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

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*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)

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Rubi [A]  time = 0.125111, antiderivative size = 89, normalized size of antiderivative = 1., 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 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])

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 (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*}

Mathematica [A]  time = 0.490405, 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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Maple [A]  time = 0.053, size = 167, normalized size = 1.9 \begin{align*}{\frac{1}{d} \left ({a}^{3} \left ({\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{6}}{\cos \left ( dx+c \right ) }}+ \left ({\frac{8}{3}}+ \left ( \sin \left ( dx+c \right ) \right ) ^{4}+{\frac{4\, \left ( \sin \left ( dx+c \right ) \right ) ^{2}}{3}} \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 ) +3\,{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 ) +{a}^{3} \left ( \tan \left ( dx+c \right ) -dx-c \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sin(d*x+c))^3*tan(d*x+c)^2,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 = 1.57021, size = 158, normalized size = 1.78 \begin{align*} -\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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(d*x+c))^3*tan(d*x+c)^2,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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Fricas [A]  time = 1.42002, size = 378, normalized size = 4.25 \begin{align*} -\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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(d*x+c))^3*tan(d*x+c)^2,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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} a^{3} \left (\int 3 \sin{\left (c + d x \right )} \tan ^{2}{\left (c + d x \right )}\, dx + \int 3 \sin ^{2}{\left (c + d x \right )} \tan ^{2}{\left (c + d x \right )}\, dx + \int \sin ^{3}{\left (c + d x \right )} \tan ^{2}{\left (c + d x \right )}\, dx + \int \tan ^{2}{\left (c + d x \right )}\, dx\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [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((a+a*sin(d*x+c))^3*tan(d*x+c)^2,x, algorithm="giac")

[Out]

Timed out