3.1545 \(\int \frac {\cos ^5(c+d x) (A+B \sin (c+d x))}{a+b \sin (c+d x)} \, dx\)

Optimal. Leaf size=202 \[ \frac {\left (a^2-b^2\right )^2 (A b-a B) \log (a+b \sin (c+d x))}{b^6 d}+\frac {\left (a^2-2 b^2\right ) (A b-a B) \sin ^2(c+d x)}{2 b^4 d}-\frac {\left (a^2 (-B)+a A b+2 b^2 B\right ) \sin ^3(c+d x)}{3 b^3 d}-\frac {\left (a^4 (-B)+a^3 A b+2 a^2 b^2 B-2 a A b^3-b^4 B\right ) \sin (c+d x)}{b^5 d}+\frac {(A b-a B) \sin ^4(c+d x)}{4 b^2 d}+\frac {B \sin ^5(c+d x)}{5 b d} \]

[Out]

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

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Rubi [A]  time = 0.25, antiderivative size = 202, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.065, Rules used = {2837, 772} \[ -\frac {\left (a^2 (-B)+a A b+2 b^2 B\right ) \sin ^3(c+d x)}{3 b^3 d}+\frac {\left (a^2-2 b^2\right ) (A b-a B) \sin ^2(c+d x)}{2 b^4 d}-\frac {\left (a^3 A b+2 a^2 b^2 B+a^4 (-B)-2 a A b^3-b^4 B\right ) \sin (c+d x)}{b^5 d}+\frac {\left (a^2-b^2\right )^2 (A b-a B) \log (a+b \sin (c+d x))}{b^6 d}+\frac {(A b-a B) \sin ^4(c+d x)}{4 b^2 d}+\frac {B \sin ^5(c+d x)}{5 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

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

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Mathematica [A]  time = 0.43, size = 148, normalized size = 0.73 \[ \frac {20 (A b-a B) \left (6 b^2 \left (a^2-b^2\right ) \sin ^2(c+d x)-12 a b \left (a^2-2 b^2\right ) \sin (c+d x)+12 \left (a^2-b^2\right )^2 \log (a+b \sin (c+d x))-4 a b^3 \sin ^3(c+d x)+3 b^4 \cos ^4(c+d x)\right )+b^5 B (150 \sin (c+d x)+25 \sin (3 (c+d x))+3 \sin (5 (c+d x)))}{240 b^6 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(20*(A*b - a*B)*(3*b^4*Cos[c + d*x]^4 + 12*(a^2 - b^2)^2*Log[a + b*Sin[c + d*x]] - 12*a*b*(a^2 - 2*b^2)*Sin[c
+ d*x] + 6*b^2*(a^2 - b^2)*Sin[c + d*x]^2 - 4*a*b^3*Sin[c + d*x]^3) + b^5*B*(150*Sin[c + d*x] + 25*Sin[3*(c +
d*x)] + 3*Sin[5*(c + d*x)]))/(240*b^6*d)

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fricas [A]  time = 0.50, size = 222, normalized size = 1.10 \[ -\frac {15 \, {\left (B a b^{4} - A b^{5}\right )} \cos \left (d x + c\right )^{4} - 30 \, {\left (B a^{3} b^{2} - A a^{2} b^{3} - B a b^{4} + A b^{5}\right )} \cos \left (d x + c\right )^{2} + 60 \, {\left (B a^{5} - A a^{4} b - 2 \, B a^{3} b^{2} + 2 \, A a^{2} b^{3} + B a b^{4} - A b^{5}\right )} \log \left (b \sin \left (d x + c\right ) + a\right ) - 4 \, {\left (3 \, B b^{5} \cos \left (d x + c\right )^{4} + 15 \, B a^{4} b - 15 \, A a^{3} b^{2} - 25 \, B a^{2} b^{3} + 25 \, A a b^{4} + 8 \, B b^{5} - {\left (5 \, B a^{2} b^{3} - 5 \, A a b^{4} - 4 \, B b^{5}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{60 \, b^{6} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.20, size = 286, normalized size = 1.42 \[ \frac {\frac {12 \, B b^{4} \sin \left (d x + c\right )^{5} - 15 \, B a b^{3} \sin \left (d x + c\right )^{4} + 15 \, A b^{4} \sin \left (d x + c\right )^{4} + 20 \, B a^{2} b^{2} \sin \left (d x + c\right )^{3} - 20 \, A a b^{3} \sin \left (d x + c\right )^{3} - 40 \, B b^{4} \sin \left (d x + c\right )^{3} - 30 \, B a^{3} b \sin \left (d x + c\right )^{2} + 30 \, A a^{2} b^{2} \sin \left (d x + c\right )^{2} + 60 \, B a b^{3} \sin \left (d x + c\right )^{2} - 60 \, A b^{4} \sin \left (d x + c\right )^{2} + 60 \, B a^{4} \sin \left (d x + c\right ) - 60 \, A a^{3} b \sin \left (d x + c\right ) - 120 \, B a^{2} b^{2} \sin \left (d x + c\right ) + 120 \, A a b^{3} \sin \left (d x + c\right ) + 60 \, B b^{4} \sin \left (d x + c\right )}{b^{5}} - \frac {60 \, {\left (B a^{5} - A a^{4} b - 2 \, B a^{3} b^{2} + 2 \, A a^{2} b^{3} + B a b^{4} - A b^{5}\right )} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{b^{6}}}{60 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/60*((12*B*b^4*sin(d*x + c)^5 - 15*B*a*b^3*sin(d*x + c)^4 + 15*A*b^4*sin(d*x + c)^4 + 20*B*a^2*b^2*sin(d*x +
c)^3 - 20*A*a*b^3*sin(d*x + c)^3 - 40*B*b^4*sin(d*x + c)^3 - 30*B*a^3*b*sin(d*x + c)^2 + 30*A*a^2*b^2*sin(d*x
+ c)^2 + 60*B*a*b^3*sin(d*x + c)^2 - 60*A*b^4*sin(d*x + c)^2 + 60*B*a^4*sin(d*x + c) - 60*A*a^3*b*sin(d*x + c)
- 120*B*a^2*b^2*sin(d*x + c) + 120*A*a*b^3*sin(d*x + c) + 60*B*b^4*sin(d*x + c))/b^5 - 60*(B*a^5 - A*a^4*b -
2*B*a^3*b^2 + 2*A*a^2*b^3 + B*a*b^4 - A*b^5)*log(abs(b*sin(d*x + c) + a))/b^6)/d

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maple [B]  time = 0.40, size = 397, normalized size = 1.97 \[ \frac {B \left (\sin ^{5}\left (d x +c \right )\right )}{5 b d}+\frac {A \left (\sin ^{4}\left (d x +c \right )\right )}{4 d b}-\frac {B \left (\sin ^{4}\left (d x +c \right )\right ) a}{4 d \,b^{2}}-\frac {A \left (\sin ^{3}\left (d x +c \right )\right ) a}{3 d \,b^{2}}+\frac {B \left (\sin ^{3}\left (d x +c \right )\right ) a^{2}}{3 d \,b^{3}}-\frac {2 B \left (\sin ^{3}\left (d x +c \right )\right )}{3 b d}+\frac {A \left (\sin ^{2}\left (d x +c \right )\right ) a^{2}}{2 d \,b^{3}}-\frac {A \left (\sin ^{2}\left (d x +c \right )\right )}{d b}-\frac {B \left (\sin ^{2}\left (d x +c \right )\right ) a^{3}}{2 d \,b^{4}}+\frac {B \left (\sin ^{2}\left (d x +c \right )\right ) a}{d \,b^{2}}-\frac {A \sin \left (d x +c \right ) a^{3}}{d \,b^{4}}+\frac {2 A \sin \left (d x +c \right ) a}{d \,b^{2}}+\frac {B \sin \left (d x +c \right ) a^{4}}{d \,b^{5}}-\frac {2 B \sin \left (d x +c \right ) a^{2}}{d \,b^{3}}+\frac {B \sin \left (d x +c \right )}{b d}+\frac {\ln \left (a +b \sin \left (d x +c \right )\right ) A \,a^{4}}{d \,b^{5}}-\frac {2 \ln \left (a +b \sin \left (d x +c \right )\right ) A \,a^{2}}{d \,b^{3}}+\frac {\ln \left (a +b \sin \left (d x +c \right )\right ) A}{d b}-\frac {\ln \left (a +b \sin \left (d x +c \right )\right ) B \,a^{5}}{d \,b^{6}}+\frac {2 \ln \left (a +b \sin \left (d x +c \right )\right ) B \,a^{3}}{d \,b^{4}}-\frac {\ln \left (a +b \sin \left (d x +c \right )\right ) B a}{d \,b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.57, size = 220, normalized size = 1.09 \[ \frac {\frac {12 \, B b^{4} \sin \left (d x + c\right )^{5} - 15 \, {\left (B a b^{3} - A b^{4}\right )} \sin \left (d x + c\right )^{4} + 20 \, {\left (B a^{2} b^{2} - A a b^{3} - 2 \, B b^{4}\right )} \sin \left (d x + c\right )^{3} - 30 \, {\left (B a^{3} b - A a^{2} b^{2} - 2 \, B a b^{3} + 2 \, A b^{4}\right )} \sin \left (d x + c\right )^{2} + 60 \, {\left (B a^{4} - A a^{3} b - 2 \, B a^{2} b^{2} + 2 \, A a b^{3} + B b^{4}\right )} \sin \left (d x + c\right )}{b^{5}} - \frac {60 \, {\left (B a^{5} - A a^{4} b - 2 \, B a^{3} b^{2} + 2 \, A a^{2} b^{3} + B a b^{4} - A b^{5}\right )} \log \left (b \sin \left (d x + c\right ) + a\right )}{b^{6}}}{60 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/60*((12*B*b^4*sin(d*x + c)^5 - 15*(B*a*b^3 - A*b^4)*sin(d*x + c)^4 + 20*(B*a^2*b^2 - A*a*b^3 - 2*B*b^4)*sin(
d*x + c)^3 - 30*(B*a^3*b - A*a^2*b^2 - 2*B*a*b^3 + 2*A*b^4)*sin(d*x + c)^2 + 60*(B*a^4 - A*a^3*b - 2*B*a^2*b^2
+ 2*A*a*b^3 + B*b^4)*sin(d*x + c))/b^5 - 60*(B*a^5 - A*a^4*b - 2*B*a^3*b^2 + 2*A*a^2*b^3 + B*a*b^4 - A*b^5)*l
og(b*sin(d*x + c) + a)/b^6)/d

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mupad [B]  time = 0.10, size = 253, normalized size = 1.25 \[ \frac {{\sin \left (c+d\,x\right )}^4\,\left (\frac {A}{4\,b}-\frac {B\,a}{4\,b^2}\right )}{d}-\frac {{\sin \left (c+d\,x\right )}^2\,\left (\frac {A}{b}-\frac {a\,\left (\frac {2\,B}{b}+\frac {a\,\left (\frac {A}{b}-\frac {B\,a}{b^2}\right )}{b}\right )}{2\,b}\right )}{d}-\frac {{\sin \left (c+d\,x\right )}^3\,\left (\frac {2\,B}{3\,b}+\frac {a\,\left (\frac {A}{b}-\frac {B\,a}{b^2}\right )}{3\,b}\right )}{d}+\frac {\sin \left (c+d\,x\right )\,\left (\frac {B}{b}+\frac {a\,\left (\frac {2\,A}{b}-\frac {a\,\left (\frac {2\,B}{b}+\frac {a\,\left (\frac {A}{b}-\frac {B\,a}{b^2}\right )}{b}\right )}{b}\right )}{b}\right )}{d}+\frac {\ln \left (a+b\,\sin \left (c+d\,x\right )\right )\,\left (-B\,a^5+A\,a^4\,b+2\,B\,a^3\,b^2-2\,A\,a^2\,b^3-B\,a\,b^4+A\,b^5\right )}{b^6\,d}+\frac {B\,{\sin \left (c+d\,x\right )}^5}{5\,b\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(sin(c + d*x)^4*(A/(4*b) - (B*a)/(4*b^2)))/d - (sin(c + d*x)^2*(A/b - (a*((2*B)/b + (a*(A/b - (B*a)/b^2))/b))/
(2*b)))/d - (sin(c + d*x)^3*((2*B)/(3*b) + (a*(A/b - (B*a)/b^2))/(3*b)))/d + (sin(c + d*x)*(B/b + (a*((2*A)/b
- (a*((2*B)/b + (a*(A/b - (B*a)/b^2))/b))/b))/b))/d + (log(a + b*sin(c + d*x))*(A*b^5 - B*a^5 - 2*A*a^2*b^3 +
2*B*a^3*b^2 + A*a^4*b - B*a*b^4))/(b^6*d) + (B*sin(c + d*x)^5)/(5*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*(A+B*sin(d*x+c))/(a+b*sin(d*x+c)),x)

[Out]

Timed out

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