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

Optimal. Leaf size=111 \[ -\frac{a^3 \cos ^3(c+d x)}{d}+\frac{7 a^3 \cos (c+d x)}{d}+\frac{a^3 \sin ^3(c+d x) \cos (c+d x)}{4 d}+\frac{19 a^3 \sin (c+d x) \cos (c+d x)}{8 d}+\frac{4 a^3 \cos (c+d x)}{d (1-\sin (c+d x))}-\frac{51 a^3 x}{8} \]

[Out]

(-51*a^3*x)/8 + (7*a^3*Cos[c + d*x])/d - (a^3*Cos[c + d*x]^3)/d + (4*a^3*Cos[c + d*x])/(d*(1 - Sin[c + d*x]))
+ (19*a^3*Cos[c + d*x]*Sin[c + d*x])/(8*d) + (a^3*Cos[c + d*x]*Sin[c + d*x]^3)/(4*d)

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Rubi [A]  time = 0.170108, antiderivative size = 111, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 6, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {2872, 2648, 2638, 2635, 8, 2633} \[ -\frac{a^3 \cos ^3(c+d x)}{d}+\frac{7 a^3 \cos (c+d x)}{d}+\frac{a^3 \sin ^3(c+d x) \cos (c+d x)}{4 d}+\frac{19 a^3 \sin (c+d x) \cos (c+d x)}{8 d}+\frac{4 a^3 \cos (c+d x)}{d (1-\sin (c+d x))}-\frac{51 a^3 x}{8} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-51*a^3*x)/8 + (7*a^3*Cos[c + d*x])/d - (a^3*Cos[c + d*x]^3)/d + (4*a^3*Cos[c + d*x])/(d*(1 - Sin[c + d*x]))
+ (19*a^3*Cos[c + d*x]*Sin[c + d*x])/(8*d) + (a^3*Cos[c + d*x]*Sin[c + d*x]^3)/(4*d)

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]))

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 2638

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

Rubi steps

\begin{align*} \int \sin (c+d x) (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)-4 a \sin ^2(c+d x)-3 a \sin ^3(c+d x)-a \sin ^4(c+d x)\right ) \, dx\\ &=-4 a^3 x-a^3 \int \sin ^4(c+d x) \, dx-\left (3 a^3\right ) \int \sin ^3(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-\left (4 a^3\right ) \int \sin ^2(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{2 a^3 \cos (c+d x) \sin (c+d x)}{d}+\frac{a^3 \cos (c+d x) \sin ^3(c+d x)}{4 d}-\frac{1}{4} \left (3 a^3\right ) \int \sin ^2(c+d x) \, dx-\left (2 a^3\right ) \int 1 \, dx+\frac{\left (3 a^3\right ) \operatorname{Subst}\left (\int \left (1-x^2\right ) \, dx,x,\cos (c+d x)\right )}{d}\\ &=-6 a^3 x+\frac{7 a^3 \cos (c+d x)}{d}-\frac{a^3 \cos ^3(c+d x)}{d}+\frac{4 a^3 \cos (c+d x)}{d (1-\sin (c+d x))}+\frac{19 a^3 \cos (c+d x) \sin (c+d x)}{8 d}+\frac{a^3 \cos (c+d x) \sin ^3(c+d x)}{4 d}-\frac{1}{8} \left (3 a^3\right ) \int 1 \, dx\\ &=-\frac{51 a^3 x}{8}+\frac{7 a^3 \cos (c+d x)}{d}-\frac{a^3 \cos ^3(c+d x)}{d}+\frac{4 a^3 \cos (c+d x)}{d (1-\sin (c+d x))}+\frac{19 a^3 \cos (c+d x) \sin (c+d x)}{8 d}+\frac{a^3 \cos (c+d x) \sin ^3(c+d x)}{4 d}\\ \end{align*}

Mathematica [A]  time = 0.802426, size = 125, normalized size = 1.13 \[ \frac{(a \sin (c+d x)+a)^3 \left (-204 (c+d x)+40 \sin (2 (c+d x))-\sin (4 (c+d x))+200 \cos (c+d x)-8 \cos (3 (c+d x))+\frac{256 \sin \left (\frac{1}{2} (c+d x)\right )}{\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )}\right )}{32 d \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^6} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a + a*Sin[c + d*x])^3*(-204*(c + d*x) + 200*Cos[c + d*x] - 8*Cos[3*(c + d*x)] + (256*Sin[(c + d*x)/2])/(Cos[
(c + d*x)/2] - Sin[(c + d*x)/2]) + 40*Sin[2*(c + d*x)] - Sin[4*(c + d*x)]))/(32*d*(Cos[(c + d*x)/2] + Sin[(c +
d*x)/2])^6)

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Maple [B]  time = 0.069, size = 212, normalized size = 1.9 \begin{align*}{\frac{1}{d} \left ({a}^{3} \left ({\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{7}}{\cos \left ( dx+c \right ) }}+ \left ( \left ( \sin \left ( dx+c \right ) \right ) ^{5}+{\frac{5\, \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{4}}+{\frac{15\,\sin \left ( dx+c \right ) }{8}} \right ) \cos \left ( dx+c \right ) -{\frac{15\,dx}{8}}-{\frac{15\,c}{8}} \right ) +3\,{a}^{3} \left ({\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{6}}{\cos \left ( dx+c \right ) }}+ \left ( 8/3+ \left ( \sin \left ( dx+c \right ) \right ) ^{4}+4/3\, \left ( \sin \left ( dx+c \right ) \right ) ^{2} \right ) \cos \left ( dx+c \right ) \right ) +3\,{a}^{3} \left ({\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{5}}{\cos \left ( dx+c \right ) }}+ \left ( \left ( \sin \left ( dx+c \right ) \right ) ^{3}+3/2\,\sin \left ( dx+c \right ) \right ) \cos \left ( dx+c \right ) -3/2\,dx-3/2\,c \right ) +{a}^{3} \left ({\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{\cos \left ( dx+c \right ) }}+ \left ( 2+ \left ( \sin \left ( dx+c \right ) \right ) ^{2} \right ) \cos \left ( dx+c \right ) \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.70465, size = 219, normalized size = 1.97 \begin{align*} -\frac{8 \,{\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} +{\left (15 \, d x + 15 \, c - \frac{9 \, \tan \left (d x + c\right )^{3} + 7 \, \tan \left (d x + c\right )}{\tan \left (d x + c\right )^{4} + 2 \, \tan \left (d x + c\right )^{2} + 1} - 8 \, \tan \left (d x + c\right )\right )} a^{3} + 12 \,{\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} - 8 \, a^{3}{\left (\frac{1}{\cos \left (d x + c\right )} + \cos \left (d x + c\right )\right )}}{8 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(8*(cos(d*x + c)^3 - 3/cos(d*x + c) - 6*cos(d*x + c))*a^3 + (15*d*x + 15*c - (9*tan(d*x + c)^3 + 7*tan(d*
x + c))/(tan(d*x + c)^4 + 2*tan(d*x + c)^2 + 1) - 8*tan(d*x + c))*a^3 + 12*(3*d*x + 3*c - tan(d*x + c)/(tan(d*
x + c)^2 + 1) - 2*tan(d*x + c))*a^3 - 8*a^3*(1/cos(d*x + c) + cos(d*x + c)))/d

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Fricas [A]  time = 1.14145, size = 440, normalized size = 3.96 \begin{align*} -\frac{2 \, a^{3} \cos \left (d x + c\right )^{5} + 8 \, a^{3} \cos \left (d x + c\right )^{4} - 15 \, a^{3} \cos \left (d x + c\right )^{3} + 51 \, a^{3} d x - 56 \, a^{3} \cos \left (d x + c\right )^{2} - 32 \, a^{3} +{\left (51 \, a^{3} d x - 67 \, a^{3}\right )} \cos \left (d x + c\right ) +{\left (2 \, a^{3} \cos \left (d x + c\right )^{4} - 6 \, a^{3} \cos \left (d x + c\right )^{3} - 51 \, a^{3} d x - 21 \, a^{3} \cos \left (d x + c\right )^{2} + 35 \, a^{3} \cos \left (d x + c\right ) - 32 \, a^{3}\right )} \sin \left (d x + c\right )}{8 \,{\left (d \cos \left (d x + c\right ) - d \sin \left (d x + c\right ) + d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(2*a^3*cos(d*x + c)^5 + 8*a^3*cos(d*x + c)^4 - 15*a^3*cos(d*x + c)^3 + 51*a^3*d*x - 56*a^3*cos(d*x + c)^2
- 32*a^3 + (51*a^3*d*x - 67*a^3)*cos(d*x + c) + (2*a^3*cos(d*x + c)^4 - 6*a^3*cos(d*x + c)^3 - 51*a^3*d*x - 2
1*a^3*cos(d*x + c)^2 + 35*a^3*cos(d*x + c) - 32*a^3)*sin(d*x + c))/(d*cos(d*x + c) - d*sin(d*x + c) + d)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]  time = 1.28266, size = 225, normalized size = 2.03 \begin{align*} -\frac{51 \,{\left (d x + c\right )} a^{3} + \frac{64 \, a^{3}}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1} + \frac{2 \,{\left (19 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 32 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{6} + 27 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 144 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 27 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 160 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 19 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 48 \, a^{3}\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )}^{4}}}{8 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(51*(d*x + c)*a^3 + 64*a^3/(tan(1/2*d*x + 1/2*c) - 1) + 2*(19*a^3*tan(1/2*d*x + 1/2*c)^7 - 32*a^3*tan(1/2
*d*x + 1/2*c)^6 + 27*a^3*tan(1/2*d*x + 1/2*c)^5 - 144*a^3*tan(1/2*d*x + 1/2*c)^4 - 27*a^3*tan(1/2*d*x + 1/2*c)
^3 - 160*a^3*tan(1/2*d*x + 1/2*c)^2 - 19*a^3*tan(1/2*d*x + 1/2*c) - 48*a^3)/(tan(1/2*d*x + 1/2*c)^2 + 1)^4)/d