3.184 \(\int \frac {\sec ^4(c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^3} \, dx\)

Optimal. Leaf size=76 \[ \frac {i \sec ^3(c+d x)}{3 a^3 d}-\frac {4 i \sec (c+d x)}{a^3 d}+\frac {5 \tanh ^{-1}(\sin (c+d x))}{2 a^3 d}-\frac {3 \tan (c+d x) \sec (c+d x)}{2 a^3 d} \]

[Out]

5/2*arctanh(sin(d*x+c))/a^3/d-4*I*sec(d*x+c)/a^3/d+1/3*I*sec(d*x+c)^3/a^3/d-3/2*sec(d*x+c)*tan(d*x+c)/a^3/d

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Rubi [A]  time = 0.18, antiderivative size = 76, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 6, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {3092, 3090, 3770, 2606, 8, 2611} \[ \frac {i \sec ^3(c+d x)}{3 a^3 d}-\frac {4 i \sec (c+d x)}{a^3 d}+\frac {5 \tanh ^{-1}(\sin (c+d x))}{2 a^3 d}-\frac {3 \tan (c+d x) \sec (c+d x)}{2 a^3 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^4/(a*Cos[c + d*x] + I*a*Sin[c + d*x])^3,x]

[Out]

(5*ArcTanh[Sin[c + d*x]])/(2*a^3*d) - ((4*I)*Sec[c + d*x])/(a^3*d) + ((I/3)*Sec[c + d*x]^3)/(a^3*d) - (3*Sec[c
+ d*x]*Tan[c + d*x])/(2*a^3*d)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2606

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a/f, Subst[
Int[(a*x)^(m - 1)*(-1 + x^2)^((n - 1)/2), x], x, Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n -
1)/2] &&  !(IntegerQ[m/2] && LtQ[0, m, n + 1])

Rule 2611

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

Rule 3090

Int[cos[(c_.) + (d_.)*(x_)]^(m_.)*(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_.), x_Sym
bol] :> Int[ExpandTrig[cos[c + d*x]^m*(a*cos[c + d*x] + b*sin[c + d*x])^n, x], x] /; FreeQ[{a, b, c, d}, x] &&
IntegerQ[m] && IGtQ[n, 0]

Rule 3092

Int[cos[(c_.) + (d_.)*(x_)]^(m_.)*(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symb
ol] :> Dist[a^n*b^n, Int[Cos[c + d*x]^m/(b*Cos[c + d*x] + a*Sin[c + d*x])^n, x], x] /; FreeQ[{a, b, c, d, m},
x] && EqQ[a^2 + b^2, 0] && ILtQ[n, 0]

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 ^4(c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^3} \, dx &=\frac {i \int \sec ^4(c+d x) (i a \cos (c+d x)+a \sin (c+d x))^3 \, dx}{a^6}\\ &=\frac {i \int \left (-i a^3 \sec (c+d x)-3 a^3 \sec (c+d x) \tan (c+d x)+3 i a^3 \sec (c+d x) \tan ^2(c+d x)+a^3 \sec (c+d x) \tan ^3(c+d x)\right ) \, dx}{a^6}\\ &=\frac {i \int \sec (c+d x) \tan ^3(c+d x) \, dx}{a^3}-\frac {(3 i) \int \sec (c+d x) \tan (c+d x) \, dx}{a^3}+\frac {\int \sec (c+d x) \, dx}{a^3}-\frac {3 \int \sec (c+d x) \tan ^2(c+d x) \, dx}{a^3}\\ &=\frac {\tanh ^{-1}(\sin (c+d x))}{a^3 d}-\frac {3 \sec (c+d x) \tan (c+d x)}{2 a^3 d}+\frac {3 \int \sec (c+d x) \, dx}{2 a^3}+\frac {i \operatorname {Subst}\left (\int \left (-1+x^2\right ) \, dx,x,\sec (c+d x)\right )}{a^3 d}-\frac {(3 i) \operatorname {Subst}(\int 1 \, dx,x,\sec (c+d x))}{a^3 d}\\ &=\frac {5 \tanh ^{-1}(\sin (c+d x))}{2 a^3 d}-\frac {4 i \sec (c+d x)}{a^3 d}+\frac {i \sec ^3(c+d x)}{3 a^3 d}-\frac {3 \sec (c+d x) \tan (c+d x)}{2 a^3 d}\\ \end {align*}

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Mathematica [A]  time = 0.47, size = 64, normalized size = 0.84 \[ \frac {i \left (\sec ^3(c+d x) (9 i \sin (2 (c+d x))-24 \cos (2 (c+d x))-20)-60 i \tanh ^{-1}\left (\cos (c) \tan \left (\frac {d x}{2}\right )+\sin (c)\right )\right )}{12 a^3 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^4/(a*Cos[c + d*x] + I*a*Sin[c + d*x])^3,x]

[Out]

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

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fricas [B]  time = 0.46, size = 182, normalized size = 2.39 \[ \frac {15 \, {\left (e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )} \log \left (e^{\left (i \, d x + i \, c\right )} + i\right ) - 15 \, {\left (e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )} \log \left (e^{\left (i \, d x + i \, c\right )} - i\right ) - 30 i \, e^{\left (5 i \, d x + 5 i \, c\right )} - 80 i \, e^{\left (3 i \, d x + 3 i \, c\right )} - 66 i \, e^{\left (i \, d x + i \, c\right )}}{6 \, {\left (a^{3} d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, a^{3} d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{3} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4/(a*cos(d*x+c)+I*a*sin(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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giac [A]  time = 0.28, size = 112, normalized size = 1.47 \[ \frac {\frac {15 \, \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{a^{3}} - \frac {15 \, \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}{a^{3}} - \frac {2 \, {\left (9 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 18 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 48 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 9 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 22 i\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3} a^{3}}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4/(a*cos(d*x+c)+I*a*sin(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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maple [B]  time = 0.29, size = 258, normalized size = 3.39 \[ -\frac {i}{3 a^{3} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {3}{2 a^{3} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {i}{2 a^{3} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {3}{2 a^{3} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {7 i}{2 a^{3} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {5 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 a^{3} d}+\frac {i}{3 a^{3} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}+\frac {3}{2 a^{3} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {i}{2 a^{3} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {3}{2 a^{3} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {7 i}{2 a^{3} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {5 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 a^{3} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^4/(a*cos(d*x+c)+I*a*sin(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 = 0.37, size = 215, normalized size = 2.83 \[ \frac {\frac {4 \, {\left (-\frac {9 i \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {48 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {18 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {9 i \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + 22\right )}}{6 i \, a^{3} - \frac {18 i \, a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {18 i \, a^{3} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {6 i \, a^{3} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}}} + \frac {5 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{3}} - \frac {5 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{3}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4/(a*cos(d*x+c)+I*a*sin(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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mupad [B]  time = 2.72, size = 135, normalized size = 1.78 \[ \frac {5\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^3\,d}+\frac {\frac {3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{a^3}-\frac {3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{a^3}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,16{}\mathrm {i}}{a^3}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,6{}\mathrm {i}}{a^3}+\frac {22{}\mathrm {i}}{3\,a^3}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(c + d*x)^4*(a*cos(c + d*x) + a*sin(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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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {\sec ^{4}{\left (c + d x \right )}}{- i \sin ^{3}{\left (c + d x \right )} - 3 \sin ^{2}{\left (c + d x \right )} \cos {\left (c + d x \right )} + 3 i \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )} + \cos ^{3}{\left (c + d x \right )}}\, dx}{a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**4/(a*cos(d*x+c)+I*a*sin(d*x+c))**3,x)

[Out]

Integral(sec(c + d*x)**4/(-I*sin(c + d*x)**3 - 3*sin(c + d*x)**2*cos(c + d*x) + 3*I*sin(c + d*x)*cos(c + d*x)*
*2 + cos(c + d*x)**3), x)/a**3

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