3.991 \(\int \frac{1}{(a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))^{5/2}} \, dx\)

Optimal. Leaf size=274 \[ -\frac{231 i}{512 a^3 c^2 f \sqrt{c-i c \tan (e+f x)}}+\frac{231 i \tanh ^{-1}\left (\frac{\sqrt{c-i c \tan (e+f x)}}{\sqrt{2} \sqrt{c}}\right )}{512 \sqrt{2} a^3 c^{5/2} f}-\frac{77 i}{256 a^3 c f (c-i c \tan (e+f x))^{3/2}}-\frac{231 i}{640 a^3 f (c-i c \tan (e+f x))^{5/2}}+\frac{33 i}{64 a^3 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}+\frac{11 i}{48 a^3 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2}}+\frac{i}{6 a^3 f (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{5/2}} \]

[Out]

(((231*I)/512)*ArcTanh[Sqrt[c - I*c*Tan[e + f*x]]/(Sqrt[2]*Sqrt[c])])/(Sqrt[2]*a^3*c^(5/2)*f) - ((231*I)/640)/
(a^3*f*(c - I*c*Tan[e + f*x])^(5/2)) + (I/6)/(a^3*f*(1 + I*Tan[e + f*x])^3*(c - I*c*Tan[e + f*x])^(5/2)) + ((1
1*I)/48)/(a^3*f*(1 + I*Tan[e + f*x])^2*(c - I*c*Tan[e + f*x])^(5/2)) + ((33*I)/64)/(a^3*f*(1 + I*Tan[e + f*x])
*(c - I*c*Tan[e + f*x])^(5/2)) - ((77*I)/256)/(a^3*c*f*(c - I*c*Tan[e + f*x])^(3/2)) - ((231*I)/512)/(a^3*c^2*
f*Sqrt[c - I*c*Tan[e + f*x]])

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Rubi [A]  time = 0.250389, antiderivative size = 274, normalized size of antiderivative = 1., number of steps used = 10, 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} \[ -\frac{231 i}{512 a^3 c^2 f \sqrt{c-i c \tan (e+f x)}}+\frac{231 i \tanh ^{-1}\left (\frac{\sqrt{c-i c \tan (e+f x)}}{\sqrt{2} \sqrt{c}}\right )}{512 \sqrt{2} a^3 c^{5/2} f}-\frac{77 i}{256 a^3 c f (c-i c \tan (e+f x))^{3/2}}-\frac{231 i}{640 a^3 f (c-i c \tan (e+f x))^{5/2}}+\frac{33 i}{64 a^3 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}+\frac{11 i}{48 a^3 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2}}+\frac{i}{6 a^3 f (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + I*a*Tan[e + f*x])^3*(c - I*c*Tan[e + f*x])^(5/2)),x]

[Out]

(((231*I)/512)*ArcTanh[Sqrt[c - I*c*Tan[e + f*x]]/(Sqrt[2]*Sqrt[c])])/(Sqrt[2]*a^3*c^(5/2)*f) - ((231*I)/640)/
(a^3*f*(c - I*c*Tan[e + f*x])^(5/2)) + (I/6)/(a^3*f*(1 + I*Tan[e + f*x])^3*(c - I*c*Tan[e + f*x])^(5/2)) + ((1
1*I)/48)/(a^3*f*(1 + I*Tan[e + f*x])^2*(c - I*c*Tan[e + f*x])^(5/2)) + ((33*I)/64)/(a^3*f*(1 + I*Tan[e + f*x])
*(c - I*c*Tan[e + f*x])^(5/2)) - ((77*I)/256)/(a^3*c*f*(c - I*c*Tan[e + f*x])^(3/2)) - ((231*I)/512)/(a^3*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))^3 (c-i c \tan (e+f x))^{5/2}} \, dx &=\frac{\int \cos ^6(e+f x) \sqrt{c-i c \tan (e+f x)} \, dx}{a^3 c^3}\\ &=\frac{\left (i c^4\right ) \operatorname{Subst}\left (\int \frac{1}{(c-x)^4 (c+x)^{7/2}} \, dx,x,-i c \tan (e+f x)\right )}{a^3 f}\\ &=\frac{i}{6 a^3 f (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{5/2}}+\frac{\left (11 i c^3\right ) \operatorname{Subst}\left (\int \frac{1}{(c-x)^3 (c+x)^{7/2}} \, dx,x,-i c \tan (e+f x)\right )}{12 a^3 f}\\ &=\frac{i}{6 a^3 f (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{5/2}}+\frac{11 i}{48 a^3 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2}}+\frac{\left (33 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 )}{32 a^3 f}\\ &=\frac{i}{6 a^3 f (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{5/2}}+\frac{11 i}{48 a^3 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2}}+\frac{33 i}{64 a^3 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}+\frac{(231 i c) \operatorname{Subst}\left (\int \frac{1}{(c-x) (c+x)^{7/2}} \, dx,x,-i c \tan (e+f x)\right )}{128 a^3 f}\\ &=-\frac{231 i}{640 a^3 f (c-i c \tan (e+f x))^{5/2}}+\frac{i}{6 a^3 f (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{5/2}}+\frac{11 i}{48 a^3 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2}}+\frac{33 i}{64 a^3 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}+\frac{(231 i) \operatorname{Subst}\left (\int \frac{1}{(c-x) (c+x)^{5/2}} \, dx,x,-i c \tan (e+f x)\right )}{256 a^3 f}\\ &=-\frac{231 i}{640 a^3 f (c-i c \tan (e+f x))^{5/2}}+\frac{i}{6 a^3 f (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{5/2}}+\frac{11 i}{48 a^3 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2}}+\frac{33 i}{64 a^3 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}-\frac{77 i}{256 a^3 c f (c-i c \tan (e+f x))^{3/2}}+\frac{(231 i) \operatorname{Subst}\left (\int \frac{1}{(c-x) (c+x)^{3/2}} \, dx,x,-i c \tan (e+f x)\right )}{512 a^3 c f}\\ &=-\frac{231 i}{640 a^3 f (c-i c \tan (e+f x))^{5/2}}+\frac{i}{6 a^3 f (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{5/2}}+\frac{11 i}{48 a^3 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2}}+\frac{33 i}{64 a^3 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}-\frac{77 i}{256 a^3 c f (c-i c \tan (e+f x))^{3/2}}-\frac{231 i}{512 a^3 c^2 f \sqrt{c-i c \tan (e+f x)}}+\frac{(231 i) \operatorname{Subst}\left (\int \frac{1}{(c-x) \sqrt{c+x}} \, dx,x,-i c \tan (e+f x)\right )}{1024 a^3 c^2 f}\\ &=-\frac{231 i}{640 a^3 f (c-i c \tan (e+f x))^{5/2}}+\frac{i}{6 a^3 f (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{5/2}}+\frac{11 i}{48 a^3 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2}}+\frac{33 i}{64 a^3 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}-\frac{77 i}{256 a^3 c f (c-i c \tan (e+f x))^{3/2}}-\frac{231 i}{512 a^3 c^2 f \sqrt{c-i c \tan (e+f x)}}+\frac{(231 i) \operatorname{Subst}\left (\int \frac{1}{2 c-x^2} \, dx,x,\sqrt{c-i c \tan (e+f x)}\right )}{512 a^3 c^2 f}\\ &=\frac{231 i \tanh ^{-1}\left (\frac{\sqrt{c-i c \tan (e+f x)}}{\sqrt{2} \sqrt{c}}\right )}{512 \sqrt{2} a^3 c^{5/2} f}-\frac{231 i}{640 a^3 f (c-i c \tan (e+f x))^{5/2}}+\frac{i}{6 a^3 f (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{5/2}}+\frac{11 i}{48 a^3 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2}}+\frac{33 i}{64 a^3 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}-\frac{77 i}{256 a^3 c f (c-i c \tan (e+f x))^{3/2}}-\frac{231 i}{512 a^3 c^2 f \sqrt{c-i c \tan (e+f x)}}\\ \end{align*}

Mathematica [A]  time = 5.60807, size = 171, normalized size = 0.62 \[ -\frac{i e^{-6 i (e+f x)} \left (-350 e^{2 i (e+f x)}-1645 e^{4 i (e+f x)}+1433 e^{6 i (e+f x)}+3184 e^{8 i (e+f x)}+464 e^{10 i (e+f x)}+48 e^{12 i (e+f x)}-3465 e^{6 i (e+f x)} \sqrt{1+e^{2 i (e+f x)}} \tanh ^{-1}\left (\sqrt{1+e^{2 i (e+f x)}}\right )-40\right ) \sqrt{c-i c \tan (e+f x)}}{15360 a^3 c^3 f} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + I*a*Tan[e + f*x])^3*(c - I*c*Tan[e + f*x])^(5/2)),x]

[Out]

((-I/15360)*(-40 - 350*E^((2*I)*(e + f*x)) - 1645*E^((4*I)*(e + f*x)) + 1433*E^((6*I)*(e + f*x)) + 3184*E^((8*
I)*(e + f*x)) + 464*E^((10*I)*(e + f*x)) + 48*E^((12*I)*(e + f*x)) - 3465*E^((6*I)*(e + f*x))*Sqrt[1 + E^((2*I
)*(e + f*x))]*ArcTanh[Sqrt[1 + E^((2*I)*(e + f*x))]])*Sqrt[c - I*c*Tan[e + f*x]])/(a^3*c^3*E^((6*I)*(e + f*x))
*f)

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Maple [A]  time = 0.044, size = 178, normalized size = 0.7 \begin{align*}{\frac{2\,i{c}^{4}}{f{a}^{3}} \left ( -{\frac{1}{32\,{c}^{6}} \left ({\frac{1}{ \left ( -c-ic\tan \left ( fx+e \right ) \right ) ^{3}} \left ({\frac{71}{32} \left ( c-ic\tan \left ( fx+e \right ) \right ) ^{{\frac{5}{2}}}}-{\frac{59\,c}{6} \left ( c-ic\tan \left ( fx+e \right ) \right ) ^{{\frac{3}{2}}}}+{\frac{89\,{c}^{2}}{8}\sqrt{c-ic\tan \left ( fx+e \right ) }} \right ) }-{\frac{231\,\sqrt{2}}{64}{\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{5}{32\,{c}^{6}}{\frac{1}{\sqrt{c-ic\tan \left ( fx+e \right ) }}}}-{\frac{1}{24\,{c}^{5}} \left ( c-ic\tan \left ( fx+e \right ) \right ) ^{-{\frac{3}{2}}}}-{\frac{1}{80\,{c}^{4}} \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))^3/(c-I*c*tan(f*x+e))^(5/2),x)

[Out]

2*I/f/a^3*c^4*(-1/32/c^6*((71/32*(c-I*c*tan(f*x+e))^(5/2)-59/6*(c-I*c*tan(f*x+e))^(3/2)*c+89/8*c^2*(c-I*c*tan(
f*x+e))^(1/2))/(-c-I*c*tan(f*x+e))^3-231/64*2^(1/2)/c^(1/2)*arctanh(1/2*(c-I*c*tan(f*x+e))^(1/2)*2^(1/2)/c^(1/
2)))-5/32/c^6/(c-I*c*tan(f*x+e))^(1/2)-1/24/c^5/(c-I*c*tan(f*x+e))^(3/2)-1/80/c^4/(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))^3/(c-I*c*tan(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.71969, size = 1062, normalized size = 3.88 \begin{align*} \frac{{\left (3465 i \, \sqrt{\frac{1}{2}} a^{3} c^{3} f \sqrt{\frac{1}{a^{6} c^{5} f^{2}}} e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (\frac{{\left (\sqrt{2} \sqrt{\frac{1}{2}}{\left (59136 i \, a^{3} c^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + 59136 i \, a^{3} c^{2} f\right )} \sqrt{\frac{c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt{\frac{1}{a^{6} c^{5} f^{2}}} + 59136 i\right )} e^{\left (-i \, f x - i \, e\right )}}{65536 \, a^{3} c^{2} f}\right ) - 3465 i \, \sqrt{\frac{1}{2}} a^{3} c^{3} f \sqrt{\frac{1}{a^{6} c^{5} f^{2}}} e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (\frac{{\left (\sqrt{2} \sqrt{\frac{1}{2}}{\left (-59136 i \, a^{3} c^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} - 59136 i \, a^{3} c^{2} f\right )} \sqrt{\frac{c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt{\frac{1}{a^{6} c^{5} f^{2}}} + 59136 i\right )} e^{\left (-i \, f x - i \, e\right )}}{65536 \, a^{3} c^{2} f}\right ) + \sqrt{2} \sqrt{\frac{c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}{\left (-48 i \, e^{\left (12 i \, f x + 12 i \, e\right )} - 464 i \, e^{\left (10 i \, f x + 10 i \, e\right )} - 3184 i \, e^{\left (8 i \, f x + 8 i \, e\right )} - 1433 i \, e^{\left (6 i \, f x + 6 i \, e\right )} + 1645 i \, e^{\left (4 i \, f x + 4 i \, e\right )} + 350 i \, e^{\left (2 i \, f x + 2 i \, e\right )} + 40 i\right )}\right )} e^{\left (-6 i \, f x - 6 i \, e\right )}}{15360 \, a^{3} 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))^3/(c-I*c*tan(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

1/15360*(3465*I*sqrt(1/2)*a^3*c^3*f*sqrt(1/(a^6*c^5*f^2))*e^(6*I*f*x + 6*I*e)*log(1/65536*(sqrt(2)*sqrt(1/2)*(
59136*I*a^3*c^2*f*e^(2*I*f*x + 2*I*e) + 59136*I*a^3*c^2*f)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(1/(a^6*c^5*f
^2)) + 59136*I)*e^(-I*f*x - I*e)/(a^3*c^2*f)) - 3465*I*sqrt(1/2)*a^3*c^3*f*sqrt(1/(a^6*c^5*f^2))*e^(6*I*f*x +
6*I*e)*log(1/65536*(sqrt(2)*sqrt(1/2)*(-59136*I*a^3*c^2*f*e^(2*I*f*x + 2*I*e) - 59136*I*a^3*c^2*f)*sqrt(c/(e^(
2*I*f*x + 2*I*e) + 1))*sqrt(1/(a^6*c^5*f^2)) + 59136*I)*e^(-I*f*x - I*e)/(a^3*c^2*f)) + sqrt(2)*sqrt(c/(e^(2*I
*f*x + 2*I*e) + 1))*(-48*I*e^(12*I*f*x + 12*I*e) - 464*I*e^(10*I*f*x + 10*I*e) - 3184*I*e^(8*I*f*x + 8*I*e) -
1433*I*e^(6*I*f*x + 6*I*e) + 1645*I*e^(4*I*f*x + 4*I*e) + 350*I*e^(2*I*f*x + 2*I*e) + 40*I))*e^(-6*I*f*x - 6*I
*e)/(a^3*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))**3/(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 )}^{3}{\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))^3/(c-I*c*tan(f*x+e))^(5/2),x, algorithm="giac")

[Out]

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