3.37 \(\int \sec ^5(c+d x) (a+a \sin (c+d x))^3 \, dx\)

Optimal. Leaf size=23 \[ \frac {a^5}{2 d (a-a \sin (c+d x))^2} \]

[Out]

1/2*a^5/d/(a-a*sin(d*x+c))^2

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Rubi [A]  time = 0.04, antiderivative size = 23, 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 = {2667, 32} \[ \frac {a^5}{2 d (a-a \sin (c+d x))^2} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^5*(a + a*Sin[c + d*x])^3,x]

[Out]

a^5/(2*d*(a - a*Sin[c + d*x])^2)

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rule 2667

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

Rubi steps

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

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Mathematica [A]  time = 0.23, size = 35, normalized size = 1.52 \[ \frac {a^3}{2 d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^4} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^5*(a + a*Sin[c + d*x])^3,x]

[Out]

a^3/(2*d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^4)

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fricas [A]  time = 0.53, size = 30, normalized size = 1.30 \[ -\frac {a^{3}}{2 \, {\left (d \cos \left (d x + c\right )^{2} + 2 \, d \sin \left (d x + c\right ) - 2 \, d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5*(a+a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.70, size = 63, normalized size = 2.74 \[ \frac {2 \, {\left (a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{d {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5*(a+a*sin(d*x+c))^3,x, algorithm="giac")

[Out]

2*(a^3*tan(1/2*d*x + 1/2*c)^3 - a^3*tan(1/2*d*x + 1/2*c)^2 + a^3*tan(1/2*d*x + 1/2*c))/(d*(tan(1/2*d*x + 1/2*c
) - 1)^4)

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maple [B]  time = 0.24, size = 146, normalized size = 6.35 \[ \frac {a^{3} \left (\sin ^{4}\left (d x +c \right )\right )}{4 d \cos \left (d x +c \right )^{4}}+\frac {3 a^{3} \left (\sin ^{3}\left (d x +c \right )\right )}{4 d \cos \left (d x +c \right )^{4}}+\frac {3 a^{3} \left (\sin ^{3}\left (d x +c \right )\right )}{8 d \cos \left (d x +c \right )^{2}}+\frac {3 a^{3} \sin \left (d x +c \right )}{8 d}+\frac {3 a^{3}}{4 d \cos \left (d x +c \right )^{4}}+\frac {a^{3} \tan \left (d x +c \right ) \left (\sec ^{3}\left (d x +c \right )\right )}{4 d}+\frac {3 a^{3} \sec \left (d x +c \right ) \tan \left (d x +c \right )}{8 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^5*(a+a*sin(d*x+c))^3,x)

[Out]

1/4/d*a^3*sin(d*x+c)^4/cos(d*x+c)^4+3/4/d*a^3*sin(d*x+c)^3/cos(d*x+c)^4+3/8/d*a^3*sin(d*x+c)^3/cos(d*x+c)^2+3/
8*a^3*sin(d*x+c)/d+3/4/d*a^3/cos(d*x+c)^4+1/4/d*a^3*tan(d*x+c)*sec(d*x+c)^3+3/8/d*a^3*sec(d*x+c)*tan(d*x+c)

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maxima [A]  time = 0.66, size = 28, normalized size = 1.22 \[ \frac {a^{3}}{2 \, {\left (\sin \left (d x + c\right )^{2} - 2 \, \sin \left (d x + c\right ) + 1\right )} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5*(a+a*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.07, size = 18, normalized size = 0.78 \[ \frac {a^3}{2\,d\,{\left (\sin \left (c+d\,x\right )-1\right )}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*sin(c + d*x))^3/cos(c + d*x)^5,x)

[Out]

a^3/(2*d*(sin(c + d*x) - 1)^2)

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

[Out]

Timed out

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