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

Optimal. Leaf size=158 \[ -\frac{b^3 \left (a^2-b^2\right )}{2 a^6 d (a \cos (c+d x)+b)^2}+\frac{b^2 \left (3 a^2-5 b^2\right )}{a^6 d (a \cos (c+d x)+b)}-\frac{\left (a^2-6 b^2\right ) \cos (c+d x)}{a^5 d}+\frac{b \left (3 a^2-10 b^2\right ) \log (a \cos (c+d x)+b)}{a^6 d}-\frac{3 b \cos ^2(c+d x)}{2 a^4 d}+\frac{\cos ^3(c+d x)}{3 a^3 d} \]

[Out]

-(((a^2 - 6*b^2)*Cos[c + d*x])/(a^5*d)) - (3*b*Cos[c + d*x]^2)/(2*a^4*d) + Cos[c + d*x]^3/(3*a^3*d) - (b^3*(a^
2 - b^2))/(2*a^6*d*(b + a*Cos[c + d*x])^2) + (b^2*(3*a^2 - 5*b^2))/(a^6*d*(b + a*Cos[c + d*x])) + (b*(3*a^2 -
10*b^2)*Log[b + a*Cos[c + d*x]])/(a^6*d)

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Rubi [A]  time = 0.271682, antiderivative size = 158, 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, 894} \[ -\frac{b^3 \left (a^2-b^2\right )}{2 a^6 d (a \cos (c+d x)+b)^2}+\frac{b^2 \left (3 a^2-5 b^2\right )}{a^6 d (a \cos (c+d x)+b)}-\frac{\left (a^2-6 b^2\right ) \cos (c+d x)}{a^5 d}+\frac{b \left (3 a^2-10 b^2\right ) \log (a \cos (c+d x)+b)}{a^6 d}-\frac{3 b \cos ^2(c+d x)}{2 a^4 d}+\frac{\cos ^3(c+d x)}{3 a^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((a^2 - 6*b^2)*Cos[c + d*x])/(a^5*d)) - (3*b*Cos[c + d*x]^2)/(2*a^4*d) + Cos[c + d*x]^3/(3*a^3*d) - (b^3*(a^
2 - b^2))/(2*a^6*d*(b + a*Cos[c + d*x])^2) + (b^2*(3*a^2 - 5*b^2))/(a^6*d*(b + a*Cos[c + d*x])) + (b*(3*a^2 -
10*b^2)*Log[b + a*Cos[c + d*x]])/(a^6*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 894

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIn
tegrand[(d + e*x)^m*(f + g*x)^n*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] &&
NeQ[c*d^2 + a*e^2, 0] && IntegerQ[p] && ((EqQ[p, 1] && IntegersQ[m, n]) || (ILtQ[m, 0] && ILtQ[n, 0]))

Rubi steps

\begin{align*} \int \frac{\sin ^3(c+d x)}{(a+b \sec (c+d x))^3} \, dx &=-\int \frac{\cos ^3(c+d x) \sin ^3(c+d x)}{(-b-a \cos (c+d x))^3} \, dx\\ &=\frac{\operatorname{Subst}\left (\int \frac{x^3 \left (a^2-x^2\right )}{a^3 (-b+x)^3} \, dx,x,-a \cos (c+d x)\right )}{a^3 d}\\ &=\frac{\operatorname{Subst}\left (\int \frac{x^3 \left (a^2-x^2\right )}{(-b+x)^3} \, dx,x,-a \cos (c+d x)\right )}{a^6 d}\\ &=\frac{\operatorname{Subst}\left (\int \left (a^2 \left (1-\frac{6 b^2}{a^2}\right )+\frac{-a^2 b^3+b^5}{(b-x)^3}+\frac{3 a^2 b^2-5 b^4}{(b-x)^2}+\frac{-3 a^2 b+10 b^3}{b-x}-3 b x-x^2\right ) \, dx,x,-a \cos (c+d x)\right )}{a^6 d}\\ &=-\frac{\left (a^2-6 b^2\right ) \cos (c+d x)}{a^5 d}-\frac{3 b \cos ^2(c+d x)}{2 a^4 d}+\frac{\cos ^3(c+d x)}{3 a^3 d}-\frac{b^3 \left (a^2-b^2\right )}{2 a^6 d (b+a \cos (c+d x))^2}+\frac{b^2 \left (3 a^2-5 b^2\right )}{a^6 d (b+a \cos (c+d x))}+\frac{b \left (3 a^2-10 b^2\right ) \log (b+a \cos (c+d x))}{a^6 d}\\ \end{align*}

Mathematica [A]  time = 0.921472, size = 208, normalized size = 1.32 \[ \frac{\sec ^3(c+d x) (a \cos (c+d x)+b) \left (9 a^4 (2 a \cos (c+d x)+b)-(a \cos (c+d x)+b)^2 \left (72 a \left (a^2-8 b^2\right ) \cos (c+d x)+\frac{6 \left (-48 a^2 b^2+3 a^4+80 b^4\right )}{a \cos (c+d x)+b}+\frac{48 a^2 b^3-9 a^4 b-48 b^5}{(a \cos (c+d x)+b)^2}+96 \left (10 b^3-3 a^2 b\right ) \log (a \cos (c+d x)+b)+72 a^2 b \cos (2 (c+d x))-8 a^3 \cos (3 (c+d x))\right )\right )}{96 a^6 d (a+b \sec (c+d x))^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((b + a*Cos[c + d*x])*(9*a^4*(b + 2*a*Cos[c + d*x]) - (b + a*Cos[c + d*x])^2*(72*a*(a^2 - 8*b^2)*Cos[c + d*x]
+ (-9*a^4*b + 48*a^2*b^3 - 48*b^5)/(b + a*Cos[c + d*x])^2 + (6*(3*a^4 - 48*a^2*b^2 + 80*b^4))/(b + a*Cos[c + d
*x]) + 72*a^2*b*Cos[2*(c + d*x)] - 8*a^3*Cos[3*(c + d*x)] + 96*(-3*a^2*b + 10*b^3)*Log[b + a*Cos[c + d*x]]))*S
ec[c + d*x]^3)/(96*a^6*d*(a + b*Sec[c + d*x])^3)

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Maple [A]  time = 0.062, size = 200, normalized size = 1.3 \begin{align*}{\frac{ \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{3\,{a}^{3}d}}-{\frac{3\,b \left ( \cos \left ( dx+c \right ) \right ) ^{2}}{2\,{a}^{4}d}}-{\frac{\cos \left ( dx+c \right ) }{{a}^{3}d}}+6\,{\frac{\cos \left ( dx+c \right ){b}^{2}}{d{a}^{5}}}+3\,{\frac{b\ln \left ( b+a\cos \left ( dx+c \right ) \right ) }{{a}^{4}d}}-10\,{\frac{{b}^{3}\ln \left ( b+a\cos \left ( dx+c \right ) \right ) }{d{a}^{6}}}-{\frac{{b}^{3}}{2\,{a}^{4}d \left ( b+a\cos \left ( dx+c \right ) \right ) ^{2}}}+{\frac{{b}^{5}}{2\,d{a}^{6} \left ( b+a\cos \left ( dx+c \right ) \right ) ^{2}}}+3\,{\frac{{b}^{2}}{{a}^{4}d \left ( b+a\cos \left ( dx+c \right ) \right ) }}-5\,{\frac{{b}^{4}}{d{a}^{6} \left ( b+a\cos \left ( dx+c \right ) \right ) }} \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))^3,x)

[Out]

1/3*cos(d*x+c)^3/a^3/d-3/2*b*cos(d*x+c)^2/a^4/d-cos(d*x+c)/a^3/d+6/d/a^5*cos(d*x+c)*b^2+3*b*ln(b+a*cos(d*x+c))
/a^4/d-10/d/a^6*b^3*ln(b+a*cos(d*x+c))-1/2*b^3/a^4/d/(b+a*cos(d*x+c))^2+1/2/d*b^5/a^6/(b+a*cos(d*x+c))^2+3*b^2
/a^4/d/(b+a*cos(d*x+c))-5/d/a^6*b^4/(b+a*cos(d*x+c))

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Maxima [A]  time = 0.968214, size = 208, normalized size = 1.32 \begin{align*} \frac{\frac{3 \,{\left (5 \, a^{2} b^{3} - 9 \, b^{5} + 2 \,{\left (3 \, a^{3} b^{2} - 5 \, a b^{4}\right )} \cos \left (d x + c\right )\right )}}{a^{8} \cos \left (d x + c\right )^{2} + 2 \, a^{7} b \cos \left (d x + c\right ) + a^{6} b^{2}} + \frac{2 \, a^{2} \cos \left (d x + c\right )^{3} - 9 \, a b \cos \left (d x + c\right )^{2} - 6 \,{\left (a^{2} - 6 \, b^{2}\right )} \cos \left (d x + c\right )}{a^{5}} + \frac{6 \,{\left (3 \, a^{2} b - 10 \, b^{3}\right )} \log \left (a \cos \left (d x + c\right ) + b\right )}{a^{6}}}{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))^3,x, algorithm="maxima")

[Out]

1/6*(3*(5*a^2*b^3 - 9*b^5 + 2*(3*a^3*b^2 - 5*a*b^4)*cos(d*x + c))/(a^8*cos(d*x + c)^2 + 2*a^7*b*cos(d*x + c) +
a^6*b^2) + (2*a^2*cos(d*x + c)^3 - 9*a*b*cos(d*x + c)^2 - 6*(a^2 - 6*b^2)*cos(d*x + c))/a^5 + 6*(3*a^2*b - 10
*b^3)*log(a*cos(d*x + c) + b)/a^6)/d

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Fricas [A]  time = 2.15444, size = 524, normalized size = 3.32 \begin{align*} \frac{4 \, a^{5} \cos \left (d x + c\right )^{5} - 10 \, a^{4} b \cos \left (d x + c\right )^{4} + 39 \, a^{2} b^{3} - 54 \, b^{5} - 4 \,{\left (3 \, a^{5} - 10 \, a^{3} b^{2}\right )} \cos \left (d x + c\right )^{3} - 3 \,{\left (5 \, a^{4} b - 42 \, a^{2} b^{3}\right )} \cos \left (d x + c\right )^{2} + 6 \,{\left (7 \, a^{3} b^{2} + 2 \, a b^{4}\right )} \cos \left (d x + c\right ) + 12 \,{\left (3 \, a^{2} b^{3} - 10 \, b^{5} +{\left (3 \, a^{4} b - 10 \, a^{2} b^{3}\right )} \cos \left (d x + c\right )^{2} + 2 \,{\left (3 \, a^{3} b^{2} - 10 \, a b^{4}\right )} \cos \left (d x + c\right )\right )} \log \left (a \cos \left (d x + c\right ) + b\right )}{12 \,{\left (a^{8} d \cos \left (d x + c\right )^{2} + 2 \, a^{7} b d \cos \left (d x + c\right ) + a^{6} b^{2} d\right )}} \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))^3,x, algorithm="fricas")

[Out]

1/12*(4*a^5*cos(d*x + c)^5 - 10*a^4*b*cos(d*x + c)^4 + 39*a^2*b^3 - 54*b^5 - 4*(3*a^5 - 10*a^3*b^2)*cos(d*x +
c)^3 - 3*(5*a^4*b - 42*a^2*b^3)*cos(d*x + c)^2 + 6*(7*a^3*b^2 + 2*a*b^4)*cos(d*x + c) + 12*(3*a^2*b^3 - 10*b^5
+ (3*a^4*b - 10*a^2*b^3)*cos(d*x + c)^2 + 2*(3*a^3*b^2 - 10*a*b^4)*cos(d*x + c))*log(a*cos(d*x + c) + b))/(a^
8*d*cos(d*x + c)^2 + 2*a^7*b*d*cos(d*x + c) + a^6*b^2*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))**3,x)

[Out]

Timed out

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Giac [A]  time = 1.44863, size = 230, normalized size = 1.46 \begin{align*} \frac{{\left (3 \, a^{2} b - 10 \, b^{3}\right )} \log \left ({\left | -a \cos \left (d x + c\right ) - b \right |}\right )}{a^{6} d} + \frac{5 \, a^{2} b^{3} - 9 \, b^{5} + \frac{2 \,{\left (3 \, a^{3} b^{2} d - 5 \, a b^{4} d\right )} \cos \left (d x + c\right )}{d}}{2 \,{\left (a \cos \left (d x + c\right ) + b\right )}^{2} a^{6} d} + \frac{2 \, a^{6} d^{8} \cos \left (d x + c\right )^{3} - 9 \, a^{5} b d^{8} \cos \left (d x + c\right )^{2} - 6 \, a^{6} d^{8} \cos \left (d x + c\right ) + 36 \, a^{4} b^{2} d^{8} \cos \left (d x + c\right )}{6 \, a^{9} d^{9}} \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))^3,x, algorithm="giac")

[Out]

(3*a^2*b - 10*b^3)*log(abs(-a*cos(d*x + c) - b))/(a^6*d) + 1/2*(5*a^2*b^3 - 9*b^5 + 2*(3*a^3*b^2*d - 5*a*b^4*d
)*cos(d*x + c)/d)/((a*cos(d*x + c) + b)^2*a^6*d) + 1/6*(2*a^6*d^8*cos(d*x + c)^3 - 9*a^5*b*d^8*cos(d*x + c)^2
- 6*a^6*d^8*cos(d*x + c) + 36*a^4*b^2*d^8*cos(d*x + c))/(a^9*d^9)