3.439 \(\int \frac {\cos ^3(c+d x)}{(a+b \sin (c+d x))^2} \, dx\)

Optimal. Leaf size=63 \[ \frac {a^2-b^2}{b^3 d (a+b \sin (c+d x))}+\frac {2 a \log (a+b \sin (c+d x))}{b^3 d}-\frac {\sin (c+d x)}{b^2 d} \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 63, 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} \[ \frac {a^2-b^2}{b^3 d (a+b \sin (c+d x))}+\frac {2 a \log (a+b \sin (c+d x))}{b^3 d}-\frac {\sin (c+d x)}{b^2 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^3/(a + b*Sin[c + d*x])^2,x]

[Out]

(2*a*Log[a + b*Sin[c + d*x]])/(b^3*d) - Sin[c + d*x]/(b^2*d) + (a^2 - b^2)/(b^3*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 ^3(c+d x)}{(a+b \sin (c+d x))^2} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {b^2-x^2}{(a+x)^2} \, dx,x,b \sin (c+d x)\right )}{b^3 d}\\ &=\frac {\operatorname {Subst}\left (\int \left (-1+\frac {-a^2+b^2}{(a+x)^2}+\frac {2 a}{a+x}\right ) \, dx,x,b \sin (c+d x)\right )}{b^3 d}\\ &=\frac {2 a \log (a+b \sin (c+d x))}{b^3 d}-\frac {\sin (c+d x)}{b^2 d}+\frac {a^2-b^2}{b^3 d (a+b \sin (c+d x))}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 52, normalized size = 0.83 \[ \frac {\frac {(a-b) (a+b)}{a+b \sin (c+d x)}+2 a \log (a+b \sin (c+d x))-b \sin (c+d x)}{b^3 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^3/(a + b*Sin[c + d*x])^2,x]

[Out]

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

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fricas [A]  time = 0.46, size = 78, normalized size = 1.24 \[ \frac {b^{2} \cos \left (d x + c\right )^{2} - a b \sin \left (d x + c\right ) + a^{2} - 2 \, b^{2} + 2 \, {\left (a b \sin \left (d x + c\right ) + a^{2}\right )} \log \left (b \sin \left (d x + c\right ) + a\right )}{b^{4} d \sin \left (d x + c\right ) + a b^{3} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3/(a+b*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 1.69, size = 91, normalized size = 1.44 \[ -\frac {\frac {2 \, a \log \left (\frac {{\left | b \sin \left (d x + c\right ) + a \right |}}{{\left (b \sin \left (d x + c\right ) + a\right )}^{2} {\left | b \right |}}\right )}{b^{3}} + \frac {b \sin \left (d x + c\right ) + a}{b^{3}} - \frac {a^{2}}{{\left (b \sin \left (d x + c\right ) + a\right )} b^{3}} + \frac {1}{{\left (b \sin \left (d x + c\right ) + a\right )} b}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3/(a+b*sin(d*x+c))^2,x, algorithm="giac")

[Out]

-(2*a*log(abs(b*sin(d*x + c) + a)/((b*sin(d*x + c) + a)^2*abs(b)))/b^3 + (b*sin(d*x + c) + a)/b^3 - a^2/((b*si
n(d*x + c) + a)*b^3) + 1/((b*sin(d*x + c) + a)*b))/d

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maple [A]  time = 0.24, size = 78, normalized size = 1.24 \[ -\frac {\sin \left (d x +c \right )}{b^{2} d}+\frac {2 a \ln \left (a +b \sin \left (d x +c \right )\right )}{b^{3} d}+\frac {a^{2}}{d \,b^{3} \left (a +b \sin \left (d x +c \right )\right )}-\frac {1}{b d \left (a +b \sin \left (d x +c \right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^3/(a+b*sin(d*x+c))^2,x)

[Out]

-sin(d*x+c)/b^2/d+2*a*ln(a+b*sin(d*x+c))/b^3/d+1/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 = 61, normalized size = 0.97 \[ \frac {\frac {a^{2} - b^{2}}{b^{4} \sin \left (d x + c\right ) + a b^{3}} + \frac {2 \, a \log \left (b \sin \left (d x + c\right ) + a\right )}{b^{3}} - \frac {\sin \left (d x + c\right )}{b^{2}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3/(a+b*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.08, size = 69, normalized size = 1.10 \[ \frac {2\,a\,\ln \left (a+b\,\sin \left (c+d\,x\right )\right )}{b^3\,d}-\frac {\sin \left (c+d\,x\right )}{b^2\,d}+\frac {a^2-b^2}{b\,d\,\left (\sin \left (c+d\,x\right )\,b^3+a\,b^2\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^3/(a + b*sin(c + d*x))^2,x)

[Out]

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

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sympy [A]  time = 1.88, size = 221, normalized size = 3.51 \[ \begin {cases} \frac {x \cos ^{3}{\relax (c )}}{a^{2}} & \text {for}\: b = 0 \wedge d = 0 \\\frac {\frac {2 \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {\sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d}}{a^{2}} & \text {for}\: b = 0 \\\frac {x \cos ^{3}{\relax (c )}}{\left (a + b \sin {\relax (c )}\right )^{2}} & \text {for}\: d = 0 \\\frac {2 a^{2} \log {\left (\frac {a}{b} + \sin {\left (c + d x \right )} \right )}}{a b^{3} d + b^{4} d \sin {\left (c + d x \right )}} + \frac {2 a^{2}}{a b^{3} d + b^{4} d \sin {\left (c + d x \right )}} + \frac {2 a b \log {\left (\frac {a}{b} + \sin {\left (c + d x \right )} \right )} \sin {\left (c + d x \right )}}{a b^{3} d + b^{4} d \sin {\left (c + d x \right )}} - \frac {2 b^{2} \sin ^{2}{\left (c + d x \right )}}{a b^{3} d + b^{4} d \sin {\left (c + d x \right )}} - \frac {b^{2} \cos ^{2}{\left (c + d x \right )}}{a b^{3} d + b^{4} d \sin {\left (c + d x \right )}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**3/(a+b*sin(d*x+c))**2,x)

[Out]

Piecewise((x*cos(c)**3/a**2, Eq(b, 0) & Eq(d, 0)), ((2*sin(c + d*x)**3/(3*d) + sin(c + d*x)*cos(c + d*x)**2/d)
/a**2, Eq(b, 0)), (x*cos(c)**3/(a + b*sin(c))**2, Eq(d, 0)), (2*a**2*log(a/b + sin(c + d*x))/(a*b**3*d + b**4*
d*sin(c + d*x)) + 2*a**2/(a*b**3*d + b**4*d*sin(c + d*x)) + 2*a*b*log(a/b + sin(c + d*x))*sin(c + d*x)/(a*b**3
*d + b**4*d*sin(c + d*x)) - 2*b**2*sin(c + d*x)**2/(a*b**3*d + b**4*d*sin(c + d*x)) - b**2*cos(c + d*x)**2/(a*
b**3*d + b**4*d*sin(c + d*x)), True))

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