3.195 \(\int \frac{\sin ^9(c+d x)}{a-b \sin ^4(c+d x)} \, dx\)
Optimal. Leaf size=177 \[ -\frac{a^{3/2} \tan ^{-1}\left (\frac{\sqrt [4]{b} \cos (c+d x)}{\sqrt{\sqrt{a}-\sqrt{b}}}\right )}{2 b^{9/4} d \sqrt{\sqrt{a}-\sqrt{b}}}-\frac{a^{3/2} \tanh ^{-1}\left (\frac{\sqrt [4]{b} \cos (c+d x)}{\sqrt{\sqrt{a}+\sqrt{b}}}\right )}{2 b^{9/4} d \sqrt{\sqrt{a}+\sqrt{b}}}+\frac{(a+b) \cos (c+d x)}{b^2 d}+\frac{\cos ^5(c+d x)}{5 b d}-\frac{2 \cos ^3(c+d x)}{3 b d} \]
[Out]
-(a^(3/2)*ArcTan[(b^(1/4)*Cos[c + d*x])/Sqrt[Sqrt[a] - Sqrt[b]]])/(2*Sqrt[Sqrt[a] - Sqrt[b]]*b^(9/4)*d) - (a^(
3/2)*ArcTanh[(b^(1/4)*Cos[c + d*x])/Sqrt[Sqrt[a] + Sqrt[b]]])/(2*Sqrt[Sqrt[a] + Sqrt[b]]*b^(9/4)*d) + ((a + b)
*Cos[c + d*x])/(b^2*d) - (2*Cos[c + d*x]^3)/(3*b*d) + Cos[c + d*x]^5/(5*b*d)
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Rubi [A] time = 0.251947, antiderivative size = 177, normalized size of antiderivative = 1.,
number of steps used = 6, number of rules used = 5, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.208, Rules used =
{3215, 1170, 1093, 205, 208} \[ -\frac{a^{3/2} \tan ^{-1}\left (\frac{\sqrt [4]{b} \cos (c+d x)}{\sqrt{\sqrt{a}-\sqrt{b}}}\right )}{2 b^{9/4} d \sqrt{\sqrt{a}-\sqrt{b}}}-\frac{a^{3/2} \tanh ^{-1}\left (\frac{\sqrt [4]{b} \cos (c+d x)}{\sqrt{\sqrt{a}+\sqrt{b}}}\right )}{2 b^{9/4} d \sqrt{\sqrt{a}+\sqrt{b}}}+\frac{(a+b) \cos (c+d x)}{b^2 d}+\frac{\cos ^5(c+d x)}{5 b d}-\frac{2 \cos ^3(c+d x)}{3 b d} \]
Antiderivative was successfully verified.
[In]
Int[Sin[c + d*x]^9/(a - b*Sin[c + d*x]^4),x]
[Out]
-(a^(3/2)*ArcTan[(b^(1/4)*Cos[c + d*x])/Sqrt[Sqrt[a] - Sqrt[b]]])/(2*Sqrt[Sqrt[a] - Sqrt[b]]*b^(9/4)*d) - (a^(
3/2)*ArcTanh[(b^(1/4)*Cos[c + d*x])/Sqrt[Sqrt[a] + Sqrt[b]]])/(2*Sqrt[Sqrt[a] + Sqrt[b]]*b^(9/4)*d) + ((a + b)
*Cos[c + d*x])/(b^2*d) - (2*Cos[c + d*x]^3)/(3*b*d) + Cos[c + d*x]^5/(5*b*d)
Rule 3215
Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^4)^(p_.), x_Symbol] :> With[{ff = Free
Factors[Cos[e + f*x], x]}, -Dist[ff/f, Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*(a + b - 2*b*ff^2*x^2 + b*ff^4*x^4
)^p, x], x, Cos[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]
Rule 1170
Int[((d_) + (e_.)*(x_)^2)^(q_)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> Int[ExpandIntegrand[(d + e*x
^2)^q/(a + b*x^2 + c*x^4), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a
*e^2, 0] && IntegerQ[q]
Rule 1093
Int[((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(-1), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Dist[c/q, Int[1/(b/
2 - q/2 + c*x^2), x], x] - Dist[c/q, Int[1/(b/2 + q/2 + c*x^2), x], x]] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*
a*c, 0] && PosQ[b^2 - 4*a*c]
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]
Rule 208
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]
Rubi steps
\begin{align*} \int \frac{\sin ^9(c+d x)}{a-b \sin ^4(c+d x)} \, dx &=-\frac{\operatorname{Subst}\left (\int \frac{\left (1-x^2\right )^4}{a-b+2 b x^2-b x^4} \, dx,x,\cos (c+d x)\right )}{d}\\ &=-\frac{\operatorname{Subst}\left (\int \left (-\frac{a+b}{b^2}+\frac{2 x^2}{b}-\frac{x^4}{b}+\frac{a^2}{b^2 \left (a-b+2 b x^2-b x^4\right )}\right ) \, dx,x,\cos (c+d x)\right )}{d}\\ &=\frac{(a+b) \cos (c+d x)}{b^2 d}-\frac{2 \cos ^3(c+d x)}{3 b d}+\frac{\cos ^5(c+d x)}{5 b d}-\frac{a^2 \operatorname{Subst}\left (\int \frac{1}{a-b+2 b x^2-b x^4} \, dx,x,\cos (c+d x)\right )}{b^2 d}\\ &=\frac{(a+b) \cos (c+d x)}{b^2 d}-\frac{2 \cos ^3(c+d x)}{3 b d}+\frac{\cos ^5(c+d x)}{5 b d}+\frac{a^{3/2} \operatorname{Subst}\left (\int \frac{1}{-\sqrt{a} \sqrt{b}+b-b x^2} \, dx,x,\cos (c+d x)\right )}{2 b^{3/2} d}-\frac{a^{3/2} \operatorname{Subst}\left (\int \frac{1}{\sqrt{a} \sqrt{b}+b-b x^2} \, dx,x,\cos (c+d x)\right )}{2 b^{3/2} d}\\ &=-\frac{a^{3/2} \tan ^{-1}\left (\frac{\sqrt [4]{b} \cos (c+d x)}{\sqrt{\sqrt{a}-\sqrt{b}}}\right )}{2 \sqrt{\sqrt{a}-\sqrt{b}} b^{9/4} d}-\frac{a^{3/2} \tanh ^{-1}\left (\frac{\sqrt [4]{b} \cos (c+d x)}{\sqrt{\sqrt{a}+\sqrt{b}}}\right )}{2 \sqrt{\sqrt{a}+\sqrt{b}} b^{9/4} d}+\frac{(a+b) \cos (c+d x)}{b^2 d}-\frac{2 \cos ^3(c+d x)}{3 b d}+\frac{\cos ^5(c+d x)}{5 b d}\\ \end{align*}
Mathematica [C] time = 0.470281, size = 228, normalized size = 1.29 \[ \frac{\cos (c+d x) (120 a-28 b \cos (2 (c+d x))+3 b \cos (4 (c+d x))+89 b)+60 i a^2 \text{RootSum}\left [-16 \text{$\#$1}^4 a+\text{$\#$1}^8 b-4 \text{$\#$1}^6 b+6 \text{$\#$1}^4 b-4 \text{$\#$1}^2 b+b\& ,\frac{-i \text{$\#$1}^3 \log \left (\text{$\#$1}^2-2 \text{$\#$1} \cos (c+d x)+1\right )+i \text{$\#$1} \log \left (\text{$\#$1}^2-2 \text{$\#$1} \cos (c+d x)+1\right )+2 \text{$\#$1}^3 \tan ^{-1}\left (\frac{\sin (c+d x)}{\cos (c+d x)-\text{$\#$1}}\right )-2 \text{$\#$1} \tan ^{-1}\left (\frac{\sin (c+d x)}{\cos (c+d x)-\text{$\#$1}}\right )}{-8 \text{$\#$1}^2 a+\text{$\#$1}^6 b-3 \text{$\#$1}^4 b+3 \text{$\#$1}^2 b-b}\& \right ]}{120 b^2 d} \]
Antiderivative was successfully verified.
[In]
Integrate[Sin[c + d*x]^9/(a - b*Sin[c + d*x]^4),x]
[Out]
(Cos[c + d*x]*(120*a + 89*b - 28*b*Cos[2*(c + d*x)] + 3*b*Cos[4*(c + d*x)]) + (60*I)*a^2*RootSum[b - 4*b*#1^2
- 16*a*#1^4 + 6*b*#1^4 - 4*b*#1^6 + b*#1^8 & , (-2*ArcTan[Sin[c + d*x]/(Cos[c + d*x] - #1)]*#1 + I*Log[1 - 2*C
os[c + d*x]*#1 + #1^2]*#1 + 2*ArcTan[Sin[c + d*x]/(Cos[c + d*x] - #1)]*#1^3 - I*Log[1 - 2*Cos[c + d*x]*#1 + #1
^2]*#1^3)/(-b - 8*a*#1^2 + 3*b*#1^2 - 3*b*#1^4 + b*#1^6) & ])/(120*b^2*d)
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Maple [A] time = 0.116, size = 159, normalized size = 0.9 \begin{align*}{\frac{ \left ( \cos \left ( dx+c \right ) \right ) ^{5}}{5\,bd}}-{\frac{2\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{3\,bd}}+{\frac{\cos \left ( dx+c \right ) a}{{b}^{2}d}}+{\frac{\cos \left ( dx+c \right ) }{bd}}-{\frac{{a}^{2}}{2\,bd}\arctan \left ({b\cos \left ( dx+c \right ){\frac{1}{\sqrt{ \left ( \sqrt{ab}-b \right ) b}}}} \right ){\frac{1}{\sqrt{ab}}}{\frac{1}{\sqrt{ \left ( \sqrt{ab}-b \right ) b}}}}-{\frac{{a}^{2}}{2\,bd}{\it Artanh} \left ({b\cos \left ( dx+c \right ){\frac{1}{\sqrt{ \left ( \sqrt{ab}+b \right ) b}}}} \right ){\frac{1}{\sqrt{ab}}}{\frac{1}{\sqrt{ \left ( \sqrt{ab}+b \right ) b}}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(sin(d*x+c)^9/(a-b*sin(d*x+c)^4),x)
[Out]
1/5*cos(d*x+c)^5/b/d-2/3*cos(d*x+c)^3/b/d+a*cos(d*x+c)/b^2/d+cos(d*x+c)/b/d-1/2/d*a^2/b/(a*b)^(1/2)/(((a*b)^(1
/2)-b)*b)^(1/2)*arctan(cos(d*x+c)*b/(((a*b)^(1/2)-b)*b)^(1/2))-1/2/d*a^2/b/(a*b)^(1/2)/(((a*b)^(1/2)+b)*b)^(1/
2)*arctanh(cos(d*x+c)*b/(((a*b)^(1/2)+b)*b)^(1/2))
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Maxima [F] time = 0., size = 0, normalized size = 0. \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)^9/(a-b*sin(d*x+c)^4),x, algorithm="maxima")
[Out]
1/240*(240*b^2*d*integrate(8*(4*a^2*b*cos(3*d*x + 3*c)*sin(2*d*x + 2*c) + 2*(8*a^3 - 3*a^2*b)*cos(3*d*x + 3*c)
*sin(4*d*x + 4*c) - 2*(8*a^3 - 3*a^2*b)*cos(4*d*x + 4*c)*sin(3*d*x + 3*c) - (a^2*b*sin(5*d*x + 5*c) - a^2*b*si
n(3*d*x + 3*c))*cos(8*d*x + 8*c) + 4*(a^2*b*sin(5*d*x + 5*c) - a^2*b*sin(3*d*x + 3*c))*cos(6*d*x + 6*c) - 2*(2
*a^2*b*sin(2*d*x + 2*c) + (8*a^3 - 3*a^2*b)*sin(4*d*x + 4*c))*cos(5*d*x + 5*c) + (a^2*b*cos(5*d*x + 5*c) - a^2
*b*cos(3*d*x + 3*c))*sin(8*d*x + 8*c) - 4*(a^2*b*cos(5*d*x + 5*c) - a^2*b*cos(3*d*x + 3*c))*sin(6*d*x + 6*c) +
(4*a^2*b*cos(2*d*x + 2*c) - a^2*b + 2*(8*a^3 - 3*a^2*b)*cos(4*d*x + 4*c))*sin(5*d*x + 5*c) - (4*a^2*b*cos(2*d
*x + 2*c) - a^2*b)*sin(3*d*x + 3*c))/(b^4*cos(8*d*x + 8*c)^2 + 16*b^4*cos(6*d*x + 6*c)^2 + 16*b^4*cos(2*d*x +
2*c)^2 + b^4*sin(8*d*x + 8*c)^2 + 16*b^4*sin(6*d*x + 6*c)^2 + 16*b^4*sin(2*d*x + 2*c)^2 - 8*b^4*cos(2*d*x + 2*
c) + b^4 + 4*(64*a^2*b^2 - 48*a*b^3 + 9*b^4)*cos(4*d*x + 4*c)^2 + 4*(64*a^2*b^2 - 48*a*b^3 + 9*b^4)*sin(4*d*x
+ 4*c)^2 + 16*(8*a*b^3 - 3*b^4)*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) - 2*(4*b^4*cos(6*d*x + 6*c) + 4*b^4*cos(2*d*
x + 2*c) - b^4 + 2*(8*a*b^3 - 3*b^4)*cos(4*d*x + 4*c))*cos(8*d*x + 8*c) + 8*(4*b^4*cos(2*d*x + 2*c) - b^4 + 2*
(8*a*b^3 - 3*b^4)*cos(4*d*x + 4*c))*cos(6*d*x + 6*c) - 4*(8*a*b^3 - 3*b^4 - 4*(8*a*b^3 - 3*b^4)*cos(2*d*x + 2*
c))*cos(4*d*x + 4*c) - 4*(2*b^4*sin(6*d*x + 6*c) + 2*b^4*sin(2*d*x + 2*c) + (8*a*b^3 - 3*b^4)*sin(4*d*x + 4*c)
)*sin(8*d*x + 8*c) + 16*(2*b^4*sin(2*d*x + 2*c) + (8*a*b^3 - 3*b^4)*sin(4*d*x + 4*c))*sin(6*d*x + 6*c)), x) +
3*b*cos(5*d*x + 5*c) - 25*b*cos(3*d*x + 3*c) + 30*(8*a + 5*b)*cos(d*x + c))/(b^2*d)
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Fricas [B] time = 3.01022, size = 1758, normalized size = 9.93 \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)^9/(a-b*sin(d*x+c)^4),x, algorithm="fricas")
[Out]
1/60*(12*b*cos(d*x + c)^5 - 15*b^2*d*sqrt(-((a*b^4 - b^5)*sqrt(a^7/((a^2*b^9 - 2*a*b^10 + b^11)*d^4))*d^2 + a^
3)/((a*b^4 - b^5)*d^2))*log(a^5*cos(d*x + c) + (a^4*b^2*d - (a*b^7 - b^8)*sqrt(a^7/((a^2*b^9 - 2*a*b^10 + b^11
)*d^4))*d^3)*sqrt(-((a*b^4 - b^5)*sqrt(a^7/((a^2*b^9 - 2*a*b^10 + b^11)*d^4))*d^2 + a^3)/((a*b^4 - b^5)*d^2)))
+ 15*b^2*d*sqrt(((a*b^4 - b^5)*sqrt(a^7/((a^2*b^9 - 2*a*b^10 + b^11)*d^4))*d^2 - a^3)/((a*b^4 - b^5)*d^2))*lo
g(a^5*cos(d*x + c) - (a^4*b^2*d + (a*b^7 - b^8)*sqrt(a^7/((a^2*b^9 - 2*a*b^10 + b^11)*d^4))*d^3)*sqrt(((a*b^4
- b^5)*sqrt(a^7/((a^2*b^9 - 2*a*b^10 + b^11)*d^4))*d^2 - a^3)/((a*b^4 - b^5)*d^2))) + 15*b^2*d*sqrt(-((a*b^4 -
b^5)*sqrt(a^7/((a^2*b^9 - 2*a*b^10 + b^11)*d^4))*d^2 + a^3)/((a*b^4 - b^5)*d^2))*log(-a^5*cos(d*x + c) + (a^4
*b^2*d - (a*b^7 - b^8)*sqrt(a^7/((a^2*b^9 - 2*a*b^10 + b^11)*d^4))*d^3)*sqrt(-((a*b^4 - b^5)*sqrt(a^7/((a^2*b^
9 - 2*a*b^10 + b^11)*d^4))*d^2 + a^3)/((a*b^4 - b^5)*d^2))) - 15*b^2*d*sqrt(((a*b^4 - b^5)*sqrt(a^7/((a^2*b^9
- 2*a*b^10 + b^11)*d^4))*d^2 - a^3)/((a*b^4 - b^5)*d^2))*log(-a^5*cos(d*x + c) - (a^4*b^2*d + (a*b^7 - b^8)*sq
rt(a^7/((a^2*b^9 - 2*a*b^10 + b^11)*d^4))*d^3)*sqrt(((a*b^4 - b^5)*sqrt(a^7/((a^2*b^9 - 2*a*b^10 + b^11)*d^4))
*d^2 - a^3)/((a*b^4 - b^5)*d^2))) - 40*b*cos(d*x + c)^3 + 60*(a + b)*cos(d*x + c))/(b^2*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)**9/(a-b*sin(d*x+c)**4),x)
[Out]
Timed out
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Giac [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: NotImplementedError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sin(d*x+c)^9/(a-b*sin(d*x+c)^4),x, algorithm="giac")
[Out]
Exception raised: NotImplementedError