3.106 \(\int \frac{\sin ^6(c+d x)}{(a+b \sin ^2(c+d x))^3} \, dx\)

Optimal. Leaf size=148 \[ -\frac{\sqrt{a} \left (8 a^2+20 a b+15 b^2\right ) \tan ^{-1}\left (\frac{\sqrt{a+b} \tan (c+d x)}{\sqrt{a}}\right )}{8 b^3 d (a+b)^{5/2}}+\frac{a (4 a+7 b) \tan (c+d x)}{8 b^2 d (a+b)^2 \left ((a+b) \tan ^2(c+d x)+a\right )}+\frac{a \tan ^3(c+d x)}{4 b d (a+b) \left ((a+b) \tan ^2(c+d x)+a\right )^2}+\frac{x}{b^3} \]

[Out]

x/b^3 - (Sqrt[a]*(8*a^2 + 20*a*b + 15*b^2)*ArcTan[(Sqrt[a + b]*Tan[c + d*x])/Sqrt[a]])/(8*b^3*(a + b)^(5/2)*d)
+ (a*Tan[c + d*x]^3)/(4*b*(a + b)*d*(a + (a + b)*Tan[c + d*x]^2)^2) + (a*(4*a + 7*b)*Tan[c + d*x])/(8*b^2*(a
+ b)^2*d*(a + (a + b)*Tan[c + d*x]^2))

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Rubi [A]  time = 0.278846, antiderivative size = 148, 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} \[ -\frac{\sqrt{a} \left (8 a^2+20 a b+15 b^2\right ) \tan ^{-1}\left (\frac{\sqrt{a+b} \tan (c+d x)}{\sqrt{a}}\right )}{8 b^3 d (a+b)^{5/2}}+\frac{a (4 a+7 b) \tan (c+d x)}{8 b^2 d (a+b)^2 \left ((a+b) \tan ^2(c+d x)+a\right )}+\frac{a \tan ^3(c+d x)}{4 b d (a+b) \left ((a+b) \tan ^2(c+d x)+a\right )^2}+\frac{x}{b^3} \]

Antiderivative was successfully verified.

[In]

Int[Sin[c + d*x]^6/(a + b*Sin[c + d*x]^2)^3,x]

[Out]

x/b^3 - (Sqrt[a]*(8*a^2 + 20*a*b + 15*b^2)*ArcTan[(Sqrt[a + b]*Tan[c + d*x])/Sqrt[a]])/(8*b^3*(a + b)^(5/2)*d)
+ (a*Tan[c + d*x]^3)/(4*b*(a + b)*d*(a + (a + b)*Tan[c + d*x]^2)^2) + (a*(4*a + 7*b)*Tan[c + d*x])/(8*b^2*(a
+ b)^2*d*(a + (a + b)*Tan[c + d*x]^2))

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)}{\left (a+b \sin ^2(c+d x)\right )^3} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^6}{\left (1+x^2\right ) \left (a+(a+b) x^2\right )^3} \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac{a \tan ^3(c+d x)}{4 b (a+b) d \left (a+(a+b) \tan ^2(c+d x)\right )^2}-\frac{\operatorname{Subst}\left (\int \frac{x^2 \left (3 a+(-a-4 b) x^2\right )}{\left (1+x^2\right ) \left (a+(a+b) x^2\right )^2} \, dx,x,\tan (c+d x)\right )}{4 b (a+b) d}\\ &=\frac{a \tan ^3(c+d x)}{4 b (a+b) d \left (a+(a+b) \tan ^2(c+d x)\right )^2}+\frac{a (4 a+7 b) \tan (c+d x)}{8 b^2 (a+b)^2 d \left (a+(a+b) \tan ^2(c+d x)\right )}-\frac{\operatorname{Subst}\left (\int \frac{a (4 a+7 b)+\left (-4 a^2-9 a b-8 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 (a+b)^2 d}\\ &=\frac{a \tan ^3(c+d x)}{4 b (a+b) d \left (a+(a+b) \tan ^2(c+d x)\right )^2}+\frac{a (4 a+7 b) \tan (c+d x)}{8 b^2 (a+b)^2 d \left (a+(a+b) \tan ^2(c+d x)\right )}+\frac{\operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\tan (c+d x)\right )}{b^3 d}-\frac{\left (a \left (8 a^2+20 a b+15 b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{a+(a+b) x^2} \, dx,x,\tan (c+d x)\right )}{8 b^3 (a+b)^2 d}\\ &=\frac{x}{b^3}-\frac{\sqrt{a} \left (8 a^2+20 a b+15 b^2\right ) \tan ^{-1}\left (\frac{\sqrt{a+b} \tan (c+d x)}{\sqrt{a}}\right )}{8 b^3 (a+b)^{5/2} d}+\frac{a \tan ^3(c+d x)}{4 b (a+b) d \left (a+(a+b) \tan ^2(c+d x)\right )^2}+\frac{a (4 a+7 b) \tan (c+d x)}{8 b^2 (a+b)^2 d \left (a+(a+b) \tan ^2(c+d x)\right )}\\ \end{align*}

Mathematica [A]  time = 2.58449, size = 134, normalized size = 0.91 \[ \frac{-\frac{\sqrt{a} \left (8 a^2+20 a b+15 b^2\right ) \tan ^{-1}\left (\frac{\sqrt{a+b} \tan (c+d x)}{\sqrt{a}}\right )}{(a+b)^{5/2}}+\frac{a b \sin (2 (c+d x)) \left (8 a^2-3 b (2 a+3 b) \cos (2 (c+d x))+20 a b+9 b^2\right )}{(a+b)^2 (2 a-b \cos (2 (c+d x))+b)^2}+8 (c+d x)}{8 b^3 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[c + d*x]^6/(a + b*Sin[c + d*x]^2)^3,x]

[Out]

(8*(c + d*x) - (Sqrt[a]*(8*a^2 + 20*a*b + 15*b^2)*ArcTan[(Sqrt[a + b]*Tan[c + d*x])/Sqrt[a]])/(a + b)^(5/2) +
(a*b*(8*a^2 + 20*a*b + 9*b^2 - 3*b*(2*a + 3*b)*Cos[2*(c + d*x)])*Sin[2*(c + d*x)])/((a + b)^2*(2*a + b - b*Cos
[2*(c + d*x)])^2))/(8*b^3*d)

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Maple [B]  time = 0.096, size = 363, normalized size = 2.5 \begin{align*}{\frac{\arctan \left ( \tan \left ( dx+c \right ) \right ) }{d{b}^{3}}}+{\frac{{a}^{2} \left ( \tan \left ( dx+c \right ) \right ) ^{3}}{2\,{b}^{2}d \left ( a \left ( \tan \left ( dx+c \right ) \right ) ^{2}+ \left ( \tan \left ( dx+c \right ) \right ) ^{2}b+a \right ) ^{2} \left ( a+b \right ) }}+{\frac{9\,a \left ( \tan \left ( dx+c \right ) \right ) ^{3}}{8\,bd \left ( a \left ( \tan \left ( dx+c \right ) \right ) ^{2}+ \left ( \tan \left ( dx+c \right ) \right ) ^{2}b+a \right ) ^{2} \left ( a+b \right ) }}+{\frac{{a}^{3}\tan \left ( dx+c \right ) }{2\,{b}^{2}d \left ( a \left ( \tan \left ( dx+c \right ) \right ) ^{2}+ \left ( \tan \left ( dx+c \right ) \right ) ^{2}b+a \right ) ^{2} \left ({a}^{2}+2\,ab+{b}^{2} \right ) }}+{\frac{7\,{a}^{2}\tan \left ( dx+c \right ) }{8\,bd \left ( a \left ( \tan \left ( dx+c \right ) \right ) ^{2}+ \left ( \tan \left ( dx+c \right ) \right ) ^{2}b+a \right ) ^{2} \left ({a}^{2}+2\,ab+{b}^{2} \right ) }}-{\frac{{a}^{3}}{d{b}^{3} \left ({a}^{2}+2\,ab+{b}^{2} \right ) }\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 ) }}}}-{\frac{5\,{a}^{2}}{2\,{b}^{2}d \left ({a}^{2}+2\,ab+{b}^{2} \right ) }\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 ) }}}}-{\frac{15\,a}{8\,bd \left ({a}^{2}+2\,ab+{b}^{2} \right ) }\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)^3,x)

[Out]

1/d/b^3*arctan(tan(d*x+c))+1/2/d*a^2/b^2/(a*tan(d*x+c)^2+tan(d*x+c)^2*b+a)^2/(a+b)*tan(d*x+c)^3+9/8/d*a/b/(a*t
an(d*x+c)^2+tan(d*x+c)^2*b+a)^2/(a+b)*tan(d*x+c)^3+1/2/d*a^3/b^2/(a*tan(d*x+c)^2+tan(d*x+c)^2*b+a)^2/(a^2+2*a*
b+b^2)*tan(d*x+c)+7/8/d*a^2/b/(a*tan(d*x+c)^2+tan(d*x+c)^2*b+a)^2/(a^2+2*a*b+b^2)*tan(d*x+c)-1/d*a^3/b^3/(a^2+
2*a*b+b^2)/(a*(a+b))^(1/2)*arctan((a+b)*tan(d*x+c)/(a*(a+b))^(1/2))-5/2/d*a^2/b^2/(a^2+2*a*b+b^2)/(a*(a+b))^(1
/2)*arctan((a+b)*tan(d*x+c)/(a*(a+b))^(1/2))-15/8/d*a/b/(a^2+2*a*b+b^2)/(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)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.22013, size = 2175, normalized size = 14.7 \begin{align*} \text{result too large to display} \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)^3,x, algorithm="fricas")

[Out]

[1/32*(32*(a^2*b^2 + 2*a*b^3 + b^4)*d*x*cos(d*x + c)^4 - 64*(a^3*b + 3*a^2*b^2 + 3*a*b^3 + b^4)*d*x*cos(d*x +
c)^2 + 32*(a^4 + 4*a^3*b + 6*a^2*b^2 + 4*a*b^3 + b^4)*d*x + ((8*a^2*b^2 + 20*a*b^3 + 15*b^4)*cos(d*x + c)^4 +
8*a^4 + 36*a^3*b + 63*a^2*b^2 + 50*a*b^3 + 15*b^4 - 2*(8*a^3*b + 28*a^2*b^2 + 35*a*b^3 + 15*b^4)*cos(d*x + c)^
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)) - 4*(3*(2*a^2*b^2 + 3*a*b
^3)*cos(d*x + c)^3 - (4*a^3*b + 13*a^2*b^2 + 9*a*b^3)*cos(d*x + c))*sin(d*x + c))/((a^2*b^5 + 2*a*b^6 + b^7)*d
*cos(d*x + c)^4 - 2*(a^3*b^4 + 3*a^2*b^5 + 3*a*b^6 + b^7)*d*cos(d*x + c)^2 + (a^4*b^3 + 4*a^3*b^4 + 6*a^2*b^5
+ 4*a*b^6 + b^7)*d), 1/16*(16*(a^2*b^2 + 2*a*b^3 + b^4)*d*x*cos(d*x + c)^4 - 32*(a^3*b + 3*a^2*b^2 + 3*a*b^3 +
b^4)*d*x*cos(d*x + c)^2 + 16*(a^4 + 4*a^3*b + 6*a^2*b^2 + 4*a*b^3 + b^4)*d*x + ((8*a^2*b^2 + 20*a*b^3 + 15*b^
4)*cos(d*x + c)^4 + 8*a^4 + 36*a^3*b + 63*a^2*b^2 + 50*a*b^3 + 15*b^4 - 2*(8*a^3*b + 28*a^2*b^2 + 35*a*b^3 + 1
5*b^4)*cos(d*x + c)^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))) - 2*(3*(2*a^2*b^2 + 3*a*b^3)*cos(d*x + c)^3 - (4*a^3*b + 13*a^2*b^2 + 9*a*b^3)*cos(d*x +
c))*sin(d*x + c))/((a^2*b^5 + 2*a*b^6 + b^7)*d*cos(d*x + c)^4 - 2*(a^3*b^4 + 3*a^2*b^5 + 3*a*b^6 + b^7)*d*cos
(d*x + c)^2 + (a^4*b^3 + 4*a^3*b^4 + 6*a^2*b^5 + 4*a*b^6 + b^7)*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)**3,x)

[Out]

Timed out

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Giac [A]  time = 1.2044, size = 302, normalized size = 2.04 \begin{align*} -\frac{\frac{{\left (8 \, a^{3} + 20 \, a^{2} b + 15 \, a b^{2}\right )}{\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 )}}{{\left (a^{2} b^{3} + 2 \, a b^{4} + b^{5}\right )} \sqrt{a^{2} + a b}} - \frac{4 \, a^{3} \tan \left (d x + c\right )^{3} + 13 \, a^{2} b \tan \left (d x + c\right )^{3} + 9 \, a b^{2} \tan \left (d x + c\right )^{3} + 4 \, a^{3} \tan \left (d x + c\right ) + 7 \, a^{2} b \tan \left (d x + c\right )}{{\left (a^{2} b^{2} + 2 \, a b^{3} + b^{4}\right )}{\left (a \tan \left (d x + c\right )^{2} + b \tan \left (d x + c\right )^{2} + a\right )}^{2}} - \frac{8 \,{\left (d x + c\right )}}{b^{3}}}{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)^3,x, algorithm="giac")

[Out]

-1/8*((8*a^3 + 20*a^2*b + 15*a*b^2)*(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^2*b^3 + 2*a*b^4 + b^5)*sqrt(a^2 + a*b)) - (4*a^3*tan(d*x + c)^3 + 13*a^2*b
*tan(d*x + c)^3 + 9*a*b^2*tan(d*x + c)^3 + 4*a^3*tan(d*x + c) + 7*a^2*b*tan(d*x + c))/((a^2*b^2 + 2*a*b^3 + b^
4)*(a*tan(d*x + c)^2 + b*tan(d*x + c)^2 + a)^2) - 8*(d*x + c)/b^3)/d