3.1311 \(\int \frac {\cos ^5(c+d x) \sin ^2(c+d x)}{a+b \sin (c+d x)} \, dx\)

Optimal. Leaf size=180 \[ \frac {a^2 \left (a^2-b^2\right )^2 \log (a+b \sin (c+d x))}{b^7 d}-\frac {a \left (a^2-b^2\right )^2 \sin (c+d x)}{b^6 d}+\frac {\left (a^2-b^2\right )^2 \sin ^2(c+d x)}{2 b^5 d}-\frac {a \left (a^2-2 b^2\right ) \sin ^3(c+d x)}{3 b^4 d}+\frac {\left (a^2-2 b^2\right ) \sin ^4(c+d x)}{4 b^3 d}-\frac {a \sin ^5(c+d x)}{5 b^2 d}+\frac {\sin ^6(c+d x)}{6 b d} \]

[Out]

a^2*(a^2-b^2)^2*ln(a+b*sin(d*x+c))/b^7/d-a*(a^2-b^2)^2*sin(d*x+c)/b^6/d+1/2*(a^2-b^2)^2*sin(d*x+c)^2/b^5/d-1/3
*a*(a^2-2*b^2)*sin(d*x+c)^3/b^4/d+1/4*(a^2-2*b^2)*sin(d*x+c)^4/b^3/d-1/5*a*sin(d*x+c)^5/b^2/d+1/6*sin(d*x+c)^6
/b/d

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Rubi [A]  time = 0.21, antiderivative size = 180, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {2837, 12, 948} \[ \frac {\left (a^2-2 b^2\right ) \sin ^4(c+d x)}{4 b^3 d}-\frac {a \left (a^2-2 b^2\right ) \sin ^3(c+d x)}{3 b^4 d}+\frac {\left (a^2-b^2\right )^2 \sin ^2(c+d x)}{2 b^5 d}-\frac {a \left (a^2-b^2\right )^2 \sin (c+d x)}{b^6 d}+\frac {a^2 \left (a^2-b^2\right )^2 \log (a+b \sin (c+d x))}{b^7 d}-\frac {a \sin ^5(c+d x)}{5 b^2 d}+\frac {\sin ^6(c+d x)}{6 b d} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]^5*Sin[c + d*x]^2)/(a + b*Sin[c + d*x]),x]

[Out]

(a^2*(a^2 - b^2)^2*Log[a + b*Sin[c + d*x]])/(b^7*d) - (a*(a^2 - b^2)^2*Sin[c + d*x])/(b^6*d) + ((a^2 - b^2)^2*
Sin[c + d*x]^2)/(2*b^5*d) - (a*(a^2 - 2*b^2)*Sin[c + d*x]^3)/(3*b^4*d) + ((a^2 - 2*b^2)*Sin[c + d*x]^4)/(4*b^3
*d) - (a*Sin[c + d*x]^5)/(5*b^2*d) + Sin[c + d*x]^6/(6*b*d)

Rule 12

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

Rule 948

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] && IGtQ[p, 0] && (IGtQ[m, 0] || (EqQ[m, -2] && EqQ[p, 1] && EqQ[d, 0]))

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]

Rubi steps

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

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Mathematica [A]  time = 0.72, size = 153, normalized size = 0.85 \[ \frac {60 \left (a^3-a b^2\right )^2 \log (a+b \sin (c+d x))+30 b^2 \left (a^2-b^2\right )^2 \sin ^2(c+d x)-60 a b \left (a^2-b^2\right )^2 \sin (c+d x)+15 b^4 \left (a^2-2 b^2\right ) \sin ^4(c+d x)-20 a b^3 \left (a^2-2 b^2\right ) \sin ^3(c+d x)-12 a b^5 \sin ^5(c+d x)+10 b^6 \sin ^6(c+d x)}{60 b^7 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]^5*Sin[c + d*x]^2)/(a + b*Sin[c + d*x]),x]

[Out]

(60*(a^3 - a*b^2)^2*Log[a + b*Sin[c + d*x]] - 60*a*b*(a^2 - b^2)^2*Sin[c + d*x] + 30*b^2*(a^2 - b^2)^2*Sin[c +
d*x]^2 - 20*a*b^3*(a^2 - 2*b^2)*Sin[c + d*x]^3 + 15*b^4*(a^2 - 2*b^2)*Sin[c + d*x]^4 - 12*a*b^5*Sin[c + d*x]^
5 + 10*b^6*Sin[c + d*x]^6)/(60*b^7*d)

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fricas [A]  time = 0.77, size = 164, normalized size = 0.91 \[ -\frac {10 \, b^{6} \cos \left (d x + c\right )^{6} - 15 \, a^{2} b^{4} \cos \left (d x + c\right )^{4} + 30 \, {\left (a^{4} b^{2} - a^{2} b^{4}\right )} \cos \left (d x + c\right )^{2} - 60 \, {\left (a^{6} - 2 \, a^{4} b^{2} + a^{2} b^{4}\right )} \log \left (b \sin \left (d x + c\right ) + a\right ) + 4 \, {\left (3 \, a b^{5} \cos \left (d x + c\right )^{4} + 15 \, a^{5} b - 25 \, a^{3} b^{3} + 8 \, a b^{5} - {\left (5 \, a^{3} b^{3} - 4 \, a b^{5}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{60 \, b^{7} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*sin(d*x+c)^2/(a+b*sin(d*x+c)),x, algorithm="fricas")

[Out]

-1/60*(10*b^6*cos(d*x + c)^6 - 15*a^2*b^4*cos(d*x + c)^4 + 30*(a^4*b^2 - a^2*b^4)*cos(d*x + c)^2 - 60*(a^6 - 2
*a^4*b^2 + a^2*b^4)*log(b*sin(d*x + c) + a) + 4*(3*a*b^5*cos(d*x + c)^4 + 15*a^5*b - 25*a^3*b^3 + 8*a*b^5 - (5
*a^3*b^3 - 4*a*b^5)*cos(d*x + c)^2)*sin(d*x + c))/(b^7*d)

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giac [A]  time = 0.18, size = 213, normalized size = 1.18 \[ \frac {\frac {10 \, b^{5} \sin \left (d x + c\right )^{6} - 12 \, a b^{4} \sin \left (d x + c\right )^{5} + 15 \, a^{2} b^{3} \sin \left (d x + c\right )^{4} - 30 \, b^{5} \sin \left (d x + c\right )^{4} - 20 \, a^{3} b^{2} \sin \left (d x + c\right )^{3} + 40 \, a b^{4} \sin \left (d x + c\right )^{3} + 30 \, a^{4} b \sin \left (d x + c\right )^{2} - 60 \, a^{2} b^{3} \sin \left (d x + c\right )^{2} + 30 \, b^{5} \sin \left (d x + c\right )^{2} - 60 \, a^{5} \sin \left (d x + c\right ) + 120 \, a^{3} b^{2} \sin \left (d x + c\right ) - 60 \, a b^{4} \sin \left (d x + c\right )}{b^{6}} + \frac {60 \, {\left (a^{6} - 2 \, a^{4} b^{2} + a^{2} b^{4}\right )} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{b^{7}}}{60 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*sin(d*x+c)^2/(a+b*sin(d*x+c)),x, algorithm="giac")

[Out]

1/60*((10*b^5*sin(d*x + c)^6 - 12*a*b^4*sin(d*x + c)^5 + 15*a^2*b^3*sin(d*x + c)^4 - 30*b^5*sin(d*x + c)^4 - 2
0*a^3*b^2*sin(d*x + c)^3 + 40*a*b^4*sin(d*x + c)^3 + 30*a^4*b*sin(d*x + c)^2 - 60*a^2*b^3*sin(d*x + c)^2 + 30*
b^5*sin(d*x + c)^2 - 60*a^5*sin(d*x + c) + 120*a^3*b^2*sin(d*x + c) - 60*a*b^4*sin(d*x + c))/b^6 + 60*(a^6 - 2
*a^4*b^2 + a^2*b^4)*log(abs(b*sin(d*x + c) + a))/b^7)/d

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maple [A]  time = 0.29, size = 273, normalized size = 1.52 \[ \frac {\sin ^{6}\left (d x +c \right )}{6 b d}-\frac {a \left (\sin ^{5}\left (d x +c \right )\right )}{5 b^{2} d}+\frac {\left (\sin ^{4}\left (d x +c \right )\right ) a^{2}}{4 d \,b^{3}}-\frac {\sin ^{4}\left (d x +c \right )}{2 b d}-\frac {\left (\sin ^{3}\left (d x +c \right )\right ) a^{3}}{3 d \,b^{4}}+\frac {2 a \left (\sin ^{3}\left (d x +c \right )\right )}{3 b^{2} d}+\frac {\left (\sin ^{2}\left (d x +c \right )\right ) a^{4}}{2 d \,b^{5}}-\frac {a^{2} \left (\sin ^{2}\left (d x +c \right )\right )}{d \,b^{3}}+\frac {\sin ^{2}\left (d x +c \right )}{2 b d}-\frac {\sin \left (d x +c \right ) a^{5}}{d \,b^{6}}+\frac {2 \sin \left (d x +c \right ) a^{3}}{d \,b^{4}}-\frac {a \sin \left (d x +c \right )}{b^{2} d}+\frac {a^{6} \ln \left (a +b \sin \left (d x +c \right )\right )}{d \,b^{7}}-\frac {2 a^{4} \ln \left (a +b \sin \left (d x +c \right )\right )}{d \,b^{5}}+\frac {\ln \left (a +b \sin \left (d x +c \right )\right ) a^{2}}{d \,b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^5*sin(d*x+c)^2/(a+b*sin(d*x+c)),x)

[Out]

1/6*sin(d*x+c)^6/b/d-1/5*a*sin(d*x+c)^5/b^2/d+1/4/d/b^3*sin(d*x+c)^4*a^2-1/2*sin(d*x+c)^4/b/d-1/3/d/b^4*sin(d*
x+c)^3*a^3+2/3*a*sin(d*x+c)^3/b^2/d+1/2/d/b^5*sin(d*x+c)^2*a^4-1/d/b^3*a^2*sin(d*x+c)^2+1/2*sin(d*x+c)^2/b/d-1
/d/b^6*sin(d*x+c)*a^5+2/d/b^4*sin(d*x+c)*a^3-a*sin(d*x+c)/b^2/d+1/d*a^6/b^7*ln(a+b*sin(d*x+c))-2/d*a^4/b^5*ln(
a+b*sin(d*x+c))+1/d/b^3*ln(a+b*sin(d*x+c))*a^2

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maxima [A]  time = 0.31, size = 172, normalized size = 0.96 \[ \frac {\frac {10 \, b^{5} \sin \left (d x + c\right )^{6} - 12 \, a b^{4} \sin \left (d x + c\right )^{5} + 15 \, {\left (a^{2} b^{3} - 2 \, b^{5}\right )} \sin \left (d x + c\right )^{4} - 20 \, {\left (a^{3} b^{2} - 2 \, a b^{4}\right )} \sin \left (d x + c\right )^{3} + 30 \, {\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} \sin \left (d x + c\right )^{2} - 60 \, {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} \sin \left (d x + c\right )}{b^{6}} + \frac {60 \, {\left (a^{6} - 2 \, a^{4} b^{2} + a^{2} b^{4}\right )} \log \left (b \sin \left (d x + c\right ) + a\right )}{b^{7}}}{60 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*sin(d*x+c)^2/(a+b*sin(d*x+c)),x, algorithm="maxima")

[Out]

1/60*((10*b^5*sin(d*x + c)^6 - 12*a*b^4*sin(d*x + c)^5 + 15*(a^2*b^3 - 2*b^5)*sin(d*x + c)^4 - 20*(a^3*b^2 - 2
*a*b^4)*sin(d*x + c)^3 + 30*(a^4*b - 2*a^2*b^3 + b^5)*sin(d*x + c)^2 - 60*(a^5 - 2*a^3*b^2 + a*b^4)*sin(d*x +
c))/b^6 + 60*(a^6 - 2*a^4*b^2 + a^2*b^4)*log(b*sin(d*x + c) + a)/b^7)/d

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mupad [B]  time = 11.78, size = 191, normalized size = 1.06 \[ \frac {{\sin \left (c+d\,x\right )}^2\,\left (\frac {1}{2\,b}-\frac {a^2\,\left (\frac {1}{b}-\frac {a^2}{2\,b^3}\right )}{b^2}\right )-{\sin \left (c+d\,x\right )}^4\,\left (\frac {1}{2\,b}-\frac {a^2}{4\,b^3}\right )+\frac {{\sin \left (c+d\,x\right )}^6}{6\,b}-\frac {a\,{\sin \left (c+d\,x\right )}^5}{5\,b^2}+\frac {\ln \left (a+b\,\sin \left (c+d\,x\right )\right )\,\left (a^6-2\,a^4\,b^2+a^2\,b^4\right )}{b^7}-\frac {a\,\sin \left (c+d\,x\right )\,\left (\frac {1}{b}-\frac {a^2\,\left (\frac {2}{b}-\frac {a^2}{b^3}\right )}{b^2}\right )}{b}+\frac {a\,{\sin \left (c+d\,x\right )}^3\,\left (\frac {2}{b}-\frac {a^2}{b^3}\right )}{3\,b}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cos(c + d*x)^5*sin(c + d*x)^2)/(a + b*sin(c + d*x)),x)

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**5*sin(d*x+c)**2/(a+b*sin(d*x+c)),x)

[Out]

Timed out

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