3.10.57 \(\int (a+i a \tan (e+f x))^2 \sqrt {c-i c \tan (e+f x)} \, dx\) [957]

Optimal. Leaf size=60 \[ \frac {4 i a^2 \sqrt {c-i c \tan (e+f x)}}{f}-\frac {2 i a^2 (c-i c \tan (e+f x))^{3/2}}{3 c f} \]

[Out]

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

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Rubi [A]
time = 0.10, antiderivative size = 60, 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 = {3603, 3568, 45} \begin {gather*} \frac {4 i a^2 \sqrt {c-i c \tan (e+f x)}}{f}-\frac {2 i a^2 (c-i c \tan (e+f x))^{3/2}}{3 c f} \end {gather*}

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*Sqrt[c - I*c*Tan[e + f*x]])/f - (((2*I)/3)*a^2*(c - I*c*Tan[e + f*x])^(3/2))/(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 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*} \int (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))^{3/2}} \, dx\\ &=\frac {\left (i a^2\right ) \text {Subst}\left (\int \frac {c-x}{\sqrt {c+x}} \, dx,x,-i c \tan (e+f x)\right )}{c f}\\ &=\frac {\left (i a^2\right ) \text {Subst}\left (\int \left (\frac {2 c}{\sqrt {c+x}}-\sqrt {c+x}\right ) \, dx,x,-i c \tan (e+f x)\right )}{c f}\\ &=\frac {4 i a^2 \sqrt {c-i c \tan (e+f x)}}{f}-\frac {2 i a^2 (c-i c \tan (e+f x))^{3/2}}{3 c f}\\ \end {align*}

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Mathematica [A]
time = 0.92, size = 37, normalized size = 0.62 \begin {gather*} -\frac {2 a^2 (-5 i+\tan (e+f x)) \sqrt {c-i c \tan (e+f x)}}{3 f} \end {gather*}

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*(-5*I + Tan[e + f*x])*Sqrt[c - I*c*Tan[e + f*x]])/(3*f)

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Maple [A]
time = 0.23, size = 47, normalized size = 0.78

method	result	size
derivativedivides	\(-\frac {2 i a^{2} \left (\frac {\left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}-2 c \sqrt {c -i c \tan \left (f x +e \right )}\right )}{f c}\)	\(47\)
default	\(-\frac {2 i a^{2} \left (\frac {\left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}-2 c \sqrt {c -i c \tan \left (f x +e \right )}\right )}{f c}\)	\(47\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c-I*c*tan(f*x+e))^(1/2)*(a+I*a*tan(f*x+e))^2,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.28, size = 47, normalized size = 0.78 \begin {gather*} -\frac {2 i \, {\left ({\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} a^{2} - 6 \, \sqrt {-i \, c \tan \left (f x + e\right ) + c} a^{2} c\right )}}{3 \, c f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-I*c*tan(f*x+e))^(1/2)*(a+I*a*tan(f*x+e))^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 1.36, size = 60, normalized size = 1.00 \begin {gather*} -\frac {4 \, \sqrt {2} {\left (-3 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}}}{3 \, {\left (f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-I*c*tan(f*x+e))^(1/2)*(a+I*a*tan(f*x+e))^2,x, algorithm="fricas")

[Out]

-4/3*sqrt(2)*(-3*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))/(f*e^(2*I*f*x + 2*I*e)
+ f)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - a^{2} \left (\int \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{2}{\left (e + f x \right )}\, dx + \int \left (- 2 i \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )}\right )\, dx + \int \left (- \sqrt {- i c \tan {\left (e + f x \right )} + c}\right )\, dx\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-I*c*tan(f*x+e))^(1/2)*(a+I*a*tan(f*x+e))^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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Mupad [B]
time = 4.79, size = 87, normalized size = 1.45 \begin {gather*} \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 )\,5{}\mathrm {i}-\sin \left (2\,e+2\,f\,x\right )+5{}\mathrm {i}\right )}{3\,f\,\left (\cos \left (2\,e+2\,f\,x\right )+1\right )} \end {gather*}

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)*5i -
sin(2*e + 2*f*x) + 5i))/(3*f*(cos(2*e + 2*f*x) + 1))

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