3.812 \(\int \sec (c+d x) (a+a \sin (c+d x))^3 \tan ^3(c+d x) \, dx\)

Optimal. Leaf size=101 \[ -\frac {3 a^3 \cos (c+d x)}{d}-\frac {a^3 \sin (c+d x) \cos (c+d x)}{2 d}-\frac {19 a^3 \cos (c+d x)}{3 d (1-\sin (c+d x))}+\frac {2 a^3 \cos (c+d x)}{3 d (1-\sin (c+d x))^2}+\frac {11 a^3 x}{2} \]

[Out]

11/2*a^3*x-3*a^3*cos(d*x+c)/d+2/3*a^3*cos(d*x+c)/d/(1-sin(d*x+c))^2-19/3*a^3*cos(d*x+c)/d/(1-sin(d*x+c))-1/2*a
^3*cos(d*x+c)*sin(d*x+c)/d

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Rubi [A]  time = 0.16, antiderivative size = 101, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {2872, 2650, 2648, 2638, 2635, 8} \[ -\frac {3 a^3 \cos (c+d x)}{d}-\frac {a^3 \sin (c+d x) \cos (c+d x)}{2 d}-\frac {19 a^3 \cos (c+d x)}{3 d (1-\sin (c+d x))}+\frac {2 a^3 \cos (c+d x)}{3 d (1-\sin (c+d x))^2}+\frac {11 a^3 x}{2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(11*a^3*x)/2 - (3*a^3*Cos[c + d*x])/d + (2*a^3*Cos[c + d*x])/(3*d*(1 - Sin[c + d*x])^2) - (19*a^3*Cos[c + d*x]
)/(3*d*(1 - Sin[c + d*x])) - (a^3*Cos[c + d*x]*Sin[c + d*x])/(2*d)

Rule 8

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

Rule 2635

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

Rule 2638

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

Rule 2648

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> -Simp[Cos[c + d*x]/(d*(b + a*Sin[c + d*x])), x]
/; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 2650

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*Cos[c + d*x]*(a + b*Sin[c + d*x])^n)/(a*
d*(2*n + 1)), x] + Dist[(n + 1)/(a*(2*n + 1)), Int[(a + b*Sin[c + d*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d},
x] && EqQ[a^2 - b^2, 0] && LtQ[n, -1] && IntegerQ[2*n]

Rule 2872

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(
m_), x_Symbol] :> Dist[1/a^p, Int[ExpandTrig[(d*sin[e + f*x])^n*(a - b*sin[e + f*x])^(p/2)*(a + b*sin[e + f*x]
)^(m + p/2), x], x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && IntegersQ[m, n, p/2] && ((GtQ[m,
0] && GtQ[p, 0] && LtQ[-m - p, n, -1]) || (GtQ[m, 2] && LtQ[p, 0] && GtQ[m + p/2, 0]))

Rubi steps

\begin {align*} \int \sec (c+d x) (a+a \sin (c+d x))^3 \tan ^3(c+d x) \, dx &=a^4 \int \left (\frac {5}{a}+\frac {2}{a (-1+\sin (c+d x))^2}+\frac {7}{a (-1+\sin (c+d x))}+\frac {3 \sin (c+d x)}{a}+\frac {\sin ^2(c+d x)}{a}\right ) \, dx\\ &=5 a^3 x+a^3 \int \sin ^2(c+d x) \, dx+\left (2 a^3\right ) \int \frac {1}{(-1+\sin (c+d x))^2} \, dx+\left (3 a^3\right ) \int \sin (c+d x) \, dx+\left (7 a^3\right ) \int \frac {1}{-1+\sin (c+d x)} \, dx\\ &=5 a^3 x-\frac {3 a^3 \cos (c+d x)}{d}+\frac {2 a^3 \cos (c+d x)}{3 d (1-\sin (c+d x))^2}-\frac {7 a^3 \cos (c+d x)}{d (1-\sin (c+d x))}-\frac {a^3 \cos (c+d x) \sin (c+d x)}{2 d}+\frac {1}{2} a^3 \int 1 \, dx-\frac {1}{3} \left (2 a^3\right ) \int \frac {1}{-1+\sin (c+d x)} \, dx\\ &=\frac {11 a^3 x}{2}-\frac {3 a^3 \cos (c+d x)}{d}+\frac {2 a^3 \cos (c+d x)}{3 d (1-\sin (c+d x))^2}-\frac {19 a^3 \cos (c+d x)}{3 d (1-\sin (c+d x))}-\frac {a^3 \cos (c+d x) \sin (c+d x)}{2 d}\\ \end {align*}

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Mathematica [A]  time = 1.48, size = 159, normalized size = 1.57 \[ -\frac {a^3 \left (-3 (132 c+132 d x+89) \cos \left (\frac {1}{2} (c+d x)\right )+(132 c+132 d x+403) \cos \left (\frac {3}{2} (c+d x)\right )+3 \left (-9 \cos \left (\frac {5}{2} (c+d x)\right )+\cos \left (\frac {7}{2} (c+d x)\right )+2 \sin \left (\frac {1}{2} (c+d x)\right ) ((44 c+44 d x-43) \cos (c+d x)-10 \cos (2 (c+d x))-\cos (3 (c+d x))+88 c+88 d x+86)\right )\right )}{48 d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/48*(a^3*(-3*(89 + 132*c + 132*d*x)*Cos[(c + d*x)/2] + (403 + 132*c + 132*d*x)*Cos[(3*(c + d*x))/2] + 3*(-9*
Cos[(5*(c + d*x))/2] + Cos[(7*(c + d*x))/2] + 2*(86 + 88*c + 88*d*x + (-43 + 44*c + 44*d*x)*Cos[c + d*x] - 10*
Cos[2*(c + d*x)] - Cos[3*(c + d*x)])*Sin[(c + d*x)/2])))/(d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^3)

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fricas [B]  time = 0.46, size = 196, normalized size = 1.94 \[ \frac {3 \, a^{3} \cos \left (d x + c\right )^{4} - 12 \, a^{3} \cos \left (d x + c\right )^{3} - 66 \, a^{3} d x - 4 \, a^{3} + {\left (33 \, a^{3} d x + 53 \, a^{3}\right )} \cos \left (d x + c\right )^{2} - {\left (33 \, a^{3} d x - 64 \, a^{3}\right )} \cos \left (d x + c\right ) - {\left (3 \, a^{3} \cos \left (d x + c\right )^{3} - 66 \, a^{3} d x + 15 \, a^{3} \cos \left (d x + c\right )^{2} + 4 \, a^{3} - {\left (33 \, a^{3} d x - 68 \, a^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{6 \, {\left (d \cos \left (d x + c\right )^{2} - d \cos \left (d x + c\right ) + {\left (d \cos \left (d x + c\right ) + 2 \, d\right )} \sin \left (d x + c\right ) - 2 \, d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*sin(d*x+c)^3*(a+a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

1/6*(3*a^3*cos(d*x + c)^4 - 12*a^3*cos(d*x + c)^3 - 66*a^3*d*x - 4*a^3 + (33*a^3*d*x + 53*a^3)*cos(d*x + c)^2
- (33*a^3*d*x - 64*a^3)*cos(d*x + c) - (3*a^3*cos(d*x + c)^3 - 66*a^3*d*x + 15*a^3*cos(d*x + c)^2 + 4*a^3 - (3
3*a^3*d*x - 68*a^3)*cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^2 - d*cos(d*x + c) + (d*cos(d*x + c) + 2*d)*si
n(d*x + c) - 2*d)

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giac [A]  time = 0.26, size = 135, normalized size = 1.34 \[ \frac {33 \, {\left (d x + c\right )} a^{3} + \frac {6 \, {\left (a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 6 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 6 \, a^{3}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2}} + \frac {4 \, {\left (15 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 36 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 17 \, a^{3}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}^{3}}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*sin(d*x+c)^3*(a+a*sin(d*x+c))^3,x, algorithm="giac")

[Out]

1/6*(33*(d*x + c)*a^3 + 6*(a^3*tan(1/2*d*x + 1/2*c)^3 - 6*a^3*tan(1/2*d*x + 1/2*c)^2 - a^3*tan(1/2*d*x + 1/2*c
) - 6*a^3)/(tan(1/2*d*x + 1/2*c)^2 + 1)^2 + 4*(15*a^3*tan(1/2*d*x + 1/2*c)^2 - 36*a^3*tan(1/2*d*x + 1/2*c) + 1
7*a^3)/(tan(1/2*d*x + 1/2*c) - 1)^3)/d

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maple [B]  time = 0.58, size = 246, normalized size = 2.44 \[ \frac {a^{3} \left (\frac {\sin ^{7}\left (d x +c \right )}{3 \cos \left (d x +c \right )^{3}}-\frac {4 \left (\sin ^{7}\left (d x +c \right )\right )}{3 \cos \left (d x +c \right )}-\frac {4 \left (\sin ^{5}\left (d x +c \right )+\frac {5 \left (\sin ^{3}\left (d x +c \right )\right )}{4}+\frac {15 \sin \left (d x +c \right )}{8}\right ) \cos \left (d x +c \right )}{3}+\frac {5 d x}{2}+\frac {5 c}{2}\right )+3 a^{3} \left (\frac {\sin ^{6}\left (d x +c \right )}{3 \cos \left (d x +c \right )^{3}}-\frac {\sin ^{6}\left (d x +c \right )}{\cos \left (d x +c \right )}-\left (\frac {8}{3}+\sin ^{4}\left (d x +c \right )+\frac {4 \left (\sin ^{2}\left (d x +c \right )\right )}{3}\right ) \cos \left (d x +c \right )\right )+3 a^{3} \left (\frac {\left (\tan ^{3}\left (d x +c \right )\right )}{3}-\tan \left (d x +c \right )+d x +c \right )+a^{3} \left (\frac {\sin ^{4}\left (d x +c \right )}{3 \cos \left (d x +c \right )^{3}}-\frac {\sin ^{4}\left (d x +c \right )}{3 \cos \left (d x +c \right )}-\frac {\left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )}{3}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^4*sin(d*x+c)^3*(a+a*sin(d*x+c))^3,x)

[Out]

1/d*(a^3*(1/3*sin(d*x+c)^7/cos(d*x+c)^3-4/3*sin(d*x+c)^7/cos(d*x+c)-4/3*(sin(d*x+c)^5+5/4*sin(d*x+c)^3+15/8*si
n(d*x+c))*cos(d*x+c)+5/2*d*x+5/2*c)+3*a^3*(1/3*sin(d*x+c)^6/cos(d*x+c)^3-sin(d*x+c)^6/cos(d*x+c)-(8/3+sin(d*x+
c)^4+4/3*sin(d*x+c)^2)*cos(d*x+c))+3*a^3*(1/3*tan(d*x+c)^3-tan(d*x+c)+d*x+c)+a^3*(1/3*sin(d*x+c)^4/cos(d*x+c)^
3-1/3*sin(d*x+c)^4/cos(d*x+c)-1/3*(2+sin(d*x+c)^2)*cos(d*x+c)))

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maxima [A]  time = 0.43, size = 145, normalized size = 1.44 \[ \frac {{\left (2 \, \tan \left (d x + c\right )^{3} + 15 \, d x + 15 \, c - \frac {3 \, \tan \left (d x + c\right )}{\tan \left (d x + c\right )^{2} + 1} - 12 \, \tan \left (d x + c\right )\right )} a^{3} + 6 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, d x + 3 \, c - 3 \, \tan \left (d x + c\right )\right )} a^{3} - 6 \, a^{3} {\left (\frac {6 \, \cos \left (d x + c\right )^{2} - 1}{\cos \left (d x + c\right )^{3}} + 3 \, \cos \left (d x + c\right )\right )} - \frac {2 \, {\left (3 \, \cos \left (d x + c\right )^{2} - 1\right )} a^{3}}{\cos \left (d x + c\right )^{3}}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*sin(d*x+c)^3*(a+a*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

1/6*((2*tan(d*x + c)^3 + 15*d*x + 15*c - 3*tan(d*x + c)/(tan(d*x + c)^2 + 1) - 12*tan(d*x + c))*a^3 + 6*(tan(d
*x + c)^3 + 3*d*x + 3*c - 3*tan(d*x + c))*a^3 - 6*a^3*((6*cos(d*x + c)^2 - 1)/cos(d*x + c)^3 + 3*cos(d*x + c))
- 2*(3*cos(d*x + c)^2 - 1)*a^3/cos(d*x + c)^3)/d

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mupad [B]  time = 14.96, size = 287, normalized size = 2.84 \[ \frac {11\,a^3\,x}{2}+\frac {\frac {11\,a^3\,\left (c+d\,x\right )}{2}-\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {33\,a^3\,\left (c+d\,x\right )}{2}-\frac {a^3\,\left (99\,c+99\,d\,x-246\right )}{6}\right )-\frac {a^3\,\left (33\,c+33\,d\,x-104\right )}{6}+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (\frac {33\,a^3\,\left (c+d\,x\right )}{2}-\frac {a^3\,\left (99\,c+99\,d\,x-66\right )}{6}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (\frac {55\,a^3\,\left (c+d\,x\right )}{2}-\frac {a^3\,\left (165\,c+165\,d\,x-198\right )}{6}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {55\,a^3\,\left (c+d\,x\right )}{2}-\frac {a^3\,\left (165\,c+165\,d\,x-322\right )}{6}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (\frac {77\,a^3\,\left (c+d\,x\right )}{2}-\frac {a^3\,\left (231\,c+231\,d\,x-308\right )}{6}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {77\,a^3\,\left (c+d\,x\right )}{2}-\frac {a^3\,\left (231\,c+231\,d\,x-420\right )}{6}\right )}{d\,{\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )}^3\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((sin(c + d*x)^3*(a + a*sin(c + d*x))^3)/cos(c + d*x)^4,x)

[Out]

(11*a^3*x)/2 + ((11*a^3*(c + d*x))/2 - tan(c/2 + (d*x)/2)*((33*a^3*(c + d*x))/2 - (a^3*(99*c + 99*d*x - 246))/
6) - (a^3*(33*c + 33*d*x - 104))/6 + tan(c/2 + (d*x)/2)^6*((33*a^3*(c + d*x))/2 - (a^3*(99*c + 99*d*x - 66))/6
) - tan(c/2 + (d*x)/2)^5*((55*a^3*(c + d*x))/2 - (a^3*(165*c + 165*d*x - 198))/6) + tan(c/2 + (d*x)/2)^2*((55*
a^3*(c + d*x))/2 - (a^3*(165*c + 165*d*x - 322))/6) + tan(c/2 + (d*x)/2)^4*((77*a^3*(c + d*x))/2 - (a^3*(231*c
+ 231*d*x - 308))/6) - tan(c/2 + (d*x)/2)^3*((77*a^3*(c + d*x))/2 - (a^3*(231*c + 231*d*x - 420))/6))/(d*(tan
(c/2 + (d*x)/2) - 1)^3*(tan(c/2 + (d*x)/2)^2 + 1)^2)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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