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*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)

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Rubi [A]  time = 0.162259, antiderivative size = 101, normalized size of antiderivative = 1., 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 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 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 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]

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*}

Mathematica [A]  time = 1.45991, 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]

-(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])))/(48*d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^3)

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Maple [B]  time = 0.088, size = 246, normalized size = 2.4 \begin{align*}{\frac{1}{d} \left ({a}^{3} \left ({\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{7}}{3\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}}}-{\frac{4\, \left ( \sin \left ( dx+c \right ) \right ) ^{7}}{3\,\cos \left ( dx+c \right ) }}-{\frac{4\,\cos \left ( dx+c \right ) }{3} \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 ) }+{\frac{5\,dx}{2}}+{\frac{5\,c}{2}} \right ) +3\,{a}^{3} \left ( 1/3\,{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{6}}{ \left ( \cos \left ( dx+c \right ) \right ) ^{3}}}-{\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 ( 1/3\, \left ( \tan \left ( dx+c \right ) \right ) ^{3}-\tan \left ( dx+c \right ) +dx+c \right ) +{a}^{3} \left ({\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{3\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}}}-{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{3\,\cos \left ( dx+c \right ) }}-{\frac{ \left ( 2+ \left ( \sin \left ( dx+c \right ) \right ) ^{2} \right ) \cos \left ( dx+c \right ) }{3}} \right ) \right ) } \end{align*}

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 = 1.70007, size = 196, normalized size = 1.94 \begin{align*} \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} \end{align*}

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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Fricas [B]  time = 1.4259, size = 471, normalized size = 4.66 \begin{align*} \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 )}} \end{align*}

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

[Out]

Timed out

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Giac [A]  time = 1.21789, size = 182, normalized size = 1.8 \begin{align*} \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} \end{align*}

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