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

Optimal. Leaf size=62 \[ \frac{a^2 \cos ^3(c+d x)}{3 d}+\frac{a^2 \cos ^2(c+d x)}{d}+\frac{a^2 \sec (c+d x)}{d}-\frac{2 a^2 \log (\cos (c+d x))}{d} \]

[Out]

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

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Rubi [A]  time = 0.123591, antiderivative size = 62, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {3872, 2836, 12, 75} \[ \frac{a^2 \cos ^3(c+d x)}{3 d}+\frac{a^2 \cos ^2(c+d x)}{d}+\frac{a^2 \sec (c+d x)}{d}-\frac{2 a^2 \log (\cos (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3872

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[((g*C
os[e + f*x])^p*(b + a*Sin[e + f*x])^m)/Sin[e + f*x]^m, x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]

Rule 2836

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)
*(x_)])^(n_.), x_Symbol] :> Dist[1/(b^p*f), Subst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2)*(c + (d*x)/b
)^n, x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, c, d, m, n}, x] && IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2,
0]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 75

Int[((d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_))*((e_) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*
x)*(d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && IGtQ[p, 0] && EqQ[b*e + a*f, 0] &&  !(ILtQ[n
+ p + 2, 0] && GtQ[n + 2*p, 0])

Rubi steps

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

Mathematica [A]  time = 0.197132, size = 65, normalized size = 1.05 \[ \frac{a^2 \sec (c+d x) (4 \cos (2 (c+d x))+6 \cos (3 (c+d x))+\cos (4 (c+d x))-6 \cos (c+d x) (8 \log (\cos (c+d x))+1)+27)}{24 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(27 + 4*Cos[2*(c + d*x)] + 6*Cos[3*(c + d*x)] + Cos[4*(c + d*x)] - 6*Cos[c + d*x]*(1 + 8*Log[Cos[c + d*x]
]))*Sec[c + d*x])/(24*d)

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Maple [A]  time = 0.04, size = 92, normalized size = 1.5 \begin{align*}{\frac{2\,{a}^{2}\cos \left ( dx+c \right ) \left ( \sin \left ( dx+c \right ) \right ) ^{2}}{3\,d}}+{\frac{4\,{a}^{2}\cos \left ( dx+c \right ) }{3\,d}}-{\frac{{a}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{2}}{d}}-2\,{\frac{{a}^{2}\ln \left ( \cos \left ( dx+c \right ) \right ) }{d}}+{\frac{{a}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{d\cos \left ( dx+c \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sec(d*x+c))^2*sin(d*x+c)^3,x)

[Out]

2/3/d*a^2*cos(d*x+c)*sin(d*x+c)^2+4/3*a^2*cos(d*x+c)/d-1/d*a^2*sin(d*x+c)^2-2*a^2*ln(cos(d*x+c))/d+1/d*a^2*sin
(d*x+c)^4/cos(d*x+c)

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Maxima [A]  time = 0.993747, size = 76, normalized size = 1.23 \begin{align*} \frac{a^{2} \cos \left (d x + c\right )^{3} + 3 \, a^{2} \cos \left (d x + c\right )^{2} - 6 \, a^{2} \log \left (\cos \left (d x + c\right )\right ) + \frac{3 \, a^{2}}{\cos \left (d x + c\right )}}{3 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^2*sin(d*x+c)^3,x, algorithm="maxima")

[Out]

1/3*(a^2*cos(d*x + c)^3 + 3*a^2*cos(d*x + c)^2 - 6*a^2*log(cos(d*x + c)) + 3*a^2/cos(d*x + c))/d

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Fricas [A]  time = 1.77674, size = 186, normalized size = 3. \begin{align*} \frac{2 \, a^{2} \cos \left (d x + c\right )^{4} + 6 \, a^{2} \cos \left (d x + c\right )^{3} - 12 \, a^{2} \cos \left (d x + c\right ) \log \left (-\cos \left (d x + c\right )\right ) - 3 \, a^{2} \cos \left (d x + c\right ) + 6 \, a^{2}}{6 \, d \cos \left (d x + c\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^2*sin(d*x+c)^3,x, algorithm="fricas")

[Out]

1/6*(2*a^2*cos(d*x + c)^4 + 6*a^2*cos(d*x + c)^3 - 12*a^2*cos(d*x + c)*log(-cos(d*x + c)) - 3*a^2*cos(d*x + c)
+ 6*a^2)/(d*cos(d*x + c))

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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((a+a*sec(d*x+c))**2*sin(d*x+c)**3,x)

[Out]

Timed out

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Giac [A]  time = 1.4018, size = 100, normalized size = 1.61 \begin{align*} -\frac{2 \, a^{2} \log \left (\frac{{\left | \cos \left (d x + c\right ) \right |}}{{\left | d \right |}}\right )}{d} + \frac{a^{2}}{d \cos \left (d x + c\right )} + \frac{a^{2} d^{5} \cos \left (d x + c\right )^{3} + 3 \, a^{2} d^{5} \cos \left (d x + c\right )^{2}}{3 \, d^{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^2*sin(d*x+c)^3,x, algorithm="giac")

[Out]

-2*a^2*log(abs(cos(d*x + c))/abs(d))/d + a^2/(d*cos(d*x + c)) + 1/3*(a^2*d^5*cos(d*x + c)^3 + 3*a^2*d^5*cos(d*
x + c)^2)/d^6