3.1.45 \(\int \sec ^3(c+d x) (a+i a \tan (c+d x))^3 \, dx\) [45]

Optimal. Leaf size=127 \[ \frac {7 a^3 \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {7 i a^3 \sec ^3(c+d x)}{12 d}+\frac {7 a^3 \sec (c+d x) \tan (c+d x)}{8 d}+\frac {i a \sec ^3(c+d x) (a+i a \tan (c+d x))^2}{5 d}+\frac {7 i \sec ^3(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{20 d} \]

[Out]

7/8*a^3*arctanh(sin(d*x+c))/d+7/12*I*a^3*sec(d*x+c)^3/d+7/8*a^3*sec(d*x+c)*tan(d*x+c)/d+1/5*I*a*sec(d*x+c)^3*(
a+I*a*tan(d*x+c))^2/d+7/20*I*sec(d*x+c)^3*(a^3+I*a^3*tan(d*x+c))/d

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Rubi [A]
time = 0.10, antiderivative size = 127, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3579, 3567, 3853, 3855} \begin {gather*} \frac {7 i a^3 \sec ^3(c+d x)}{12 d}+\frac {7 a^3 \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {7 i \sec ^3(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{20 d}+\frac {7 a^3 \tan (c+d x) \sec (c+d x)}{8 d}+\frac {i a \sec ^3(c+d x) (a+i a \tan (c+d x))^2}{5 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(7*a^3*ArcTanh[Sin[c + d*x]])/(8*d) + (((7*I)/12)*a^3*Sec[c + d*x]^3)/d + (7*a^3*Sec[c + d*x]*Tan[c + d*x])/(8
*d) + ((I/5)*a*Sec[c + d*x]^3*(a + I*a*Tan[c + d*x])^2)/d + (((7*I)/20)*Sec[c + d*x]^3*(a^3 + I*a^3*Tan[c + d*
x]))/d

Rule 3567

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[b*((d*Sec[
e + f*x])^m/(f*m)), x] + Dist[a, Int[(d*Sec[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e, f, m}, x] && (IntegerQ[2
*m] || NeQ[a^2 + b^2, 0])

Rule 3579

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[b*(d*
Sec[e + f*x])^m*((a + b*Tan[e + f*x])^(n - 1)/(f*(m + n - 1))), x] + Dist[a*((m + 2*n - 2)/(m + n - 1)), Int[(
d*Sec[e + f*x])^m*(a + b*Tan[e + f*x])^(n - 1), x], x] /; FreeQ[{a, b, d, e, f, m}, x] && EqQ[a^2 + b^2, 0] &&
GtQ[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 \sec ^3(c+d x) (a+i a \tan (c+d x))^3 \, dx &=\frac {i a \sec ^3(c+d x) (a+i a \tan (c+d x))^2}{5 d}+\frac {1}{5} (7 a) \int \sec ^3(c+d x) (a+i a \tan (c+d x))^2 \, dx\\ &=\frac {i a \sec ^3(c+d x) (a+i a \tan (c+d x))^2}{5 d}+\frac {7 i \sec ^3(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{20 d}+\frac {1}{4} \left (7 a^2\right ) \int \sec ^3(c+d x) (a+i a \tan (c+d x)) \, dx\\ &=\frac {7 i a^3 \sec ^3(c+d x)}{12 d}+\frac {i a \sec ^3(c+d x) (a+i a \tan (c+d x))^2}{5 d}+\frac {7 i \sec ^3(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{20 d}+\frac {1}{4} \left (7 a^3\right ) \int \sec ^3(c+d x) \, dx\\ &=\frac {7 i a^3 \sec ^3(c+d x)}{12 d}+\frac {7 a^3 \sec (c+d x) \tan (c+d x)}{8 d}+\frac {i a \sec ^3(c+d x) (a+i a \tan (c+d x))^2}{5 d}+\frac {7 i \sec ^3(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{20 d}+\frac {1}{8} \left (7 a^3\right ) \int \sec (c+d x) \, dx\\ &=\frac {7 a^3 \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {7 i a^3 \sec ^3(c+d x)}{12 d}+\frac {7 a^3 \sec (c+d x) \tan (c+d x)}{8 d}+\frac {i a \sec ^3(c+d x) (a+i a \tan (c+d x))^2}{5 d}+\frac {7 i \sec ^3(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{20 d}\\ \end {align*}

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Mathematica [A]
time = 0.64, size = 102, normalized size = 0.80 \begin {gather*} \frac {a^3 (\cos (3 d x)+i \sin (3 d x)) \left (1680 \tanh ^{-1}\left (\sin (c)+\cos (c) \tan \left (\frac {d x}{2}\right )\right )+\sec ^5(c+d x) (448 i+640 i \cos (2 (c+d x))-150 \sin (2 (c+d x))+105 \sin (4 (c+d x)))\right )}{960 d (\cos (d x)+i \sin (d x))^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*(Cos[3*d*x] + I*Sin[3*d*x])*(1680*ArcTanh[Sin[c] + Cos[c]*Tan[(d*x)/2]] + Sec[c + d*x]^5*(448*I + (640*I)
*Cos[2*(c + d*x)] - 150*Sin[2*(c + d*x)] + 105*Sin[4*(c + d*x)])))/(960*d*(Cos[d*x] + I*Sin[d*x])^3)

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Maple [A]
time = 0.30, size = 200, normalized size = 1.57

method	result	size
risch	\(-\frac {i a^{3} \left (105 \,{\mathrm e}^{9 i \left (d x +c \right )}-790 \,{\mathrm e}^{7 i \left (d x +c \right )}-896 \,{\mathrm e}^{5 i \left (d x +c \right )}-490 \,{\mathrm e}^{3 i \left (d x +c \right )}-105 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{60 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{5}}+\frac {7 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{8 d}-\frac {7 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{8 d}\)	\(122\)
derivativedivides	\(\frac {-i a^{3} \left (\frac {\sin ^{4}\left (d x +c \right )}{5 \cos \left (d x +c \right )^{5}}+\frac {\sin ^{4}\left (d x +c \right )}{15 \cos \left (d x +c \right )^{3}}-\frac {\sin ^{4}\left (d x +c \right )}{15 \cos \left (d x +c \right )}-\frac {\left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )}{15}\right )-3 a^{3} \left (\frac {\sin ^{3}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}+\frac {\sin ^{3}\left (d x +c \right )}{8 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{8}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+\frac {i a^{3}}{\cos \left (d x +c \right )^{3}}+a^{3} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\)	\(200\)
default	\(\frac {-i a^{3} \left (\frac {\sin ^{4}\left (d x +c \right )}{5 \cos \left (d x +c \right )^{5}}+\frac {\sin ^{4}\left (d x +c \right )}{15 \cos \left (d x +c \right )^{3}}-\frac {\sin ^{4}\left (d x +c \right )}{15 \cos \left (d x +c \right )}-\frac {\left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )}{15}\right )-3 a^{3} \left (\frac {\sin ^{3}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}+\frac {\sin ^{3}\left (d x +c \right )}{8 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{8}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+\frac {i a^{3}}{\cos \left (d x +c \right )^{3}}+a^{3} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\)	\(200\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^3*(a+I*a*tan(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

1/d*(-I*a^3*(1/5*sin(d*x+c)^4/cos(d*x+c)^5+1/15*sin(d*x+c)^4/cos(d*x+c)^3-1/15*sin(d*x+c)^4/cos(d*x+c)-1/15*(2
+sin(d*x+c)^2)*cos(d*x+c))-3*a^3*(1/4*sin(d*x+c)^3/cos(d*x+c)^4+1/8*sin(d*x+c)^3/cos(d*x+c)^2+1/8*sin(d*x+c)-1
/8*ln(sec(d*x+c)+tan(d*x+c)))+I*a^3/cos(d*x+c)^3+a^3*(1/2*sec(d*x+c)*tan(d*x+c)+1/2*ln(sec(d*x+c)+tan(d*x+c)))
)

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Maxima [A]
time = 0.29, size = 155, normalized size = 1.22 \begin {gather*} -\frac {45 \, a^{3} {\left (\frac {2 \, {\left (\sin \left (d x + c\right )^{3} + \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 60 \, a^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - \frac {240 i \, a^{3}}{\cos \left (d x + c\right )^{3}} - \frac {16 i \, {\left (5 \, \cos \left (d x + c\right )^{2} - 3\right )} a^{3}}{\cos \left (d x + c\right )^{5}}}{240 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/240*(45*a^3*(2*(sin(d*x + c)^3 + sin(d*x + c))/(sin(d*x + c)^4 - 2*sin(d*x + c)^2 + 1) - log(sin(d*x + c) +
1) + log(sin(d*x + c) - 1)) + 60*a^3*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d
*x + c) - 1)) - 240*I*a^3/cos(d*x + c)^3 - 16*I*(5*cos(d*x + c)^2 - 3)*a^3/cos(d*x + c)^5)/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. 310 vs. \(2 (107) = 214\).
time = 0.36, size = 310, normalized size = 2.44 \begin {gather*} \frac {-210 i \, a^{3} e^{\left (9 i \, d x + 9 i \, c\right )} + 1580 i \, a^{3} e^{\left (7 i \, d x + 7 i \, c\right )} + 1792 i \, a^{3} e^{\left (5 i \, d x + 5 i \, c\right )} + 980 i \, a^{3} e^{\left (3 i \, d x + 3 i \, c\right )} + 210 i \, a^{3} e^{\left (i \, d x + i \, c\right )} + 105 \, {\left (a^{3} e^{\left (10 i \, d x + 10 i \, c\right )} + 5 \, a^{3} e^{\left (8 i \, d x + 8 i \, c\right )} + 10 \, a^{3} e^{\left (6 i \, d x + 6 i \, c\right )} + 10 \, a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} + 5 \, a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{3}\right )} \log \left (e^{\left (i \, d x + i \, c\right )} + i\right ) - 105 \, {\left (a^{3} e^{\left (10 i \, d x + 10 i \, c\right )} + 5 \, a^{3} e^{\left (8 i \, d x + 8 i \, c\right )} + 10 \, a^{3} e^{\left (6 i \, d x + 6 i \, c\right )} + 10 \, a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} + 5 \, a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{3}\right )} \log \left (e^{\left (i \, d x + i \, c\right )} - i\right )}{120 \, {\left (d e^{\left (10 i \, d x + 10 i \, c\right )} + 5 \, d e^{\left (8 i \, d x + 8 i \, c\right )} + 10 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 10 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 5 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/120*(-210*I*a^3*e^(9*I*d*x + 9*I*c) + 1580*I*a^3*e^(7*I*d*x + 7*I*c) + 1792*I*a^3*e^(5*I*d*x + 5*I*c) + 980*
I*a^3*e^(3*I*d*x + 3*I*c) + 210*I*a^3*e^(I*d*x + I*c) + 105*(a^3*e^(10*I*d*x + 10*I*c) + 5*a^3*e^(8*I*d*x + 8*
I*c) + 10*a^3*e^(6*I*d*x + 6*I*c) + 10*a^3*e^(4*I*d*x + 4*I*c) + 5*a^3*e^(2*I*d*x + 2*I*c) + a^3)*log(e^(I*d*x
+ I*c) + I) - 105*(a^3*e^(10*I*d*x + 10*I*c) + 5*a^3*e^(8*I*d*x + 8*I*c) + 10*a^3*e^(6*I*d*x + 6*I*c) + 10*a^
3*e^(4*I*d*x + 4*I*c) + 5*a^3*e^(2*I*d*x + 2*I*c) + a^3)*log(e^(I*d*x + I*c) - I))/(d*e^(10*I*d*x + 10*I*c) +
5*d*e^(8*I*d*x + 8*I*c) + 10*d*e^(6*I*d*x + 6*I*c) + 10*d*e^(4*I*d*x + 4*I*c) + 5*d*e^(2*I*d*x + 2*I*c) + d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.84, size = 189, normalized size = 1.49 \begin {gather*} \frac {105 \, a^{3} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right ) - 105 \, a^{3} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right ) + \frac {2 \, {\left (15 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 360 i \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} - 390 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 960 i \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 400 i \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 390 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 320 i \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 15 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 136 i \, a^{3}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{5}}}{120 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/120*(105*a^3*log(tan(1/2*d*x + 1/2*c) + 1) - 105*a^3*log(tan(1/2*d*x + 1/2*c) - 1) + 2*(15*a^3*tan(1/2*d*x +
1/2*c)^9 - 360*I*a^3*tan(1/2*d*x + 1/2*c)^8 - 390*a^3*tan(1/2*d*x + 1/2*c)^7 + 960*I*a^3*tan(1/2*d*x + 1/2*c)
^6 - 400*I*a^3*tan(1/2*d*x + 1/2*c)^4 + 390*a^3*tan(1/2*d*x + 1/2*c)^3 + 320*I*a^3*tan(1/2*d*x + 1/2*c)^2 - 15
*a^3*tan(1/2*d*x + 1/2*c) - 136*I*a^3)/(tan(1/2*d*x + 1/2*c)^2 - 1)^5)/d

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Mupad [B]
time = 6.98, size = 228, normalized size = 1.80 \begin {gather*} \frac {7\,a^3\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{4\,d}-\frac {-\frac {a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{4}+a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,6{}\mathrm {i}+\frac {13\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{2}-a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,16{}\mathrm {i}+\frac {a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,20{}\mathrm {i}}{3}-\frac {13\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{2}-\frac {a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,16{}\mathrm {i}}{3}+\frac {a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}+\frac {a^3\,34{}\mathrm {i}}{15}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\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((a + a*tan(c + d*x)*1i)^3/cos(c + d*x)^3,x)

[Out]

(7*a^3*atanh(tan(c/2 + (d*x)/2)))/(4*d) - ((a^3*tan(c/2 + (d*x)/2)^4*20i)/3 - (13*a^3*tan(c/2 + (d*x)/2)^3)/2
- (a^3*tan(c/2 + (d*x)/2)^2*16i)/3 - a^3*tan(c/2 + (d*x)/2)^6*16i + (13*a^3*tan(c/2 + (d*x)/2)^7)/2 + a^3*tan(
c/2 + (d*x)/2)^8*6i - (a^3*tan(c/2 + (d*x)/2)^9)/4 + (a^3*34i)/15 + (a^3*tan(c/2 + (d*x)/2))/4)/(d*(5*tan(c/2
+ (d*x)/2)^2 - 10*tan(c/2 + (d*x)/2)^4 + 10*tan(c/2 + (d*x)/2)^6 - 5*tan(c/2 + (d*x)/2)^8 + tan(c/2 + (d*x)/2)
^10 - 1))

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