3.141 \(\int \frac{\sec ^7(c+d x)}{(a+i a \tan (c+d x))^3} \, dx\)

Optimal. Leaf size=93 \[ -\frac{5 i \sec ^3(c+d x)}{3 a^3 d}+\frac{5 \tanh ^{-1}(\sin (c+d x))}{2 a^3 d}+\frac{5 \tan (c+d x) \sec (c+d x)}{2 a^3 d}-\frac{2 i \sec ^5(c+d x)}{a d (a+i a \tan (c+d x))^2} \]

[Out]

(5*ArcTanh[Sin[c + d*x]])/(2*a^3*d) - (((5*I)/3)*Sec[c + d*x]^3)/(a^3*d) + (5*Sec[c + d*x]*Tan[c + d*x])/(2*a^
3*d) - ((2*I)*Sec[c + d*x]^5)/(a*d*(a + I*a*Tan[c + d*x])^2)

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Rubi [A]  time = 0.0981361, antiderivative size = 93, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {3500, 3501, 3768, 3770} \[ -\frac{5 i \sec ^3(c+d x)}{3 a^3 d}+\frac{5 \tanh ^{-1}(\sin (c+d x))}{2 a^3 d}+\frac{5 \tan (c+d x) \sec (c+d x)}{2 a^3 d}-\frac{2 i \sec ^5(c+d x)}{a d (a+i a \tan (c+d x))^2} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^7/(a + I*a*Tan[c + d*x])^3,x]

[Out]

(5*ArcTanh[Sin[c + d*x]])/(2*a^3*d) - (((5*I)/3)*Sec[c + d*x]^3)/(a^3*d) + (5*Sec[c + d*x]*Tan[c + d*x])/(2*a^
3*d) - ((2*I)*Sec[c + d*x]^5)/(a*d*(a + I*a*Tan[c + d*x])^2)

Rule 3500

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(2*d^2
*(d*Sec[e + f*x])^(m - 2)*(a + b*Tan[e + f*x])^(n + 1))/(b*f*(m + 2*n)), x] - Dist[(d^2*(m - 2))/(b^2*(m + 2*n
)), Int[(d*Sec[e + f*x])^(m - 2)*(a + b*Tan[e + f*x])^(n + 2), x], x] /; FreeQ[{a, b, d, e, f, m}, x] && EqQ[a
^2 + b^2, 0] && LtQ[n, -1] && ((ILtQ[n/2, 0] && IGtQ[m - 1/2, 0]) || EqQ[n, -2] || IGtQ[m + n, 0] || (Integers
Q[n, m + 1/2] && GtQ[2*m + n + 1, 0])) && IntegerQ[2*m]

Rule 3501

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(d^2*
(d*Sec[e + f*x])^(m - 2)*(a + b*Tan[e + f*x])^(n + 1))/(b*f*(m + n - 1)), x] + Dist[(d^2*(m - 2))/(a*(m + n -
1)), Int[(d*Sec[e + f*x])^(m - 2)*(a + b*Tan[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2
+ b^2, 0] && LtQ[n, 0] && GtQ[m, 1] &&  !ILtQ[m + n, 0] && NeQ[m + n - 1, 0] && IntegersQ[2*m, 2*n]

Rule 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[2*n]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin{align*} \int \frac{\sec ^7(c+d x)}{(a+i a \tan (c+d x))^3} \, dx &=-\frac{2 i \sec ^5(c+d x)}{a d (a+i a \tan (c+d x))^2}+\frac{5 \int \frac{\sec ^5(c+d x)}{a+i a \tan (c+d x)} \, dx}{a^2}\\ &=-\frac{5 i \sec ^3(c+d x)}{3 a^3 d}-\frac{2 i \sec ^5(c+d x)}{a d (a+i a \tan (c+d x))^2}+\frac{5 \int \sec ^3(c+d x) \, dx}{a^3}\\ &=-\frac{5 i \sec ^3(c+d x)}{3 a^3 d}+\frac{5 \sec (c+d x) \tan (c+d x)}{2 a^3 d}-\frac{2 i \sec ^5(c+d x)}{a d (a+i a \tan (c+d x))^2}+\frac{5 \int \sec (c+d x) \, dx}{2 a^3}\\ &=\frac{5 \tanh ^{-1}(\sin (c+d x))}{2 a^3 d}-\frac{5 i \sec ^3(c+d x)}{3 a^3 d}+\frac{5 \sec (c+d x) \tan (c+d x)}{2 a^3 d}-\frac{2 i \sec ^5(c+d x)}{a d (a+i a \tan (c+d x))^2}\\ \end{align*}

Mathematica [A]  time = 0.379595, size = 63, normalized size = 0.68 \[ \frac{60 \tanh ^{-1}\left (\cos (c) \tan \left (\frac{d x}{2}\right )+\sin (c)\right )-i \sec ^3(c+d x) (-9 i \sin (2 (c+d x))+24 \cos (2 (c+d x))+20)}{12 a^3 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^7/(a + I*a*Tan[c + d*x])^3,x]

[Out]

(60*ArcTanh[Sin[c] + Cos[c]*Tan[(d*x)/2]] - I*Sec[c + d*x]^3*(20 + 24*Cos[2*(c + d*x)] - (9*I)*Sin[2*(c + d*x)
]))/(12*a^3*d)

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Maple [B]  time = 0.098, size = 258, normalized size = 2.8 \begin{align*}{\frac{{\frac{i}{3}}}{d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-3}}+{\frac{3}{2\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-2}}-{\frac{{\frac{i}{2}}}{d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-2}}-{\frac{3}{2\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}-{\frac{{\frac{7\,i}{2}}}{d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}+{\frac{5}{2\,d{a}^{3}}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) }-{\frac{{\frac{i}{3}}}{d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-3}}-{\frac{3}{2\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-2}}-{\frac{{\frac{i}{2}}}{d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-2}}-{\frac{3}{2\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}+{\frac{{\frac{7\,i}{2}}}{d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}-{\frac{5}{2\,d{a}^{3}}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^7/(a+I*a*tan(d*x+c))^3,x)

[Out]

1/3*I/a^3/d/(tan(1/2*d*x+1/2*c)+1)^3+3/2/a^3/d/(tan(1/2*d*x+1/2*c)+1)^2-1/2*I/a^3/d/(tan(1/2*d*x+1/2*c)+1)^2-3
/2/a^3/d/(tan(1/2*d*x+1/2*c)+1)-7/2*I/a^3/d/(tan(1/2*d*x+1/2*c)+1)+5/2/a^3/d*ln(tan(1/2*d*x+1/2*c)+1)-1/3*I/a^
3/d/(tan(1/2*d*x+1/2*c)-1)^3-3/2/a^3/d/(tan(1/2*d*x+1/2*c)-1)^2-1/2*I/a^3/d/(tan(1/2*d*x+1/2*c)-1)^2-3/2/a^3/d
/(tan(1/2*d*x+1/2*c)-1)+7/2*I/a^3/d/(tan(1/2*d*x+1/2*c)-1)-5/2/a^3/d*ln(tan(1/2*d*x+1/2*c)-1)

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Maxima [B]  time = 1.00669, size = 290, normalized size = 3.12 \begin{align*} \frac{\frac{4 \,{\left (-\frac{9 i \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{48 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{18 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac{9 i \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + 22\right )}}{6 i \, a^{3} - \frac{18 i \, a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{18 i \, a^{3} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac{6 i \, a^{3} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}}} + \frac{5 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{3}} - \frac{5 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{3}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^7/(a+I*a*tan(d*x+c))^3,x, algorithm="maxima")

[Out]

1/2*(4*(-9*I*sin(d*x + c)/(cos(d*x + c) + 1) - 48*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 18*sin(d*x + c)^4/(cos
(d*x + c) + 1)^4 + 9*I*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 22)/(6*I*a^3 - 18*I*a^3*sin(d*x + c)^2/(cos(d*x +
c) + 1)^2 + 18*I*a^3*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 6*I*a^3*sin(d*x + c)^6/(cos(d*x + c) + 1)^6) + 5*l
og(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a^3 - 5*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a^3)/d

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Fricas [B]  time = 2.49059, size = 521, normalized size = 5.6 \begin{align*} \frac{15 \,{\left (e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )} \log \left (e^{\left (i \, d x + i \, c\right )} + i\right ) - 15 \,{\left (e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )} \log \left (e^{\left (i \, d x + i \, c\right )} - i\right ) - 30 i \, e^{\left (5 i \, d x + 5 i \, c\right )} - 80 i \, e^{\left (3 i \, d x + 3 i \, c\right )} - 66 i \, e^{\left (i \, d x + i \, c\right )}}{6 \,{\left (a^{3} d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, a^{3} d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{3} d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^7/(a+I*a*tan(d*x+c))^3,x, algorithm="fricas")

[Out]

1/6*(15*(e^(6*I*d*x + 6*I*c) + 3*e^(4*I*d*x + 4*I*c) + 3*e^(2*I*d*x + 2*I*c) + 1)*log(e^(I*d*x + I*c) + I) - 1
5*(e^(6*I*d*x + 6*I*c) + 3*e^(4*I*d*x + 4*I*c) + 3*e^(2*I*d*x + 2*I*c) + 1)*log(e^(I*d*x + I*c) - I) - 30*I*e^
(5*I*d*x + 5*I*c) - 80*I*e^(3*I*d*x + 3*I*c) - 66*I*e^(I*d*x + I*c))/(a^3*d*e^(6*I*d*x + 6*I*c) + 3*a^3*d*e^(4
*I*d*x + 4*I*c) + 3*a^3*d*e^(2*I*d*x + 2*I*c) + a^3*d)

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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(sec(d*x+c)**7/(a+I*a*tan(d*x+c))**3,x)

[Out]

Exception raised: AttributeError

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Giac [A]  time = 1.2489, size = 154, normalized size = 1.66 \begin{align*} \frac{\frac{15 \, \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right )}{a^{3}} - \frac{15 \, \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right )}{a^{3}} - \frac{2 \,{\left (9 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 18 i \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 48 i \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 9 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 22 i\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )}^{3} a^{3}}}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^7/(a+I*a*tan(d*x+c))^3,x, algorithm="giac")

[Out]

1/6*(15*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^3 - 15*log(abs(tan(1/2*d*x + 1/2*c) - 1))/a^3 - 2*(9*tan(1/2*d*x
+ 1/2*c)^5 - 18*I*tan(1/2*d*x + 1/2*c)^4 + 48*I*tan(1/2*d*x + 1/2*c)^2 - 9*tan(1/2*d*x + 1/2*c) - 22*I)/((tan(
1/2*d*x + 1/2*c)^2 - 1)^3*a^3))/d