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

Optimal. Leaf size=43 \[ -\frac{a^2 \cos (c+d x)}{d}+\frac{a^2 \sec (c+d x)}{d}-\frac{2 a^2 \log (\cos (c+d x))}{d} \]

[Out]

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

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Rubi [A]  time = 0.0767932, antiderivative size = 43, 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} \[ -\frac{a^2 \cos (c+d x)}{d}+\frac{a^2 \sec (c+d x)}{d}-\frac{2 a^2 \log (\cos (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.112216, size = 31, normalized size = 0.72 \[ \frac{a^2 (\sin (c+d x) \tan (c+d x)-2 \log (\cos (c+d x))+1)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.02, size = 46, normalized size = 1.1 \begin{align*}{\frac{{a}^{2}\sec \left ( dx+c \right ) }{d}}+2\,{\frac{{a}^{2}\ln \left ( \sec \left ( dx+c \right ) \right ) }{d}}-{\frac{{a}^{2}}{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))^2*sin(d*x+c),x)

[Out]

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

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Maxima [A]  time = 1.00563, size = 55, normalized size = 1.28 \begin{align*} -\frac{a^{2} \cos \left (d x + c\right ) + 2 \, a^{2} \log \left (\cos \left (d x + c\right )\right ) - \frac{a^{2}}{\cos \left (d x + c\right )}}{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),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.75952, size = 116, normalized size = 2.7 \begin{align*} -\frac{a^{2} \cos \left (d x + c\right )^{2} + 2 \, a^{2} \cos \left (d x + c\right ) \log \left (-\cos \left (d x + c\right )\right ) - a^{2}}{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),x, algorithm="fricas")

[Out]

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

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

[Out]

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

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Giac [A]  time = 1.38293, size = 69, normalized size = 1.6 \begin{align*} -\frac{a^{2} \cos \left (d x + c\right )}{d} - \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 )} \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),x, algorithm="giac")

[Out]

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