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

Optimal. Leaf size=183 \[ -\frac{\log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} \sin (c+d x)+b^{2/3} \sin ^2(c+d x)\right )}{9 a^{5/3} \sqrt [3]{b} d}+\frac{2 \log \left (\sqrt [3]{a}+\sqrt [3]{b} \sin (c+d x)\right )}{9 a^{5/3} \sqrt [3]{b} d}-\frac{2 \tan ^{-1}\left (\frac{\sqrt [3]{a}-2 \sqrt [3]{b} \sin (c+d x)}{\sqrt{3} \sqrt [3]{a}}\right )}{3 \sqrt{3} a^{5/3} \sqrt [3]{b} d}+\frac{a+b \sin (c+d x)}{3 a b d \left (a+b \sin ^3(c+d x)\right )} \]

[Out]

(-2*ArcTan[(a^(1/3) - 2*b^(1/3)*Sin[c + d*x])/(Sqrt[3]*a^(1/3))])/(3*Sqrt[3]*a^(5/3)*b^(1/3)*d) + (2*Log[a^(1/
3) + b^(1/3)*Sin[c + d*x]])/(9*a^(5/3)*b^(1/3)*d) - Log[a^(2/3) - a^(1/3)*b^(1/3)*Sin[c + d*x] + b^(2/3)*Sin[c
+ d*x]^2]/(9*a^(5/3)*b^(1/3)*d) + (a + b*Sin[c + d*x])/(3*a*b*d*(a + b*Sin[c + d*x]^3))

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Rubi [A]  time = 0.162416, antiderivative size = 183, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 9, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.391, Rules used = {3223, 1854, 12, 200, 31, 634, 617, 204, 628} \[ -\frac{\log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} \sin (c+d x)+b^{2/3} \sin ^2(c+d x)\right )}{9 a^{5/3} \sqrt [3]{b} d}+\frac{2 \log \left (\sqrt [3]{a}+\sqrt [3]{b} \sin (c+d x)\right )}{9 a^{5/3} \sqrt [3]{b} d}-\frac{2 \tan ^{-1}\left (\frac{\sqrt [3]{a}-2 \sqrt [3]{b} \sin (c+d x)}{\sqrt{3} \sqrt [3]{a}}\right )}{3 \sqrt{3} a^{5/3} \sqrt [3]{b} d}+\frac{a+b \sin (c+d x)}{3 a b d \left (a+b \sin ^3(c+d x)\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*ArcTan[(a^(1/3) - 2*b^(1/3)*Sin[c + d*x])/(Sqrt[3]*a^(1/3))])/(3*Sqrt[3]*a^(5/3)*b^(1/3)*d) + (2*Log[a^(1/
3) + b^(1/3)*Sin[c + d*x]])/(9*a^(5/3)*b^(1/3)*d) - Log[a^(2/3) - a^(1/3)*b^(1/3)*Sin[c + d*x] + b^(2/3)*Sin[c
+ d*x]^2]/(9*a^(5/3)*b^(1/3)*d) + (a + b*Sin[c + d*x])/(3*a*b*d*(a + b*Sin[c + d*x]^3))

Rule 3223

Int[cos[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*((c_.)*sin[(e_.) + (f_.)*(x_)])^(n_))^(p_.), x_Symbol] :> With
[{ff = FreeFactors[Sin[e + f*x], x]}, Dist[ff/f, Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*(a + b*(c*ff*x)^n)^p, x]
, x, Sin[e + f*x]/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && IntegerQ[(m - 1)/2] && (EqQ[n, 4] || GtQ[m, 0
] || IGtQ[p, 0] || IntegersQ[m, p])

Rule 1854

Int[(Pq_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Module[{q = Expon[Pq, x], i}, Simp[((a*Coeff[Pq, x, q] -
b*x*ExpandToSum[Pq - Coeff[Pq, x, q]*x^q, x])*(a + b*x^n)^(p + 1))/(a*b*n*(p + 1)), x] + Dist[1/(a*n*(p + 1))
, Int[Sum[(n*(p + 1) + i + 1)*Coeff[Pq, x, i]*x^i, {i, 0, q - 1}]*(a + b*x^n)^(p + 1), x], x] /; q == n - 1] /
; FreeQ[{a, b}, x] && PolyQ[Pq, x] && IGtQ[n, 0] && LtQ[p, -1]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 200

Int[((a_) + (b_.)*(x_)^3)^(-1), x_Symbol] :> Dist[1/(3*Rt[a, 3]^2), Int[1/(Rt[a, 3] + Rt[b, 3]*x), x], x] + Di
st[1/(3*Rt[a, 3]^2), Int[(2*Rt[a, 3] - Rt[b, 3]*x)/(Rt[a, 3]^2 - Rt[a, 3]*Rt[b, 3]*x + Rt[b, 3]^2*x^2), x], x]
/; FreeQ[{a, b}, x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 634

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rubi steps

\begin{align*} \int \frac{\cos ^3(c+d x)}{\left (a+b \sin ^3(c+d x)\right )^2} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1-x^2}{\left (a+b x^3\right )^2} \, dx,x,\sin (c+d x)\right )}{d}\\ &=\frac{a+b \sin (c+d x)}{3 a b d \left (a+b \sin ^3(c+d x)\right )}-\frac{\operatorname{Subst}\left (\int -\frac{2}{a+b x^3} \, dx,x,\sin (c+d x)\right )}{3 a d}\\ &=\frac{a+b \sin (c+d x)}{3 a b d \left (a+b \sin ^3(c+d x)\right )}+\frac{2 \operatorname{Subst}\left (\int \frac{1}{a+b x^3} \, dx,x,\sin (c+d x)\right )}{3 a d}\\ &=\frac{a+b \sin (c+d x)}{3 a b d \left (a+b \sin ^3(c+d x)\right )}+\frac{2 \operatorname{Subst}\left (\int \frac{1}{\sqrt [3]{a}+\sqrt [3]{b} x} \, dx,x,\sin (c+d x)\right )}{9 a^{5/3} d}+\frac{2 \operatorname{Subst}\left (\int \frac{2 \sqrt [3]{a}-\sqrt [3]{b} x}{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2} \, dx,x,\sin (c+d x)\right )}{9 a^{5/3} d}\\ &=\frac{2 \log \left (\sqrt [3]{a}+\sqrt [3]{b} \sin (c+d x)\right )}{9 a^{5/3} \sqrt [3]{b} d}+\frac{a+b \sin (c+d x)}{3 a b d \left (a+b \sin ^3(c+d x)\right )}+\frac{\operatorname{Subst}\left (\int \frac{1}{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2} \, dx,x,\sin (c+d x)\right )}{3 a^{4/3} d}-\frac{\operatorname{Subst}\left (\int \frac{-\sqrt [3]{a} \sqrt [3]{b}+2 b^{2/3} x}{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2} \, dx,x,\sin (c+d x)\right )}{9 a^{5/3} \sqrt [3]{b} d}\\ &=\frac{2 \log \left (\sqrt [3]{a}+\sqrt [3]{b} \sin (c+d x)\right )}{9 a^{5/3} \sqrt [3]{b} d}-\frac{\log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} \sin (c+d x)+b^{2/3} \sin ^2(c+d x)\right )}{9 a^{5/3} \sqrt [3]{b} d}+\frac{a+b \sin (c+d x)}{3 a b d \left (a+b \sin ^3(c+d x)\right )}+\frac{2 \operatorname{Subst}\left (\int \frac{1}{-3-x^2} \, dx,x,1-\frac{2 \sqrt [3]{b} \sin (c+d x)}{\sqrt [3]{a}}\right )}{3 a^{5/3} \sqrt [3]{b} d}\\ &=-\frac{2 \tan ^{-1}\left (\frac{1-\frac{2 \sqrt [3]{b} \sin (c+d x)}{\sqrt [3]{a}}}{\sqrt{3}}\right )}{3 \sqrt{3} a^{5/3} \sqrt [3]{b} d}+\frac{2 \log \left (\sqrt [3]{a}+\sqrt [3]{b} \sin (c+d x)\right )}{9 a^{5/3} \sqrt [3]{b} d}-\frac{\log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} \sin (c+d x)+b^{2/3} \sin ^2(c+d x)\right )}{9 a^{5/3} \sqrt [3]{b} d}+\frac{a+b \sin (c+d x)}{3 a b d \left (a+b \sin ^3(c+d x)\right )}\\ \end{align*}

Mathematica [A]  time = 0.842328, size = 184, normalized size = 1.01 \[ \frac{\frac{-\frac{b^{2/3} \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} \sin (c+d x)+b^{2/3} \sin ^2(c+d x)\right )}{a^{5/3}}+\frac{2 b^{2/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} \sin (c+d x)\right )}{a^{5/3}}+\frac{3}{a+b \sin ^3(c+d x)}}{b}-\frac{2 \sqrt{3} \tan ^{-1}\left (\frac{\sqrt [3]{a}-2 \sqrt [3]{b} \sin (c+d x)}{\sqrt{3} \sqrt [3]{a}}\right )}{a^{5/3} \sqrt [3]{b}}+\frac{3 \sin (c+d x)}{a \left (a+b \sin ^3(c+d x)\right )}}{9 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-2*Sqrt[3]*ArcTan[(a^(1/3) - 2*b^(1/3)*Sin[c + d*x])/(Sqrt[3]*a^(1/3))])/(a^(5/3)*b^(1/3)) + (3*Sin[c + d*x]
)/(a*(a + b*Sin[c + d*x]^3)) + ((2*b^(2/3)*Log[a^(1/3) + b^(1/3)*Sin[c + d*x]])/a^(5/3) - (b^(2/3)*Log[a^(2/3)
- a^(1/3)*b^(1/3)*Sin[c + d*x] + b^(2/3)*Sin[c + d*x]^2])/a^(5/3) + 3/(a + b*Sin[c + d*x]^3))/b)/(9*d)

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Maple [A]  time = 0.141, size = 179, normalized size = 1. \begin{align*}{\frac{\sin \left ( dx+c \right ) }{3\,da \left ( a+b \left ( \sin \left ( dx+c \right ) \right ) ^{3} \right ) }}+{\frac{2}{9\,abd}\ln \left ( \sin \left ( dx+c \right ) +\sqrt [3]{{\frac{a}{b}}} \right ) \left ({\frac{a}{b}} \right ) ^{-{\frac{2}{3}}}}-{\frac{1}{9\,abd}\ln \left ( \left ( \sin \left ( dx+c \right ) \right ) ^{2}-\sqrt [3]{{\frac{a}{b}}}\sin \left ( dx+c \right ) + \left ({\frac{a}{b}} \right ) ^{{\frac{2}{3}}} \right ) \left ({\frac{a}{b}} \right ) ^{-{\frac{2}{3}}}}+{\frac{2\,\sqrt{3}}{9\,abd}\arctan \left ({\frac{\sqrt{3}}{3} \left ( 2\,{\sin \left ( dx+c \right ){\frac{1}{\sqrt [3]{{\frac{a}{b}}}}}}-1 \right ) } \right ) \left ({\frac{a}{b}} \right ) ^{-{\frac{2}{3}}}}+{\frac{1}{3\,bd \left ( a+b \left ( \sin \left ( dx+c \right ) \right ) ^{3} \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*sin(d*x+c)/a/d/(a+b*sin(d*x+c)^3)+2/9/d/b/a/(a/b)^(2/3)*ln(sin(d*x+c)+(a/b)^(1/3))-1/9/d/b/a/(a/b)^(2/3)*l
n(sin(d*x+c)^2-(a/b)^(1/3)*sin(d*x+c)+(a/b)^(2/3))+2/9/d/b/a/(a/b)^(2/3)*3^(1/2)*arctan(1/3*3^(1/2)*(2/(a/b)^(
1/3)*sin(d*x+c)-1))+1/3/d/b/(a+b*sin(d*x+c)^3)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.70672, size = 1609, normalized size = 8.79 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/9*(3*a^2*b*sin(d*x + c) + 3*a^3 + 3*sqrt(1/3)*(a^2*b - (a*b^2*cos(d*x + c)^2 - a*b^2)*sin(d*x + c))*sqrt(-(
a^2*b)^(1/3)/b)*log(-(3*(a^2*b)^(1/3)*a*sin(d*x + c) + a^2 + 3*sqrt(1/3)*(2*a*b*cos(d*x + c)^2 - 2*a*b - (a^2*
b)^(2/3)*sin(d*x + c) + (a^2*b)^(1/3)*a)*sqrt(-(a^2*b)^(1/3)/b) + 2*(a*b*cos(d*x + c)^2 - a*b)*sin(d*x + c))/(
(b*cos(d*x + c)^2 - b)*sin(d*x + c) - a)) + (a^2*b)^(2/3)*((b*cos(d*x + c)^2 - b)*sin(d*x + c) - a)*log(-a*b*c
os(d*x + c)^2 + a*b - (a^2*b)^(2/3)*sin(d*x + c) + (a^2*b)^(1/3)*a) - 2*(a^2*b)^(2/3)*((b*cos(d*x + c)^2 - b)*
sin(d*x + c) - a)*log(a*b*sin(d*x + c) + (a^2*b)^(2/3)))/(a^4*b*d - (a^3*b^2*d*cos(d*x + c)^2 - a^3*b^2*d)*sin
(d*x + c)), 1/9*(3*a^2*b*sin(d*x + c) + 3*a^3 + 6*sqrt(1/3)*(a^2*b - (a*b^2*cos(d*x + c)^2 - a*b^2)*sin(d*x +
c))*sqrt((a^2*b)^(1/3)/b)*arctan(sqrt(1/3)*(2*(a^2*b)^(2/3)*sin(d*x + c) - (a^2*b)^(1/3)*a)*sqrt((a^2*b)^(1/3)
/b)/a^2) + (a^2*b)^(2/3)*((b*cos(d*x + c)^2 - b)*sin(d*x + c) - a)*log(-a*b*cos(d*x + c)^2 + a*b - (a^2*b)^(2/
3)*sin(d*x + c) + (a^2*b)^(1/3)*a) - 2*(a^2*b)^(2/3)*((b*cos(d*x + c)^2 - b)*sin(d*x + c) - a)*log(a*b*sin(d*x
+ c) + (a^2*b)^(2/3)))/(a^4*b*d - (a^3*b^2*d*cos(d*x + c)^2 - a^3*b^2*d)*sin(d*x + c))]

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

[Out]

Timed out

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Giac [A]  time = 1.18884, size = 228, normalized size = 1.25 \begin{align*} -\frac{\frac{2 \, \left (-\frac{a}{b}\right )^{\frac{1}{3}} \log \left ({\left | -\left (-\frac{a}{b}\right )^{\frac{1}{3}} + \sin \left (d x + c\right ) \right |}\right )}{a^{2}} - \frac{2 \, \sqrt{3} \left (-a b^{2}\right )^{\frac{1}{3}} \arctan \left (\frac{\sqrt{3}{\left (\left (-\frac{a}{b}\right )^{\frac{1}{3}} + 2 \, \sin \left (d x + c\right )\right )}}{3 \, \left (-\frac{a}{b}\right )^{\frac{1}{3}}}\right )}{a^{2} b} - \frac{\left (-a b^{2}\right )^{\frac{1}{3}} \log \left (\sin \left (d x + c\right )^{2} + \left (-\frac{a}{b}\right )^{\frac{1}{3}} \sin \left (d x + c\right ) + \left (-\frac{a}{b}\right )^{\frac{2}{3}}\right )}{a^{2} b} - \frac{3 \,{\left (b \sin \left (d x + c\right ) + a\right )}}{{\left (b \sin \left (d x + c\right )^{3} + a\right )} a b}}{9 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/9*(2*(-a/b)^(1/3)*log(abs(-(-a/b)^(1/3) + sin(d*x + c)))/a^2 - 2*sqrt(3)*(-a*b^2)^(1/3)*arctan(1/3*sqrt(3)*
((-a/b)^(1/3) + 2*sin(d*x + c))/(-a/b)^(1/3))/(a^2*b) - (-a*b^2)^(1/3)*log(sin(d*x + c)^2 + (-a/b)^(1/3)*sin(d
*x + c) + (-a/b)^(2/3))/(a^2*b) - 3*(b*sin(d*x + c) + a)/((b*sin(d*x + c)^3 + a)*a*b))/d