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

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

[Out]

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

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Rubi [A]  time = 0.0914808, antiderivative size = 62, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, Rules used = {3872, 2833, 12, 43} \[ -\frac{a^3 \cos (c+d x)}{d}+\frac{a^3 \sec ^2(c+d x)}{2 d}+\frac{3 a^3 \sec (c+d x)}{d}-\frac{3 a^3 \log (\cos (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a^3*Cos[c + d*x])/d) - (3*a^3*Log[Cos[c + d*x]])/d + (3*a^3*Sec[c + d*x])/d + (a^3*Sec[c + d*x]^2)/(2*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 2833

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

Rule 12

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

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

Mathematica [A]  time = 0.228813, size = 65, normalized size = 1.05 \[ -\frac{a^3 \sec ^2(c+d x) (-9 \cos (c+d x)+\cos (3 (c+d x))+6 \log (\cos (c+d x))+\cos (2 (c+d x)) (6 \log (\cos (c+d x))-2)-4)}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a^3*(-4 - 9*Cos[c + d*x] + Cos[3*(c + d*x)] + 6*Log[Cos[c + d*x]] + Cos[2*(c + d*x)]*(-2 + 6*Log[Cos[c + d*x
]]))*Sec[c + d*x]^2)/(4*d)

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Maple [A]  time = 0.021, size = 63, normalized size = 1. \begin{align*}{\frac{{a}^{3} \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{2\,d}}+3\,{\frac{{a}^{3}\sec \left ( dx+c \right ) }{d}}+3\,{\frac{{a}^{3}\ln \left ( \sec \left ( dx+c \right ) \right ) }{d}}-{\frac{{a}^{3}}{d\sec \left ( dx+c \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.986018, size = 74, normalized size = 1.19 \begin{align*} -\frac{2 \, a^{3} \cos \left (d x + c\right ) + 6 \, a^{3} \log \left (\cos \left (d x + c\right )\right ) - \frac{6 \, a^{3}}{\cos \left (d x + c\right )} - \frac{a^{3}}{\cos \left (d x + c\right )^{2}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.84729, size = 158, normalized size = 2.55 \begin{align*} -\frac{2 \, a^{3} \cos \left (d x + c\right )^{3} + 6 \, a^{3} \cos \left (d x + c\right )^{2} \log \left (-\cos \left (d x + c\right )\right ) - 6 \, a^{3} \cos \left (d x + c\right ) - a^{3}}{2 \, d \cos \left (d x + c\right )^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} a^{3} \left (\int 3 \sin{\left (c + d x \right )} \sec{\left (c + d x \right )}\, dx + \int 3 \sin{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int \sin{\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}\, dx + \int \sin{\left (c + d x \right )}\, dx\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.29489, size = 86, normalized size = 1.39 \begin{align*} -\frac{a^{3} \cos \left (d x + c\right )}{d} - \frac{3 \, a^{3} \log \left (\frac{{\left | \cos \left (d x + c\right ) \right |}}{{\left | d \right |}}\right )}{d} + \frac{6 \, a^{3} \cos \left (d x + c\right ) + a^{3}}{2 \, d \cos \left (d x + c\right )^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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