3.10.72 \(\int \frac {(c-i c \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^2} \, dx\) [972]

Optimal. Leaf size=146 \[ \frac {3 i c^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{4 \sqrt {2} a^2 f}+\frac {i c^3 (c-i c \tan (e+f x))^{3/2}}{2 a^2 f (c+i c \tan (e+f x))^2}-\frac {3 i c^3 \sqrt {c-i c \tan (e+f x)}}{4 a^2 f (c+i c \tan (e+f x))} \]

[Out]

3/8*I*c^(5/2)*arctanh(1/2*(c-I*c*tan(f*x+e))^(1/2)*2^(1/2)/c^(1/2))/a^2/f*2^(1/2)+1/2*I*c^3*(c-I*c*tan(f*x+e))
^(3/2)/a^2/f/(c+I*c*tan(f*x+e))^2-3/4*I*c^3*(c-I*c*tan(f*x+e))^(1/2)/a^2/f/(c+I*c*tan(f*x+e))

________________________________________________________________________________________

Rubi [A]
time = 0.13, antiderivative size = 146, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {3603, 3568, 43, 65, 212} \begin {gather*} \frac {3 i c^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{4 \sqrt {2} a^2 f}-\frac {3 i c^3 \sqrt {c-i c \tan (e+f x)}}{4 a^2 f (c+i c \tan (e+f x))}+\frac {i c^3 (c-i c \tan (e+f x))^{3/2}}{2 a^2 f (c+i c \tan (e+f x))^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 43

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, n
}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] &&  !IntegerQ[n] && GtQ[n, 0]

Rule 65

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 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 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 \frac {(c-i c \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^2} \, dx &=\frac {\int \cos ^4(e+f x) (c-i c \tan (e+f x))^{9/2} \, dx}{a^2 c^2}\\ &=\frac {\left (i c^3\right ) \text {Subst}\left (\int \frac {(c+x)^{3/2}}{(c-x)^3} \, dx,x,-i c \tan (e+f x)\right )}{a^2 f}\\ &=\frac {i c^3 (c-i c \tan (e+f x))^{3/2}}{2 a^2 f (c+i c \tan (e+f x))^2}-\frac {\left (3 i c^3\right ) \text {Subst}\left (\int \frac {\sqrt {c+x}}{(c-x)^2} \, dx,x,-i c \tan (e+f x)\right )}{4 a^2 f}\\ &=\frac {i c^3 (c-i c \tan (e+f x))^{3/2}}{2 a^2 f (c+i c \tan (e+f x))^2}-\frac {3 i c^3 \sqrt {c-i c \tan (e+f x)}}{4 a^2 f (c+i c \tan (e+f x))}+\frac {\left (3 i c^3\right ) \text {Subst}\left (\int \frac {1}{(c-x) \sqrt {c+x}} \, dx,x,-i c \tan (e+f x)\right )}{8 a^2 f}\\ &=\frac {i c^3 (c-i c \tan (e+f x))^{3/2}}{2 a^2 f (c+i c \tan (e+f x))^2}-\frac {3 i c^3 \sqrt {c-i c \tan (e+f x)}}{4 a^2 f (c+i c \tan (e+f x))}+\frac {\left (3 i c^3\right ) \text {Subst}\left (\int \frac {1}{2 c-x^2} \, dx,x,\sqrt {c-i c \tan (e+f x)}\right )}{4 a^2 f}\\ &=\frac {3 i c^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{4 \sqrt {2} a^2 f}+\frac {i c^3 (c-i c \tan (e+f x))^{3/2}}{2 a^2 f (c+i c \tan (e+f x))^2}-\frac {3 i c^3 \sqrt {c-i c \tan (e+f x)}}{4 a^2 f (c+i c \tan (e+f x))}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 1.55, size = 138, normalized size = 0.95 \begin {gather*} \frac {c^2 (i \cos (2 (e+f x))+\sin (2 (e+f x))) \left (3 \sqrt {2} \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right ) (\cos (2 (e+f x))+i \sin (2 (e+f x)))-(1+\cos (2 (e+f x))+5 i \sin (2 (e+f x))) \sqrt {c-i c \tan (e+f x)}\right )}{8 a^2 f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(c^2*(I*Cos[2*(e + f*x)] + Sin[2*(e + f*x)])*(3*Sqrt[2]*Sqrt[c]*ArcTanh[Sqrt[c - I*c*Tan[e + f*x]]/(Sqrt[2]*Sq
rt[c])]*(Cos[2*(e + f*x)] + I*Sin[2*(e + f*x)]) - (1 + Cos[2*(e + f*x)] + (5*I)*Sin[2*(e + f*x)])*Sqrt[c - I*c
*Tan[e + f*x]]))/(8*a^2*f)

________________________________________________________________________________________

Maple [A]
time = 0.29, size = 95, normalized size = 0.65

method	result	size
derivativedivides	\(-\frac {2 i c^{3} \left (-\frac {4 \left (\frac {5 \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{32}-\frac {3 c \sqrt {c -i c \tan \left (f x +e \right )}}{16}\right )}{\left (c +i c \tan \left (f x +e \right )\right )^{2}}-\frac {3 \sqrt {2}\, \arctanh \left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{16 \sqrt {c}}\right )}{f \,a^{2}}\)	\(95\)
default	\(-\frac {2 i c^{3} \left (-\frac {4 \left (\frac {5 \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{32}-\frac {3 c \sqrt {c -i c \tan \left (f x +e \right )}}{16}\right )}{\left (c +i c \tan \left (f x +e \right )\right )^{2}}-\frac {3 \sqrt {2}\, \arctanh \left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{16 \sqrt {c}}\right )}{f \,a^{2}}\)	\(95\)

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))^2,x,method=_RETURNVERBOSE)

[Out]

-2*I/f/a^2*c^3*(-4*(5/32*(c-I*c*tan(f*x+e))^(3/2)-3/16*c*(c-I*c*tan(f*x+e))^(1/2))/(c+I*c*tan(f*x+e))^2-3/16*2
^(1/2)/c^(1/2)*arctanh(1/2*(c-I*c*tan(f*x+e))^(1/2)*2^(1/2)/c^(1/2)))

________________________________________________________________________________________

Maxima [A]
time = 0.52, size = 159, normalized size = 1.09 \begin {gather*} -\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^{2}} - \frac {4 \, {\left (5 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} c^{4} - 6 \, \sqrt {-i \, c \tan \left (f x + e\right ) + c} c^{5}\right )}}{{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{2} a^{2} - 4 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )} a^{2} c + 4 \, a^{2} c^{2}}\right )}}{16 \, c f} \end {gather*}

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))^2,x, algorithm="maxima")

[Out]

-1/16*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*t
an(f*x + e) + c)))/a^2 - 4*(5*(-I*c*tan(f*x + e) + c)^(3/2)*c^4 - 6*sqrt(-I*c*tan(f*x + e) + c)*c^5)/((-I*c*ta
n(f*x + e) + c)^2*a^2 - 4*(-I*c*tan(f*x + e) + c)*a^2*c + 4*a^2*c^2))/(c*f)

________________________________________________________________________________________

Fricas [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 311 vs. \(2 (116) = 232\).
time = 1.79, size = 311, normalized size = 2.13 \begin {gather*} -\frac {{\left (3 \, \sqrt {\frac {1}{2}} a^{2} f \sqrt {-\frac {c^{5}}{a^{4} f^{2}}} e^{\left (4 i \, f x + 4 i \, e\right )} \log \left (-\frac {3 \, {\left (-i \, c^{3} + \sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{2} f\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {-\frac {c^{5}}{a^{4} f^{2}}}\right )} e^{\left (-i \, f x - i \, e\right )}}{2 \, a^{2} f}\right ) - 3 \, \sqrt {\frac {1}{2}} a^{2} f \sqrt {-\frac {c^{5}}{a^{4} f^{2}}} e^{\left (4 i \, f x + 4 i \, e\right )} \log \left (-\frac {3 \, {\left (-i \, c^{3} - \sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{2} f\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {-\frac {c^{5}}{a^{4} f^{2}}}\right )} e^{\left (-i \, f x - i \, e\right )}}{2 \, a^{2} f}\right ) - \sqrt {2} {\left (-3 i \, c^{2} e^{\left (4 i \, f x + 4 i \, e\right )} - i \, c^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + 2 i \, c^{2}\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-4 i \, f x - 4 i \, e\right )}}{8 \, a^{2} f} \end {gather*}

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))^2,x, algorithm="fricas")

[Out]

-1/8*(3*sqrt(1/2)*a^2*f*sqrt(-c^5/(a^4*f^2))*e^(4*I*f*x + 4*I*e)*log(-3/2*(-I*c^3 + sqrt(2)*sqrt(1/2)*(a^2*f*e
^(2*I*f*x + 2*I*e) + a^2*f)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(-c^5/(a^4*f^2)))*e^(-I*f*x - I*e)/(a^2*f))
- 3*sqrt(1/2)*a^2*f*sqrt(-c^5/(a^4*f^2))*e^(4*I*f*x + 4*I*e)*log(-3/2*(-I*c^3 - sqrt(2)*sqrt(1/2)*(a^2*f*e^(2*
I*f*x + 2*I*e) + a^2*f)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(-c^5/(a^4*f^2)))*e^(-I*f*x - I*e)/(a^2*f)) - sq
rt(2)*(-3*I*c^2*e^(4*I*f*x + 4*I*e) - I*c^2*e^(2*I*f*x + 2*I*e) + 2*I*c^2)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1)))*
e^(-4*I*f*x - 4*I*e)/(a^2*f)

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \frac {\int \frac {c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c}}{\tan ^{2}{\left (e + f x \right )} - 2 i \tan {\left (e + f x \right )} - 1}\, dx + \int \left (- \frac {c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{2}{\left (e + f x \right )}}{\tan ^{2}{\left (e + f x \right )} - 2 i \tan {\left (e + f x \right )} - 1}\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 ^{2}{\left (e + f x \right )} - 2 i \tan {\left (e + f x \right )} - 1}\right )\, dx}{a^{2}} \end {gather*}

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))**2,x)

[Out]

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

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

Mupad [B]
time = 0.28, size = 135, normalized size = 0.92 \begin {gather*} -\frac {\frac {c^4\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}\,3{}\mathrm {i}}{2\,a^2\,f}-\frac {c^3\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{3/2}\,5{}\mathrm {i}}{4\,a^2\,f}}{{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^2-4\,c\,\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )+4\,c^2}+\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}}{8\,a^2\,f} \end {gather*}

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)^2,x)

[Out]

(2^(1/2)*(-c)^(5/2)*atan((2^(1/2)*(c - c*tan(e + f*x)*1i)^(1/2))/(2*(-c)^(1/2)))*3i)/(8*a^2*f) - ((c^4*(c - c*
tan(e + f*x)*1i)^(1/2)*3i)/(2*a^2*f) - (c^3*(c - c*tan(e + f*x)*1i)^(3/2)*5i)/(4*a^2*f))/((c - c*tan(e + f*x)*
1i)^2 - 4*c*(c - c*tan(e + f*x)*1i) + 4*c^2)

________________________________________________________________________________________