3.1010 \(\int \frac{(c-i c \tan (e+f x))^{5/2}}{\sqrt{a+i a \tan (e+f x)}} \, dx\)

Optimal. Leaf size=153 \[ \frac{6 i c^{5/2} \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{a+i a \tan (e+f x)}}{\sqrt{a} \sqrt{c-i c \tan (e+f x)}}\right )}{\sqrt{a} f}+\frac{3 i c^2 \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{a f}+\frac{2 i c (c-i c \tan (e+f x))^{3/2}}{f \sqrt{a+i a \tan (e+f x)}} \]

[Out]

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

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Rubi [A]  time = 0.159191, antiderivative size = 153, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.171, Rules used = {3523, 47, 50, 63, 217, 203} \[ \frac{6 i c^{5/2} \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{a+i a \tan (e+f x)}}{\sqrt{a} \sqrt{c-i c \tan (e+f x)}}\right )}{\sqrt{a} f}+\frac{3 i c^2 \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{a f}+\frac{2 i c (c-i c \tan (e+f x))^{3/2}}{f \sqrt{a+i a \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3523

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

Rule 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{(c-i c \tan (e+f x))^{5/2}}{\sqrt{a+i a \tan (e+f x)}} \, dx &=\frac{(a c) \operatorname{Subst}\left (\int \frac{(c-i c x)^{3/2}}{(a+i a x)^{3/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{2 i c (c-i c \tan (e+f x))^{3/2}}{f \sqrt{a+i a \tan (e+f x)}}-\frac{\left (3 c^2\right ) \operatorname{Subst}\left (\int \frac{\sqrt{c-i c x}}{\sqrt{a+i a x}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{3 i c^2 \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{a f}+\frac{2 i c (c-i c \tan (e+f x))^{3/2}}{f \sqrt{a+i a \tan (e+f x)}}-\frac{\left (3 c^3\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+i a x} \sqrt{c-i c x}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{3 i c^2 \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{a f}+\frac{2 i c (c-i c \tan (e+f x))^{3/2}}{f \sqrt{a+i a \tan (e+f x)}}+\frac{\left (6 i c^3\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{2 c-\frac{c x^2}{a}}} \, dx,x,\sqrt{a+i a \tan (e+f x)}\right )}{a f}\\ &=\frac{3 i c^2 \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{a f}+\frac{2 i c (c-i c \tan (e+f x))^{3/2}}{f \sqrt{a+i a \tan (e+f x)}}+\frac{\left (6 i c^3\right ) \operatorname{Subst}\left (\int \frac{1}{1+\frac{c x^2}{a}} \, dx,x,\frac{\sqrt{a+i a \tan (e+f x)}}{\sqrt{c-i c \tan (e+f x)}}\right )}{a f}\\ &=\frac{6 i c^{5/2} \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{a+i a \tan (e+f x)}}{\sqrt{a} \sqrt{c-i c \tan (e+f x)}}\right )}{\sqrt{a} f}+\frac{3 i c^2 \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{a f}+\frac{2 i c (c-i c \tan (e+f x))^{3/2}}{f \sqrt{a+i a \tan (e+f x)}}\\ \end{align*}

Mathematica [A]  time = 3.47265, size = 113, normalized size = 0.74 \[ \frac{c^3 (\cos (f x)+i \sin (f x)) (\sin (f x)+i \cos (f x)) \left (\tan ^2(e+f x)-4 i \tan (e+f x)+6 \sec (e+f x) \tan ^{-1}(\cos (e+f x)+i \sin (e+f x))+5\right )}{f \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c^3*(Cos[f*x] + I*Sin[f*x])*(I*Cos[f*x] + Sin[f*x])*(5 + 6*ArcTan[Cos[e + f*x] + I*Sin[e + f*x]]*Sec[e + f*x]
- (4*I)*Tan[e + f*x] + Tan[e + f*x]^2))/(f*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c - I*c*Tan[e + f*x]])

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Maple [B]  time = 0.08, size = 300, normalized size = 2. \begin{align*}{\frac{i{c}^{2}}{fa \left ( -\tan \left ( fx+e \right ) +i \right ) ^{2}}\sqrt{-c \left ( -1+i\tan \left ( fx+e \right ) \right ) }\sqrt{a \left ( 1+i\tan \left ( fx+e \right ) \right ) } \left ( 3\,i\ln \left ({ \left ( ac\tan \left ( fx+e \right ) +\sqrt{ac \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }\sqrt{ac} \right ){\frac{1}{\sqrt{ac}}}} \right ) \left ( \tan \left ( fx+e \right ) \right ) ^{2}ac-3\,i\ln \left ({ \left ( ac\tan \left ( fx+e \right ) +\sqrt{ac \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }\sqrt{ac} \right ){\frac{1}{\sqrt{ac}}}} \right ) ac-6\,i\sqrt{ac}\sqrt{ac \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }\tan \left ( fx+e \right ) +6\,\ln \left ({\frac{ac\tan \left ( fx+e \right ) +\sqrt{ac \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }\sqrt{ac}}{\sqrt{ac}}} \right ) \tan \left ( fx+e \right ) ac+\sqrt{ac}\sqrt{ac \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) } \left ( \tan \left ( fx+e \right ) \right ) ^{2}-5\,\sqrt{ac \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }\sqrt{ac} \right ){\frac{1}{\sqrt{ac \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }}}{\frac{1}{\sqrt{ac}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

I/f*(-c*(-1+I*tan(f*x+e)))^(1/2)*(a*(1+I*tan(f*x+e)))^(1/2)*c^2/a*(3*I*ln((a*c*tan(f*x+e)+(a*c*(1+tan(f*x+e)^2
))^(1/2)*(a*c)^(1/2))/(a*c)^(1/2))*tan(f*x+e)^2*a*c-3*I*ln((a*c*tan(f*x+e)+(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^
(1/2))/(a*c)^(1/2))*a*c-6*I*(a*c)^(1/2)*(a*c*(1+tan(f*x+e)^2))^(1/2)*tan(f*x+e)+6*ln((a*c*tan(f*x+e)+(a*c*(1+t
an(f*x+e)^2))^(1/2)*(a*c)^(1/2))/(a*c)^(1/2))*tan(f*x+e)*a*c+(a*c)^(1/2)*(a*c*(1+tan(f*x+e)^2))^(1/2)*tan(f*x+
e)^2-5*(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/2))/(a*c*(1+tan(f*x+e)^2))^(1/2)/(a*c)^(1/2)/(-tan(f*x+e)+I)^2

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Maxima [A]  time = 2.01967, size = 250, normalized size = 1.63 \begin{align*} -\frac{{\left (6 i \, c^{2} \arctan \left (\cos \left (f x + e\right ), \sin \left (f x + e\right ) + 1\right ) \cos \left (f x + e\right ) + 6 i \, c^{2} \arctan \left (\cos \left (f x + e\right ), -\sin \left (f x + e\right ) + 1\right ) \cos \left (f x + e\right ) - 8 i \, c^{2} \cos \left (f x + e\right )^{2} + 3 \, c^{2} \cos \left (f x + e\right ) \log \left (\cos \left (f x + e\right )^{2} + \sin \left (f x + e\right )^{2} + 2 \, \sin \left (f x + e\right ) + 1\right ) - 3 \, c^{2} \cos \left (f x + e\right ) \log \left (\cos \left (f x + e\right )^{2} + \sin \left (f x + e\right )^{2} - 2 \, \sin \left (f x + e\right ) + 1\right ) - 8 \, c^{2} \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 2 i \, c^{2}\right )} \sqrt{c}}{2 \, \sqrt{a} f \cos \left (f x + e\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-I*c*tan(f*x+e))^(5/2)/(a+I*a*tan(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

-1/2*(6*I*c^2*arctan2(cos(f*x + e), sin(f*x + e) + 1)*cos(f*x + e) + 6*I*c^2*arctan2(cos(f*x + e), -sin(f*x +
e) + 1)*cos(f*x + e) - 8*I*c^2*cos(f*x + e)^2 + 3*c^2*cos(f*x + e)*log(cos(f*x + e)^2 + sin(f*x + e)^2 + 2*sin
(f*x + e) + 1) - 3*c^2*cos(f*x + e)*log(cos(f*x + e)^2 + sin(f*x + e)^2 - 2*sin(f*x + e) + 1) - 8*c^2*cos(f*x
+ e)*sin(f*x + e) - 2*I*c^2)*sqrt(c)/(sqrt(a)*f*cos(f*x + e))

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Fricas [B]  time = 1.66494, size = 1010, normalized size = 6.6 \begin{align*} -\frac{{\left (3 \, \sqrt{\frac{c^{5}}{a f^{2}}} a f e^{\left (2 i \, f x + 2 i \, e\right )} \log \left (-\frac{2 \,{\left (c^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + c^{2}\right )} \sqrt{\frac{a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt{\frac{c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} e^{\left (i \, f x + i \, e\right )} -{\left (i \, a f e^{\left (2 i \, f x + 2 i \, e\right )} - i \, a f\right )} \sqrt{\frac{c^{5}}{a f^{2}}}}{2 \,{\left (c^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + c^{2}\right )}}\right ) - 3 \, \sqrt{\frac{c^{5}}{a f^{2}}} a f e^{\left (2 i \, f x + 2 i \, e\right )} \log \left (-\frac{2 \,{\left (c^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + c^{2}\right )} \sqrt{\frac{a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt{\frac{c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} e^{\left (i \, f x + i \, e\right )} -{\left (-i \, a f e^{\left (2 i \, f x + 2 i \, e\right )} + i \, a f\right )} \sqrt{\frac{c^{5}}{a f^{2}}}}{2 \,{\left (c^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + c^{2}\right )}}\right ) -{\left (-10 i \, c^{2} e^{\left (3 i \, f x + 3 i \, e\right )} + 12 i \, c^{2} e^{\left (2 i \, f x + 2 i \, e\right )} - 10 i \, c^{2} e^{\left (i \, f x + i \, e\right )} + 8 i \, c^{2}\right )} \sqrt{\frac{a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt{\frac{c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} e^{\left (i \, f x + i \, e\right )}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{2 \, a f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-I*c*tan(f*x+e))^(5/2)/(a+I*a*tan(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

-1/2*(3*sqrt(c^5/(a*f^2))*a*f*e^(2*I*f*x + 2*I*e)*log(-1/2*(2*(c^2*e^(2*I*f*x + 2*I*e) + c^2)*sqrt(a/(e^(2*I*f
*x + 2*I*e) + 1))*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*e^(I*f*x + I*e) - (I*a*f*e^(2*I*f*x + 2*I*e) - I*a*f)*sqrt
(c^5/(a*f^2)))/(c^2*e^(2*I*f*x + 2*I*e) + c^2)) - 3*sqrt(c^5/(a*f^2))*a*f*e^(2*I*f*x + 2*I*e)*log(-1/2*(2*(c^2
*e^(2*I*f*x + 2*I*e) + c^2)*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*e^(I*f*x + I*e
) - (-I*a*f*e^(2*I*f*x + 2*I*e) + I*a*f)*sqrt(c^5/(a*f^2)))/(c^2*e^(2*I*f*x + 2*I*e) + c^2)) - (-10*I*c^2*e^(3
*I*f*x + 3*I*e) + 12*I*c^2*e^(2*I*f*x + 2*I*e) - 10*I*c^2*e^(I*f*x + I*e) + 8*I*c^2)*sqrt(a/(e^(2*I*f*x + 2*I*
e) + 1))*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*e^(I*f*x + I*e))*e^(-2*I*f*x - 2*I*e)/(a*f)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac{5}{2}}}{\sqrt{i \, a \tan \left (f x + e\right ) + a}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-I*c*tan(f*x+e))^(5/2)/(a+I*a*tan(f*x+e))^(1/2),x, algorithm="giac")

[Out]

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