3.198 \(\int \frac{\sin ^3(c+d x)}{a+b \sec (c+d x)} \, dx\)

Optimal. Leaf size=89 \[ -\frac{\left (a^2-b^2\right ) \cos (c+d x)}{a^3 d}+\frac{b \left (a^2-b^2\right ) \log (a \cos (c+d x)+b)}{a^4 d}-\frac{b \cos ^2(c+d x)}{2 a^2 d}+\frac{\cos ^3(c+d x)}{3 a d} \]

[Out]

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

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Rubi [A]  time = 0.15573, antiderivative size = 89, 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, 2837, 12, 772} \[ -\frac{\left (a^2-b^2\right ) \cos (c+d x)}{a^3 d}+\frac{b \left (a^2-b^2\right ) \log (a \cos (c+d x)+b)}{a^4 d}-\frac{b \cos ^2(c+d x)}{2 a^2 d}+\frac{\cos ^3(c+d x)}{3 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((a^2 - b^2)*Cos[c + d*x])/(a^3*d)) - (b*Cos[c + d*x]^2)/(2*a^2*d) + Cos[c + d*x]^3/(3*a*d) + (b*(a^2 - b^2)
*Log[b + a*Cos[c + d*x]])/(a^4*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 2837

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*(c + (d*x)/b)^n*(b^2 - x^2)^((p - 1)/2), x], x
, b*Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IntegerQ[(p - 1)/2] && NeQ[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 772

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegr
and[(d + e*x)^m*(f + g*x)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && IGtQ[p, 0]

Rubi steps

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

Mathematica [A]  time = 0.187618, size = 89, normalized size = 1. \[ \frac{\left (12 a b^2-9 a^3\right ) \cos (c+d x)-3 a^2 b \cos (2 (c+d x))+12 a^2 b \log (a \cos (c+d x)+b)+a^3 \cos (3 (c+d x))-12 b^3 \log (a \cos (c+d x)+b)}{12 a^4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-9*a^3 + 12*a*b^2)*Cos[c + d*x] - 3*a^2*b*Cos[2*(c + d*x)] + a^3*Cos[3*(c + d*x)] + 12*a^2*b*Log[b + a*Cos[c
+ d*x]] - 12*b^3*Log[b + a*Cos[c + d*x]])/(12*a^4*d)

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Maple [A]  time = 0.041, size = 106, normalized size = 1.2 \begin{align*}{\frac{ \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{3\,ad}}-{\frac{b \left ( \cos \left ( dx+c \right ) \right ) ^{2}}{2\,{a}^{2}d}}-{\frac{\cos \left ( dx+c \right ) }{ad}}+{\frac{{b}^{2}\cos \left ( dx+c \right ) }{d{a}^{3}}}+{\frac{b\ln \left ( b+a\cos \left ( dx+c \right ) \right ) }{{a}^{2}d}}-{\frac{{b}^{3}\ln \left ( b+a\cos \left ( dx+c \right ) \right ) }{d{a}^{4}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.97901, size = 108, normalized size = 1.21 \begin{align*} \frac{\frac{2 \, a^{2} \cos \left (d x + c\right )^{3} - 3 \, a b \cos \left (d x + c\right )^{2} - 6 \,{\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )}{a^{3}} + \frac{6 \,{\left (a^{2} b - b^{3}\right )} \log \left (a \cos \left (d x + c\right ) + b\right )}{a^{4}}}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.80522, size = 181, normalized size = 2.03 \begin{align*} \frac{2 \, a^{3} \cos \left (d x + c\right )^{3} - 3 \, a^{2} b \cos \left (d x + c\right )^{2} - 6 \,{\left (a^{3} - a b^{2}\right )} \cos \left (d x + c\right ) + 6 \,{\left (a^{2} b - b^{3}\right )} \log \left (a \cos \left (d x + c\right ) + b\right )}{6 \, a^{4} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(2*a^3*cos(d*x + c)^3 - 3*a^2*b*cos(d*x + c)^2 - 6*(a^3 - a*b^2)*cos(d*x + c) + 6*(a^2*b - b^3)*log(a*cos(
d*x + c) + b))/(a^4*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(sin(d*x+c)**3/(a+b*sec(d*x+c)),x)

[Out]

Timed out

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Giac [A]  time = 1.30075, size = 138, normalized size = 1.55 \begin{align*} \frac{{\left (a^{2} b - b^{3}\right )} \log \left ({\left | -a \cos \left (d x + c\right ) - b \right |}\right )}{a^{4} d} + \frac{2 \, a^{2} d^{2} \cos \left (d x + c\right )^{3} - 3 \, a b d^{2} \cos \left (d x + c\right )^{2} - 6 \, a^{2} d^{2} \cos \left (d x + c\right ) + 6 \, b^{2} d^{2} \cos \left (d x + c\right )}{6 \, a^{3} d^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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