3.973 \(\int \frac{(c-i c \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^3} \, dx\)

Optimal. Leaf size=193 \[ -\frac{i c^4 \sqrt{c-i c \tan (e+f x)}}{4 a^3 f (c+i c \tan (e+f x))^2}+\frac{i c^4 (c-i c \tan (e+f x))^{3/2}}{3 a^3 f (c+i c \tan (e+f x))^3}+\frac{i c^3 \sqrt{c-i c \tan (e+f x)}}{16 a^3 f (c+i c \tan (e+f x))}+\frac{i c^{5/2} \tanh ^{-1}\left (\frac{\sqrt{c-i c \tan (e+f x)}}{\sqrt{2} \sqrt{c}}\right )}{16 \sqrt{2} a^3 f} \]

[Out]

((I/16)*c^(5/2)*ArcTanh[Sqrt[c - I*c*Tan[e + f*x]]/(Sqrt[2]*Sqrt[c])])/(Sqrt[2]*a^3*f) + ((I/3)*c^4*(c - I*c*T
an[e + f*x])^(3/2))/(a^3*f*(c + I*c*Tan[e + f*x])^3) - ((I/4)*c^4*Sqrt[c - I*c*Tan[e + f*x]])/(a^3*f*(c + I*c*
Tan[e + f*x])^2) + ((I/16)*c^3*Sqrt[c - I*c*Tan[e + f*x]])/(a^3*f*(c + I*c*Tan[e + f*x]))

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Rubi [A]  time = 0.204621, antiderivative size = 193, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {3522, 3487, 47, 51, 63, 206} \[ -\frac{i c^4 \sqrt{c-i c \tan (e+f x)}}{4 a^3 f (c+i c \tan (e+f x))^2}+\frac{i c^4 (c-i c \tan (e+f x))^{3/2}}{3 a^3 f (c+i c \tan (e+f x))^3}+\frac{i c^3 \sqrt{c-i c \tan (e+f x)}}{16 a^3 f (c+i c \tan (e+f x))}+\frac{i c^{5/2} \tanh ^{-1}\left (\frac{\sqrt{c-i c \tan (e+f x)}}{\sqrt{2} \sqrt{c}}\right )}{16 \sqrt{2} a^3 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((I/16)*c^(5/2)*ArcTanh[Sqrt[c - I*c*Tan[e + f*x]]/(Sqrt[2]*Sqrt[c])])/(Sqrt[2]*a^3*f) + ((I/3)*c^4*(c - I*c*T
an[e + f*x])^(3/2))/(a^3*f*(c + I*c*Tan[e + f*x])^3) - ((I/4)*c^4*Sqrt[c - I*c*Tan[e + f*x]])/(a^3*f*(c + I*c*
Tan[e + f*x])^2) + ((I/16)*c^3*Sqrt[c - I*c*Tan[e + f*x]])/(a^3*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 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && 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))^3} \, dx &=\frac{\int \cos ^6(e+f x) (c-i c \tan (e+f x))^{11/2} \, dx}{a^3 c^3}\\ &=\frac{\left (i c^4\right ) \operatorname{Subst}\left (\int \frac{(c+x)^{3/2}}{(c-x)^4} \, dx,x,-i c \tan (e+f x)\right )}{a^3 f}\\ &=\frac{i c^4 (c-i c \tan (e+f x))^{3/2}}{3 a^3 f (c+i c \tan (e+f x))^3}-\frac{\left (i c^4\right ) \operatorname{Subst}\left (\int \frac{\sqrt{c+x}}{(c-x)^3} \, dx,x,-i c \tan (e+f x)\right )}{2 a^3 f}\\ &=\frac{i c^4 (c-i c \tan (e+f x))^{3/2}}{3 a^3 f (c+i c \tan (e+f x))^3}-\frac{i c^4 \sqrt{c-i c \tan (e+f x)}}{4 a^3 f (c+i c \tan (e+f x))^2}+\frac{\left (i c^4\right ) \operatorname{Subst}\left (\int \frac{1}{(c-x)^2 \sqrt{c+x}} \, dx,x,-i c \tan (e+f x)\right )}{8 a^3 f}\\ &=\frac{i c^4 (c-i c \tan (e+f x))^{3/2}}{3 a^3 f (c+i c \tan (e+f x))^3}-\frac{i c^4 \sqrt{c-i c \tan (e+f x)}}{4 a^3 f (c+i c \tan (e+f x))^2}+\frac{i c^3 \sqrt{c-i c \tan (e+f x)}}{16 a^3 f (c+i c \tan (e+f x))}+\frac{\left (i c^3\right ) \operatorname{Subst}\left (\int \frac{1}{(c-x) \sqrt{c+x}} \, dx,x,-i c \tan (e+f x)\right )}{32 a^3 f}\\ &=\frac{i c^4 (c-i c \tan (e+f x))^{3/2}}{3 a^3 f (c+i c \tan (e+f x))^3}-\frac{i c^4 \sqrt{c-i c \tan (e+f x)}}{4 a^3 f (c+i c \tan (e+f x))^2}+\frac{i c^3 \sqrt{c-i c \tan (e+f x)}}{16 a^3 f (c+i c \tan (e+f x))}+\frac{\left (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 )}{16 a^3 f}\\ &=\frac{i c^{5/2} \tanh ^{-1}\left (\frac{\sqrt{c-i c \tan (e+f x)}}{\sqrt{2} \sqrt{c}}\right )}{16 \sqrt{2} a^3 f}+\frac{i c^4 (c-i c \tan (e+f x))^{3/2}}{3 a^3 f (c+i c \tan (e+f x))^3}-\frac{i c^4 \sqrt{c-i c \tan (e+f x)}}{4 a^3 f (c+i c \tan (e+f x))^2}+\frac{i c^3 \sqrt{c-i c \tan (e+f x)}}{16 a^3 f (c+i c \tan (e+f x))}\\ \end{align*}

Mathematica [A]  time = 3.65219, size = 152, normalized size = 0.79 \[ \frac{c^2 (\sin (3 (e+f x))+i \cos (3 (e+f x))) \left (\sqrt{c-i c \tan (e+f x)} \left (9 \cos (e+f x)+5 \cos (3 (e+f x))-44 i \sin (e+f x) \cos ^2(e+f x)\right )+3 \sqrt{2} \sqrt{c} (\cos (3 (e+f x))+i \sin (3 (e+f x))) \tanh ^{-1}\left (\frac{\sqrt{c-i c \tan (e+f x)}}{\sqrt{2} \sqrt{c}}\right )\right )}{96 a^3 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c^2*(I*Cos[3*(e + f*x)] + Sin[3*(e + f*x)])*(3*Sqrt[2]*Sqrt[c]*ArcTanh[Sqrt[c - I*c*Tan[e + f*x]]/(Sqrt[2]*Sq
rt[c])]*(Cos[3*(e + f*x)] + I*Sin[3*(e + f*x)]) + (9*Cos[e + f*x] + 5*Cos[3*(e + f*x)] - (44*I)*Cos[e + f*x]^2
*Sin[e + f*x])*Sqrt[c - I*c*Tan[e + f*x]]))/(96*a^3*f)

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Maple [A]  time = 0.04, size = 115, normalized size = 0.6 \begin{align*}{\frac{2\,i{c}^{4}}{f{a}^{3}} \left ({\frac{1}{ \left ( -c-ic\tan \left ( fx+e \right ) \right ) ^{3}} \left ( -{\frac{1}{32\,c} \left ( c-ic\tan \left ( fx+e \right ) \right ) ^{{\frac{5}{2}}}}-{\frac{1}{6} \left ( c-ic\tan \left ( fx+e \right ) \right ) ^{{\frac{3}{2}}}}+{\frac{c}{8}\sqrt{c-ic\tan \left ( fx+e \right ) }} \right ) }+{\frac{\sqrt{2}}{64}{\it Artanh} \left ({\frac{\sqrt{2}}{2}\sqrt{c-ic\tan \left ( fx+e \right ) }{\frac{1}{\sqrt{c}}}} \right ){c}^{-{\frac{3}{2}}}} \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))^3,x)

[Out]

2*I/f/a^3*c^4*((-1/32/c*(c-I*c*tan(f*x+e))^(5/2)-1/6*(c-I*c*tan(f*x+e))^(3/2)+1/8*c*(c-I*c*tan(f*x+e))^(1/2))/
(-c-I*c*tan(f*x+e))^3+1/64/c^(3/2)*2^(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))^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.34733, size = 819, normalized size = 4.24 \begin{align*} \frac{{\left (3 \, \sqrt{\frac{1}{2}} a^{3} f \sqrt{-\frac{c^{5}}{a^{6} f^{2}}} e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (\frac{{\left (i \, c^{3} + \sqrt{2} \sqrt{\frac{1}{2}}{\left (a^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{3} f\right )} \sqrt{\frac{c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt{-\frac{c^{5}}{a^{6} f^{2}}}\right )} e^{\left (-i \, f x - i \, e\right )}}{8 \, a^{3} f}\right ) - 3 \, \sqrt{\frac{1}{2}} a^{3} f \sqrt{-\frac{c^{5}}{a^{6} f^{2}}} e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (\frac{{\left (i \, c^{3} - \sqrt{2} \sqrt{\frac{1}{2}}{\left (a^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{3} f\right )} \sqrt{\frac{c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt{-\frac{c^{5}}{a^{6} f^{2}}}\right )} e^{\left (-i \, f x - i \, e\right )}}{8 \, a^{3} f}\right ) + \sqrt{2}{\left (-3 i \, c^{2} e^{\left (6 i \, f x + 6 i \, e\right )} - i \, c^{2} e^{\left (4 i \, f x + 4 i \, e\right )} + 10 i \, c^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + 8 i \, c^{2}\right )} \sqrt{\frac{c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-6 i \, f x - 6 i \, e\right )}}{96 \, a^{3} 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))^3,x, algorithm="fricas")

[Out]

1/96*(3*sqrt(1/2)*a^3*f*sqrt(-c^5/(a^6*f^2))*e^(6*I*f*x + 6*I*e)*log(1/8*(I*c^3 + sqrt(2)*sqrt(1/2)*(a^3*f*e^(
2*I*f*x + 2*I*e) + a^3*f)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(-c^5/(a^6*f^2)))*e^(-I*f*x - I*e)/(a^3*f)) -
3*sqrt(1/2)*a^3*f*sqrt(-c^5/(a^6*f^2))*e^(6*I*f*x + 6*I*e)*log(1/8*(I*c^3 - sqrt(2)*sqrt(1/2)*(a^3*f*e^(2*I*f*
x + 2*I*e) + a^3*f)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(-c^5/(a^6*f^2)))*e^(-I*f*x - I*e)/(a^3*f)) + sqrt(2
)*(-3*I*c^2*e^(6*I*f*x + 6*I*e) - I*c^2*e^(4*I*f*x + 4*I*e) + 10*I*c^2*e^(2*I*f*x + 2*I*e) + 8*I*c^2)*sqrt(c/(
e^(2*I*f*x + 2*I*e) + 1)))*e^(-6*I*f*x - 6*I*e)/(a^3*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))**3,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}}}{{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{3}}\,{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))^3,x, algorithm="giac")

[Out]

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