3.86 \(\int \frac{\sin ^6(c+d x)}{a+b \sin ^2(c+d x)} \, dx\)

Optimal. Leaf size=117 \[ -\frac{a^{5/2} \tan ^{-1}\left (\frac{\sqrt{a+b} \tan (c+d x)}{\sqrt{a}}\right )}{b^3 d \sqrt{a+b}}+\frac{x \left (8 a^2-4 a b+3 b^2\right )}{8 b^3}+\frac{(4 a-3 b) \sin (c+d x) \cos (c+d x)}{8 b^2 d}-\frac{\sin ^3(c+d x) \cos (c+d x)}{4 b d} \]

[Out]

((8*a^2 - 4*a*b + 3*b^2)*x)/(8*b^3) - (a^(5/2)*ArcTan[(Sqrt[a + b]*Tan[c + d*x])/Sqrt[a]])/(b^3*Sqrt[a + b]*d)
+ ((4*a - 3*b)*Cos[c + d*x]*Sin[c + d*x])/(8*b^2*d) - (Cos[c + d*x]*Sin[c + d*x]^3)/(4*b*d)

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Rubi [A]  time = 0.223681, antiderivative size = 117, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.261, Rules used = {3187, 470, 578, 522, 203, 205} \[ -\frac{a^{5/2} \tan ^{-1}\left (\frac{\sqrt{a+b} \tan (c+d x)}{\sqrt{a}}\right )}{b^3 d \sqrt{a+b}}+\frac{x \left (8 a^2-4 a b+3 b^2\right )}{8 b^3}+\frac{(4 a-3 b) \sin (c+d x) \cos (c+d x)}{8 b^2 d}-\frac{\sin ^3(c+d x) \cos (c+d x)}{4 b d} \]

Antiderivative was successfully verified.

[In]

Int[Sin[c + d*x]^6/(a + b*Sin[c + d*x]^2),x]

[Out]

((8*a^2 - 4*a*b + 3*b^2)*x)/(8*b^3) - (a^(5/2)*ArcTan[(Sqrt[a + b]*Tan[c + d*x])/Sqrt[a]])/(b^3*Sqrt[a + b]*d)
+ ((4*a - 3*b)*Cos[c + d*x]*Sin[c + d*x])/(8*b^2*d) - (Cos[c + d*x]*Sin[c + d*x]^3)/(4*b*d)

Rule 3187

Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.), x_Symbol] :> With[{ff = FreeF
actors[Tan[e + f*x], x]}, Dist[ff^(m + 1)/f, Subst[Int[(x^m*(a + (a + b)*ff^2*x^2)^p)/(1 + ff^2*x^2)^(m/2 + p
+ 1), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x] && IntegerQ[m/2] && IntegerQ[p]

Rule 470

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> -Simp[(a*e^(2
*n - 1)*(e*x)^(m - 2*n + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(b*n*(b*c - a*d)*(p + 1)), x] + Dist[e^(2
*n)/(b*n*(b*c - a*d)*(p + 1)), Int[(e*x)^(m - 2*n)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[a*c*(m - 2*n + 1) +
(a*d*(m - n + n*q + 1) + b*c*n*(p + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b*c - a*d, 0] &
& IGtQ[n, 0] && LtQ[p, -1] && GtQ[m - n + 1, n] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x]

Rule 578

Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_)*((e_) + (f_.)*(x_)^(n_)), x
_Symbol] :> Simp[(g^(n - 1)*(b*e - a*f)*(g*x)^(m - n + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(b*n*(b*c -
a*d)*(p + 1)), x] - Dist[g^n/(b*n*(b*c - a*d)*(p + 1)), Int[(g*x)^(m - n)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*S
imp[c*(b*e - a*f)*(m - n + 1) + (d*(b*e - a*f)*(m + n*q + 1) - b*n*(c*f - d*e)*(p + 1))*x^n, x], x], x] /; Fre
eQ[{a, b, c, d, e, f, g, q}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m - n + 1, 0]

Rule 522

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*((c_) + (d_.)*(x_)^(n_))), x_Symbol] :> Dist[(b*e - a*f
)/(b*c - a*d), Int[1/(a + b*x^n), x], x] - Dist[(d*e - c*f)/(b*c - a*d), Int[1/(c + d*x^n), x], x] /; FreeQ[{a
, b, c, d, e, f, n}, x]

Rule 203

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

Rule 205

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

Rubi steps

\begin{align*} \int \frac{\sin ^6(c+d x)}{a+b \sin ^2(c+d x)} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^6}{\left (1+x^2\right )^3 \left (a+(a+b) x^2\right )} \, dx,x,\tan (c+d x)\right )}{d}\\ &=-\frac{\cos (c+d x) \sin ^3(c+d x)}{4 b d}+\frac{\operatorname{Subst}\left (\int \frac{x^2 \left (3 a+(-a+3 b) x^2\right )}{\left (1+x^2\right )^2 \left (a+(a+b) x^2\right )} \, dx,x,\tan (c+d x)\right )}{4 b d}\\ &=\frac{(4 a-3 b) \cos (c+d x) \sin (c+d x)}{8 b^2 d}-\frac{\cos (c+d x) \sin ^3(c+d x)}{4 b d}-\frac{\operatorname{Subst}\left (\int \frac{a (4 a-3 b)+\left (-4 a^2+a b-3 b^2\right ) x^2}{\left (1+x^2\right ) \left (a+(a+b) x^2\right )} \, dx,x,\tan (c+d x)\right )}{8 b^2 d}\\ &=\frac{(4 a-3 b) \cos (c+d x) \sin (c+d x)}{8 b^2 d}-\frac{\cos (c+d x) \sin ^3(c+d x)}{4 b d}-\frac{a^3 \operatorname{Subst}\left (\int \frac{1}{a+(a+b) x^2} \, dx,x,\tan (c+d x)\right )}{b^3 d}+\frac{\left (8 a^2-4 a b+3 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\tan (c+d x)\right )}{8 b^3 d}\\ &=\frac{\left (8 a^2-4 a b+3 b^2\right ) x}{8 b^3}-\frac{a^{5/2} \tan ^{-1}\left (\frac{\sqrt{a+b} \tan (c+d x)}{\sqrt{a}}\right )}{b^3 \sqrt{a+b} d}+\frac{(4 a-3 b) \cos (c+d x) \sin (c+d x)}{8 b^2 d}-\frac{\cos (c+d x) \sin ^3(c+d x)}{4 b d}\\ \end{align*}

Mathematica [A]  time = 0.445376, size = 95, normalized size = 0.81 \[ \frac{4 \left (8 a^2-4 a b+3 b^2\right ) (c+d x)-\frac{32 a^{5/2} \tan ^{-1}\left (\frac{\sqrt{a+b} \tan (c+d x)}{\sqrt{a}}\right )}{\sqrt{a+b}}+8 b (a-b) \sin (2 (c+d x))+b^2 \sin (4 (c+d x))}{32 b^3 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[c + d*x]^6/(a + b*Sin[c + d*x]^2),x]

[Out]

(4*(8*a^2 - 4*a*b + 3*b^2)*(c + d*x) - (32*a^(5/2)*ArcTan[(Sqrt[a + b]*Tan[c + d*x])/Sqrt[a]])/Sqrt[a + b] + 8
*(a - b)*b*Sin[2*(c + d*x)] + b^2*Sin[4*(c + d*x)])/(32*b^3*d)

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Maple [A]  time = 0.092, size = 196, normalized size = 1.7 \begin{align*}{\frac{ \left ( \tan \left ( dx+c \right ) \right ) ^{3}a}{2\,{b}^{2}d \left ( \left ( \tan \left ( dx+c \right ) \right ) ^{2}+1 \right ) ^{2}}}-{\frac{5\, \left ( \tan \left ( dx+c \right ) \right ) ^{3}}{8\,bd \left ( \left ( \tan \left ( dx+c \right ) \right ) ^{2}+1 \right ) ^{2}}}+{\frac{a\tan \left ( dx+c \right ) }{2\,{b}^{2}d \left ( \left ( \tan \left ( dx+c \right ) \right ) ^{2}+1 \right ) ^{2}}}-{\frac{3\,\tan \left ( dx+c \right ) }{8\,bd \left ( \left ( \tan \left ( dx+c \right ) \right ) ^{2}+1 \right ) ^{2}}}+{\frac{\arctan \left ( \tan \left ( dx+c \right ) \right ){a}^{2}}{d{b}^{3}}}-{\frac{\arctan \left ( \tan \left ( dx+c \right ) \right ) a}{2\,{b}^{2}d}}+{\frac{3\,\arctan \left ( \tan \left ( dx+c \right ) \right ) }{8\,bd}}-{\frac{{a}^{3}}{d{b}^{3}}\arctan \left ({ \left ( a+b \right ) \tan \left ( dx+c \right ){\frac{1}{\sqrt{a \left ( a+b \right ) }}}} \right ){\frac{1}{\sqrt{a \left ( a+b \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(d*x+c)^6/(a+sin(d*x+c)^2*b),x)

[Out]

1/2/d/b^2/(tan(d*x+c)^2+1)^2*tan(d*x+c)^3*a-5/8/d/b/(tan(d*x+c)^2+1)^2*tan(d*x+c)^3+1/2/d/b^2/(tan(d*x+c)^2+1)
^2*tan(d*x+c)*a-3/8/d/b/(tan(d*x+c)^2+1)^2*tan(d*x+c)+1/d/b^3*arctan(tan(d*x+c))*a^2-1/2/d/b^2*arctan(tan(d*x+
c))*a+3/8/d/b*arctan(tan(d*x+c))-1/d*a^3/b^3/(a*(a+b))^(1/2)*arctan((a+b)*tan(d*x+c)/(a*(a+b))^(1/2))

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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(sin(d*x+c)^6/(a+b*sin(d*x+c)^2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.92646, size = 884, normalized size = 7.56 \begin{align*} \left [\frac{2 \, a^{2} \sqrt{-\frac{a}{a + b}} \log \left (\frac{{\left (8 \, a^{2} + 8 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{4} - 2 \,{\left (4 \, a^{2} + 5 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{2} + 4 \,{\left ({\left (2 \, a^{2} + 3 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{3} -{\left (a^{2} + 2 \, a b + b^{2}\right )} \cos \left (d x + c\right )\right )} \sqrt{-\frac{a}{a + b}} \sin \left (d x + c\right ) + a^{2} + 2 \, a b + b^{2}}{b^{2} \cos \left (d x + c\right )^{4} - 2 \,{\left (a b + b^{2}\right )} \cos \left (d x + c\right )^{2} + a^{2} + 2 \, a b + b^{2}}\right ) +{\left (8 \, a^{2} - 4 \, a b + 3 \, b^{2}\right )} d x +{\left (2 \, b^{2} \cos \left (d x + c\right )^{3} +{\left (4 \, a b - 5 \, b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{8 \, b^{3} d}, \frac{4 \, a^{2} \sqrt{\frac{a}{a + b}} \arctan \left (\frac{{\left ({\left (2 \, a + b\right )} \cos \left (d x + c\right )^{2} - a - b\right )} \sqrt{\frac{a}{a + b}}}{2 \, a \cos \left (d x + c\right ) \sin \left (d x + c\right )}\right ) +{\left (8 \, a^{2} - 4 \, a b + 3 \, b^{2}\right )} d x +{\left (2 \, b^{2} \cos \left (d x + c\right )^{3} +{\left (4 \, a b - 5 \, b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{8 \, b^{3} d}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^6/(a+b*sin(d*x+c)^2),x, algorithm="fricas")

[Out]

[1/8*(2*a^2*sqrt(-a/(a + b))*log(((8*a^2 + 8*a*b + b^2)*cos(d*x + c)^4 - 2*(4*a^2 + 5*a*b + b^2)*cos(d*x + c)^
2 + 4*((2*a^2 + 3*a*b + b^2)*cos(d*x + c)^3 - (a^2 + 2*a*b + b^2)*cos(d*x + c))*sqrt(-a/(a + b))*sin(d*x + c)
+ a^2 + 2*a*b + b^2)/(b^2*cos(d*x + c)^4 - 2*(a*b + b^2)*cos(d*x + c)^2 + a^2 + 2*a*b + b^2)) + (8*a^2 - 4*a*b
+ 3*b^2)*d*x + (2*b^2*cos(d*x + c)^3 + (4*a*b - 5*b^2)*cos(d*x + c))*sin(d*x + c))/(b^3*d), 1/8*(4*a^2*sqrt(a
/(a + b))*arctan(1/2*((2*a + b)*cos(d*x + c)^2 - a - b)*sqrt(a/(a + b))/(a*cos(d*x + c)*sin(d*x + c))) + (8*a^
2 - 4*a*b + 3*b^2)*d*x + (2*b^2*cos(d*x + c)^3 + (4*a*b - 5*b^2)*cos(d*x + c))*sin(d*x + c))/(b^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(sin(d*x+c)**6/(a+b*sin(d*x+c)**2),x)

[Out]

Timed out

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Giac [A]  time = 1.2068, size = 212, normalized size = 1.81 \begin{align*} -\frac{\frac{8 \,{\left (\pi \left \lfloor \frac{d x + c}{\pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (2 \, a + 2 \, b\right ) + \arctan \left (\frac{a \tan \left (d x + c\right ) + b \tan \left (d x + c\right )}{\sqrt{a^{2} + a b}}\right )\right )} a^{3}}{\sqrt{a^{2} + a b} b^{3}} - \frac{{\left (8 \, a^{2} - 4 \, a b + 3 \, b^{2}\right )}{\left (d x + c\right )}}{b^{3}} - \frac{4 \, a \tan \left (d x + c\right )^{3} - 5 \, b \tan \left (d x + c\right )^{3} + 4 \, a \tan \left (d x + c\right ) - 3 \, b \tan \left (d x + c\right )}{{\left (\tan \left (d x + c\right )^{2} + 1\right )}^{2} b^{2}}}{8 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^6/(a+b*sin(d*x+c)^2),x, algorithm="giac")

[Out]

-1/8*(8*(pi*floor((d*x + c)/pi + 1/2)*sgn(2*a + 2*b) + arctan((a*tan(d*x + c) + b*tan(d*x + c))/sqrt(a^2 + a*b
)))*a^3/(sqrt(a^2 + a*b)*b^3) - (8*a^2 - 4*a*b + 3*b^2)*(d*x + c)/b^3 - (4*a*tan(d*x + c)^3 - 5*b*tan(d*x + c)
^3 + 4*a*tan(d*x + c) - 3*b*tan(d*x + c))/((tan(d*x + c)^2 + 1)^2*b^2))/d