tan (e+f x))^{3/2}}{3 f}-\frac {8 i a^3 (c-i c \tan (e+f x))^{5/2}}{5 c f}+\frac {2 i a^3 (c-i c \tan (e+f x))^{7/2}}{7 c^2 f} \]

[Out]

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

Rubi [A] (verified)

Time = 0.21 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {3603, 3568, 45} \[ \int (a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))^{3/2} \, dx=\frac {2 i a^3 (c-i c \tan (e+f x))^{7/2}}{7 c^2 f}-\frac {8 i a^3 (c-i c \tan (e+f x))^{5/2}}{5 c f}+\frac {8 i a^3 (c-i c \tan (e+f x))^{3/2}}{3 f} \]

[In]

Int[(a + I*a*Tan[e + f*x])^3*(c - I*c*Tan[e + f*x])^(3/2),x]

[Out]

(((8*I)/3)*a^3*(c - I*c*Tan[e + f*x])^(3/2))/f - (((8*I)/5)*a^3*(c - I*c*Tan[e + f*x])^(5/2))/(c*f) + (((2*I)/
7)*a^3*(c - I*c*Tan[e + f*x])^(7/2))/(c^2*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 3568

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[1/(a^(m - 2)*b
*f), Subst[Int[(a - x)^(m/2 - 1)*(a + x)^(n + m/2 - 1), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, e, f, n}, x
] && EqQ[a^2 + b^2, 0] && IntegerQ[m/2]

Rule 3603

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Di
st[a^m*c^m, Int[Sec[e + f*x]^(2*m)*(c + d*Tan[e + f*x])^(n - m), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] &&
EqQ[b*c + a*d, 0] && EqQ[a^2 + b^2, 0] && IntegerQ[m] &&  !(IGtQ[n, 0] && (LtQ[m, 0] || GtQ[m, n]))

Rubi steps \begin{align*} \text {integral}& = \left (a^3 c^3\right ) \int \frac {\sec ^6(e+f x)}{(c-i c \tan (e+f x))^{3/2}} \, dx \\ & = \frac {\left (i a^3\right ) \text {Subst}\left (\int (c-x)^2 \sqrt {c+x} \, dx,x,-i c \tan (e+f x)\right )}{c^2 f} \\ & = \frac {\left (i a^3\right ) \text {Subst}\left (\int \left (4 c^2 \sqrt {c+x}-4 c (c+x)^{3/2}+(c+x)^{5/2}\right ) \, dx,x,-i c \tan (e+f x)\right )}{c^2 f} \\ & = \frac {8 i a^3 (c-i c \tan (e+f x))^{3/2}}{3 f}-\frac {8 i a^3 (c-i c \tan (e+f x))^{5/2}}{5 c f}+\frac {2 i a^3 (c-i c \tan (e+f x))^{7/2}}{7 c^2 f} \\ \end{align*}

Mathematica [A] (verified)

Time = 2.10 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.64 \[ \int (a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))^{3/2} \, dx=-\frac {2 a^3 c (i+\tan (e+f x)) \sqrt {c-i c \tan (e+f x)} \left (-71-54 i \tan (e+f x)+15 \tan ^2(e+f x)\right )}{105 f} \]

[In]

Integrate[(a + I*a*Tan[e + f*x])^3*(c - I*c*Tan[e + f*x])^(3/2),x]

[Out]

(-2*a^3*c*(I + Tan[e + f*x])*Sqrt[c - I*c*Tan[e + f*x]]*(-71 - (54*I)*Tan[e + f*x] + 15*Tan[e + f*x]^2))/(105*
f)

Maple [A] (verified)

Time = 0.66 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.70

method	result	size
derivativedivides	\(\frac {2 i a^{3} \left (\frac {\left (c -i c \tan \left (f x +e \right )\right )^{\frac {7}{2}}}{7}-\frac {4 c \left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{5}+\frac {4 c^{2} \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}\right )}{f \,c^{2}}\)	\(66\)
default	\(\frac {2 i a^{3} \left (\frac {\left (c -i c \tan \left (f x +e \right )\right )^{\frac {7}{2}}}{7}-\frac {4 c \left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{5}+\frac {4 c^{2} \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}\right )}{f \,c^{2}}\)	\(66\)
parts	\(\frac {2 i a^{3} c \left (-\sqrt {c -i c \tan \left (f x +e \right )}+\sqrt {c}\, \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )\right )}{f}-\frac {2 i a^{3} \left (-\frac {\left (c -i c \tan \left (f x +e \right )\right )^{\frac {7}{2}}}{7}+\frac {c \left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{5}-\frac {c^{2} \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}-c^{3} \sqrt {c -i c \tan \left (f x +e \right )}+c^{\frac {7}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )\right )}{f \,c^{2}}+\frac {3 i a^{3} \left (\frac {2 \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}+2 c \sqrt {c -i c \tan \left (f x +e \right )}-2 c^{\frac {3}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )\right )}{f}-\frac {6 i a^{3} \left (\frac {\left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{5}+c^{2} \sqrt {c -i c \tan \left (f x +e \right )}-c^{\frac {5}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )\right )}{f c}\)	\(325\)

[In]

int((a+I*a*tan(f*x+e))^3*(c-I*c*tan(f*x+e))^(3/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.04 \[ \int (a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))^{3/2} \, dx=-\frac {16 \, \sqrt {2} {\left (-35 i \, a^{3} c e^{\left (4 i \, f x + 4 i \, e\right )} - 28 i \, a^{3} c e^{\left (2 i \, f x + 2 i \, e\right )} - 8 i \, a^{3} c\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{105 \, {\left (f e^{\left (6 i \, f x + 6 i \, e\right )} + 3 \, f e^{\left (4 i \, f x + 4 i \, e\right )} + 3 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )}} \]

[In]

integrate((a+I*a*tan(f*x+e))^3*(c-I*c*tan(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

-16/105*sqrt(2)*(-35*I*a^3*c*e^(4*I*f*x + 4*I*e) - 28*I*a^3*c*e^(2*I*f*x + 2*I*e) - 8*I*a^3*c)*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]

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

[In]

integrate((a+I*a*tan(f*x+e))**3*(c-I*c*tan(f*x+e))**(3/2),x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.71 \[ \int (a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))^{3/2} \, dx=\frac {2 i \, {\left (15 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {7}{2}} a^{3} - 84 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}} a^{3} c + 140 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} a^{3} c^{2}\right )}}{105 \, c^{2} f} \]

[In]

integrate((a+I*a*tan(f*x+e))^3*(c-I*c*tan(f*x+e))^(3/2),x, algorithm="maxima")

[Out]

2/105*I*(15*(-I*c*tan(f*x + e) + c)^(7/2)*a^3 - 84*(-I*c*tan(f*x + e) + c)^(5/2)*a^3*c + 140*(-I*c*tan(f*x + e
) + c)^(3/2)*a^3*c^2)/(c^2*f)

Giac [F]

\[ \int (a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))^{3/2} \, dx=\int { {\left (i \, a \tan \left (f x + e\right ) + a\right )}^{3} {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} \,d x } \]

[In]

integrate((a+I*a*tan(f*x+e))^3*(c-I*c*tan(f*x+e))^(3/2),x, algorithm="giac")

[Out]

integrate((I*a*tan(f*x + e) + a)^3*(-I*c*tan(f*x + e) + c)^(3/2), x)

Mupad [B] (verification not implemented)

Time = 10.56 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.01 \[ \int (a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))^{3/2} \, dx=\frac {16\,a^3\,c\,\sqrt {c+\frac {c\,\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1}}\,\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}\,28{}\mathrm {i}+{\mathrm {e}}^{e\,4{}\mathrm {i}+f\,x\,4{}\mathrm {i}}\,35{}\mathrm {i}+8{}\mathrm {i}\right )}{105\,f\,{\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1\right )}^3} \]

[In]

int((a + a*tan(e + f*x)*1i)^3*(c - c*tan(e + f*x)*1i)^(3/2),x)

[Out]

(16*a^3*c*(c + (c*(exp(e*2i + f*x*2i)*1i - 1i)*1i)/(exp(e*2i + f*x*2i) + 1))^(1/2)*(exp(e*2i + f*x*2i)*28i + e
xp(e*4i + f*x*4i)*35i + 8i))/(105*f*(exp(e*2i + f*x*2i) + 1)^3)