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

Optimal. Leaf size=40 \[ \frac{2 a^4}{d (a-a \sin (c+d x))}+\frac{a^3 \log (1-\sin (c+d x))}{d} \]

[Out]

(a^3*Log[1 - Sin[c + d*x]])/d + (2*a^4)/(d*(a - a*Sin[c + d*x]))

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Rubi [A]  time = 0.0522718, antiderivative size = 40, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.095, Rules used = {2667, 43} \[ \frac{2 a^4}{d (a-a \sin (c+d x))}+\frac{a^3 \log (1-\sin (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*Log[1 - Sin[c + d*x]])/d + (2*a^4)/(d*(a - a*Sin[c + d*x]))

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])

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \sec ^3(c+d x) (a+a \sin (c+d x))^3 \, dx &=\frac{a^3 \operatorname{Subst}\left (\int \frac{a+x}{(a-x)^2} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac{a^3 \operatorname{Subst}\left (\int \left (\frac{2 a}{(a-x)^2}+\frac{1}{-a+x}\right ) \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac{a^3 \log (1-\sin (c+d x))}{d}+\frac{2 a^4}{d (a-a \sin (c+d x))}\\ \end{align*}

Mathematica [A]  time = 0.0410921, size = 59, normalized size = 1.48 \[ \frac{a^3 (1-\sin (c+d x)) (\sin (c+d x)+1) \sec ^2(c+d x) \left (\frac{2}{1-\sin (c+d x)}+\log (1-\sin (c+d x))\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*Sec[c + d*x]^2*(Log[1 - Sin[c + d*x]] + 2/(1 - Sin[c + d*x]))*(1 - Sin[c + d*x])*(1 + Sin[c + d*x]))/d

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Maple [B]  time = 0.069, size = 128, normalized size = 3.2 \begin{align*}{\frac{{a}^{3} \left ( \tan \left ( dx+c \right ) \right ) ^{2}}{2\,d}}+{\frac{{a}^{3}\ln \left ( \cos \left ( dx+c \right ) \right ) }{d}}+{\frac{3\,{a}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{2\,d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}+{\frac{3\,{a}^{3}\sin \left ( dx+c \right ) }{2\,d}}-{\frac{{a}^{3}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{\frac{3\,{a}^{3}}{2\,d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}+{\frac{{a}^{3}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.95084, size = 45, normalized size = 1.12 \begin{align*} \frac{a^{3} \log \left (\sin \left (d x + c\right ) - 1\right ) - \frac{2 \, a^{3}}{\sin \left (d x + c\right ) - 1}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.6929, size = 109, normalized size = 2.72 \begin{align*} -\frac{2 \, a^{3} -{\left (a^{3} \sin \left (d x + c\right ) - a^{3}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right )}{d \sin \left (d x + c\right ) - d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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Giac [B]  time = 1.20148, size = 124, normalized size = 3.1 \begin{align*} -\frac{a^{3} \log \left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right ) - 2 \, a^{3} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) + \frac{3 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 10 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 3 \, a^{3}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1\right )}^{2}}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(a^3*log(tan(1/2*d*x + 1/2*c)^2 + 1) - 2*a^3*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + (3*a^3*tan(1/2*d*x + 1/2*c)
^2 - 10*a^3*tan(1/2*d*x + 1/2*c) + 3*a^3)/(tan(1/2*d*x + 1/2*c) - 1)^2)/d