3.21 \(\int \sec ^3(c+d x) (a+a \sin (c+d x))^2 \, dx\)

Optimal. Leaf size=20 \[ \frac{a^3}{d (a-a \sin (c+d x))} \]

[Out]

a^3/(d*(a - a*Sin[c + d*x]))

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Rubi [A]  time = 0.0376204, antiderivative size = 20, 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 = {2667, 32} \[ \frac{a^3}{d (a-a \sin (c+d x))} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^3*(a + a*Sin[c + d*x])^2,x]

[Out]

a^3/(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 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]

Rubi steps

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

Mathematica [A]  time = 0.138102, size = 32, normalized size = 1.6 \[ \frac{a^2}{d \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^3*(a + a*Sin[c + d*x])^2,x]

[Out]

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

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Maple [B]  time = 0.062, size = 75, normalized size = 3.8 \begin{align*}{\frac{{a}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{2\,d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}+{\frac{{a}^{2}\sin \left ( dx+c \right ) }{2\,d}}+{\frac{{a}^{2}}{d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}+{\frac{{a}^{2}\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))^2,x)

[Out]

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

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Maxima [A]  time = 0.932716, size = 24, normalized size = 1.2 \begin{align*} -\frac{a^{2}}{d{\left (\sin \left (d x + c\right ) - 1\right )}} \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))^2,x, algorithm="maxima")

[Out]

-a^2/(d*(sin(d*x + c) - 1))

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Fricas [A]  time = 1.54091, size = 36, normalized size = 1.8 \begin{align*} -\frac{a^{2}}{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))^2,x, algorithm="fricas")

[Out]

-a^2/(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))**2,x)

[Out]

Timed out

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Giac [A]  time = 1.17787, size = 41, normalized size = 2.05 \begin{align*} \frac{2 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{d{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1\right )}^{2}} \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))^2,x, algorithm="giac")

[Out]

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