3.766 \(\int (a+a \sin (c+d x))^3 \tan ^2(c+d x) \, dx\)

Optimal. Leaf size=89 \[ -\frac {a^3 \cos ^3(c+d x)}{3 d}+\frac {5 a^3 \cos (c+d x)}{d}+\frac {3 a^3 \sin (c+d x) \cos (c+d x)}{2 d}+\frac {4 a^3 \cos (c+d x)}{d (1-\sin (c+d x))}-\frac {11 a^3 x}{2} \]

[Out]

-11/2*a^3*x+5*a^3*cos(d*x+c)/d-1/3*a^3*cos(d*x+c)^3/d+4*a^3*cos(d*x+c)/d/(1-sin(d*x+c))+3/2*a^3*cos(d*x+c)*sin
(d*x+c)/d

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Rubi [A]  time = 0.12, antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {2709, 2648, 2638, 2635, 8, 2633} \[ -\frac {a^3 \cos ^3(c+d x)}{3 d}+\frac {5 a^3 \cos (c+d x)}{d}+\frac {3 a^3 \sin (c+d x) \cos (c+d x)}{2 d}+\frac {4 a^3 \cos (c+d x)}{d (1-\sin (c+d x))}-\frac {11 a^3 x}{2} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sin[c + d*x])^3*Tan[c + d*x]^2,x]

[Out]

(-11*a^3*x)/2 + (5*a^3*Cos[c + d*x])/d - (a^3*Cos[c + d*x]^3)/(3*d) + (4*a^3*Cos[c + d*x])/(d*(1 - Sin[c + d*x
])) + (3*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 2633

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]

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 2709

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*tan[(e_.) + (f_.)*(x_)]^(p_), x_Symbol] :> Dist[a^p, Int[Expan
dIntegrand[(Sin[e + f*x]^p*(a + b*Sin[e + f*x])^(m - p/2))/(a - b*Sin[e + f*x])^(p/2), x], x], x] /; FreeQ[{a,
b, e, f}, x] && EqQ[a^2 - b^2, 0] && IntegersQ[m, p/2] && (LtQ[p, 0] || GtQ[m - p/2, 0])

Rubi steps

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

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Mathematica [A]  time = 0.48, size = 115, normalized size = 1.29 \[ \frac {(a \sin (c+d x)+a)^3 \left (-66 (c+d x)+9 \sin (2 (c+d x))+57 \cos (c+d x)-\cos (3 (c+d x))+\frac {96 \sin \left (\frac {1}{2} (c+d x)\right )}{\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )}\right )}{12 d \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^6} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sin[c + d*x])^3*Tan[c + d*x]^2,x]

[Out]

((a + a*Sin[c + d*x])^3*(-66*(c + d*x) + 57*Cos[c + d*x] - Cos[3*(c + d*x)] + (96*Sin[(c + d*x)/2])/(Cos[(c +
d*x)/2] - Sin[(c + d*x)/2]) + 9*Sin[2*(c + d*x)]))/(12*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^6)

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fricas [A]  time = 0.45, size = 154, normalized size = 1.73 \[ -\frac {2 \, a^{3} \cos \left (d x + c\right )^{4} - 7 \, a^{3} \cos \left (d x + c\right )^{3} + 33 \, a^{3} d x - 30 \, a^{3} \cos \left (d x + c\right )^{2} - 24 \, a^{3} + 3 \, {\left (11 \, a^{3} d x - 15 \, a^{3}\right )} \cos \left (d x + c\right ) - {\left (2 \, a^{3} \cos \left (d x + c\right )^{3} + 33 \, a^{3} d x + 9 \, a^{3} \cos \left (d x + c\right )^{2} - 21 \, a^{3} \cos \left (d x + c\right ) + 24 \, a^{3}\right )} \sin \left (d x + c\right )}{6 \, {\left (d \cos \left (d x + c\right ) - d \sin \left (d x + c\right ) + d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(2*a^3*cos(d*x + c)^4 - 7*a^3*cos(d*x + c)^3 + 33*a^3*d*x - 30*a^3*cos(d*x + c)^2 - 24*a^3 + 3*(11*a^3*d*
x - 15*a^3)*cos(d*x + c) - (2*a^3*cos(d*x + c)^3 + 33*a^3*d*x + 9*a^3*cos(d*x + c)^2 - 21*a^3*cos(d*x + c) + 2
4*a^3)*sin(d*x + c))/(d*cos(d*x + c) - d*sin(d*x + c) + d)

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giac [A]  time = 0.22, size = 119, normalized size = 1.34 \[ -\frac {33 \, {\left (d x + c\right )} a^{3} + \frac {48 \, a^{3}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1} + \frac {2 \, {\left (9 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 24 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 60 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 9 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 28 \, a^{3}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3}}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(33*(d*x + c)*a^3 + 48*a^3/(tan(1/2*d*x + 1/2*c) - 1) + 2*(9*a^3*tan(1/2*d*x + 1/2*c)^5 - 24*a^3*tan(1/2*
d*x + 1/2*c)^4 - 60*a^3*tan(1/2*d*x + 1/2*c)^2 - 9*a^3*tan(1/2*d*x + 1/2*c) - 28*a^3)/(tan(1/2*d*x + 1/2*c)^2
+ 1)^3)/d

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maple [A]  time = 0.51, size = 167, normalized size = 1.88 \[ \frac {a^{3} \left (\frac {\sin ^{6}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (\frac {8}{3}+\sin ^{4}\left (d x +c \right )+\frac {4 \left (\sin ^{2}\left (d x +c \right )\right )}{3}\right ) \cos \left (d x +c \right )\right )+3 a^{3} \left (\frac {\sin ^{5}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (\sin ^{3}\left (d x +c \right )+\frac {3 \sin \left (d x +c \right )}{2}\right ) \cos \left (d x +c \right )-\frac {3 d x}{2}-\frac {3 c}{2}\right )+3 a^{3} \left (\frac {\sin ^{4}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )\right )+a^{3} \left (\tan \left (d x +c \right )-d x -c \right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.42, size = 117, normalized size = 1.31 \[ -\frac {2 \, {\left (\cos \left (d x + c\right )^{3} - \frac {3}{\cos \left (d x + c\right )} - 6 \, \cos \left (d x + c\right )\right )} a^{3} + 9 \, {\left (3 \, d x + 3 \, c - \frac {\tan \left (d x + c\right )}{\tan \left (d x + c\right )^{2} + 1} - 2 \, \tan \left (d x + c\right )\right )} a^{3} + 6 \, {\left (d x + c - \tan \left (d x + c\right )\right )} a^{3} - 18 \, a^{3} {\left (\frac {1}{\cos \left (d x + c\right )} + \cos \left (d x + c\right )\right )}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(2*(cos(d*x + c)^3 - 3/cos(d*x + c) - 6*cos(d*x + c))*a^3 + 9*(3*d*x + 3*c - tan(d*x + c)/(tan(d*x + c)^2
+ 1) - 2*tan(d*x + c))*a^3 + 6*(d*x + c - tan(d*x + c))*a^3 - 18*a^3*(1/cos(d*x + c) + cos(d*x + c)))/d

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mupad [B]  time = 14.76, size = 288, normalized size = 3.24 \[ -\frac {11\,a^3\,x}{2}-\frac {\frac {11\,a^3\,\left (c+d\,x\right )}{2}-\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {11\,a^3\,\left (c+d\,x\right )}{2}-\frac {a^3\,\left (33\,c+33\,d\,x-38\right )}{6}\right )-\frac {a^3\,\left (33\,c+33\,d\,x-104\right )}{6}+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (\frac {11\,a^3\,\left (c+d\,x\right )}{2}-\frac {a^3\,\left (33\,c+33\,d\,x-66\right )}{6}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (\frac {33\,a^3\,\left (c+d\,x\right )}{2}-\frac {a^3\,\left (99\,c+99\,d\,x-66\right )}{6}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {33\,a^3\,\left (c+d\,x\right )}{2}-\frac {a^3\,\left (99\,c+99\,d\,x-120\right )}{6}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (\frac {33\,a^3\,\left (c+d\,x\right )}{2}-\frac {a^3\,\left (99\,c+99\,d\,x-192\right )}{6}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {33\,a^3\,\left (c+d\,x\right )}{2}-\frac {a^3\,\left (99\,c+99\,d\,x-246\right )}{6}\right )}{d\,\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- (11*a^3*x)/2 - ((11*a^3*(c + d*x))/2 - tan(c/2 + (d*x)/2)*((11*a^3*(c + d*x))/2 - (a^3*(33*c + 33*d*x - 38))
/6) - (a^3*(33*c + 33*d*x - 104))/6 + tan(c/2 + (d*x)/2)^6*((11*a^3*(c + d*x))/2 - (a^3*(33*c + 33*d*x - 66))/
6) - tan(c/2 + (d*x)/2)^5*((33*a^3*(c + d*x))/2 - (a^3*(99*c + 99*d*x - 66))/6) - tan(c/2 + (d*x)/2)^3*((33*a^
3*(c + d*x))/2 - (a^3*(99*c + 99*d*x - 120))/6) + tan(c/2 + (d*x)/2)^4*((33*a^3*(c + d*x))/2 - (a^3*(99*c + 99
*d*x - 192))/6) + tan(c/2 + (d*x)/2)^2*((33*a^3*(c + d*x))/2 - (a^3*(99*c + 99*d*x - 246))/6))/(d*(tan(c/2 + (
d*x)/2) - 1)*(tan(c/2 + (d*x)/2)^2 + 1)^3)

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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)**2*sin(d*x+c)**2*(a+a*sin(d*x+c))**3,x)

[Out]

Timed out

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