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

[Out]

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

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Rubi [A]  time = 0.18, antiderivative size = 125, normalized size of antiderivative = 1.00, 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 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])

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 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]))

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*}

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Mathematica [F]  time = 180.01, size = 0, normalized size = 0.00 \[ \text {\$Aborted} \]

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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fricas [B]  time = 0.63, size = 259, normalized size = 2.07 \[ \frac {{\left (6 \, \sqrt {2} a \sqrt {-\frac {c^{5}}{a^{2} f^{2}}} f e^{\left (2 i \, f x + 2 i \, e\right )} \log \left (\frac {{\left (-12 i \, c^{3} + 12 \, {\left (a f e^{\left (2 i \, f x + 2 i \, e\right )} + a f\right )} \sqrt {-\frac {c^{5}}{a^{2} f^{2}}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-i \, f x - i \, e\right )}}{a f}\right ) - 6 \, \sqrt {2} a \sqrt {-\frac {c^{5}}{a^{2} f^{2}}} f e^{\left (2 i \, f x + 2 i \, e\right )} \log \left (\frac {{\left (-12 i \, c^{3} - 12 \, {\left (a f e^{\left (2 i \, f x + 2 i \, e\right )} + a f\right )} \sqrt {-\frac {c^{5}}{a^{2} f^{2}}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-i \, f x - i \, e\right )}}{a f}\right ) + \sqrt {2} {\left (12 i \, c^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + 4 i \, c^{2}\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{4 \, a f} \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}}}{i \, a \tan \left (f x + e\right ) + a}\,{d x} \]

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)

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maple [A]  time = 0.35, size = 95, normalized size = 0.76 \[ \frac {2 i c^{2} \left (\sqrt {c -i c \tan \left (f x +e \right )}+4 c \left (-\frac {\sqrt {c -i c \tan \left (f x +e \right )}}{4 \left (-c -i c \tan \left (f x +e \right )\right )}-\frac {3 \sqrt {2}\, \arctanh \left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{8 \sqrt {c}}\right )\right )}{f a} \]

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 [A]  time = 0.92, size = 128, normalized size = 1.02 \[ \frac {i \, {\left (\frac {3 \, \sqrt {2} c^{\frac {7}{2}} \log \left (-\frac {\sqrt {2} \sqrt {c} - \sqrt {-i \, c \tan \left (f x + e\right ) + c}}{\sqrt {2} \sqrt {c} + \sqrt {-i \, c \tan \left (f x + e\right ) + c}}\right )}{a} - \frac {4 \, \sqrt {-i \, c \tan \left (f x + e\right ) + c} c^{4}}{{\left (-i \, c \tan \left (f x + e\right ) + c\right )} a - 2 \, a c} + \frac {4 \, \sqrt {-i \, c \tan \left (f x + e\right ) + c} c^{3}}{a}\right )}}{2 \, c f} \]

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]

1/2*I*(3*sqrt(2)*c^(7/2)*log(-(sqrt(2)*sqrt(c) - sqrt(-I*c*tan(f*x + e) + c))/(sqrt(2)*sqrt(c) + sqrt(-I*c*tan
(f*x + e) + c)))/a - 4*sqrt(-I*c*tan(f*x + e) + c)*c^4/((-I*c*tan(f*x + e) + c)*a - 2*a*c) + 4*sqrt(-I*c*tan(f
*x + e) + c)*c^3/a)/(c*f)

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mupad [B]  time = 0.35, size = 109, normalized size = 0.87 \[ \frac {c^2\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}\,2{}\mathrm {i}}{a\,f}-\frac {\sqrt {2}\,{\left (-c\right )}^{5/2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {-c}}\right )\,3{}\mathrm {i}}{a\,f}+\frac {c^3\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}\,2{}\mathrm {i}}{a\,f\,\left (c+c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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]

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

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