3.394 \(\int \frac{\cos ^5(c+d x)}{(a+b \sin ^3(c+d x))^2} \, dx\)

Optimal. Leaf size=238 \[ \frac{\left (a^{4/3}-b^{4/3}\right ) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} \sin (c+d x)+b^{2/3} \sin ^2(c+d x)\right )}{9 a^{5/3} b^{5/3} d}-\frac{2 \left (a^{4/3}-b^{4/3}\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} \sin (c+d x)\right )}{9 a^{5/3} b^{5/3} d}-\frac{2 \left (a^{4/3}+b^{4/3}\right ) \tan ^{-1}\left (\frac{\sqrt [3]{a}-2 \sqrt [3]{b} \sin (c+d x)}{\sqrt{3} \sqrt [3]{a}}\right )}{3 \sqrt{3} a^{5/3} b^{5/3} d}+\frac{\sin (c+d x) \left (-a \sin (c+d x)-2 b \sin ^2(c+d x)+b\right )}{3 a b d \left (a+b \sin ^3(c+d x)\right )} \]

[Out]

(-2*(a^(4/3) + b^(4/3))*ArcTan[(a^(1/3) - 2*b^(1/3)*Sin[c + d*x])/(Sqrt[3]*a^(1/3))])/(3*Sqrt[3]*a^(5/3)*b^(5/
3)*d) - (2*(a^(4/3) - b^(4/3))*Log[a^(1/3) + b^(1/3)*Sin[c + d*x]])/(9*a^(5/3)*b^(5/3)*d) + ((a^(4/3) - b^(4/3
))*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^(5/3)*d) + (Sin[c + d*x]
*(b - a*Sin[c + d*x] - 2*b*Sin[c + d*x]^2))/(3*a*b*d*(a + b*Sin[c + d*x]^3))

________________________________________________________________________________________

Rubi [A]  time = 0.224561, antiderivative size = 238, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.348, Rules used = {3223, 1858, 1860, 31, 634, 617, 204, 628} \[ \frac{\left (a^{4/3}-b^{4/3}\right ) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} \sin (c+d x)+b^{2/3} \sin ^2(c+d x)\right )}{9 a^{5/3} b^{5/3} d}-\frac{2 \left (a^{4/3}-b^{4/3}\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} \sin (c+d x)\right )}{9 a^{5/3} b^{5/3} d}-\frac{2 \left (a^{4/3}+b^{4/3}\right ) \tan ^{-1}\left (\frac{\sqrt [3]{a}-2 \sqrt [3]{b} \sin (c+d x)}{\sqrt{3} \sqrt [3]{a}}\right )}{3 \sqrt{3} a^{5/3} b^{5/3} d}+\frac{\sin (c+d x) \left (-a \sin (c+d x)-2 b \sin ^2(c+d x)+b\right )}{3 a b d \left (a+b \sin ^3(c+d x)\right )} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^5/(a + b*Sin[c + d*x]^3)^2,x]

[Out]

(-2*(a^(4/3) + b^(4/3))*ArcTan[(a^(1/3) - 2*b^(1/3)*Sin[c + d*x])/(Sqrt[3]*a^(1/3))])/(3*Sqrt[3]*a^(5/3)*b^(5/
3)*d) - (2*(a^(4/3) - b^(4/3))*Log[a^(1/3) + b^(1/3)*Sin[c + d*x]])/(9*a^(5/3)*b^(5/3)*d) + ((a^(4/3) - b^(4/3
))*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^(5/3)*d) + (Sin[c + d*x]
*(b - a*Sin[c + d*x] - 2*b*Sin[c + d*x]^2))/(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 1858

Int[(Pq_)*((a_) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> With[{q = Expon[Pq, x]}, Module[{Q = PolynomialQuotient
[b^(Floor[(q - 1)/n] + 1)*Pq, a + b*x^n, x], R = PolynomialRemainder[b^(Floor[(q - 1)/n] + 1)*Pq, a + b*x^n, x
]}, Dist[1/(a*n*(p + 1)*b^(Floor[(q - 1)/n] + 1)), Int[(a + b*x^n)^(p + 1)*ExpandToSum[a*n*(p + 1)*Q + n*(p +
1)*R + D[x*R, x], x], x], x] - Simp[(x*R*(a + b*x^n)^(p + 1))/(a*n*(p + 1)*b^(Floor[(q - 1)/n] + 1)), x]] /; G
eQ[q, n]] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && IGtQ[n, 0] && LtQ[p, -1]

Rule 1860

Int[((A_) + (B_.)*(x_))/((a_) + (b_.)*(x_)^3), x_Symbol] :> With[{r = Numerator[Rt[a/b, 3]], s = Denominator[R
t[a/b, 3]]}, -Dist[(r*(B*r - A*s))/(3*a*s), Int[1/(r + s*x), x], x] + Dist[r/(3*a*s), Int[(r*(B*r + 2*A*s) + s
*(B*r - A*s)*x)/(r^2 - r*s*x + s^2*x^2), x], x]] /; FreeQ[{a, b, A, B}, x] && NeQ[a*B^3 - b*A^3, 0] && PosQ[a/
b]

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 ^5(c+d x)}{\left (a+b \sin ^3(c+d x)\right )^2} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (1-x^2\right )^2}{\left (a+b x^3\right )^2} \, dx,x,\sin (c+d x)\right )}{d}\\ &=\frac{\sin (c+d x) \left (b-a \sin (c+d x)-2 b \sin ^2(c+d x)\right )}{3 a b d \left (a+b \sin ^3(c+d x)\right )}-\frac{\operatorname{Subst}\left (\int \frac{-2 b^2-2 a b x}{a+b x^3} \, dx,x,\sin (c+d x)\right )}{3 a b^2 d}\\ &=\frac{\sin (c+d x) \left (b-a \sin (c+d x)-2 b \sin ^2(c+d x)\right )}{3 a b d \left (a+b \sin ^3(c+d x)\right )}-\frac{\operatorname{Subst}\left (\int \frac{\sqrt [3]{a} \left (-2 a^{4/3} b-4 b^{7/3}\right )+\sqrt [3]{b} \left (-2 a^{4/3} b+2 b^{7/3}\right ) 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} b^{7/3} d}-\frac{\left (2 \left (a^{4/3}-b^{4/3}\right )\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt [3]{a}+\sqrt [3]{b} x} \, dx,x,\sin (c+d x)\right )}{9 a^{5/3} b^{4/3} d}\\ &=-\frac{2 \left (a^{4/3}-b^{4/3}\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} \sin (c+d x)\right )}{9 a^{5/3} b^{5/3} d}+\frac{\sin (c+d x) \left (b-a \sin (c+d x)-2 b \sin ^2(c+d x)\right )}{3 a b d \left (a+b \sin ^3(c+d x)\right )}+\frac{\left (\frac{1}{a^{4/3}}+\frac{1}{b^{4/3}}\right ) \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 d}+\frac{\left (a^{4/3}-b^{4/3}\right ) \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} b^{5/3} d}\\ &=-\frac{2 \left (a^{4/3}-b^{4/3}\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} \sin (c+d x)\right )}{9 a^{5/3} b^{5/3} d}+\frac{\left (a^{4/3}-b^{4/3}\right ) \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} b^{5/3} d}+\frac{\sin (c+d x) \left (b-a \sin (c+d x)-2 b \sin ^2(c+d x)\right )}{3 a b d \left (a+b \sin ^3(c+d x)\right )}+\frac{\left (2 \left (a^{4/3}+b^{4/3}\right )\right ) \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} b^{5/3} d}\\ &=-\frac{2 \left (a^{4/3}+b^{4/3}\right ) \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} b^{5/3} d}-\frac{2 \left (a^{4/3}-b^{4/3}\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} \sin (c+d x)\right )}{9 a^{5/3} b^{5/3} d}+\frac{\left (a^{4/3}-b^{4/3}\right ) \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} b^{5/3} d}+\frac{\sin (c+d x) \left (b-a \sin (c+d x)-2 b \sin ^2(c+d x)\right )}{3 a b d \left (a+b \sin ^3(c+d x)\right )}\\ \end{align*}

Mathematica [C]  time = 1.03697, size = 258, normalized size = 1.08 \[ \frac{-\frac{2 \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} \sin (c+d x)+b^{2/3} \sin ^2(c+d x)\right )}{a^{5/3} \sqrt [3]{b}}+\frac{4 \log \left (\sqrt [3]{a}+\sqrt [3]{b} \sin (c+d x)\right )}{a^{5/3} \sqrt [3]{b}}-\frac{4 \sqrt{3} \tan ^{-1}\left (\frac{\sqrt [3]{a}-2 \sqrt [3]{b} \sin (c+d x)}{\sqrt{3} \sqrt [3]{a}}\right )}{a^{5/3} \sqrt [3]{b}}+\frac{9 \sin ^2(c+d x) \, _2F_1\left (\frac{2}{3},1;\frac{5}{3};-\frac{b \sin ^3(c+d x)}{a}\right )}{a b}-\frac{9 \sin ^2(c+d x) \, _2F_1\left (\frac{2}{3},2;\frac{5}{3};-\frac{b \sin ^3(c+d x)}{a}\right )}{a b}+\frac{6 \sin (c+d x)}{a \left (a+b \sin ^3(c+d x)\right )}+\frac{12}{b \left (a+b \sin ^3(c+d x)\right )}}{18 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^5/(a + b*Sin[c + d*x]^3)^2,x]

[Out]

((-4*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)) + (4*Log[a^(1/3)
+ b^(1/3)*Sin[c + d*x]])/(a^(5/3)*b^(1/3)) - (2*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)*b^(1/3)) + (9*Hypergeometric2F1[2/3, 1, 5/3, -((b*Sin[c + d*x]^3)/a)]*Sin[c + d*x]^2)/(a*b) -
(9*Hypergeometric2F1[2/3, 2, 5/3, -((b*Sin[c + d*x]^3)/a)]*Sin[c + d*x]^2)/(a*b) + 12/(b*(a + b*Sin[c + d*x]^
3)) + (6*Sin[c + d*x])/(a*(a + b*Sin[c + d*x]^3)))/(18*d)

________________________________________________________________________________________

Maple [A]  time = 0.145, size = 327, normalized size = 1.4 \begin{align*} -{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{2}}{3\,bd \left ( a+b \left ( \sin \left ( dx+c \right ) \right ) ^{3} \right ) }}+{\frac{\sin \left ( dx+c \right ) }{3\,da \left ( a+b \left ( \sin \left ( dx+c \right ) \right ) ^{3} \right ) }}+{\frac{2}{3\,bd \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{2}{9\,{b}^{2}d}\ln \left ( \sin \left ( dx+c \right ) +\sqrt [3]{{\frac{a}{b}}} \right ){\frac{1}{\sqrt [3]{{\frac{a}{b}}}}}}+{\frac{1}{9\,{b}^{2}d}\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 ){\frac{1}{\sqrt [3]{{\frac{a}{b}}}}}}+{\frac{2\,\sqrt{3}}{9\,{b}^{2}d}\arctan \left ({\frac{\sqrt{3}}{3} \left ( 2\,{\sin \left ( dx+c \right ){\frac{1}{\sqrt [3]{{\frac{a}{b}}}}}}-1 \right ) } \right ){\frac{1}{\sqrt [3]{{\frac{a}{b}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^5/(a+b*sin(d*x+c)^3)^2,x)

[Out]

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

________________________________________________________________________________________

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)^5/(a+b*sin(d*x+c)^3)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [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)^5/(a+b*sin(d*x+c)^3)^2,x, algorithm="fricas")

[Out]

Timed out

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 1.19461, size = 308, normalized size = 1.29 \begin{align*} -\frac{\frac{2 \,{\left (a \left (-\frac{a}{b}\right )^{\frac{1}{3}} + b\right )} \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} b} + \frac{3 \,{\left (a \sin \left (d x + c\right )^{2} - b \sin \left (d x + c\right ) - 2 \, a\right )}}{{\left (b \sin \left (d x + c\right )^{3} + a\right )} a b} - \frac{2 \, \sqrt{3}{\left (\left (-a b^{2}\right )^{\frac{1}{3}} b^{2} - \left (-a b^{2}\right )^{\frac{2}{3}} a\right )} \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^{3}} - \frac{{\left (\left (-a b^{2}\right )^{\frac{1}{3}} b^{2} + \left (-a b^{2}\right )^{\frac{2}{3}} a\right )} \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^{3}}}{9 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5/(a+b*sin(d*x+c)^3)^2,x, algorithm="giac")

[Out]

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