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

Optimal. Leaf size=58 \[ -\frac {2 i a^2 \sqrt {c-i c \tan (e+f x)}}{c f}-\frac {4 i a^2}{f \sqrt {c-i c \tan (e+f x)}} \]

[Out]

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

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Rubi [A]  time = 0.15, antiderivative size = 58, 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} \[ -\frac {2 i a^2 \sqrt {c-i c \tan (e+f x)}}{c f}-\frac {4 i a^2}{f \sqrt {c-i c \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]  time = 2.36, size = 91, normalized size = 1.57 \[ \frac {2 a^2 \sqrt {c-i c \tan (e+f x)} (-2 \sin (2 e)-2 i \cos (2 e)+\sin (2 f x)-i \cos (2 f x)) (\cos (e+f x)+i \sin (e+f x))^2}{c f (\cos (f x)+i \sin (f x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.49, size = 47, normalized size = 0.81 \[ \frac {\sqrt {2} {\left (-2 i \, a^{2} e^{\left (2 i \, f x + 2 i \, e\right )} - 4 i \, a^{2}\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{c f} \]

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))^(1/2),x, algorithm="fricas")

[Out]

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

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

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))^(1/2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.23, size = 45, normalized size = 0.78 \[ -\frac {2 i a^{2} \left (\sqrt {c -i c \tan \left (f x +e \right )}+\frac {2 c}{\sqrt {c -i c \tan \left (f x +e \right )}}\right )}{f c} \]

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))^(1/2),x)

[Out]

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

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maxima [A]  time = 0.52, size = 45, normalized size = 0.78 \[ -\frac {2 i \, {\left (\sqrt {-i \, c \tan \left (f x + e\right ) + c} a^{2} + \frac {2 \, a^{2} c}{\sqrt {-i \, c \tan \left (f x + e\right ) + c}}\right )}}{c f} \]

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))^(1/2),x, algorithm="maxima")

[Out]

-2*I*(sqrt(-I*c*tan(f*x + e) + c)*a^2 + 2*a^2*c/sqrt(-I*c*tan(f*x + e) + c))/(c*f)

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mupad [B]  time = 4.86, size = 77, normalized size = 1.33 \[ -\frac {2\,a^2\,\sqrt {\frac {c\,\left (\cos \left (2\,e+2\,f\,x\right )+1-\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}\,\left (\cos \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}-\sin \left (2\,e+2\,f\,x\right )+2{}\mathrm {i}\right )}{c\,f} \]

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)^(1/2),x)

[Out]

-(2*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 -
sin(2*e + 2*f*x) + 2i))/(c*f)

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

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))**(1/2),x)

[Out]

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

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