3.989 \(\int \frac{1}{(a+i a \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}} \, dx\)

Optimal. Leaf size=188 \[ -\frac{7 i}{16 a c^2 f \sqrt{c-i c \tan (e+f x)}}+\frac{7 i \tanh ^{-1}\left (\frac{\sqrt{c-i c \tan (e+f x)}}{\sqrt{2} \sqrt{c}}\right )}{16 \sqrt{2} a c^{5/2} f}-\frac{7 i}{24 a c f (c-i c \tan (e+f x))^{3/2}}-\frac{7 i}{20 a f (c-i c \tan (e+f x))^{5/2}}+\frac{i}{2 a f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}} \]

[Out]

(((7*I)/16)*ArcTanh[Sqrt[c - I*c*Tan[e + f*x]]/(Sqrt[2]*Sqrt[c])])/(Sqrt[2]*a*c^(5/2)*f) - ((7*I)/20)/(a*f*(c
- I*c*Tan[e + f*x])^(5/2)) + (I/2)/(a*f*(1 + I*Tan[e + f*x])*(c - I*c*Tan[e + f*x])^(5/2)) - ((7*I)/24)/(a*c*f
*(c - I*c*Tan[e + f*x])^(3/2)) - ((7*I)/16)/(a*c^2*f*Sqrt[c - I*c*Tan[e + f*x]])

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Rubi [A]  time = 0.223694, antiderivative size = 188, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 5, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.152, Rules used = {3522, 3487, 51, 63, 206} \[ -\frac{7 i}{16 a c^2 f \sqrt{c-i c \tan (e+f x)}}+\frac{7 i \tanh ^{-1}\left (\frac{\sqrt{c-i c \tan (e+f x)}}{\sqrt{2} \sqrt{c}}\right )}{16 \sqrt{2} a c^{5/2} f}-\frac{7 i}{24 a c f (c-i c \tan (e+f x))^{3/2}}-\frac{7 i}{20 a f (c-i c \tan (e+f x))^{5/2}}+\frac{i}{2 a f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + I*a*Tan[e + f*x])*(c - I*c*Tan[e + f*x])^(5/2)),x]

[Out]

(((7*I)/16)*ArcTanh[Sqrt[c - I*c*Tan[e + f*x]]/(Sqrt[2]*Sqrt[c])])/(Sqrt[2]*a*c^(5/2)*f) - ((7*I)/20)/(a*f*(c
- I*c*Tan[e + f*x])^(5/2)) + (I/2)/(a*f*(1 + I*Tan[e + f*x])*(c - I*c*Tan[e + f*x])^(5/2)) - ((7*I)/24)/(a*c*f
*(c - I*c*Tan[e + f*x])^(3/2)) - ((7*I)/16)/(a*c^2*f*Sqrt[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 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{1}{(a+i a \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}} \, dx &=\frac{\int \frac{\cos ^2(e+f x)}{(c-i c \tan (e+f x))^{3/2}} \, dx}{a c}\\ &=\frac{\left (i c^2\right ) \operatorname{Subst}\left (\int \frac{1}{(c-x)^2 (c+x)^{7/2}} \, dx,x,-i c \tan (e+f x)\right )}{a f}\\ &=\frac{i}{2 a f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}+\frac{(7 i c) \operatorname{Subst}\left (\int \frac{1}{(c-x) (c+x)^{7/2}} \, dx,x,-i c \tan (e+f x)\right )}{4 a f}\\ &=-\frac{7 i}{20 a f (c-i c \tan (e+f x))^{5/2}}+\frac{i}{2 a f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}+\frac{(7 i) \operatorname{Subst}\left (\int \frac{1}{(c-x) (c+x)^{5/2}} \, dx,x,-i c \tan (e+f x)\right )}{8 a f}\\ &=-\frac{7 i}{20 a f (c-i c \tan (e+f x))^{5/2}}+\frac{i}{2 a f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}-\frac{7 i}{24 a c f (c-i c \tan (e+f x))^{3/2}}+\frac{(7 i) \operatorname{Subst}\left (\int \frac{1}{(c-x) (c+x)^{3/2}} \, dx,x,-i c \tan (e+f x)\right )}{16 a c f}\\ &=-\frac{7 i}{20 a f (c-i c \tan (e+f x))^{5/2}}+\frac{i}{2 a f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}-\frac{7 i}{24 a c f (c-i c \tan (e+f x))^{3/2}}-\frac{7 i}{16 a c^2 f \sqrt{c-i c \tan (e+f x)}}+\frac{(7 i) \operatorname{Subst}\left (\int \frac{1}{(c-x) \sqrt{c+x}} \, dx,x,-i c \tan (e+f x)\right )}{32 a c^2 f}\\ &=-\frac{7 i}{20 a f (c-i c \tan (e+f x))^{5/2}}+\frac{i}{2 a f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}-\frac{7 i}{24 a c f (c-i c \tan (e+f x))^{3/2}}-\frac{7 i}{16 a c^2 f \sqrt{c-i c \tan (e+f x)}}+\frac{(7 i) \operatorname{Subst}\left (\int \frac{1}{2 c-x^2} \, dx,x,\sqrt{c-i c \tan (e+f x)}\right )}{16 a c^2 f}\\ &=\frac{7 i \tanh ^{-1}\left (\frac{\sqrt{c-i c \tan (e+f x)}}{\sqrt{2} \sqrt{c}}\right )}{16 \sqrt{2} a c^{5/2} f}-\frac{7 i}{20 a f (c-i c \tan (e+f x))^{5/2}}+\frac{i}{2 a f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}-\frac{7 i}{24 a c f (c-i c \tan (e+f x))^{3/2}}-\frac{7 i}{16 a c^2 f \sqrt{c-i c \tan (e+f x)}}\\ \end{align*}

Mathematica [A]  time = 3.4298, size = 149, normalized size = 0.79 \[ \frac{\sqrt{c-i c \tan (e+f x)} (\cos (2 (e+f x))+i \sin (2 (e+f x))) \left (-63 \sin (2 (e+f x))+21 \sin (4 (e+f x))-139 i \cos (2 (e+f x))+9 i \cos (4 (e+f x))+105 i e^{-2 i (e+f x)} \sqrt{1+e^{2 i (e+f x)}} \tanh ^{-1}\left (\sqrt{1+e^{2 i (e+f x)}}\right )-148 i\right )}{480 a c^3 f} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + I*a*Tan[e + f*x])*(c - I*c*Tan[e + f*x])^(5/2)),x]

[Out]

((Cos[2*(e + f*x)] + I*Sin[2*(e + f*x)])*(-148*I + ((105*I)*Sqrt[1 + E^((2*I)*(e + f*x))]*ArcTanh[Sqrt[1 + E^(
(2*I)*(e + f*x))]])/E^((2*I)*(e + f*x)) - (139*I)*Cos[2*(e + f*x)] + (9*I)*Cos[4*(e + f*x)] - 63*Sin[2*(e + f*
x)] + 21*Sin[4*(e + f*x)])*Sqrt[c - I*c*Tan[e + f*x]])/(480*a*c^3*f)

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Maple [A]  time = 0.036, size = 140, normalized size = 0.7 \begin{align*}{\frac{2\,i{c}^{2}}{fa} \left ( -{\frac{1}{16\,{c}^{4}} \left ({\frac{1}{-2\,c-2\,ic\tan \left ( fx+e \right ) }\sqrt{c-ic\tan \left ( fx+e \right ) }}-{\frac{7\,\sqrt{2}}{4}{\it Artanh} \left ({\frac{\sqrt{2}}{2}\sqrt{c-ic\tan \left ( fx+e \right ) }{\frac{1}{\sqrt{c}}}} \right ){\frac{1}{\sqrt{c}}}} \right ) }-{\frac{3}{16\,{c}^{4}}{\frac{1}{\sqrt{c-ic\tan \left ( fx+e \right ) }}}}-{\frac{1}{12\,{c}^{3}} \left ( c-ic\tan \left ( fx+e \right ) \right ) ^{-{\frac{3}{2}}}}-{\frac{1}{20\,{c}^{2}} \left ( c-ic\tan \left ( fx+e \right ) \right ) ^{-{\frac{5}{2}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*I/f/a*c^2*(-1/16/c^4*(1/2*(c-I*c*tan(f*x+e))^(1/2)/(-c-I*c*tan(f*x+e))-7/4*2^(1/2)/c^(1/2)*arctanh(1/2*(c-I*
c*tan(f*x+e))^(1/2)*2^(1/2)/c^(1/2)))-3/16/c^4/(c-I*c*tan(f*x+e))^(1/2)-1/12/c^3/(c-I*c*tan(f*x+e))^(3/2)-1/20
/c^2/(c-I*c*tan(f*x+e))^(5/2))

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.60638, size = 913, normalized size = 4.86 \begin{align*} \frac{{\left (105 i \, \sqrt{\frac{1}{2}} a c^{3} f \sqrt{\frac{1}{a^{2} c^{5} f^{2}}} e^{\left (2 i \, f x + 2 i \, e\right )} \log \left (\frac{{\left (\sqrt{2} \sqrt{\frac{1}{2}}{\left (56 i \, a c^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + 56 i \, a c^{2} f\right )} \sqrt{\frac{c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt{\frac{1}{a^{2} c^{5} f^{2}}} + 56 i\right )} e^{\left (-i \, f x - i \, e\right )}}{64 \, a c^{2} f}\right ) - 105 i \, \sqrt{\frac{1}{2}} a c^{3} f \sqrt{\frac{1}{a^{2} c^{5} f^{2}}} e^{\left (2 i \, f x + 2 i \, e\right )} \log \left (\frac{{\left (\sqrt{2} \sqrt{\frac{1}{2}}{\left (-56 i \, a c^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} - 56 i \, a c^{2} f\right )} \sqrt{\frac{c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt{\frac{1}{a^{2} c^{5} f^{2}}} + 56 i\right )} e^{\left (-i \, f x - i \, e\right )}}{64 \, a c^{2} f}\right ) + \sqrt{2} \sqrt{\frac{c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}{\left (-6 i \, e^{\left (8 i \, f x + 8 i \, e\right )} - 38 i \, e^{\left (6 i \, f x + 6 i \, e\right )} - 148 i \, e^{\left (4 i \, f x + 4 i \, e\right )} - 101 i \, e^{\left (2 i \, f x + 2 i \, e\right )} + 15 i\right )}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{480 \, a c^{3} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/480*(105*I*sqrt(1/2)*a*c^3*f*sqrt(1/(a^2*c^5*f^2))*e^(2*I*f*x + 2*I*e)*log(1/64*(sqrt(2)*sqrt(1/2)*(56*I*a*c
^2*f*e^(2*I*f*x + 2*I*e) + 56*I*a*c^2*f)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(1/(a^2*c^5*f^2)) + 56*I)*e^(-I
*f*x - I*e)/(a*c^2*f)) - 105*I*sqrt(1/2)*a*c^3*f*sqrt(1/(a^2*c^5*f^2))*e^(2*I*f*x + 2*I*e)*log(1/64*(sqrt(2)*s
qrt(1/2)*(-56*I*a*c^2*f*e^(2*I*f*x + 2*I*e) - 56*I*a*c^2*f)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(1/(a^2*c^5*
f^2)) + 56*I)*e^(-I*f*x - I*e)/(a*c^2*f)) + sqrt(2)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*(-6*I*e^(8*I*f*x + 8*I*e
) - 38*I*e^(6*I*f*x + 6*I*e) - 148*I*e^(4*I*f*x + 4*I*e) - 101*I*e^(2*I*f*x + 2*I*e) + 15*I))*e^(-2*I*f*x - 2*
I*e)/(a*c^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(1/(a+I*a*tan(f*x+e))/(c-I*c*tan(f*x+e))**(5/2),x)

[Out]

Exception raised: AttributeError

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (i \, a \tan \left (f x + e\right ) + a\right )}{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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