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

Optimal. Leaf size=120 \[ -\frac {\left (a^2-b^2\right )^2}{b^5 d (a+b \sin (c+d x))}-\frac {4 a \left (a^2-b^2\right ) \log (a+b \sin (c+d x))}{b^5 d}+\frac {\left (3 a^2-2 b^2\right ) \sin (c+d x)}{b^4 d}-\frac {a \sin ^2(c+d x)}{b^3 d}+\frac {\sin ^3(c+d x)}{3 b^2 d} \]

[Out]

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

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Rubi [A]  time = 0.10, antiderivative size = 120, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {2668, 697} \[ \frac {\left (3 a^2-2 b^2\right ) \sin (c+d x)}{b^4 d}-\frac {\left (a^2-b^2\right )^2}{b^5 d (a+b \sin (c+d x))}-\frac {4 a \left (a^2-b^2\right ) \log (a+b \sin (c+d x))}{b^5 d}-\frac {a \sin ^2(c+d x)}{b^3 d}+\frac {\sin ^3(c+d x)}{3 b^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 697

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*(a + c*
x^2)^p, x], x] /; FreeQ[{a, c, d, e, m}, x] && NeQ[c*d^2 + a*e^2, 0] && IGtQ[p, 0]

Rule 2668

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^m*(b^2 - x^2)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && Integer
Q[(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))^2} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\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 (3 a^2 \left (1-\frac {2 b^2}{3 a^2}\right )-2 a x+x^2+\frac {\left (a^2-b^2\right )^2}{(a+x)^2}-\frac {4 \left (a^3-a b^2\right )}{a+x}\right ) \, dx,x,b \sin (c+d x)\right )}{b^5 d}\\ &=-\frac {4 a \left (a^2-b^2\right ) \log (a+b \sin (c+d x))}{b^5 d}+\frac {\left (3 a^2-2 b^2\right ) \sin (c+d x)}{b^4 d}-\frac {a \sin ^2(c+d x)}{b^3 d}+\frac {\sin ^3(c+d x)}{3 b^2 d}-\frac {\left (a^2-b^2\right )^2}{b^5 d (a+b \sin (c+d x))}\\ \end {align*}

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Mathematica [A]  time = 0.63, size = 127, normalized size = 1.06 \[ \frac {\left (8 a^2 b-4 b^3\right ) \sin (c+d x)+\frac {b^4 \cos ^4(c+d x)-4 \left (a^2-b^2\right ) \left (3 a^2 \log (a+b \sin (c+d x))+a^2+3 a b \sin (c+d x) \log (a+b \sin (c+d x))-b^2\right )}{a+b \sin (c+d x)}-2 a b^2 \sin ^2(c+d x)}{3 b^5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((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*Log[a + b*Sin[c + d*x]]*Sin[c + d*x]))/(a + b*Sin[c + d*x]))/(3*b^5*d)

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fricas [A]  time = 0.49, size = 156, normalized size = 1.30 \[ \frac {2 \, b^{4} \cos \left (d x + c\right )^{4} - 6 \, a^{4} + 27 \, a^{2} b^{2} - 16 \, b^{4} - 4 \, {\left (3 \, a^{2} b^{2} - 2 \, b^{4}\right )} \cos \left (d x + c\right )^{2} - 24 \, {\left (a^{4} - a^{2} b^{2} + {\left (a^{3} b - a b^{3}\right )} \sin \left (d x + c\right )\right )} \log \left (b \sin \left (d x + c\right ) + a\right ) + {\left (4 \, a b^{3} \cos \left (d x + c\right )^{2} + 18 \, a^{3} b - 13 \, a b^{3}\right )} \sin \left (d x + c\right )}{6 \, {\left (b^{6} d \sin \left (d x + c\right ) + a b^{5} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(2*b^4*cos(d*x + c)^4 - 6*a^4 + 27*a^2*b^2 - 16*b^4 - 4*(3*a^2*b^2 - 2*b^4)*cos(d*x + c)^2 - 24*(a^4 - a^2
*b^2 + (a^3*b - a*b^3)*sin(d*x + c))*log(b*sin(d*x + c) + a) + (4*a*b^3*cos(d*x + c)^2 + 18*a^3*b - 13*a*b^3)*
sin(d*x + c))/(b^6*d*sin(d*x + c) + a*b^5*d)

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giac [A]  time = 0.84, size = 150, normalized size = 1.25 \[ -\frac {\frac {12 \, {\left (a^{3} - a b^{2}\right )} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{b^{5}} - \frac {b^{4} \sin \left (d x + c\right )^{3} - 3 \, a b^{3} \sin \left (d x + c\right )^{2} + 9 \, a^{2} b^{2} \sin \left (d x + c\right ) - 6 \, b^{4} \sin \left (d x + c\right )}{b^{6}} - \frac {3 \, {\left (4 \, a^{3} b \sin \left (d x + c\right ) - 4 \, a b^{3} \sin \left (d x + c\right ) + 3 \, a^{4} - 2 \, a^{2} b^{2} - b^{4}\right )}}{{\left (b \sin \left (d x + c\right ) + a\right )} b^{5}}}{3 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(12*(a^3 - a*b^2)*log(abs(b*sin(d*x + c) + a))/b^5 - (b^4*sin(d*x + c)^3 - 3*a*b^3*sin(d*x + c)^2 + 9*a^2
*b^2*sin(d*x + c) - 6*b^4*sin(d*x + c))/b^6 - 3*(4*a^3*b*sin(d*x + c) - 4*a*b^3*sin(d*x + c) + 3*a^4 - 2*a^2*b
^2 - b^4)/((b*sin(d*x + c) + a)*b^5))/d

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maple [A]  time = 0.25, size = 174, normalized size = 1.45 \[ \frac {\sin ^{3}\left (d x +c \right )}{3 b^{2} d}-\frac {a \left (\sin ^{2}\left (d x +c \right )\right )}{b^{3} d}+\frac {3 a^{2} \sin \left (d x +c \right )}{d \,b^{4}}-\frac {2 \sin \left (d x +c \right )}{b^{2} d}-\frac {4 a^{3} \ln \left (a +b \sin \left (d x +c \right )\right )}{d \,b^{5}}+\frac {4 a \ln \left (a +b \sin \left (d x +c \right )\right )}{b^{3} d}-\frac {a^{4}}{d \,b^{5} \left (a +b \sin \left (d x +c \right )\right )}+\frac {2 a^{2}}{d \,b^{3} \left (a +b \sin \left (d x +c \right )\right )}-\frac {1}{b d \left (a +b \sin \left (d x +c \right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.31, size = 116, normalized size = 0.97 \[ -\frac {\frac {3 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )}}{b^{6} \sin \left (d x + c\right ) + a b^{5}} - \frac {b^{2} \sin \left (d x + c\right )^{3} - 3 \, a b \sin \left (d x + c\right )^{2} + 3 \, {\left (3 \, a^{2} - 2 \, b^{2}\right )} \sin \left (d x + c\right )}{b^{4}} + \frac {12 \, {\left (a^{3} - a b^{2}\right )} \log \left (b \sin \left (d x + c\right ) + a\right )}{b^{5}}}{3 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.08, size = 118, normalized size = 0.98 \[ -\frac {\sin \left (c+d\,x\right )\,\left (\frac {2}{b^2}-\frac {3\,a^2}{b^4}\right )-\frac {{\sin \left (c+d\,x\right )}^3}{3\,b^2}+\frac {a\,{\sin \left (c+d\,x\right )}^2}{b^3}-\frac {\ln \left (a+b\,\sin \left (c+d\,x\right )\right )\,\left (4\,a\,b^2-4\,a^3\right )}{b^5}+\frac {a^4-2\,a^2\,b^2+b^4}{b\,\left (\sin \left (c+d\,x\right )\,b^5+a\,b^4\right )}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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