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

Optimal. Leaf size=64 \[ -\frac{2 a^5 \cos (c+d x)}{d \left (a^2-a^2 \sin (c+d x)\right )}+a^3 x+\frac{\sec ^3(c+d x) (a \sin (c+d x)+a)^3}{3 d} \]

[Out]

a^3*x + (Sec[c + d*x]^3*(a + a*Sin[c + d*x])^3)/(3*d) - (2*a^5*Cos[c + d*x])/(d*(a^2 - a^2*Sin[c + d*x]))

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Rubi [A]  time = 0.138811, antiderivative size = 64, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.148, Rules used = {2855, 2670, 2680, 8} \[ -\frac{2 a^5 \cos (c+d x)}{d \left (a^2-a^2 \sin (c+d x)\right )}+a^3 x+\frac{\sec ^3(c+d x) (a \sin (c+d x)+a)^3}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

a^3*x + (Sec[c + d*x]^3*(a + a*Sin[c + d*x])^3)/(3*d) - (2*a^5*Cos[c + d*x])/(d*(a^2 - a^2*Sin[c + d*x]))

Rule 2855

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> -Simp[((b*c + a*d)*(g*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^m)/(a*f*g*(p +
1)), x] + Dist[(b*(a*d*m + b*c*(m + p + 1)))/(a*g^2*(p + 1)), Int[(g*Cos[e + f*x])^(p + 2)*(a + b*Sin[e + f*x]
)^(m - 1), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && EqQ[a^2 - b^2, 0] && GtQ[m, -1] && LtQ[p, -1]

Rule 2670

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Dist[(a/g)^
(2*m), Int[(g*Cos[e + f*x])^(2*m + p)/(a - b*Sin[e + f*x])^m, x], x] /; FreeQ[{a, b, e, f, g}, x] && EqQ[a^2 -
b^2, 0] && IntegerQ[m] && LtQ[p, -1] && GeQ[2*m + p, 0]

Rule 2680

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(2*g*(
g*Cos[e + f*x])^(p - 1)*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(2*m + p + 1)), x] + Dist[(g^2*(p - 1))/(b^2*(2*m +
p + 1)), Int[(g*Cos[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^(m + 2), x], x] /; FreeQ[{a, b, e, f, g}, x] && Eq
Q[a^2 - b^2, 0] && LeQ[m, -2] && GtQ[p, 1] && NeQ[2*m + p + 1, 0] &&  !ILtQ[m + p + 1, 0] && IntegersQ[2*m, 2*
p]

Rule 8

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

Rubi steps

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

Mathematica [A]  time = 0.719077, size = 107, normalized size = 1.67 \[ -\frac{a^3 \left (-9 (c+d x+2) \cos \left (\frac{1}{2} (c+d x)\right )+(3 c+3 d x+14) \cos \left (\frac{3}{2} (c+d x)\right )+6 \sin \left (\frac{1}{2} (c+d x)\right ) (2 (c+d x+2)+(c+d x) \cos (c+d x))\right )}{6 d \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a^3*(-9*(2 + c + d*x)*Cos[(c + d*x)/2] + (14 + 3*c + 3*d*x)*Cos[(3*(c + d*x))/2] + 6*(2*(2 + c + d*x) + (c +
d*x)*Cos[c + d*x])*Sin[(c + d*x)/2]))/(6*d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^3)

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Maple [B]  time = 0.071, size = 126, normalized size = 2. \begin{align*}{\frac{1}{d} \left ({a}^{3} \left ({\frac{ \left ( \tan \left ( dx+c \right ) \right ) ^{3}}{3}}-\tan \left ( dx+c \right ) +dx+c \right ) +3\,{a}^{3} \left ( 1/3\,{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{ \left ( \cos \left ( dx+c \right ) \right ) ^{3}}}-1/3\,{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{\cos \left ( dx+c \right ) }}-1/3\, \left ( 2+ \left ( \sin \left ( dx+c \right ) \right ) ^{2} \right ) \cos \left ( dx+c \right ) \right ) +{\frac{{a}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{ \left ( \cos \left ( dx+c \right ) \right ) ^{3}}}+{\frac{{a}^{3}}{3\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.69052, size = 113, normalized size = 1.77 \begin{align*} \frac{3 \, a^{3} \tan \left (d x + c\right )^{3} +{\left (\tan \left (d x + c\right )^{3} + 3 \, d x + 3 \, c - 3 \, \tan \left (d x + c\right )\right )} a^{3} - \frac{3 \,{\left (3 \, \cos \left (d x + c\right )^{2} - 1\right )} a^{3}}{\cos \left (d x + c\right )^{3}} + \frac{a^{3}}{\cos \left (d x + c\right )^{3}}}{3 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 1.35476, size = 335, normalized size = 5.23 \begin{align*} -\frac{6 \, a^{3} d x + 2 \, a^{3} -{\left (3 \, a^{3} d x + 7 \, a^{3}\right )} \cos \left (d x + c\right )^{2} +{\left (3 \, a^{3} d x - 5 \, a^{3}\right )} \cos \left (d x + c\right ) -{\left (6 \, a^{3} d x - 2 \, a^{3} +{\left (3 \, a^{3} d x - 7 \, a^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{3 \,{\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)*(a+a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

-1/3*(6*a^3*d*x + 2*a^3 - (3*a^3*d*x + 7*a^3)*cos(d*x + c)^2 + (3*a^3*d*x - 5*a^3)*cos(d*x + c) - (6*a^3*d*x -
2*a^3 + (3*a^3*d*x - 7*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)*sin(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)*(a+a*sin(d*x+c))**3,x)

[Out]

Timed out

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Giac [A]  time = 1.25368, size = 90, normalized size = 1.41 \begin{align*} \frac{3 \,{\left (d x + c\right )} a^{3} + \frac{2 \,{\left (3 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 12 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 5 \, a^{3}\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1\right )}^{3}}}{3 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(3*(d*x + c)*a^3 + 2*(3*a^3*tan(1/2*d*x + 1/2*c)^2 - 12*a^3*tan(1/2*d*x + 1/2*c) + 5*a^3)/(tan(1/2*d*x + 1
/2*c) - 1)^3)/d