3.1553 \(\int \frac {\cos ^5(c+d x) (A+B \sin (c+d x))}{(a+b \sin (c+d x))^2} \, dx\)

Optimal. Leaf size=206 \[ -\frac {\left (a^2-b^2\right )^2 (A b-a B)}{b^6 d (a+b \sin (c+d x))}-\frac {\left (a^2-b^2\right ) \left (-5 a^2 B+4 a A b+b^2 B\right ) \log (a+b \sin (c+d x))}{b^6 d}-\frac {\left (-3 a^2 B+2 a A b+2 b^2 B\right ) \sin ^2(c+d x)}{2 b^4 d}+\frac {\left (-4 a^3 B+3 a^2 A b+4 a b^2 B-2 A b^3\right ) \sin (c+d x)}{b^5 d}+\frac {(A b-2 a B) \sin ^3(c+d x)}{3 b^3 d}+\frac {B \sin ^4(c+d x)}{4 b^2 d} \]

[Out]

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

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Rubi [A]  time = 0.27, antiderivative size = 206, 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} \[ -\frac {\left (-3 a^2 B+2 a A b+2 b^2 B\right ) \sin ^2(c+d x)}{2 b^4 d}+\frac {\left (3 a^2 A b-4 a^3 B+4 a b^2 B-2 A b^3\right ) \sin (c+d x)}{b^5 d}-\frac {\left (a^2-b^2\right )^2 (A b-a B)}{b^6 d (a+b \sin (c+d x))}-\frac {\left (a^2-b^2\right ) \left (-5 a^2 B+4 a A b+b^2 B\right ) \log (a+b \sin (c+d x))}{b^6 d}+\frac {(A b-2 a B) \sin ^3(c+d x)}{3 b^3 d}+\frac {B \sin ^4(c+d x)}{4 b^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]  time = 2.16, size = 234, normalized size = 1.14 \[ \frac {4 \left (A-\frac {a B}{b}\right ) \left (\left (8 a^2 b-4 b^3\right ) \sin (c+d x)+\frac {b^4 \cos ^4(c+d x)-4 \left (a^2-b^2\right ) \left (3 a^2 \log (a+b \sin (c+d x))+a^2+3 a b \sin (c+d x) \log (a+b \sin (c+d x))-b^2\right )}{a+b \sin (c+d x)}-2 a b^2 \sin ^2(c+d x)\right )+B \left (6 b \left (a^2-b^2\right ) \sin ^2(c+d x)-12 a \left (a^2-2 b^2\right ) \sin (c+d x)+\frac {12 \left (a^2-b^2\right )^2 \log (a+b \sin (c+d x))}{b}-4 a b^2 \sin ^3(c+d x)+3 b^3 \cos ^4(c+d x)\right )}{12 b^5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(B*(3*b^3*Cos[c + d*x]^4 + (12*(a^2 - b^2)^2*Log[a + b*Sin[c + d*x]])/b - 12*a*(a^2 - 2*b^2)*Sin[c + d*x] + 6*
b*(a^2 - b^2)*Sin[c + d*x]^2 - 4*a*b^2*Sin[c + d*x]^3) + 4*(A - (a*B)/b)*((8*a^2*b - 4*b^3)*Sin[c + d*x] - 2*a
*b^2*Sin[c + d*x]^2 + (b^4*Cos[c + d*x]^4 - 4*(a^2 - b^2)*(a^2 - b^2 + 3*a^2*Log[a + b*Sin[c + d*x]] + 3*a*b*L
og[a + b*Sin[c + d*x]]*Sin[c + d*x]))/(a + b*Sin[c + d*x])))/(12*b^5*d)

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fricas [A]  time = 0.54, size = 322, normalized size = 1.56 \[ \frac {96 \, B a^{5} - 96 \, A a^{4} b - 504 \, B a^{3} b^{2} + 432 \, A a^{2} b^{3} + 383 \, B a b^{4} - 256 \, A b^{5} - 8 \, {\left (5 \, B a b^{4} - 4 \, A b^{5}\right )} \cos \left (d x + c\right )^{4} + 16 \, {\left (15 \, B a^{3} b^{2} - 12 \, A a^{2} b^{3} - 13 \, B a b^{4} + 8 \, A b^{5}\right )} \cos \left (d x + c\right )^{2} + 96 \, {\left (5 \, B a^{5} - 4 \, A a^{4} b - 6 \, B a^{3} b^{2} + 4 \, A a^{2} b^{3} + B a b^{4} + {\left (5 \, B a^{4} b - 4 \, A a^{3} b^{2} - 6 \, B a^{2} b^{3} + 4 \, A a b^{4} + B b^{5}\right )} \sin \left (d x + c\right )\right )} \log \left (b \sin \left (d x + c\right ) + a\right ) + {\left (24 \, B b^{5} \cos \left (d x + c\right )^{4} - 384 \, B a^{4} b + 288 \, A a^{3} b^{2} + 392 \, B a^{2} b^{3} - 208 \, A a b^{4} - 33 \, B b^{5} - 16 \, {\left (5 \, B a^{2} b^{3} - 4 \, A a b^{4} - 3 \, B b^{5}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{96 \, {\left (b^{7} d \sin \left (d x + c\right ) + a b^{6} d\right )}} \]

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))^2,x, algorithm="fricas")

[Out]

1/96*(96*B*a^5 - 96*A*a^4*b - 504*B*a^3*b^2 + 432*A*a^2*b^3 + 383*B*a*b^4 - 256*A*b^5 - 8*(5*B*a*b^4 - 4*A*b^5
)*cos(d*x + c)^4 + 16*(15*B*a^3*b^2 - 12*A*a^2*b^3 - 13*B*a*b^4 + 8*A*b^5)*cos(d*x + c)^2 + 96*(5*B*a^5 - 4*A*
a^4*b - 6*B*a^3*b^2 + 4*A*a^2*b^3 + B*a*b^4 + (5*B*a^4*b - 4*A*a^3*b^2 - 6*B*a^2*b^3 + 4*A*a*b^4 + B*b^5)*sin(
d*x + c))*log(b*sin(d*x + c) + a) + (24*B*b^5*cos(d*x + c)^4 - 384*B*a^4*b + 288*A*a^3*b^2 + 392*B*a^2*b^3 - 2
08*A*a*b^4 - 33*B*b^5 - 16*(5*B*a^2*b^3 - 4*A*a*b^4 - 3*B*b^5)*cos(d*x + c)^2)*sin(d*x + c))/(b^7*d*sin(d*x +
c) + a*b^6*d)

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giac [A]  time = 0.25, size = 328, normalized size = 1.59 \[ \frac {\frac {12 \, {\left (5 \, B a^{4} - 4 \, A a^{3} b - 6 \, B a^{2} b^{2} + 4 \, A a b^{3} + B b^{4}\right )} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{b^{6}} - \frac {12 \, {\left (5 \, B a^{4} b \sin \left (d x + c\right ) - 4 \, A a^{3} b^{2} \sin \left (d x + c\right ) - 6 \, B a^{2} b^{3} \sin \left (d x + c\right ) + 4 \, A a b^{4} \sin \left (d x + c\right ) + B b^{5} \sin \left (d x + c\right ) + 4 \, B a^{5} - 3 \, A a^{4} b - 4 \, B a^{3} b^{2} + 2 \, A a^{2} b^{3} + A b^{5}\right )}}{{\left (b \sin \left (d x + c\right ) + a\right )} b^{6}} + \frac {3 \, B b^{6} \sin \left (d x + c\right )^{4} - 8 \, B a b^{5} \sin \left (d x + c\right )^{3} + 4 \, A b^{6} \sin \left (d x + c\right )^{3} + 18 \, B a^{2} b^{4} \sin \left (d x + c\right )^{2} - 12 \, A a b^{5} \sin \left (d x + c\right )^{2} - 12 \, B b^{6} \sin \left (d x + c\right )^{2} - 48 \, B a^{3} b^{3} \sin \left (d x + c\right ) + 36 \, A a^{2} b^{4} \sin \left (d x + c\right ) + 48 \, B a b^{5} \sin \left (d x + c\right ) - 24 \, A b^{6} \sin \left (d x + c\right )}{b^{8}}}{12 \, d} \]

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))^2,x, algorithm="giac")

[Out]

1/12*(12*(5*B*a^4 - 4*A*a^3*b - 6*B*a^2*b^2 + 4*A*a*b^3 + B*b^4)*log(abs(b*sin(d*x + c) + a))/b^6 - 12*(5*B*a^
4*b*sin(d*x + c) - 4*A*a^3*b^2*sin(d*x + c) - 6*B*a^2*b^3*sin(d*x + c) + 4*A*a*b^4*sin(d*x + c) + B*b^5*sin(d*
x + c) + 4*B*a^5 - 3*A*a^4*b - 4*B*a^3*b^2 + 2*A*a^2*b^3 + A*b^5)/((b*sin(d*x + c) + a)*b^6) + (3*B*b^6*sin(d*
x + c)^4 - 8*B*a*b^5*sin(d*x + c)^3 + 4*A*b^6*sin(d*x + c)^3 + 18*B*a^2*b^4*sin(d*x + c)^2 - 12*A*a*b^5*sin(d*
x + c)^2 - 12*B*b^6*sin(d*x + c)^2 - 48*B*a^3*b^3*sin(d*x + c) + 36*A*a^2*b^4*sin(d*x + c) + 48*B*a*b^5*sin(d*
x + c) - 24*A*b^6*sin(d*x + c))/b^8)/d

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maple [B]  time = 0.66, size = 422, normalized size = 2.05 \[ \frac {B \left (\sin ^{4}\left (d x +c \right )\right )}{4 b^{2} d}+\frac {A \left (\sin ^{3}\left (d x +c \right )\right )}{3 d \,b^{2}}-\frac {2 B \left (\sin ^{3}\left (d x +c \right )\right ) a}{3 d \,b^{3}}-\frac {A \left (\sin ^{2}\left (d x +c \right )\right ) a}{d \,b^{3}}+\frac {3 B \left (\sin ^{2}\left (d x +c \right )\right ) a^{2}}{2 d \,b^{4}}-\frac {B \left (\sin ^{2}\left (d x +c \right )\right )}{b^{2} d}+\frac {3 A \,a^{2} \sin \left (d x +c \right )}{d \,b^{4}}-\frac {2 A \sin \left (d x +c \right )}{d \,b^{2}}-\frac {4 B \,a^{3} \sin \left (d x +c \right )}{d \,b^{5}}+\frac {4 B a \sin \left (d x +c \right )}{d \,b^{3}}-\frac {4 \ln \left (a +b \sin \left (d x +c \right )\right ) A \,a^{3}}{d \,b^{5}}+\frac {4 \ln \left (a +b \sin \left (d x +c \right )\right ) A a}{d \,b^{3}}+\frac {5 \ln \left (a +b \sin \left (d x +c \right )\right ) B \,a^{4}}{d \,b^{6}}-\frac {6 \ln \left (a +b \sin \left (d x +c \right )\right ) B \,a^{2}}{d \,b^{4}}+\frac {B \ln \left (a +b \sin \left (d x +c \right )\right )}{b^{2} d}-\frac {A \,a^{4}}{d \,b^{5} \left (a +b \sin \left (d x +c \right )\right )}+\frac {2 A \,a^{2}}{d \,b^{3} \left (a +b \sin \left (d x +c \right )\right )}-\frac {A}{d b \left (a +b \sin \left (d x +c \right )\right )}+\frac {B \,a^{5}}{d \,b^{6} \left (a +b \sin \left (d x +c \right )\right )}-\frac {2 B \,a^{3}}{d \,b^{4} \left (a +b \sin \left (d x +c \right )\right )}+\frac {a B}{d \,b^{2} \left (a +b \sin \left (d x +c \right )\right )} \]

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))^2,x)

[Out]

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

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maxima [A]  time = 0.39, size = 229, normalized size = 1.11 \[ \frac {\frac {12 \, {\left (B a^{5} - A a^{4} b - 2 \, B a^{3} b^{2} + 2 \, A a^{2} b^{3} + B a b^{4} - A b^{5}\right )}}{b^{7} \sin \left (d x + c\right ) + a b^{6}} + \frac {3 \, B b^{3} \sin \left (d x + c\right )^{4} - 4 \, {\left (2 \, B a b^{2} - A b^{3}\right )} \sin \left (d x + c\right )^{3} + 6 \, {\left (3 \, B a^{2} b - 2 \, A a b^{2} - 2 \, B b^{3}\right )} \sin \left (d x + c\right )^{2} - 12 \, {\left (4 \, B a^{3} - 3 \, A a^{2} b - 4 \, B a b^{2} + 2 \, A b^{3}\right )} \sin \left (d x + c\right )}{b^{5}} + \frac {12 \, {\left (5 \, B a^{4} - 4 \, A a^{3} b - 6 \, B a^{2} b^{2} + 4 \, A a b^{3} + B b^{4}\right )} \log \left (b \sin \left (d x + c\right ) + a\right )}{b^{6}}}{12 \, d} \]

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))^2,x, algorithm="maxima")

[Out]

1/12*(12*(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)/(b^7*sin(d*x + c) + a*b^6) + (3*B*b^3
*sin(d*x + c)^4 - 4*(2*B*a*b^2 - A*b^3)*sin(d*x + c)^3 + 6*(3*B*a^2*b - 2*A*a*b^2 - 2*B*b^3)*sin(d*x + c)^2 -
12*(4*B*a^3 - 3*A*a^2*b - 4*B*a*b^2 + 2*A*b^3)*sin(d*x + c))/b^5 + 12*(5*B*a^4 - 4*A*a^3*b - 6*B*a^2*b^2 + 4*A
*a*b^3 + B*b^4)*log(b*sin(d*x + c) + a)/b^6)/d

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mupad [B]  time = 0.12, size = 290, normalized size = 1.41 \[ \frac {{\sin \left (c+d\,x\right )}^3\,\left (\frac {A}{3\,b^2}-\frac {2\,B\,a}{3\,b^3}\right )}{d}-\frac {\sin \left (c+d\,x\right )\,\left (\frac {2\,A}{b^2}+\frac {a^2\,\left (\frac {A}{b^2}-\frac {2\,B\,a}{b^3}\right )}{b^2}-\frac {2\,a\,\left (\frac {2\,B}{b^2}+\frac {2\,a\,\left (\frac {A}{b^2}-\frac {2\,B\,a}{b^3}\right )}{b}+\frac {B\,a^2}{b^4}\right )}{b}\right )}{d}-\frac {{\sin \left (c+d\,x\right )}^2\,\left (\frac {B}{b^2}+\frac {a\,\left (\frac {A}{b^2}-\frac {2\,B\,a}{b^3}\right )}{b}+\frac {B\,a^2}{2\,b^4}\right )}{d}+\frac {\ln \left (a+b\,\sin \left (c+d\,x\right )\right )\,\left (5\,B\,a^4-4\,A\,a^3\,b-6\,B\,a^2\,b^2+4\,A\,a\,b^3+B\,b^4\right )}{b^6\,d}-\frac {-B\,a^5+A\,a^4\,b+2\,B\,a^3\,b^2-2\,A\,a^2\,b^3-B\,a\,b^4+A\,b^5}{b\,d\,\left (\sin \left (c+d\,x\right )\,b^6+a\,b^5\right )}+\frac {B\,{\sin \left (c+d\,x\right )}^4}{4\,b^2\,d} \]

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))^2,x)

[Out]

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

[Out]

Timed out

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