3.182 \(\int \frac {\sec ^2(c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^3} \, dx\)

Optimal. Leaf size=62 \[ \frac {4 \sin (c+d x)}{a^3 d}+\frac {4 i \cos (c+d x)}{a^3 d}+\frac {i \sec (c+d x)}{a^3 d}-\frac {3 \tanh ^{-1}(\sin (c+d x))}{a^3 d} \]

[Out]

-3*arctanh(sin(d*x+c))/a^3/d+4*I*cos(d*x+c)/a^3/d+I*sec(d*x+c)/a^3/d+4*sin(d*x+c)/a^3/d

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Rubi [A]  time = 0.16, antiderivative size = 62, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 9, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.290, Rules used = {3092, 3090, 2637, 2638, 2592, 321, 206, 2590, 14} \[ \frac {4 \sin (c+d x)}{a^3 d}+\frac {4 i \cos (c+d x)}{a^3 d}+\frac {i \sec (c+d x)}{a^3 d}-\frac {3 \tanh ^{-1}(\sin (c+d x))}{a^3 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^2/(a*Cos[c + d*x] + I*a*Sin[c + d*x])^3,x]

[Out]

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

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

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])

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 2590

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*tan[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> -Dist[f^(-1), Subst[Int[(1 - x^2
)^((m + n - 1)/2)/x^n, x], x, Cos[e + f*x]], x] /; FreeQ[{e, f}, x] && IntegersQ[m, n, (m + n - 1)/2]

Rule 2592

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> With[{ff = FreeFactors[S
in[e + f*x], x]}, Dist[ff/f, Subst[Int[(ff*x)^(m + n)/(a^2 - ff^2*x^2)^((n + 1)/2), x], x, (a*Sin[e + f*x])/ff
], x]] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n + 1)/2]

Rule 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 2638

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

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]

Rubi steps

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

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Mathematica [A]  time = 0.33, size = 109, normalized size = 1.76 \[ -\frac {i \sec ^3(c+d x) (\cos (d x)+i \sin (d x))^3 \left ((\tan (c+d x)-5 i) (\cos (2 c-d x)+i \sin (2 c-d x))+6 (\cos (3 c)+i \sin (3 c)) \tanh ^{-1}\left (\cos (c) \tan \left (\frac {d x}{2}\right )+\sin (c)\right )\right )}{a^3 d (\tan (c+d x)-i)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^2/(a*Cos[c + d*x] + I*a*Sin[c + d*x])^3,x]

[Out]

((-I)*Sec[c + d*x]^3*(Cos[d*x] + I*Sin[d*x])^3*(6*ArcTanh[Sin[c] + Cos[c]*Tan[(d*x)/2]]*(Cos[3*c] + I*Sin[3*c]
) + (Cos[2*c - d*x] + I*Sin[2*c - d*x])*(-5*I + Tan[c + d*x])))/(a^3*d*(-I + Tan[c + d*x])^3)

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fricas [A]  time = 0.61, size = 112, normalized size = 1.81 \[ -\frac {3 \, {\left (e^{\left (3 i \, d x + 3 i \, c\right )} + e^{\left (i \, d x + i \, c\right )}\right )} \log \left (e^{\left (i \, d x + i \, c\right )} + i\right ) - 3 \, {\left (e^{\left (3 i \, d x + 3 i \, c\right )} + e^{\left (i \, d x + i \, c\right )}\right )} \log \left (e^{\left (i \, d x + i \, c\right )} - i\right ) - 6 i \, e^{\left (2 i \, d x + 2 i \, c\right )} - 4 i}{a^{3} d e^{\left (3 i \, d x + 3 i \, c\right )} + a^{3} d e^{\left (i \, d x + i \, c\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2/(a*cos(d*x+c)+I*a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

-(3*(e^(3*I*d*x + 3*I*c) + e^(I*d*x + I*c))*log(e^(I*d*x + I*c) + I) - 3*(e^(3*I*d*x + 3*I*c) + e^(I*d*x + I*c
))*log(e^(I*d*x + I*c) - I) - 6*I*e^(2*I*d*x + 2*I*c) - 4*I)/(a^3*d*e^(3*I*d*x + 3*I*c) + a^3*d*e^(I*d*x + I*c
))

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giac [A]  time = 0.62, size = 110, normalized size = 1.77 \[ -\frac {\frac {3 \, \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{a^{3}} - \frac {3 \, \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}{a^{3}} - \frac {2 \, {\left (4 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 5\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + i\right )} a^{3}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2/(a*cos(d*x+c)+I*a*sin(d*x+c))^3,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.28, size = 108, normalized size = 1.74 \[ -\frac {i}{a^{3} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {3 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a^{3} d}+\frac {i}{a^{3} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {3 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a^{3} d}+\frac {8}{a^{3} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^2/(a*cos(d*x+c)+I*a*sin(d*x+c))^3,x)

[Out]

-I/a^3/d/(tan(1/2*d*x+1/2*c)-1)+3/a^3/d*ln(tan(1/2*d*x+1/2*c)-1)+I/a^3/d/(tan(1/2*d*x+1/2*c)+1)-3/a^3/d*ln(tan
(1/2*d*x+1/2*c)+1)+8/a^3/d/(tan(1/2*d*x+1/2*c)-I)

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maxima [B]  time = 0.74, size = 329, normalized size = 5.31 \[ \frac {{\left (6 \, \cos \left (3 \, d x + 3 \, c\right ) + 6 \, \cos \left (d x + c\right ) + 6 i \, \sin \left (3 \, d x + 3 \, c\right ) + 6 i \, \sin \left (d x + c\right )\right )} \arctan \left (\cos \left (d x + c\right ), \sin \left (d x + c\right ) + 1\right ) + {\left (6 \, \cos \left (3 \, d x + 3 \, c\right ) + 6 \, \cos \left (d x + c\right ) + 6 i \, \sin \left (3 \, d x + 3 \, c\right ) + 6 i \, \sin \left (d x + c\right )\right )} \arctan \left (\cos \left (d x + c\right ), -\sin \left (d x + c\right ) + 1\right ) - {\left (-3 i \, \cos \left (3 \, d x + 3 \, c\right ) - 3 i \, \cos \left (d x + c\right ) + 3 \, \sin \left (3 \, d x + 3 \, c\right ) + 3 \, \sin \left (d x + c\right )\right )} \log \left (\cos \left (d x + c\right )^{2} + \sin \left (d x + c\right )^{2} + 2 \, \sin \left (d x + c\right ) + 1\right ) - {\left (3 i \, \cos \left (3 \, d x + 3 \, c\right ) + 3 i \, \cos \left (d x + c\right ) - 3 \, \sin \left (3 \, d x + 3 \, c\right ) - 3 \, \sin \left (d x + c\right )\right )} \log \left (\cos \left (d x + c\right )^{2} + \sin \left (d x + c\right )^{2} - 2 \, \sin \left (d x + c\right ) + 1\right ) + 12 \, \cos \left (2 \, d x + 2 \, c\right ) + 12 i \, \sin \left (2 \, d x + 2 \, c\right ) + 8}{{\left (-2 i \, a^{3} \cos \left (3 \, d x + 3 \, c\right ) - 2 i \, a^{3} \cos \left (d x + c\right ) + 2 \, a^{3} \sin \left (3 \, d x + 3 \, c\right ) + 2 \, a^{3} \sin \left (d x + c\right )\right )} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2/(a*cos(d*x+c)+I*a*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

((6*cos(3*d*x + 3*c) + 6*cos(d*x + c) + 6*I*sin(3*d*x + 3*c) + 6*I*sin(d*x + c))*arctan2(cos(d*x + c), sin(d*x
+ c) + 1) + (6*cos(3*d*x + 3*c) + 6*cos(d*x + c) + 6*I*sin(3*d*x + 3*c) + 6*I*sin(d*x + c))*arctan2(cos(d*x +
c), -sin(d*x + c) + 1) - (-3*I*cos(3*d*x + 3*c) - 3*I*cos(d*x + c) + 3*sin(3*d*x + 3*c) + 3*sin(d*x + c))*log
(cos(d*x + c)^2 + sin(d*x + c)^2 + 2*sin(d*x + c) + 1) - (3*I*cos(3*d*x + 3*c) + 3*I*cos(d*x + c) - 3*sin(3*d*
x + 3*c) - 3*sin(d*x + c))*log(cos(d*x + c)^2 + sin(d*x + c)^2 - 2*sin(d*x + c) + 1) + 12*cos(2*d*x + 2*c) + 1
2*I*sin(2*d*x + 2*c) + 8)/((-2*I*a^3*cos(3*d*x + 3*c) - 2*I*a^3*cos(d*x + c) + 2*a^3*sin(3*d*x + 3*c) + 2*a^3*
sin(d*x + c))*d)

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mupad [B]  time = 0.99, size = 105, normalized size = 1.69 \[ -\frac {6\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^3\,d}-\frac {\frac {2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{a^3}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,8{}\mathrm {i}}{a^3}-\frac {10{}\mathrm {i}}{a^3}}{d\,\left (-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,1{}\mathrm {i}-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(c + d*x)^2*(a*cos(c + d*x) + a*sin(c + d*x)*1i)^3),x)

[Out]

- (6*atanh(tan(c/2 + (d*x)/2)))/(a^3*d) - ((tan(c/2 + (d*x)/2)^2*8i)/a^3 - 10i/a^3 + (2*tan(c/2 + (d*x)/2))/a^
3)/(d*(tan(c/2 + (d*x)/2)*1i - tan(c/2 + (d*x)/2)^2 - tan(c/2 + (d*x)/2)^3*1i + 1))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sec(c + d*x)**2/(-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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