3.79 \(\int \frac{\cos ^4(c+d x)}{(a+a \sin (c+d x))^3} \, dx\)

Optimal. Leaf size=49 \[ -\frac{3 \cos (c+d x)}{a^3 d}-\frac{3 x}{a^3}-\frac{2 \cos ^3(c+d x)}{a d (a \sin (c+d x)+a)^2} \]

[Out]

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

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Rubi [A]  time = 0.0850498, antiderivative size = 49, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {2680, 2682, 8} \[ -\frac{3 \cos (c+d x)}{a^3 d}-\frac{3 x}{a^3}-\frac{2 \cos ^3(c+d x)}{a d (a \sin (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^4/(a + a*Sin[c + d*x])^3,x]

[Out]

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

Rule 2680

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(2*g*(
g*Cos[e + f*x])^(p - 1)*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(2*m + p + 1)), x] + Dist[(g^2*(p - 1))/(b^2*(2*m +
p + 1)), Int[(g*Cos[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^(m + 2), x], x] /; FreeQ[{a, b, e, f, g}, x] && Eq
Q[a^2 - b^2, 0] && LeQ[m, -2] && GtQ[p, 1] && NeQ[2*m + p + 1, 0] &&  !ILtQ[m + p + 1, 0] && IntegersQ[2*m, 2*
p]

Rule 2682

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(g*(g*Cos[e
+ f*x])^(p - 1))/(b*f*(p - 1)), x] + Dist[g^2/a, Int[(g*Cos[e + f*x])^(p - 2), x], x] /; FreeQ[{a, b, e, f, g
}, x] && EqQ[a^2 - b^2, 0] && GtQ[p, 1] && IntegerQ[2*p]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

\begin{align*} \int \frac{\cos ^4(c+d x)}{(a+a \sin (c+d x))^3} \, dx &=-\frac{2 \cos ^3(c+d x)}{a d (a+a \sin (c+d x))^2}-\frac{3 \int \frac{\cos ^2(c+d x)}{a+a \sin (c+d x)} \, dx}{a^2}\\ &=-\frac{3 \cos (c+d x)}{a^3 d}-\frac{2 \cos ^3(c+d x)}{a d (a+a \sin (c+d x))^2}-\frac{3 \int 1 \, dx}{a^3}\\ &=-\frac{3 x}{a^3}-\frac{3 \cos (c+d x)}{a^3 d}-\frac{2 \cos ^3(c+d x)}{a d (a+a \sin (c+d x))^2}\\ \end{align*}

Mathematica [C]  time = 0.0383827, size = 59, normalized size = 1.2 \[ -\frac{\cos ^5(c+d x) \, _2F_1\left (\frac{3}{2},\frac{5}{2};\frac{7}{2};\frac{1}{2} (1-\sin (c+d x))\right )}{5 \sqrt{2} a^3 d (\sin (c+d x)+1)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^4/(a + a*Sin[c + d*x])^3,x]

[Out]

-(Cos[c + d*x]^5*Hypergeometric2F1[3/2, 5/2, 7/2, (1 - Sin[c + d*x])/2])/(5*Sqrt[2]*a^3*d*(1 + Sin[c + d*x])^(
5/2))

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Maple [A]  time = 0.084, size = 64, normalized size = 1.3 \begin{align*} -2\,{\frac{1}{d{a}^{3} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) }}-6\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) }{d{a}^{3}}}-8\,{\frac{1}{d{a}^{3} \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^4/(a+a*sin(d*x+c))^3,x)

[Out]

-2/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)-6/d/a^3*arctan(tan(1/2*d*x+1/2*c))-8/d/a^3/(tan(1/2*d*x+1/2*c)+1)

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Maxima [B]  time = 1.43639, size = 188, normalized size = 3.84 \begin{align*} -\frac{2 \,{\left (\frac{\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{4 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + 5}{a^{3} + \frac{a^{3} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{a^{3} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}} + \frac{3 \, \arctan \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}}\right )}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4/(a+a*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.95341, size = 203, normalized size = 4.14 \begin{align*} -\frac{3 \, d x +{\left (3 \, d x + 5\right )} \cos \left (d x + c\right ) + \cos \left (d x + c\right )^{2} +{\left (3 \, d x + \cos \left (d x + c\right ) - 4\right )} \sin \left (d x + c\right ) + 4}{a^{3} d \cos \left (d x + c\right ) + a^{3} d \sin \left (d x + c\right ) + a^{3} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4/(a+a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**4/(a+a*sin(d*x+c))**3,x)

[Out]

Timed out

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Giac [A]  time = 1.18172, size = 108, normalized size = 2.2 \begin{align*} -\frac{\frac{3 \,{\left (d x + c\right )}}{a^{3}} + \frac{2 \,{\left (4 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 5\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )} a^{3}}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4/(a+a*sin(d*x+c))^3,x, algorithm="giac")

[Out]

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