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

Optimal. Leaf size=77 \[ -\frac{a^2 \tanh ^{-1}\left (\frac{\sqrt{b} \cos (c+d x)}{\sqrt{a+b}}\right )}{b^{5/2} d \sqrt{a+b}}+\frac{(a-b) \cos (c+d x)}{b^2 d}+\frac{\cos ^3(c+d x)}{3 b d} \]

[Out]

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

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Rubi [A]  time = 0.0916907, antiderivative size = 77, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {3186, 390, 208} \[ -\frac{a^2 \tanh ^{-1}\left (\frac{\sqrt{b} \cos (c+d x)}{\sqrt{a+b}}\right )}{b^{5/2} d \sqrt{a+b}}+\frac{(a-b) \cos (c+d x)}{b^2 d}+\frac{\cos ^3(c+d x)}{3 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [C]  time = 0.507072, size = 150, normalized size = 1.95 \[ \frac{6 a^2 \tan ^{-1}\left (\frac{\sqrt{b}-i \sqrt{a} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{-a-b}}\right )+6 a^2 \tan ^{-1}\left (\frac{\sqrt{b}+i \sqrt{a} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{-a-b}}\right )+\sqrt{b} \sqrt{-a-b} \cos (c+d x) (6 a+b \cos (2 (c+d x))-5 b)}{6 b^{5/2} d \sqrt{-a-b}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(6*a^2*ArcTan[(Sqrt[b] - I*Sqrt[a]*Tan[(c + d*x)/2])/Sqrt[-a - b]] + 6*a^2*ArcTan[(Sqrt[b] + I*Sqrt[a]*Tan[(c
+ d*x)/2])/Sqrt[-a - b]] + Sqrt[-a - b]*Sqrt[b]*Cos[c + d*x]*(6*a - 5*b + b*Cos[2*(c + d*x)]))/(6*Sqrt[-a - b]
*b^(5/2)*d)

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Maple [A]  time = 0.075, size = 70, normalized size = 0.9 \begin{align*}{\frac{1}{d} \left ({\frac{1}{{b}^{2}} \left ({\frac{b \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{3}}+\cos \left ( dx+c \right ) a-b\cos \left ( dx+c \right ) \right ) }-{\frac{{a}^{2}}{{b}^{2}}{\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 ) } \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),x)

[Out]

1/d*(1/b^2*(1/3*b*cos(d*x+c)^3+cos(d*x+c)*a-b*cos(d*x+c))-a^2/b^2/((a+b)*b)^(1/2)*arctanh(cos(d*x+c)*b/((a+b)*
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)^5/(a+b*sin(d*x+c)^2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.8727, size = 491, normalized size = 6.38 \begin{align*} \left [\frac{2 \,{\left (a b^{2} + b^{3}\right )} \cos \left (d x + c\right )^{3} + 3 \, \sqrt{a b + b^{2}} a^{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 ) + 6 \,{\left (a^{2} b - b^{3}\right )} \cos \left (d x + c\right )}{6 \,{\left (a b^{3} + b^{4}\right )} d}, \frac{{\left (a b^{2} + b^{3}\right )} \cos \left (d x + c\right )^{3} + 3 \, \sqrt{-a b - b^{2}} a^{2} \arctan \left (\frac{\sqrt{-a b - b^{2}} \cos \left (d x + c\right )}{a + b}\right ) + 3 \,{\left (a^{2} b - b^{3}\right )} \cos \left (d x + c\right )}{3 \,{\left (a b^{3} + b^{4}\right )} d}\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),x, algorithm="fricas")

[Out]

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

[Out]

Timed out

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Giac [B]  time = 1.17106, size = 234, normalized size = 3.04 \begin{align*} \frac{\frac{3 \, a^{2} \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 )}{\sqrt{-a b - b^{2}} b^{2}} - \frac{2 \,{\left (3 \, a - 2 \, b - \frac{6 \, a{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac{6 \, 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}}\right )}}{b^{2}{\left (\frac{\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} - 1\right )}^{3}}}{3 \, 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),x, algorithm="giac")

[Out]

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