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+b)*cos(d*x+c)/b^2/d-2/3*cos(d*x+c)^3/b/d+1/5*cos(d*x+c)^5/b/d-1/2*a^(3/2)*arctan(b^(1/4)*cos(d*x+c)/(a^(1/2
)-b^(1/2))^(1/2))/b^(9/4)/d/(a^(1/2)-b^(1/2))^(1/2)-1/2*a^(3/2)*arctanh(b^(1/4)*cos(d*x+c)/(a^(1/2)+b^(1/2))^(
1/2))/b^(9/4)/d/(a^(1/2)+b^(1/2))^(1/2)

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Rubi [A]  time = 0.25, antiderivative size = 177, normalized size of antiderivative = 1.00, 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 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]

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 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 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]

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*}

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Mathematica [C]  time = 0.48, size = 228, normalized size = 1.29 \[ \frac {\cos (c+d x) (120 a-28 b \cos (2 (c+d x))+3 b \cos (4 (c+d x))+89 b)+60 i a^2 \text {RootSum}\left [\text {$\#$1}^8 b-4 \text {$\#$1}^6 b-16 \text {$\#$1}^4 a+6 \text {$\#$1}^4 b-4 \text {$\#$1}^2 b+b\& ,\frac {2 \text {$\#$1}^3 \tan ^{-1}\left (\frac {\sin (c+d x)}{\cos (c+d x)-\text {$\#$1}}\right )+i \text {$\#$1} \log \left (\text {$\#$1}^2-2 \text {$\#$1} \cos (c+d x)+1\right )-i \text {$\#$1}^3 \log \left (\text {$\#$1}^2-2 \text {$\#$1} \cos (c+d x)+1\right )-2 \text {$\#$1} \tan ^{-1}\left (\frac {\sin (c+d x)}{\cos (c+d x)-\text {$\#$1}}\right )}{\text {$\#$1}^6 b-3 \text {$\#$1}^4 b-8 \text {$\#$1}^2 a+3 \text {$\#$1}^2 b-b}\& \right ]}{120 b^2 d} \]

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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fricas [B]  time = 0.58, size = 872, normalized size = 4.93 \[ \frac {12 \, b \cos \left (d x + c\right )^{5} - 15 \, b^{2} d \sqrt {-\frac {{\left (a b^{4} - b^{5}\right )} \sqrt {\frac {a^{7}}{{\left (a^{2} b^{9} - 2 \, a b^{10} + b^{11}\right )} d^{4}}} d^{2} + a^{3}}{{\left (a b^{4} - b^{5}\right )} d^{2}}} \log \left (a^{5} \cos \left (d x + c\right ) + {\left (a^{4} b^{2} d - {\left (a b^{7} - b^{8}\right )} \sqrt {\frac {a^{7}}{{\left (a^{2} b^{9} - 2 \, a b^{10} + b^{11}\right )} d^{4}}} d^{3}\right )} \sqrt {-\frac {{\left (a b^{4} - b^{5}\right )} \sqrt {\frac {a^{7}}{{\left (a^{2} b^{9} - 2 \, a b^{10} + b^{11}\right )} d^{4}}} d^{2} + a^{3}}{{\left (a b^{4} - b^{5}\right )} d^{2}}}\right ) + 15 \, b^{2} d \sqrt {\frac {{\left (a b^{4} - b^{5}\right )} \sqrt {\frac {a^{7}}{{\left (a^{2} b^{9} - 2 \, a b^{10} + b^{11}\right )} d^{4}}} d^{2} - a^{3}}{{\left (a b^{4} - b^{5}\right )} d^{2}}} \log \left (a^{5} \cos \left (d x + c\right ) - {\left (a^{4} b^{2} d + {\left (a b^{7} - b^{8}\right )} \sqrt {\frac {a^{7}}{{\left (a^{2} b^{9} - 2 \, a b^{10} + b^{11}\right )} d^{4}}} d^{3}\right )} \sqrt {\frac {{\left (a b^{4} - b^{5}\right )} \sqrt {\frac {a^{7}}{{\left (a^{2} b^{9} - 2 \, a b^{10} + b^{11}\right )} d^{4}}} d^{2} - a^{3}}{{\left (a b^{4} - b^{5}\right )} d^{2}}}\right ) + 15 \, b^{2} d \sqrt {-\frac {{\left (a b^{4} - b^{5}\right )} \sqrt {\frac {a^{7}}{{\left (a^{2} b^{9} - 2 \, a b^{10} + b^{11}\right )} d^{4}}} d^{2} + a^{3}}{{\left (a b^{4} - b^{5}\right )} d^{2}}} \log \left (-a^{5} \cos \left (d x + c\right ) + {\left (a^{4} b^{2} d - {\left (a b^{7} - b^{8}\right )} \sqrt {\frac {a^{7}}{{\left (a^{2} b^{9} - 2 \, a b^{10} + b^{11}\right )} d^{4}}} d^{3}\right )} \sqrt {-\frac {{\left (a b^{4} - b^{5}\right )} \sqrt {\frac {a^{7}}{{\left (a^{2} b^{9} - 2 \, a b^{10} + b^{11}\right )} d^{4}}} d^{2} + a^{3}}{{\left (a b^{4} - b^{5}\right )} d^{2}}}\right ) - 15 \, b^{2} d \sqrt {\frac {{\left (a b^{4} - b^{5}\right )} \sqrt {\frac {a^{7}}{{\left (a^{2} b^{9} - 2 \, a b^{10} + b^{11}\right )} d^{4}}} d^{2} - a^{3}}{{\left (a b^{4} - b^{5}\right )} d^{2}}} \log \left (-a^{5} \cos \left (d x + c\right ) - {\left (a^{4} b^{2} d + {\left (a b^{7} - b^{8}\right )} \sqrt {\frac {a^{7}}{{\left (a^{2} b^{9} - 2 \, a b^{10} + b^{11}\right )} d^{4}}} d^{3}\right )} \sqrt {\frac {{\left (a b^{4} - b^{5}\right )} \sqrt {\frac {a^{7}}{{\left (a^{2} b^{9} - 2 \, a b^{10} + b^{11}\right )} d^{4}}} d^{2} - a^{3}}{{\left (a b^{4} - b^{5}\right )} d^{2}}}\right ) - 40 \, b \cos \left (d x + c\right )^{3} + 60 \, {\left (a + b\right )} \cos \left (d x + c\right )}{60 \, b^{2} d} \]

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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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: NotImplementedError} \]

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 >> Unable to parse Giac output: Warning, need to choose a branch for the
root of a polynomial with parameters. This might be wrong.The choice was done assuming [a,b]=[-42,-63]Warning
, need to choose a branch for the root of a polynomial with parameters. This might be wrong.The choice was don
e assuming [a,b]=[-2,-75]-2/d*(-15*((1-cos(c+d*x))/(1+cos(c+d*x)))^4*a-60*((1-cos(c+d*x))/(1+cos(c+d*x)))^3*a-
90*((1-cos(c+d*x))/(1+cos(c+d*x)))^2*a-80*((1-cos(c+d*x))/(1+cos(c+d*x)))^2*b-60*(1-cos(c+d*x))/(1+cos(c+d*x))
*a-40*(1-cos(c+d*x))/(1+cos(c+d*x))*b-15*a-8*b)*1/15/b^2/((1-cos(c+d*x))/(1+cos(c+d*x))+1)^5-2/d/b^2*2/d*((-2*
a^4*b+12*a^3*b^2-6*a^3*b*sqrt(a^2-a*b+sqrt(a*b)*(-a+b))+2*a^3*a*b-3*a^3*sqrt(a*b)*sqrt(a^2-a*b+sqrt(a*b)*(-a+b
))-10*a^2*b^3+12*a^2*b^2*sqrt(a^2-a*b+sqrt(a*b)*(-a+b))-12*a^2*b*a*b+6*a^2*b*sqrt(a*b)*sqrt(a^2-a*b+sqrt(a*b)*
(-a+b))+2*a*b^3*sqrt(a^2-a*b+sqrt(a*b)*(-a+b))+10*a*b^2*a*b+a*b^2*sqrt(a*b)*sqrt(a^2-a*b+sqrt(a*b)*(-a+b)))*ab
s(a-b)/(24*a^4*b-96*a^3*b^2+112*a^2*b^3-32*a*b^4-8*b^5)*(atan(tan(c+d*x)/sqrt(-(-8*a+sqrt(8*a*8*a-4*(4*a-4*b)*
4*a))/2/(4*a-4*b)))+pi*floor((c+d*x)/pi+1/2))-(-2*a^4*b+12*a^3*b^2+6*a^3*b*sqrt(a^2-a*b+sqrt(a*b)*(a-b))+2*a^3
*a*b-3*a^3*sqrt(a*b)*sqrt(a^2-a*b+sqrt(a*b)*(a-b))-10*a^2*b^3-12*a^2*b^2*sqrt(a^2-a*b+sqrt(a*b)*(a-b))-12*a^2*
b*a*b+6*a^2*b*sqrt(a*b)*sqrt(a^2-a*b+sqrt(a*b)*(a-b))-2*a*b^3*sqrt(a^2-a*b+sqrt(a*b)*(a-b))+10*a*b^2*a*b+a*b^2
*sqrt(a*b)*sqrt(a^2-a*b+sqrt(a*b)*(a-b)))*abs(a-b)/(24*a^4*b-96*a^3*b^2+112*a^2*b^3-32*a*b^4-8*b^5)*(atan(tan(
c+d*x)/sqrt(-(-8*a-sqrt(8*a*8*a-4*(4*a-4*b)*4*a))/2/(4*a-4*b)))+pi*floor((c+d*x)/pi+1/2)))

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maple [A]  time = 0.37, size = 159, normalized size = 0.90 \[ \frac {\cos ^{5}\left (d x +c \right )}{5 b d}-\frac {2 \left (\cos ^{3}\left (d x +c \right )\right )}{3 b d}+\frac {a \cos \left (d x +c \right )}{b^{2} d}+\frac {\cos \left (d x +c \right )}{b d}-\frac {a^{2} \arctanh \left (\frac {\cos \left (d x +c \right ) b}{\sqrt {\left (\sqrt {a b}+b \right ) b}}\right )}{2 d b \sqrt {a b}\, \sqrt {\left (\sqrt {a b}+b \right ) b}}-\frac {a^{2} \arctan \left (\frac {\cos \left (d x +c \right ) b}{\sqrt {\left (\sqrt {a b}-b \right ) b}}\right )}{2 d b \sqrt {a b}\, \sqrt {\left (\sqrt {a b}-b \right ) b}} \]

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)*arctanh(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)*arctan(cos(d*x+c)*b/(((a*b)^(1/2)-b)*b)^(1/2))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {result too large to display} \]

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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mupad [B]  time = 14.44, size = 1067, normalized size = 6.03 \[ \frac {{\cos \left (c+d\,x\right )}^5}{5\,b\,d}-\frac {2\,{\cos \left (c+d\,x\right )}^3}{3\,b\,d}+\frac {\cos \left (c+d\,x\right )\,\left (\frac {a-b}{b^2}+\frac {2}{b}\right )}{d}+\frac {\mathrm {atan}\left (-\frac {a^4\,\cos \left (c+d\,x\right )\,\sqrt {-\frac {\sqrt {a^7\,b^9}}{16\,\left (a\,b^9-b^{10}\right )}-\frac {a^3\,b^5}{16\,\left (a\,b^9-b^{10}\right )}}\,8{}\mathrm {i}}{\frac {2\,a^6\,b^7}{a\,b^9-b^{10}}+\frac {2\,a^3\,b^2\,\sqrt {a^7\,b^9}}{a\,b^9-b^{10}}}+\frac {a^4\,b^9\,\cos \left (c+d\,x\right )\,\sqrt {-\frac {\sqrt {a^7\,b^9}}{16\,\left (a\,b^9-b^{10}\right )}-\frac {a^3\,b^5}{16\,\left (a\,b^9-b^{10}\right )}}\,8{}\mathrm {i}}{\frac {2\,a^6\,b^{16}}{a\,b^9-b^{10}}-\frac {2\,a^7\,b^{15}}{a\,b^9-b^{10}}+\frac {2\,a^3\,b^{11}\,\sqrt {a^7\,b^9}}{a\,b^9-b^{10}}-\frac {2\,a^4\,b^{10}\,\sqrt {a^7\,b^9}}{a\,b^9-b^{10}}}+\frac {a\,b^4\,\cos \left (c+d\,x\right )\,\sqrt {-\frac {\sqrt {a^7\,b^9}}{16\,\left (a\,b^9-b^{10}\right )}-\frac {a^3\,b^5}{16\,\left (a\,b^9-b^{10}\right )}}\,\sqrt {a^7\,b^9}\,8{}\mathrm {i}}{\frac {2\,a^6\,b^{16}}{a\,b^9-b^{10}}-\frac {2\,a^7\,b^{15}}{a\,b^9-b^{10}}+\frac {2\,a^3\,b^{11}\,\sqrt {a^7\,b^9}}{a\,b^9-b^{10}}-\frac {2\,a^4\,b^{10}\,\sqrt {a^7\,b^9}}{a\,b^9-b^{10}}}\right )\,\sqrt {-\frac {\sqrt {a^7\,b^9}+a^3\,b^5}{16\,\left (a\,b^9-b^{10}\right )}}\,2{}\mathrm {i}}{d}-\frac {\mathrm {atan}\left (\frac {a^4\,\cos \left (c+d\,x\right )\,\sqrt {\frac {\sqrt {a^7\,b^9}}{16\,\left (a\,b^9-b^{10}\right )}-\frac {a^3\,b^5}{16\,\left (a\,b^9-b^{10}\right )}}\,8{}\mathrm {i}}{\frac {2\,a^6\,b^7}{a\,b^9-b^{10}}-\frac {2\,a^3\,b^2\,\sqrt {a^7\,b^9}}{a\,b^9-b^{10}}}-\frac {a^4\,b^9\,\cos \left (c+d\,x\right )\,\sqrt {\frac {\sqrt {a^7\,b^9}}{16\,\left (a\,b^9-b^{10}\right )}-\frac {a^3\,b^5}{16\,\left (a\,b^9-b^{10}\right )}}\,8{}\mathrm {i}}{\frac {2\,a^6\,b^{16}}{a\,b^9-b^{10}}-\frac {2\,a^7\,b^{15}}{a\,b^9-b^{10}}-\frac {2\,a^3\,b^{11}\,\sqrt {a^7\,b^9}}{a\,b^9-b^{10}}+\frac {2\,a^4\,b^{10}\,\sqrt {a^7\,b^9}}{a\,b^9-b^{10}}}+\frac {a\,b^4\,\cos \left (c+d\,x\right )\,\sqrt {\frac {\sqrt {a^7\,b^9}}{16\,\left (a\,b^9-b^{10}\right )}-\frac {a^3\,b^5}{16\,\left (a\,b^9-b^{10}\right )}}\,\sqrt {a^7\,b^9}\,8{}\mathrm {i}}{\frac {2\,a^6\,b^{16}}{a\,b^9-b^{10}}-\frac {2\,a^7\,b^{15}}{a\,b^9-b^{10}}-\frac {2\,a^3\,b^{11}\,\sqrt {a^7\,b^9}}{a\,b^9-b^{10}}+\frac {2\,a^4\,b^{10}\,\sqrt {a^7\,b^9}}{a\,b^9-b^{10}}}\right )\,\sqrt {\frac {\sqrt {a^7\,b^9}-a^3\,b^5}{16\,\left (a\,b^9-b^{10}\right )}}\,2{}\mathrm {i}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(c + d*x)^9/(a - b*sin(c + d*x)^4),x)

[Out]

(atan((a^4*b^9*cos(c + d*x)*(- (a^7*b^9)^(1/2)/(16*(a*b^9 - b^10)) - (a^3*b^5)/(16*(a*b^9 - b^10)))^(1/2)*8i)/
((2*a^6*b^16)/(a*b^9 - b^10) - (2*a^7*b^15)/(a*b^9 - b^10) + (2*a^3*b^11*(a^7*b^9)^(1/2))/(a*b^9 - b^10) - (2*
a^4*b^10*(a^7*b^9)^(1/2))/(a*b^9 - b^10)) - (a^4*cos(c + d*x)*(- (a^7*b^9)^(1/2)/(16*(a*b^9 - b^10)) - (a^3*b^
5)/(16*(a*b^9 - b^10)))^(1/2)*8i)/((2*a^6*b^7)/(a*b^9 - b^10) + (2*a^3*b^2*(a^7*b^9)^(1/2))/(a*b^9 - b^10)) +
(a*b^4*cos(c + d*x)*(- (a^7*b^9)^(1/2)/(16*(a*b^9 - b^10)) - (a^3*b^5)/(16*(a*b^9 - b^10)))^(1/2)*(a^7*b^9)^(1
/2)*8i)/((2*a^6*b^16)/(a*b^9 - b^10) - (2*a^7*b^15)/(a*b^9 - b^10) + (2*a^3*b^11*(a^7*b^9)^(1/2))/(a*b^9 - b^1
0) - (2*a^4*b^10*(a^7*b^9)^(1/2))/(a*b^9 - b^10)))*(-((a^7*b^9)^(1/2) + a^3*b^5)/(16*(a*b^9 - b^10)))^(1/2)*2i
)/d - (atan((a^4*cos(c + d*x)*((a^7*b^9)^(1/2)/(16*(a*b^9 - b^10)) - (a^3*b^5)/(16*(a*b^9 - b^10)))^(1/2)*8i)/
((2*a^6*b^7)/(a*b^9 - b^10) - (2*a^3*b^2*(a^7*b^9)^(1/2))/(a*b^9 - b^10)) - (a^4*b^9*cos(c + d*x)*((a^7*b^9)^(
1/2)/(16*(a*b^9 - b^10)) - (a^3*b^5)/(16*(a*b^9 - b^10)))^(1/2)*8i)/((2*a^6*b^16)/(a*b^9 - b^10) - (2*a^7*b^15
)/(a*b^9 - b^10) - (2*a^3*b^11*(a^7*b^9)^(1/2))/(a*b^9 - b^10) + (2*a^4*b^10*(a^7*b^9)^(1/2))/(a*b^9 - b^10))
+ (a*b^4*cos(c + d*x)*((a^7*b^9)^(1/2)/(16*(a*b^9 - b^10)) - (a^3*b^5)/(16*(a*b^9 - b^10)))^(1/2)*(a^7*b^9)^(1
/2)*8i)/((2*a^6*b^16)/(a*b^9 - b^10) - (2*a^7*b^15)/(a*b^9 - b^10) - (2*a^3*b^11*(a^7*b^9)^(1/2))/(a*b^9 - b^1
0) + (2*a^4*b^10*(a^7*b^9)^(1/2))/(a*b^9 - b^10)))*(((a^7*b^9)^(1/2) - a^3*b^5)/(16*(a*b^9 - b^10)))^(1/2)*2i)
/d - (2*cos(c + d*x)^3)/(3*b*d) + cos(c + d*x)^5/(5*b*d) + (cos(c + d*x)*((a - b)/b^2 + 2/b))/d

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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