3.646 \(\int \frac{\cos ^6(c+d x) \sin (c+d x)}{(a+a \sin (c+d x))^3} \, dx\)

Optimal. Leaf size=84 \[ \frac{\cos ^3(c+d x)}{a^3 d}-\frac{4 \cos (c+d x)}{a^3 d}+\frac{\sin ^3(c+d x) \cos (c+d x)}{4 a^3 d}+\frac{15 \sin (c+d x) \cos (c+d x)}{8 a^3 d}-\frac{15 x}{8 a^3} \]

[Out]

(-15*x)/(8*a^3) - (4*Cos[c + d*x])/(a^3*d) + Cos[c + d*x]^3/(a^3*d) + (15*Cos[c + d*x]*Sin[c + d*x])/(8*a^3*d)
+ (Cos[c + d*x]*Sin[c + d*x]^3)/(4*a^3*d)

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Rubi [A]  time = 0.165566, antiderivative size = 105, normalized size of antiderivative = 1.25, number of steps used = 5, number of rules used = 5, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.185, Rules used = {2859, 2679, 2682, 2635, 8} \[ -\frac{5 \cos ^3(c+d x)}{4 a^3 d}-\frac{3 \cos ^5(c+d x)}{4 d \left (a^3 \sin (c+d x)+a^3\right )}-\frac{15 \sin (c+d x) \cos (c+d x)}{8 a^3 d}-\frac{15 x}{8 a^3}-\frac{\cos ^7(c+d x)}{d (a \sin (c+d x)+a)^3} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]^6*Sin[c + d*x])/(a + a*Sin[c + d*x])^3,x]

[Out]

(-15*x)/(8*a^3) - (5*Cos[c + d*x]^3)/(4*a^3*d) - (15*Cos[c + d*x]*Sin[c + d*x])/(8*a^3*d) - Cos[c + d*x]^7/(d*
(a + a*Sin[c + d*x])^3) - (3*Cos[c + d*x]^5)/(4*d*(a^3 + a^3*Sin[c + d*x]))

Rule 2859

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> Simp[((b*c - a*d)*(g*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^m)/(a*f*g*(2*m +
p + 1)), x] + Dist[(a*d*m + b*c*(m + p + 1))/(a*b*(2*m + p + 1)), Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^
(m + 1), x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0] && (LtQ[m, -1] || ILtQ[Simplify[
m + p], 0]) && NeQ[2*m + p + 1, 0]

Rule 2679

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(g*(g*
Cos[e + f*x])^(p - 1)*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m + p)), x] + Dist[(g^2*(p - 1))/(a*(m + p)), Int[(g
*Cos[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{a, b, e, f, g}, x] && EqQ[a^2 - b^2, 0]
&& LtQ[m, -1] && GtQ[p, 1] && (GtQ[m, -2] || EqQ[2*m + p + 1, 0] || (EqQ[m, -2] && IntegerQ[p])) && NeQ[m + p,
0] && IntegersQ[2*m, 2*p]

Rule 2682

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(g*(g*Cos[e
+ f*x])^(p - 1))/(b*f*(p - 1)), x] + Dist[g^2/a, Int[(g*Cos[e + f*x])^(p - 2), x], x] /; FreeQ[{a, b, e, f, g
}, x] && EqQ[a^2 - b^2, 0] && GtQ[p, 1] && IntegerQ[2*p]

Rule 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

\begin{align*} \int \frac{\cos ^6(c+d x) \sin (c+d x)}{(a+a \sin (c+d x))^3} \, dx &=-\frac{\cos ^7(c+d x)}{d (a+a \sin (c+d x))^3}-\frac{3 \int \frac{\cos ^6(c+d x)}{(a+a \sin (c+d x))^2} \, dx}{a}\\ &=-\frac{\cos ^7(c+d x)}{d (a+a \sin (c+d x))^3}-\frac{3 \cos ^5(c+d x)}{4 d \left (a^3+a^3 \sin (c+d x)\right )}-\frac{15 \int \frac{\cos ^4(c+d x)}{a+a \sin (c+d x)} \, dx}{4 a^2}\\ &=-\frac{5 \cos ^3(c+d x)}{4 a^3 d}-\frac{\cos ^7(c+d x)}{d (a+a \sin (c+d x))^3}-\frac{3 \cos ^5(c+d x)}{4 d \left (a^3+a^3 \sin (c+d x)\right )}-\frac{15 \int \cos ^2(c+d x) \, dx}{4 a^3}\\ &=-\frac{5 \cos ^3(c+d x)}{4 a^3 d}-\frac{15 \cos (c+d x) \sin (c+d x)}{8 a^3 d}-\frac{\cos ^7(c+d x)}{d (a+a \sin (c+d x))^3}-\frac{3 \cos ^5(c+d x)}{4 d \left (a^3+a^3 \sin (c+d x)\right )}-\frac{15 \int 1 \, dx}{8 a^3}\\ &=-\frac{15 x}{8 a^3}-\frac{5 \cos ^3(c+d x)}{4 a^3 d}-\frac{15 \cos (c+d x) \sin (c+d x)}{8 a^3 d}-\frac{\cos ^7(c+d x)}{d (a+a \sin (c+d x))^3}-\frac{3 \cos ^5(c+d x)}{4 d \left (a^3+a^3 \sin (c+d x)\right )}\\ \end{align*}

Mathematica [B]  time = 1.4017, size = 255, normalized size = 3.04 \[ -\frac{120 d x \sin \left (\frac{c}{2}\right )-104 \sin \left (\frac{c}{2}+d x\right )+104 \sin \left (\frac{3 c}{2}+d x\right )-32 \sin \left (\frac{3 c}{2}+2 d x\right )-32 \sin \left (\frac{5 c}{2}+2 d x\right )+8 \sin \left (\frac{5 c}{2}+3 d x\right )-8 \sin \left (\frac{7 c}{2}+3 d x\right )+\sin \left (\frac{7 c}{2}+4 d x\right )+\sin \left (\frac{9 c}{2}+4 d x\right )+\cos \left (\frac{c}{2}\right ) (120 d x+1)+104 \cos \left (\frac{c}{2}+d x\right )+104 \cos \left (\frac{3 c}{2}+d x\right )-32 \cos \left (\frac{3 c}{2}+2 d x\right )+32 \cos \left (\frac{5 c}{2}+2 d x\right )-8 \cos \left (\frac{5 c}{2}+3 d x\right )-8 \cos \left (\frac{7 c}{2}+3 d x\right )+\cos \left (\frac{7 c}{2}+4 d x\right )-\cos \left (\frac{9 c}{2}+4 d x\right )-\sin \left (\frac{c}{2}\right )}{64 a^3 d \left (\sin \left (\frac{c}{2}\right )+\cos \left (\frac{c}{2}\right )\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]^6*Sin[c + d*x])/(a + a*Sin[c + d*x])^3,x]

[Out]

-((1 + 120*d*x)*Cos[c/2] + 104*Cos[c/2 + d*x] + 104*Cos[(3*c)/2 + d*x] - 32*Cos[(3*c)/2 + 2*d*x] + 32*Cos[(5*c
)/2 + 2*d*x] - 8*Cos[(5*c)/2 + 3*d*x] - 8*Cos[(7*c)/2 + 3*d*x] + Cos[(7*c)/2 + 4*d*x] - Cos[(9*c)/2 + 4*d*x] -
Sin[c/2] + 120*d*x*Sin[c/2] - 104*Sin[c/2 + d*x] + 104*Sin[(3*c)/2 + d*x] - 32*Sin[(3*c)/2 + 2*d*x] - 32*Sin[
(5*c)/2 + 2*d*x] + 8*Sin[(5*c)/2 + 3*d*x] - 8*Sin[(7*c)/2 + 3*d*x] + Sin[(7*c)/2 + 4*d*x] + Sin[(9*c)/2 + 4*d*
x])/(64*a^3*d*(Cos[c/2] + Sin[c/2]))

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Maple [B]  time = 0.109, size = 279, normalized size = 3.3 \begin{align*} -{\frac{15}{4\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{7} \left ( 1+ \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) ^{-4}}-2\,{\frac{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{6}}{d{a}^{3} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{4}}}-{\frac{23}{4\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5} \left ( 1+ \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) ^{-4}}-18\,{\frac{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}}{d{a}^{3} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{4}}}+{\frac{23}{4\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3} \left ( 1+ \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) ^{-4}}-22\,{\frac{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}{d{a}^{3} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{4}}}+{\frac{15}{4\,d{a}^{3}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \left ( 1+ \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) ^{-4}}-6\,{\frac{1}{d{a}^{3} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{4}}}-{\frac{15}{4\,d{a}^{3}}\arctan \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^6*sin(d*x+c)/(a+a*sin(d*x+c))^3,x)

[Out]

-15/4/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^7-2/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2
*c)^6-23/4/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^5-18/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d
*x+1/2*c)^4+23/4/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^3-22/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^4*tan
(1/2*d*x+1/2*c)^2+15/4/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)-6/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^4-
15/4/d/a^3*arctan(tan(1/2*d*x+1/2*c))

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Maxima [B]  time = 1.53987, size = 360, normalized size = 4.29 \begin{align*} \frac{\frac{\frac{15 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{88 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{23 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac{72 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac{23 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac{8 \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} - \frac{15 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} - 24}{a^{3} + \frac{4 \, a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{6 \, a^{3} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac{4 \, a^{3} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac{a^{3} \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}}} - \frac{15 \, \arctan \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}}}{4 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*sin(d*x+c)/(a+a*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

1/4*((15*sin(d*x + c)/(cos(d*x + c) + 1) - 88*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 23*sin(d*x + c)^3/(cos(d*x
+ c) + 1)^3 - 72*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 23*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 8*sin(d*x + c
)^6/(cos(d*x + c) + 1)^6 - 15*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 - 24)/(a^3 + 4*a^3*sin(d*x + c)^2/(cos(d*x +
c) + 1)^2 + 6*a^3*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 4*a^3*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + a^3*sin(d
*x + c)^8/(cos(d*x + c) + 1)^8) - 15*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^3)/d

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Fricas [A]  time = 1.06535, size = 150, normalized size = 1.79 \begin{align*} \frac{8 \, \cos \left (d x + c\right )^{3} - 15 \, d x -{\left (2 \, \cos \left (d x + c\right )^{3} - 17 \, \cos \left (d x + c\right )\right )} \sin \left (d x + c\right ) - 32 \, \cos \left (d x + c\right )}{8 \, a^{3} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*sin(d*x+c)/(a+a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

1/8*(8*cos(d*x + c)^3 - 15*d*x - (2*cos(d*x + c)^3 - 17*cos(d*x + c))*sin(d*x + c) - 32*cos(d*x + c))/(a^3*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(cos(d*x+c)**6*sin(d*x+c)/(a+a*sin(d*x+c))**3,x)

[Out]

Timed out

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Giac [A]  time = 1.37456, size = 171, normalized size = 2.04 \begin{align*} -\frac{\frac{15 \,{\left (d x + c\right )}}{a^{3}} + \frac{2 \,{\left (15 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 8 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{6} + 23 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 72 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 23 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 88 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 15 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 24\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )}^{4} a^{3}}}{8 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*sin(d*x+c)/(a+a*sin(d*x+c))^3,x, algorithm="giac")

[Out]

-1/8*(15*(d*x + c)/a^3 + 2*(15*tan(1/2*d*x + 1/2*c)^7 + 8*tan(1/2*d*x + 1/2*c)^6 + 23*tan(1/2*d*x + 1/2*c)^5 +
72*tan(1/2*d*x + 1/2*c)^4 - 23*tan(1/2*d*x + 1/2*c)^3 + 88*tan(1/2*d*x + 1/2*c)^2 - 15*tan(1/2*d*x + 1/2*c) +
24)/((tan(1/2*d*x + 1/2*c)^2 + 1)^4*a^3))/d