3.197 \(\int \frac{\sin ^5(c+d x)}{a-b \sin ^4(c+d x)} \, dx\)
Optimal. Leaf size=138 \[ -\frac{\sqrt{a} \tan ^{-1}\left (\frac{\sqrt [4]{b} \cos (c+d x)}{\sqrt{\sqrt{a}-\sqrt{b}}}\right )}{2 b^{5/4} d \sqrt{\sqrt{a}-\sqrt{b}}}-\frac{\sqrt{a} \tanh ^{-1}\left (\frac{\sqrt [4]{b} \cos (c+d x)}{\sqrt{\sqrt{a}+\sqrt{b}}}\right )}{2 b^{5/4} d \sqrt{\sqrt{a}+\sqrt{b}}}+\frac{\cos (c+d x)}{b d} \]
[Out]
-(Sqrt[a]*ArcTan[(b^(1/4)*Cos[c + d*x])/Sqrt[Sqrt[a] - Sqrt[b]]])/(2*Sqrt[Sqrt[a] - Sqrt[b]]*b^(5/4)*d) - (Sqr
t[a]*ArcTanh[(b^(1/4)*Cos[c + d*x])/Sqrt[Sqrt[a] + Sqrt[b]]])/(2*Sqrt[Sqrt[a] + Sqrt[b]]*b^(5/4)*d) + Cos[c +
d*x]/(b*d)
________________________________________________________________________________________
Rubi [A] time = 0.178527, antiderivative size = 138, 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} \[ -\frac{\sqrt{a} \tan ^{-1}\left (\frac{\sqrt [4]{b} \cos (c+d x)}{\sqrt{\sqrt{a}-\sqrt{b}}}\right )}{2 b^{5/4} d \sqrt{\sqrt{a}-\sqrt{b}}}-\frac{\sqrt{a} \tanh ^{-1}\left (\frac{\sqrt [4]{b} \cos (c+d x)}{\sqrt{\sqrt{a}+\sqrt{b}}}\right )}{2 b^{5/4} d \sqrt{\sqrt{a}+\sqrt{b}}}+\frac{\cos (c+d x)}{b d} \]
Antiderivative was successfully verified.
[In]
Int[Sin[c + d*x]^5/(a - b*Sin[c + d*x]^4),x]
[Out]
-(Sqrt[a]*ArcTan[(b^(1/4)*Cos[c + d*x])/Sqrt[Sqrt[a] - Sqrt[b]]])/(2*Sqrt[Sqrt[a] - Sqrt[b]]*b^(5/4)*d) - (Sqr
t[a]*ArcTanh[(b^(1/4)*Cos[c + d*x])/Sqrt[Sqrt[a] + Sqrt[b]]])/(2*Sqrt[Sqrt[a] + Sqrt[b]]*b^(5/4)*d) + Cos[c +
d*x]/(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 ^5(c+d x)}{a-b \sin ^4(c+d x)} \, dx &=-\frac{\operatorname{Subst}\left (\int \frac{\left (1-x^2\right )^2}{a-b+2 b x^2-b x^4} \, dx,x,\cos (c+d x)\right )}{d}\\ &=-\frac{\operatorname{Subst}\left (\int \left (-\frac{1}{b}+\frac{a}{b \left (a-b+2 b x^2-b x^4\right )}\right ) \, dx,x,\cos (c+d x)\right )}{d}\\ &=\frac{\cos (c+d x)}{b d}-\frac{a \operatorname{Subst}\left (\int \frac{1}{a-b+2 b x^2-b x^4} \, dx,x,\cos (c+d x)\right )}{b d}\\ &=\frac{\cos (c+d x)}{b d}+\frac{\sqrt{a} \operatorname{Subst}\left (\int \frac{1}{-\sqrt{a} \sqrt{b}+b-b x^2} \, dx,x,\cos (c+d x)\right )}{2 \sqrt{b} d}-\frac{\sqrt{a} \operatorname{Subst}\left (\int \frac{1}{\sqrt{a} \sqrt{b}+b-b x^2} \, dx,x,\cos (c+d x)\right )}{2 \sqrt{b} d}\\ &=-\frac{\sqrt{a} \tan ^{-1}\left (\frac{\sqrt [4]{b} \cos (c+d x)}{\sqrt{\sqrt{a}-\sqrt{b}}}\right )}{2 \sqrt{\sqrt{a}-\sqrt{b}} b^{5/4} d}-\frac{\sqrt{a} \tanh ^{-1}\left (\frac{\sqrt [4]{b} \cos (c+d x)}{\sqrt{\sqrt{a}+\sqrt{b}}}\right )}{2 \sqrt{\sqrt{a}+\sqrt{b}} b^{5/4} d}+\frac{\cos (c+d x)}{b d}\\ \end{align*}
Mathematica [C] time = 0.250087, size = 198, normalized size = 1.43 \[ \frac{2 \cos (c+d x)+i a \text{RootSum}\left [-16 \text{$\#$1}^4 a+\text{$\#$1}^8 b-4 \text{$\#$1}^6 b+6 \text{$\#$1}^4 b-4 \text{$\#$1}^2 b+b\& ,\frac{-i \text{$\#$1}^3 \log \left (\text{$\#$1}^2-2 \text{$\#$1} \cos (c+d x)+1\right )+i \text{$\#$1} \log \left (\text{$\#$1}^2-2 \text{$\#$1} \cos (c+d x)+1\right )+2 \text{$\#$1}^3 \tan ^{-1}\left (\frac{\sin (c+d x)}{\cos (c+d x)-\text{$\#$1}}\right )-2 \text{$\#$1} \tan ^{-1}\left (\frac{\sin (c+d x)}{\cos (c+d x)-\text{$\#$1}}\right )}{-8 \text{$\#$1}^2 a+\text{$\#$1}^6 b-3 \text{$\#$1}^4 b+3 \text{$\#$1}^2 b-b}\& \right ]}{2 b d} \]
Antiderivative was successfully verified.
[In]
Integrate[Sin[c + d*x]^5/(a - b*Sin[c + d*x]^4),x]
[Out]
(2*Cos[c + d*x] + I*a*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*Cos[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) & ])/(2*b*
d)
________________________________________________________________________________________
Maple [A] time = 0.085, size = 103, normalized size = 0.8 \begin{align*}{\frac{\cos \left ( dx+c \right ) }{bd}}-{\frac{a}{2\,d}\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\,d}{\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)^5/(a-b*sin(d*x+c)^4),x)
[Out]
cos(d*x+c)/b/d-1/2/d*a/(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/(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))
________________________________________________________________________________________
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)^5/(a-b*sin(d*x+c)^4),x, algorithm="maxima")
[Out]
(b*d*integrate(8*(4*a*b*cos(3*d*x + 3*c)*sin(2*d*x + 2*c) + 2*(8*a^2 - 3*a*b)*cos(3*d*x + 3*c)*sin(4*d*x + 4*c
) - 2*(8*a^2 - 3*a*b)*cos(4*d*x + 4*c)*sin(3*d*x + 3*c) - (a*b*sin(5*d*x + 5*c) - a*b*sin(3*d*x + 3*c))*cos(8*
d*x + 8*c) + 4*(a*b*sin(5*d*x + 5*c) - a*b*sin(3*d*x + 3*c))*cos(6*d*x + 6*c) - 2*(2*a*b*sin(2*d*x + 2*c) + (8
*a^2 - 3*a*b)*sin(4*d*x + 4*c))*cos(5*d*x + 5*c) + (a*b*cos(5*d*x + 5*c) - a*b*cos(3*d*x + 3*c))*sin(8*d*x + 8
*c) - 4*(a*b*cos(5*d*x + 5*c) - a*b*cos(3*d*x + 3*c))*sin(6*d*x + 6*c) + (4*a*b*cos(2*d*x + 2*c) - a*b + 2*(8*
a^2 - 3*a*b)*cos(4*d*x + 4*c))*sin(5*d*x + 5*c) - (4*a*b*cos(2*d*x + 2*c) - a*b)*sin(3*d*x + 3*c))/(b^3*cos(8*
d*x + 8*c)^2 + 16*b^3*cos(6*d*x + 6*c)^2 + 16*b^3*cos(2*d*x + 2*c)^2 + b^3*sin(8*d*x + 8*c)^2 + 16*b^3*sin(6*d
*x + 6*c)^2 + 16*b^3*sin(2*d*x + 2*c)^2 - 8*b^3*cos(2*d*x + 2*c) + b^3 + 4*(64*a^2*b - 48*a*b^2 + 9*b^3)*cos(4
*d*x + 4*c)^2 + 4*(64*a^2*b - 48*a*b^2 + 9*b^3)*sin(4*d*x + 4*c)^2 + 16*(8*a*b^2 - 3*b^3)*sin(4*d*x + 4*c)*sin
(2*d*x + 2*c) - 2*(4*b^3*cos(6*d*x + 6*c) + 4*b^3*cos(2*d*x + 2*c) - b^3 + 2*(8*a*b^2 - 3*b^3)*cos(4*d*x + 4*c
))*cos(8*d*x + 8*c) + 8*(4*b^3*cos(2*d*x + 2*c) - b^3 + 2*(8*a*b^2 - 3*b^3)*cos(4*d*x + 4*c))*cos(6*d*x + 6*c)
- 4*(8*a*b^2 - 3*b^3 - 4*(8*a*b^2 - 3*b^3)*cos(2*d*x + 2*c))*cos(4*d*x + 4*c) - 4*(2*b^3*sin(6*d*x + 6*c) + 2
*b^3*sin(2*d*x + 2*c) + (8*a*b^2 - 3*b^3)*sin(4*d*x + 4*c))*sin(8*d*x + 8*c) + 16*(2*b^3*sin(2*d*x + 2*c) + (8
*a*b^2 - 3*b^3)*sin(4*d*x + 4*c))*sin(6*d*x + 6*c)), x) + cos(d*x + c))/(b*d)
________________________________________________________________________________________
Fricas [B] time = 2.91648, size = 1592, normalized size = 11.54 \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)^5/(a-b*sin(d*x+c)^4),x, algorithm="fricas")
[Out]
-1/4*(b*d*sqrt(-((a*b^2 - b^3)*d^2*sqrt(a^3/((a^2*b^5 - 2*a*b^6 + b^7)*d^4)) + a)/((a*b^2 - b^3)*d^2))*log(a^2
*cos(d*x + c) - ((a*b^4 - b^5)*d^3*sqrt(a^3/((a^2*b^5 - 2*a*b^6 + b^7)*d^4)) - a^2*b*d)*sqrt(-((a*b^2 - b^3)*d
^2*sqrt(a^3/((a^2*b^5 - 2*a*b^6 + b^7)*d^4)) + a)/((a*b^2 - b^3)*d^2))) - b*d*sqrt(((a*b^2 - b^3)*d^2*sqrt(a^3
/((a^2*b^5 - 2*a*b^6 + b^7)*d^4)) - a)/((a*b^2 - b^3)*d^2))*log(a^2*cos(d*x + c) - ((a*b^4 - b^5)*d^3*sqrt(a^3
/((a^2*b^5 - 2*a*b^6 + b^7)*d^4)) + a^2*b*d)*sqrt(((a*b^2 - b^3)*d^2*sqrt(a^3/((a^2*b^5 - 2*a*b^6 + b^7)*d^4))
- a)/((a*b^2 - b^3)*d^2))) - b*d*sqrt(-((a*b^2 - b^3)*d^2*sqrt(a^3/((a^2*b^5 - 2*a*b^6 + b^7)*d^4)) + a)/((a*
b^2 - b^3)*d^2))*log(-a^2*cos(d*x + c) - ((a*b^4 - b^5)*d^3*sqrt(a^3/((a^2*b^5 - 2*a*b^6 + b^7)*d^4)) - a^2*b*
d)*sqrt(-((a*b^2 - b^3)*d^2*sqrt(a^3/((a^2*b^5 - 2*a*b^6 + b^7)*d^4)) + a)/((a*b^2 - b^3)*d^2))) + b*d*sqrt(((
a*b^2 - b^3)*d^2*sqrt(a^3/((a^2*b^5 - 2*a*b^6 + b^7)*d^4)) - a)/((a*b^2 - b^3)*d^2))*log(-a^2*cos(d*x + c) - (
(a*b^4 - b^5)*d^3*sqrt(a^3/((a^2*b^5 - 2*a*b^6 + b^7)*d^4)) + a^2*b*d)*sqrt(((a*b^2 - b^3)*d^2*sqrt(a^3/((a^2*
b^5 - 2*a*b^6 + b^7)*d^4)) - a)/((a*b^2 - b^3)*d^2))) - 4*cos(d*x + c))/(b*d)
________________________________________________________________________________________
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)**5/(a-b*sin(d*x+c)**4),x)
[Out]
Timed out
________________________________________________________________________________________
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)^5/(a-b*sin(d*x+c)^4),x, algorithm="giac")
[Out]
Exception raised: NotImplementedError