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

Optimal. Leaf size=125 \[ \frac{3 i c^2 \sqrt{c-i c \tan (e+f x)}}{a f}+\frac{i c^2 (c-i c \tan (e+f x))^{3/2}}{a f (c+i c \tan (e+f x))}-\frac{3 i \sqrt{2} c^{5/2} \tanh ^{-1}\left (\frac{\sqrt{c-i c \tan (e+f x)}}{\sqrt{2} \sqrt{c}}\right )}{a f} \]

[Out]

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

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Rubi [A]  time = 0.18495, antiderivative size = 125, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {3522, 3487, 47, 50, 63, 206} \[ \frac{3 i c^2 \sqrt{c-i c \tan (e+f x)}}{a f}+\frac{i c^2 (c-i c \tan (e+f x))^{3/2}}{a f (c+i c \tan (e+f x))}-\frac{3 i \sqrt{2} c^{5/2} \tanh ^{-1}\left (\frac{\sqrt{c-i c \tan (e+f x)}}{\sqrt{2} \sqrt{c}}\right )}{a f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]))

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 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 206

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

Rubi steps

\begin{align*} \int \frac{(c-i c \tan (e+f x))^{5/2}}{a+i a \tan (e+f x)} \, dx &=\frac{\int \cos ^2(e+f x) (c-i c \tan (e+f x))^{7/2} \, dx}{a c}\\ &=\frac{\left (i c^2\right ) \operatorname{Subst}\left (\int \frac{(c+x)^{3/2}}{(c-x)^2} \, dx,x,-i c \tan (e+f x)\right )}{a f}\\ &=\frac{i c^2 (c-i c \tan (e+f x))^{3/2}}{a f (c+i c \tan (e+f x))}-\frac{\left (3 i c^2\right ) \operatorname{Subst}\left (\int \frac{\sqrt{c+x}}{c-x} \, dx,x,-i c \tan (e+f x)\right )}{2 a f}\\ &=\frac{3 i c^2 \sqrt{c-i c \tan (e+f x)}}{a f}+\frac{i c^2 (c-i c \tan (e+f x))^{3/2}}{a f (c+i c \tan (e+f x))}-\frac{\left (3 i c^3\right ) \operatorname{Subst}\left (\int \frac{1}{(c-x) \sqrt{c+x}} \, dx,x,-i c \tan (e+f x)\right )}{a f}\\ &=\frac{3 i c^2 \sqrt{c-i c \tan (e+f x)}}{a f}+\frac{i c^2 (c-i c \tan (e+f x))^{3/2}}{a f (c+i c \tan (e+f x))}-\frac{\left (6 i c^3\right ) \operatorname{Subst}\left (\int \frac{1}{2 c-x^2} \, dx,x,\sqrt{c-i c \tan (e+f x)}\right )}{a f}\\ &=-\frac{3 i \sqrt{2} c^{5/2} \tanh ^{-1}\left (\frac{\sqrt{c-i c \tan (e+f x)}}{\sqrt{2} \sqrt{c}}\right )}{a f}+\frac{3 i c^2 \sqrt{c-i c \tan (e+f x)}}{a f}+\frac{i c^2 (c-i c \tan (e+f x))^{3/2}}{a f (c+i c \tan (e+f x))}\\ \end{align*}

Mathematica [F]  time = 180.007, size = 0, normalized size = 0. \[ \text{\$Aborted} \]

Verification is Not applicable to the result.

[In]

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

[Out]

$Aborted

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Maple [A]  time = 0.033, size = 95, normalized size = 0.8 \begin{align*}{\frac{2\,i{c}^{2}}{fa} \left ( \sqrt{c-ic\tan \left ( fx+e \right ) }+4\,c \left ( -1/4\,{\frac{\sqrt{c-ic\tan \left ( fx+e \right ) }}{-c-ic\tan \left ( fx+e \right ) }}-3/8\,{\frac{\sqrt{2}}{\sqrt{c}}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{c-ic\tan \left ( fx+e \right ) }\sqrt{2}}{\sqrt{c}}} \right ) } \right ) \right ) } \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)),x)

[Out]

2*I/f/a*c^2*((c-I*c*tan(f*x+e))^(1/2)+4*c*(-1/4*(c-I*c*tan(f*x+e))^(1/2)/(-c-I*c*tan(f*x+e))-3/8*2^(1/2)/c^(1/
2)*arctanh(1/2*(c-I*c*tan(f*x+e))^(1/2)*2^(1/2)/c^(1/2))))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \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)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.43491, size = 668, normalized size = 5.34 \begin{align*} \frac{{\left (6 \, \sqrt{2} a \sqrt{-\frac{c^{5}}{a^{2} f^{2}}} f e^{\left (2 i \, f x + 2 i \, e\right )} \log \left (\frac{{\left (-12 i \, c^{3} + 12 \,{\left (a f e^{\left (2 i \, f x + 2 i \, e\right )} + a f\right )} \sqrt{-\frac{c^{5}}{a^{2} f^{2}}} \sqrt{\frac{c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-i \, f x - i \, e\right )}}{a f}\right ) - 6 \, \sqrt{2} a \sqrt{-\frac{c^{5}}{a^{2} f^{2}}} f e^{\left (2 i \, f x + 2 i \, e\right )} \log \left (\frac{{\left (-12 i \, c^{3} - 12 \,{\left (a f e^{\left (2 i \, f x + 2 i \, e\right )} + a f\right )} \sqrt{-\frac{c^{5}}{a^{2} f^{2}}} \sqrt{\frac{c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-i \, f x - i \, e\right )}}{a f}\right ) + \sqrt{2}{\left (12 i \, c^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + 4 i \, c^{2}\right )} \sqrt{\frac{c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{4 \, 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)),x, algorithm="fricas")

[Out]

1/4*(6*sqrt(2)*a*sqrt(-c^5/(a^2*f^2))*f*e^(2*I*f*x + 2*I*e)*log((-12*I*c^3 + 12*(a*f*e^(2*I*f*x + 2*I*e) + a*f
)*sqrt(-c^5/(a^2*f^2))*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1)))*e^(-I*f*x - I*e)/(a*f)) - 6*sqrt(2)*a*sqrt(-c^5/(a^2
*f^2))*f*e^(2*I*f*x + 2*I*e)*log((-12*I*c^3 - 12*(a*f*e^(2*I*f*x + 2*I*e) + a*f)*sqrt(-c^5/(a^2*f^2))*sqrt(c/(
e^(2*I*f*x + 2*I*e) + 1)))*e^(-I*f*x - I*e)/(a*f)) + sqrt(2)*(12*I*c^2*e^(2*I*f*x + 2*I*e) + 4*I*c^2)*sqrt(c/(
e^(2*I*f*x + 2*I*e) + 1)))*e^(-2*I*f*x - 2*I*e)/(a*f)

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Sympy [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: AttributeError} \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)),x)

[Out]

Exception raised: AttributeError

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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}}}{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)),x, algorithm="giac")

[Out]

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