3.987 \(\int \frac {(a+i a \tan (e+f x))^2}{(c-i c \tan (e+f x))^{5/2}} \, dx\)

Optimal. Leaf size=62 \[ \frac {2 i a^2}{3 c f (c-i c \tan (e+f x))^{3/2}}-\frac {4 i a^2}{5 f (c-i c \tan (e+f x))^{5/2}} \]

[Out]

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

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Rubi [A]  time = 0.15, antiderivative size = 62, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {3522, 3487, 43} \[ \frac {2 i a^2}{3 c f (c-i c \tan (e+f x))^{3/2}}-\frac {4 i a^2}{5 f (c-i c \tan (e+f x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + I*a*Tan[e + f*x])^2/(c - I*c*Tan[e + f*x])^(5/2),x]

[Out]

(((-4*I)/5)*a^2)/(f*(c - I*c*Tan[e + f*x])^(5/2)) + (((2*I)/3)*a^2)/(c*f*(c - I*c*Tan[e + f*x])^(3/2))

Rule 43

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 3487

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 3522

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*} \int \frac {(a+i a \tan (e+f x))^2}{(c-i c \tan (e+f x))^{5/2}} \, dx &=\left (a^2 c^2\right ) \int \frac {\sec ^4(e+f x)}{(c-i c \tan (e+f x))^{9/2}} \, dx\\ &=\frac {\left (i a^2\right ) \operatorname {Subst}\left (\int \frac {c-x}{(c+x)^{7/2}} \, dx,x,-i c \tan (e+f x)\right )}{c f}\\ &=\frac {\left (i a^2\right ) \operatorname {Subst}\left (\int \left (\frac {2 c}{(c+x)^{7/2}}-\frac {1}{(c+x)^{5/2}}\right ) \, dx,x,-i c \tan (e+f x)\right )}{c f}\\ &=-\frac {4 i a^2}{5 f (c-i c \tan (e+f x))^{5/2}}+\frac {2 i a^2}{3 c f (c-i c \tan (e+f x))^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 6.58, size = 95, normalized size = 1.53 \[ \frac {2 a^2 \cos ^2(e+f x) \sqrt {c-i c \tan (e+f x)} (5 \sin (e+f x)-i \cos (e+f x)) (\cos (3 e+5 f x)+i \sin (3 e+5 f x))}{15 c^3 f (\cos (f x)+i \sin (f x))^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + I*a*Tan[e + f*x])^2/(c - I*c*Tan[e + f*x])^(5/2),x]

[Out]

(2*a^2*Cos[e + f*x]^2*((-I)*Cos[e + f*x] + 5*Sin[e + f*x])*(Cos[3*e + 5*f*x] + I*Sin[3*e + 5*f*x])*Sqrt[c - I*
c*Tan[e + f*x]])/(15*c^3*f*(Cos[f*x] + I*Sin[f*x])^2)

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fricas [A]  time = 0.46, size = 76, normalized size = 1.23 \[ \frac {\sqrt {2} {\left (-3 i \, a^{2} e^{\left (6 i \, f x + 6 i \, e\right )} - 4 i \, a^{2} e^{\left (4 i \, f x + 4 i \, e\right )} + i \, a^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + 2 i \, a^{2}\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{30 \, c^{3} f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^2/(c-I*c*tan(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

1/30*sqrt(2)*(-3*I*a^2*e^(6*I*f*x + 6*I*e) - 4*I*a^2*e^(4*I*f*x + 4*I*e) + I*a^2*e^(2*I*f*x + 2*I*e) + 2*I*a^2
)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))/(c^3*f)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{2}}{{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^2/(c-I*c*tan(f*x+e))^(5/2),x, algorithm="giac")

[Out]

integrate((I*a*tan(f*x + e) + a)^2/(-I*c*tan(f*x + e) + c)^(5/2), x)

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maple [A]  time = 0.20, size = 47, normalized size = 0.76 \[ -\frac {2 i a^{2} \left (-\frac {1}{3 \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}+\frac {2 c}{5 \left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}}\right )}{f c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+I*a*tan(f*x+e))^2/(c-I*c*tan(f*x+e))^(5/2),x)

[Out]

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

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maxima [A]  time = 0.70, size = 44, normalized size = 0.71 \[ \frac {2 i \, {\left (5 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )} a^{2} - 6 \, a^{2} c\right )}}{15 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}} c f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^2/(c-I*c*tan(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

2/15*I*(5*(-I*c*tan(f*x + e) + c)*a^2 - 6*a^2*c)/((-I*c*tan(f*x + e) + c)^(5/2)*c*f)

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mupad [B]  time = 5.55, size = 123, normalized size = 1.98 \[ \frac {a^2\,\sqrt {\frac {c\,\left (\cos \left (2\,e+2\,f\,x\right )+1-\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}\,\left (\cos \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}-\cos \left (4\,e+4\,f\,x\right )\,4{}\mathrm {i}-\cos \left (6\,e+6\,f\,x\right )\,3{}\mathrm {i}-\sin \left (2\,e+2\,f\,x\right )+4\,\sin \left (4\,e+4\,f\,x\right )+3\,\sin \left (6\,e+6\,f\,x\right )+2{}\mathrm {i}\right )}{30\,c^3\,f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*tan(e + f*x)*1i)^2/(c - c*tan(e + f*x)*1i)^(5/2),x)

[Out]

(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)*(cos(2*e + 2*f*x)*1i - co
s(4*e + 4*f*x)*4i - cos(6*e + 6*f*x)*3i - sin(2*e + 2*f*x) + 4*sin(4*e + 4*f*x) + 3*sin(6*e + 6*f*x) + 2i))/(3
0*c^3*f)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - a^{2} \left (\int \frac {\tan ^{2}{\left (e + f x \right )}}{- c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{2}{\left (e + f x \right )} - 2 i c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )} + c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c}}\, dx + \int \left (- \frac {2 i \tan {\left (e + f x \right )}}{- c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{2}{\left (e + f x \right )} - 2 i c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )} + c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c}}\right )\, dx + \int \left (- \frac {1}{- c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{2}{\left (e + f x \right )} - 2 i c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )} + c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c}}\right )\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))**2/(c-I*c*tan(f*x+e))**(5/2),x)

[Out]

-a**2*(Integral(tan(e + f*x)**2/(-c**2*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x)**2 - 2*I*c**2*sqrt(-I*c*tan(e
+ f*x) + c)*tan(e + f*x) + c**2*sqrt(-I*c*tan(e + f*x) + c)), x) + Integral(-2*I*tan(e + f*x)/(-c**2*sqrt(-I*c
*tan(e + f*x) + c)*tan(e + f*x)**2 - 2*I*c**2*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x) + c**2*sqrt(-I*c*tan(e
+ f*x) + c)), x) + Integral(-1/(-c**2*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x)**2 - 2*I*c**2*sqrt(-I*c*tan(e +
f*x) + c)*tan(e + f*x) + c**2*sqrt(-I*c*tan(e + f*x) + c)), x))

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