3.95 \(\int \frac{\sin ^5(c+d x)}{(a+b \sin ^2(c+d x))^2} \, dx\)

Optimal. Leaf size=102 \[ -\frac{a^2 \cos (c+d x)}{2 b^2 d (a+b) \left (a-b \cos ^2(c+d x)+b\right )}+\frac{a (3 a+4 b) \tanh ^{-1}\left (\frac{\sqrt{b} \cos (c+d x)}{\sqrt{a+b}}\right )}{2 b^{5/2} d (a+b)^{3/2}}-\frac{\cos (c+d x)}{b^2 d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.145686, antiderivative size = 102, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {3186, 390, 385, 208} \[ -\frac{a^2 \cos (c+d x)}{2 b^2 d (a+b) \left (a-b \cos ^2(c+d x)+b\right )}+\frac{a (3 a+4 b) \tanh ^{-1}\left (\frac{\sqrt{b} \cos (c+d x)}{\sqrt{a+b}}\right )}{2 b^{5/2} d (a+b)^{3/2}}-\frac{\cos (c+d x)}{b^2 d} \]

Antiderivative was successfully verified.

[In]

Int[Sin[c + d*x]^5/(a + b*Sin[c + d*x]^2)^2,x]

[Out]

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

Rule 3186

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(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 - b*ff^2*x^2)^p, x], x, Cos
[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]

Rule 390

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Int[PolynomialDivide[(a + b*x^n)
^p, (c + d*x^n)^(-q), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && IGtQ[p, 0] && ILt
Q[q, 0] && GeQ[p, -q]

Rule 385

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> -Simp[((b*c - a*d)*x*(a + b*x^n)^(p +
1))/(a*b*n*(p + 1)), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(a*b*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /
; FreeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/n + p, 0])

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

Mathematica [C]  time = 0.904157, size = 172, normalized size = 1.69 \[ \frac{2 \sqrt{b} \cos (c+d x) \left (-\frac{a^2}{(a+b) (2 a-b \cos (2 (c+d x))+b)}-1\right )+\frac{a (3 a+4 b) \tan ^{-1}\left (\frac{\sqrt{b}-i \sqrt{a} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{-a-b}}\right )}{(-a-b)^{3/2}}+\frac{a (3 a+4 b) \tan ^{-1}\left (\frac{\sqrt{b}+i \sqrt{a} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{-a-b}}\right )}{(-a-b)^{3/2}}}{2 b^{5/2} d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[c + d*x]^5/(a + b*Sin[c + d*x]^2)^2,x]

[Out]

((a*(3*a + 4*b)*ArcTan[(Sqrt[b] - I*Sqrt[a]*Tan[(c + d*x)/2])/Sqrt[-a - b]])/(-a - b)^(3/2) + (a*(3*a + 4*b)*A
rcTan[(Sqrt[b] + I*Sqrt[a]*Tan[(c + d*x)/2])/Sqrt[-a - b]])/(-a - b)^(3/2) + 2*Sqrt[b]*Cos[c + d*x]*(-1 - a^2/
((a + b)*(2*a + b - b*Cos[2*(c + d*x)]))))/(2*b^(5/2)*d)

________________________________________________________________________________________

Maple [A]  time = 0.083, size = 94, normalized size = 0.9 \begin{align*}{\frac{1}{d} \left ( -{\frac{\cos \left ( dx+c \right ) }{{b}^{2}}}-{\frac{a}{{b}^{2}} \left ( -{\frac{\cos \left ( dx+c \right ) a}{ \left ( 2\,a+2\,b \right ) \left ( b \left ( \cos \left ( dx+c \right ) \right ) ^{2}-a-b \right ) }}-{\frac{3\,a+4\,b}{2\,a+2\,b}{\it Artanh} \left ({b\cos \left ( dx+c \right ){\frac{1}{\sqrt{ \left ( a+b \right ) b}}}} \right ){\frac{1}{\sqrt{ \left ( a+b \right ) b}}}} \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(d*x+c)^5/(a+sin(d*x+c)^2*b)^2,x)

[Out]

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

________________________________________________________________________________________

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)^5/(a+b*sin(d*x+c)^2)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [B]  time = 1.97237, size = 941, normalized size = 9.23 \begin{align*} \left [-\frac{4 \,{\left (a^{2} b^{2} + 2 \, a b^{3} + b^{4}\right )} \cos \left (d x + c\right )^{3} +{\left (3 \, a^{3} + 7 \, a^{2} b + 4 \, a b^{2} -{\left (3 \, a^{2} b + 4 \, a b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt{a b + b^{2}} \log \left (\frac{b \cos \left (d x + c\right )^{2} + 2 \, \sqrt{a b + b^{2}} \cos \left (d x + c\right ) + a + b}{b \cos \left (d x + c\right )^{2} - a - b}\right ) - 2 \,{\left (3 \, a^{3} b + 7 \, a^{2} b^{2} + 6 \, a b^{3} + 2 \, b^{4}\right )} \cos \left (d x + c\right )}{4 \,{\left ({\left (a^{2} b^{4} + 2 \, a b^{5} + b^{6}\right )} d \cos \left (d x + c\right )^{2} -{\left (a^{3} b^{3} + 3 \, a^{2} b^{4} + 3 \, a b^{5} + b^{6}\right )} d\right )}}, -\frac{2 \,{\left (a^{2} b^{2} + 2 \, a b^{3} + b^{4}\right )} \cos \left (d x + c\right )^{3} -{\left (3 \, a^{3} + 7 \, a^{2} b + 4 \, a b^{2} -{\left (3 \, a^{2} b + 4 \, a b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt{-a b - b^{2}} \arctan \left (\frac{\sqrt{-a b - b^{2}} \cos \left (d x + c\right )}{a + b}\right ) -{\left (3 \, a^{3} b + 7 \, a^{2} b^{2} + 6 \, a b^{3} + 2 \, b^{4}\right )} \cos \left (d x + c\right )}{2 \,{\left ({\left (a^{2} b^{4} + 2 \, a b^{5} + b^{6}\right )} d \cos \left (d x + c\right )^{2} -{\left (a^{3} b^{3} + 3 \, a^{2} b^{4} + 3 \, a b^{5} + b^{6}\right )} d\right )}}\right ] \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)^2)^2,x, algorithm="fricas")

[Out]

[-1/4*(4*(a^2*b^2 + 2*a*b^3 + b^4)*cos(d*x + c)^3 + (3*a^3 + 7*a^2*b + 4*a*b^2 - (3*a^2*b + 4*a*b^2)*cos(d*x +
c)^2)*sqrt(a*b + b^2)*log((b*cos(d*x + c)^2 + 2*sqrt(a*b + b^2)*cos(d*x + c) + a + b)/(b*cos(d*x + c)^2 - a -
b)) - 2*(3*a^3*b + 7*a^2*b^2 + 6*a*b^3 + 2*b^4)*cos(d*x + c))/((a^2*b^4 + 2*a*b^5 + b^6)*d*cos(d*x + c)^2 - (
a^3*b^3 + 3*a^2*b^4 + 3*a*b^5 + b^6)*d), -1/2*(2*(a^2*b^2 + 2*a*b^3 + b^4)*cos(d*x + c)^3 - (3*a^3 + 7*a^2*b +
4*a*b^2 - (3*a^2*b + 4*a*b^2)*cos(d*x + c)^2)*sqrt(-a*b - b^2)*arctan(sqrt(-a*b - b^2)*cos(d*x + c)/(a + b))
- (3*a^3*b + 7*a^2*b^2 + 6*a*b^3 + 2*b^4)*cos(d*x + c))/((a^2*b^4 + 2*a*b^5 + b^6)*d*cos(d*x + c)^2 - (a^3*b^3
+ 3*a^2*b^4 + 3*a*b^5 + b^6)*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)**2)**2,x)

[Out]

Timed out

________________________________________________________________________________________

Giac [B]  time = 1.16373, size = 462, normalized size = 4.53 \begin{align*} -\frac{\frac{{\left (3 \, a^{2} + 4 \, a b\right )} \arctan \left (\frac{b \cos \left (d x + c\right ) + a + b}{\sqrt{-a b - b^{2}} \cos \left (d x + c\right ) + \sqrt{-a b - b^{2}}}\right )}{{\left (a b^{2} + b^{3}\right )} \sqrt{-a b - b^{2}}} + \frac{2 \,{\left (3 \, a^{2} + 2 \, a b - \frac{6 \, a^{2}{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac{14 \, a b{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac{8 \, b^{2}{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac{3 \, a^{2}{\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{4 \, a b{\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )}}{{\left (a b^{2} + b^{3}\right )}{\left (a - \frac{3 \, a{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac{4 \, b{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac{3 \, a{\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{4 \, b{\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac{a{\left (\cos \left (d x + c\right ) - 1\right )}^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}}{2 \, d} \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)^2)^2,x, algorithm="giac")

[Out]

-1/2*((3*a^2 + 4*a*b)*arctan((b*cos(d*x + c) + a + b)/(sqrt(-a*b - b^2)*cos(d*x + c) + sqrt(-a*b - b^2)))/((a*
b^2 + b^3)*sqrt(-a*b - b^2)) + 2*(3*a^2 + 2*a*b - 6*a^2*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 14*a*b*(cos(d*
x + c) - 1)/(cos(d*x + c) + 1) - 8*b^2*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 3*a^2*(cos(d*x + c) - 1)^2/(cos
(d*x + c) + 1)^2 + 4*a*b*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2)/((a*b^2 + b^3)*(a - 3*a*(cos(d*x + c) - 1)
/(cos(d*x + c) + 1) - 4*b*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 3*a*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^
2 + 4*b*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 - a*(cos(d*x + c) - 1)^3/(cos(d*x + c) + 1)^3)))/d