3.36 \(\int \sec ^4(c+d x) (a+a \sin (c+d x))^3 \, dx\)

Optimal. Leaf size=31 \[ \frac{a^6 \cos ^3(c+d x)}{3 d (a-a \sin (c+d x))^3} \]

[Out]

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

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Rubi [A]  time = 0.0847633, antiderivative size = 31, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.095, Rules used = {2670, 2671} \[ \frac{a^6 \cos ^3(c+d x)}{3 d (a-a \sin (c+d x))^3} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^4*(a + a*Sin[c + d*x])^3,x]

[Out]

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

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 2671

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

Rubi steps

\begin{align*} \int \sec ^4(c+d x) (a+a \sin (c+d x))^3 \, dx &=a^6 \int \frac{\cos ^2(c+d x)}{(a-a \sin (c+d x))^3} \, dx\\ &=\frac{a^6 \cos ^3(c+d x)}{3 d (a-a \sin (c+d x))^3}\\ \end{align*}

Mathematica [A]  time = 0.0248598, size = 28, normalized size = 0.9 \[ \frac{a^3 (\sin (c+d x)+1)^3 \sec ^3(c+d x)}{3 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^4*(a + a*Sin[c + d*x])^3,x]

[Out]

(a^3*Sec[c + d*x]^3*(1 + Sin[c + d*x])^3)/(3*d)

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Maple [B]  time = 0.07, size = 120, normalized size = 3.9 \begin{align*}{\frac{1}{d} \left ({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 ) +{\frac{{a}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{ \left ( \cos \left ( dx+c \right ) \right ) ^{3}}}+{\frac{{a}^{3}}{ \left ( \cos \left ( dx+c \right ) \right ) ^{3}}}-{a}^{3} \left ( -{\frac{2}{3}}-{\frac{ \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{3}} \right ) \tan \left ( dx+c \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 0.967453, size = 105, normalized size = 3.39 \begin{align*} \frac{3 \, a^{3} \tan \left (d x + c\right )^{3} +{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} a^{3} - \frac{{\left (3 \, \cos \left (d x + c\right )^{2} - 1\right )} a^{3}}{\cos \left (d x + c\right )^{3}} + \frac{3 \, 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*(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*tan(d*x + c))*a^3 - (3*cos(d*x + c)^2 - 1)*a^3/cos(d*x + c)^3
+ 3*a^3/cos(d*x + c)^3)/d

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Fricas [B]  time = 1.60187, size = 236, normalized size = 7.61 \begin{align*} \frac{a^{3} \cos \left (d x + c\right )^{2} - a^{3} \cos \left (d x + c\right ) - 2 \, a^{3} -{\left (a^{3} \cos \left (d x + c\right ) + 2 \, a^{3}\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*(a+a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

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

[Out]

Timed out

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Giac [A]  time = 1.16137, size = 51, normalized size = 1.65 \begin{align*} -\frac{2 \,{\left (3 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a^{3}\right )}}{3 \, d{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1\right )}^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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