3.2.41 \(\int \frac {\sec ^7(c+d x)}{(a+i a \tan (c+d x))^3} \, dx\) [141]

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

[Out]

5/2*arctanh(sin(d*x+c))/a^3/d-5/3*I*sec(d*x+c)^3/a^3/d+5/2*sec(d*x+c)*tan(d*x+c)/a^3/d-2*I*sec(d*x+c)^5/a/d/(a
+I*a*tan(d*x+c))^2

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Rubi [A]
time = 0.08, antiderivative size = 93, normalized size of antiderivative = 1.00, 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 = {3581, 3582, 3853, 3855} \begin {gather*} -\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} \end {gather*}

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 3581

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 3582

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 3853

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 3855

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*}

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Mathematica [A]
time = 0.49, size = 63, normalized size = 0.68 \begin {gather*} \frac {60 \tanh ^{-1}\left (\sin (c)+\cos (c) \tan \left (\frac {d x}{2}\right )\right )-i \sec ^3(c+d x) (20+24 \cos (2 (c+d x))-9 i \sin (2 (c+d x)))}{12 a^3 d} \end {gather*}

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 [A]
time = 0.28, size = 138, normalized size = 1.48

method	result	size
risch	\(-\frac {i \left (15 \,{\mathrm e}^{5 i \left (d x +c \right )}+40 \,{\mathrm e}^{3 i \left (d x +c \right )}+33 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{3 d \,a^{3} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3}}+\frac {5 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 a^{3} d}-\frac {5 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 a^{3} d}\)	\(100\)
derivativedivides	\(\frac {-\frac {i}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}+\frac {2 \left (-\frac {3}{4}-\frac {i}{4}\right )}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {2 \left (-\frac {3}{4}+\frac {7 i}{4}\right )}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}-\frac {5 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2}+\frac {i}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}+\frac {2 \left (\frac {3}{4}-\frac {i}{4}\right )}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {2 \left (-\frac {3}{4}-\frac {7 i}{4}\right )}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}+\frac {5 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2}}{a^{3} d}\)	\(138\)
default	\(\frac {-\frac {i}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}+\frac {2 \left (-\frac {3}{4}-\frac {i}{4}\right )}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {2 \left (-\frac {3}{4}+\frac {7 i}{4}\right )}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}-\frac {5 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2}+\frac {i}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}+\frac {2 \left (\frac {3}{4}-\frac {i}{4}\right )}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {2 \left (-\frac {3}{4}-\frac {7 i}{4}\right )}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}+\frac {5 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2}}{a^{3} d}\)	\(138\)

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,method=_RETURNVERBOSE)

[Out]

2/d/a^3*(-1/6*I/(tan(1/2*d*x+1/2*c)-1)^3-(3/4+1/4*I)/(tan(1/2*d*x+1/2*c)-1)^2+(-3/4+7/4*I)/(tan(1/2*d*x+1/2*c)
-1)-5/4*ln(tan(1/2*d*x+1/2*c)-1)+1/6*I/(tan(1/2*d*x+1/2*c)+1)^3+(3/4-1/4*I)/(tan(1/2*d*x+1/2*c)+1)^2-(3/4+7/4*
I)/(tan(1/2*d*x+1/2*c)+1)+5/4*ln(tan(1/2*d*x+1/2*c)+1))

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Maxima [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 215 vs. \(2 (81) = 162\).
time = 0.30, size = 215, normalized size = 2.31 \begin {gather*} \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 {gather*}

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] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 182 vs. \(2 (81) = 162\).
time = 0.38, size = 182, normalized size = 1.96 \begin {gather*} \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 {gather*}

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]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {i \int \frac {\sec ^{7}{\left (c + d x \right )}}{\tan ^{3}{\left (c + d x \right )} - 3 i \tan ^{2}{\left (c + d x \right )} - 3 \tan {\left (c + d x \right )} + i}\, dx}{a^{3}} \end {gather*}

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]

I*Integral(sec(c + d*x)**7/(tan(c + d*x)**3 - 3*I*tan(c + d*x)**2 - 3*tan(c + d*x) + I), x)/a**3

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Giac [A]
time = 0.72, size = 112, normalized size = 1.20 \begin {gather*} \frac {\frac {15 \, \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{a^{3}} - \frac {15 \, \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\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 {gather*}

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(tan(1/2*d*x + 1/2*c) + 1)/a^3 - 15*log(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

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Mupad [B]
time = 5.44, size = 135, normalized size = 1.45 \begin {gather*} \frac {5\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^3\,d}+\frac {\frac {3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{a^3}-\frac {3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{a^3}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,16{}\mathrm {i}}{a^3}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,6{}\mathrm {i}}{a^3}+\frac {22{}\mathrm {i}}{3\,a^3}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(c + d*x)^7*(a + a*tan(c + d*x)*1i)^3),x)

[Out]

(5*atanh(tan(c/2 + (d*x)/2)))/(a^3*d) + ((tan(c/2 + (d*x)/2)^4*6i)/a^3 - (tan(c/2 + (d*x)/2)^2*16i)/a^3 - (3*t
an(c/2 + (d*x)/2)^5)/a^3 + 22i/(3*a^3) + (3*tan(c/2 + (d*x)/2))/a^3)/(d*(3*tan(c/2 + (d*x)/2)^2 - 3*tan(c/2 +
(d*x)/2)^4 + tan(c/2 + (d*x)/2)^6 - 1))

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