3.10.56 \(\int (a+i a \tan (e+f x))^3 \sqrt {c-i c \tan (e+f x)} \, dx\) [956]

Optimal. Leaf size=92 \[ \frac {8 i a^3 \sqrt {c-i c \tan (e+f x)}}{f}-\frac {8 i a^3 (c-i c \tan (e+f x))^{3/2}}{3 c f}+\frac {2 i a^3 (c-i c \tan (e+f x))^{5/2}}{5 c^2 f} \]

[Out]

8*I*a^3*(c-I*c*tan(f*x+e))^(1/2)/f-8/3*I*a^3*(c-I*c*tan(f*x+e))^(3/2)/c/f+2/5*I*a^3*(c-I*c*tan(f*x+e))^(5/2)/c
^2/f

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Rubi [A]
time = 0.12, antiderivative size = 92, 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 {2 i a^3 (c-i c \tan (e+f x))^{5/2}}{5 c^2 f}-\frac {8 i a^3 (c-i c \tan (e+f x))^{3/2}}{3 c f}+\frac {8 i a^3 \sqrt {c-i c \tan (e+f x)}}{f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((8*I)*a^3*Sqrt[c - I*c*Tan[e + f*x]])/f - (((8*I)/3)*a^3*(c - I*c*Tan[e + f*x])^(3/2))/(c*f) + (((2*I)/5)*a^3
*(c - I*c*Tan[e + f*x])^(5/2))/(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*} \int (a+i a \tan (e+f x))^3 \sqrt {c-i c \tan (e+f x)} \, dx &=\left (a^3 c^3\right ) \int \frac {\sec ^6(e+f x)}{(c-i c \tan (e+f x))^{5/2}} \, dx\\ &=\frac {\left (i a^3\right ) \text {Subst}\left (\int \frac {(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 (\frac {4 c^2}{\sqrt {c+x}}-4 c \sqrt {c+x}+(c+x)^{3/2}\right ) \, dx,x,-i c \tan (e+f x)\right )}{c^2 f}\\ &=\frac {8 i a^3 \sqrt {c-i c \tan (e+f x)}}{f}-\frac {8 i a^3 (c-i c \tan (e+f x))^{3/2}}{3 c f}+\frac {2 i a^3 (c-i c \tan (e+f x))^{5/2}}{5 c^2 f}\\ \end {align*}

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Mathematica [A]
time = 1.32, size = 61, normalized size = 0.66 \begin {gather*} \frac {2 i a^3 \sec ^2(e+f x) (20+23 \cos (2 (e+f x))+7 i \sin (2 (e+f x))) \sqrt {c-i c \tan (e+f x)}}{15 f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(((2*I)/15)*a^3*Sec[e + f*x]^2*(20 + 23*Cos[2*(e + f*x)] + (7*I)*Sin[2*(e + f*x)])*Sqrt[c - I*c*Tan[e + f*x]])
/f

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Maple [A]
time = 0.36, size = 66, normalized size = 0.72

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

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))^3,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.28, size = 70, normalized size = 0.76 \begin {gather*} \frac {2 i \, {\left (3 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}} a^{3} - 20 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} a^{3} c + 60 \, \sqrt {-i \, c \tan \left (f x + e\right ) + c} a^{3} c^{2}\right )}}{15 \, c^{2} 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))^3,x, algorithm="maxima")

[Out]

2/15*I*(3*(-I*c*tan(f*x + e) + c)^(5/2)*a^3 - 20*(-I*c*tan(f*x + e) + c)^(3/2)*a^3*c + 60*sqrt(-I*c*tan(f*x +
e) + c)*a^3*c^2)/(c^2*f)

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Fricas [A]
time = 1.22, size = 88, normalized size = 0.96 \begin {gather*} -\frac {8 \, \sqrt {2} {\left (-15 i \, a^{3} e^{\left (4 i \, f x + 4 i \, e\right )} - 20 i \, a^{3} e^{\left (2 i \, f x + 2 i \, e\right )} - 8 i \, a^{3}\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{15 \, {\left (f e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, 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))^3,x, algorithm="fricas")

[Out]

-8/15*sqrt(2)*(-15*I*a^3*e^(4*I*f*x + 4*I*e) - 20*I*a^3*e^(2*I*f*x + 2*I*e) - 8*I*a^3)*sqrt(c/(e^(2*I*f*x + 2*
I*e) + 1))/(f*e^(4*I*f*x + 4*I*e) + 2*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*} - i a^{3} \left (\int i \sqrt {- i c \tan {\left (e + f x \right )} + c}\, dx + \int \left (- 3 \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )}\right )\, dx + \int \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{3}{\left (e + f x \right )}\, dx + \int \left (- 3 i \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{2}{\left (e + f x \right )}\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))**3,x)

[Out]

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

[Out]

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

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Mupad [B]
time = 6.21, size = 155, normalized size = 1.68 \begin {gather*} \frac {4\,a^3\,\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 )\,321{}\mathrm {i}+\cos \left (4\,e+4\,f\,x\right )\,132{}\mathrm {i}+\cos \left (6\,e+6\,f\,x\right )\,23{}\mathrm {i}-35\,\sin \left (2\,e+2\,f\,x\right )-28\,\sin \left (4\,e+4\,f\,x\right )-7\,\sin \left (6\,e+6\,f\,x\right )+212{}\mathrm {i}\right )}{15\,f\,\left (15\,\cos \left (2\,e+2\,f\,x\right )+6\,\cos \left (4\,e+4\,f\,x\right )+\cos \left (6\,e+6\,f\,x\right )+10\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(4*a^3*((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)*321i
+ cos(4*e + 4*f*x)*132i + cos(6*e + 6*f*x)*23i - 35*sin(2*e + 2*f*x) - 28*sin(4*e + 4*f*x) - 7*sin(6*e + 6*f*x
) + 212i))/(15*f*(15*cos(2*e + 2*f*x) + 6*cos(4*e + 4*f*x) + cos(6*e + 6*f*x) + 10))

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