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]

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

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Rubi [A]  time = 0.08, antiderivative size = 31, normalized size of antiderivative = 1.00, 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*}

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Mathematica [A]  time = 0.03, size = 28, normalized size = 0.90 \[ \frac {a^3 (\sin (c+d x)+1)^3 \sec ^3(c+d x)}{3 d} \]

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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fricas [B]  time = 0.67, size = 99, normalized size = 3.19 \[ \frac {a^{3} \cos \left (d x + c\right )^{2} - a^{3} \cos \left (d x + c\right ) - 2 \, a^{3} - {\left (a^{3} \cos \left (d x + c\right ) + 2 \, a^{3}\right )} \sin \left (d x + c\right )}{3 \, {\left (d \cos \left (d x + c\right )^{2} - d \cos \left (d x + c\right ) + {\left (d \cos \left (d x + c\right ) + 2 \, d\right )} \sin \left (d x + c\right ) - 2 \, d\right )}} \]

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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giac [A]  time = 1.52, size = 38, normalized size = 1.23 \[ -\frac {2 \, {\left (3 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a^{3}\right )}}{3 \, d {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}^{3}} \]

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)

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maple [B]  time = 0.26, size = 120, normalized size = 3.87 \[ \frac {a^{3} \left (\frac {\sin ^{4}\left (d x +c \right )}{3 \cos \left (d x +c \right )^{3}}-\frac {\sin ^{4}\left (d x +c \right )}{3 \cos \left (d x +c \right )}-\frac {\left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )}{3}\right )+\frac {a^{3} \left (\sin ^{3}\left (d x +c \right )\right )}{\cos \left (d x +c \right )^{3}}+\frac {a^{3}}{\cos \left (d x +c \right )^{3}}-a^{3} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )}{d} \]

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/cos(d
*x+c)^3*sin(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 = 0.48, size = 78, normalized size = 2.52 \[ \frac {3 \, a^{3} \tan \left (d x + c\right )^{3} + {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} a^{3} - \frac {{\left (3 \, \cos \left (d x + c\right )^{2} - 1\right )} a^{3}}{\cos \left (d x + c\right )^{3}} + \frac {3 \, a^{3}}{\cos \left (d x + c\right )^{3}}}{3 \, d} \]

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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mupad [B]  time = 4.59, size = 55, normalized size = 1.77 \[ -\frac {2\,a^3\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-3\right )}{3\,d\,{\left (\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )-\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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