3.80 \(\int \frac{\sin ^3(c+d x)}{a+b \sin ^2(c+d x)} \, dx\)

Optimal. Leaf size=52 \[ \frac{a \tanh ^{-1}\left (\frac{\sqrt{b} \cos (c+d x)}{\sqrt{a+b}}\right )}{b^{3/2} d \sqrt{a+b}}-\frac{\cos (c+d x)}{b d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0696408, antiderivative size = 52, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {3186, 388, 208} \[ \frac{a \tanh ^{-1}\left (\frac{\sqrt{b} \cos (c+d x)}{\sqrt{a+b}}\right )}{b^{3/2} d \sqrt{a+b}}-\frac{\cos (c+d x)}{b d} \]

Antiderivative was successfully verified.

[In]

Int[Sin[c + d*x]^3/(a + b*Sin[c + d*x]^2),x]

[Out]

(a*ArcTanh[(Sqrt[b]*Cos[c + d*x])/Sqrt[a + b]])/(b^(3/2)*Sqrt[a + b]*d) - Cos[c + d*x]/(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 388

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(d*x*(a + b*x^n)^(p + 1))/(b*(n*
(p + 1) + 1)), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(b*(n*(p + 1) + 1)), Int[(a + b*x^n)^p, x], x] /; FreeQ[{
a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && NeQ[n*(p + 1) + 1, 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 ^3(c+d x)}{a+b \sin ^2(c+d x)} \, dx &=-\frac{\operatorname{Subst}\left (\int \frac{1-x^2}{a+b-b x^2} \, 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-b x^2} \, dx,x,\cos (c+d x)\right )}{b d}\\ &=\frac{a \tanh ^{-1}\left (\frac{\sqrt{b} \cos (c+d x)}{\sqrt{a+b}}\right )}{b^{3/2} \sqrt{a+b} d}-\frac{\cos (c+d x)}{b d}\\ \end{align*}

Mathematica [C]  time = 0.257717, size = 125, normalized size = 2.4 \[ -\frac{\sqrt{b} \sqrt{-a-b} \cos (c+d x)+a \tan ^{-1}\left (\frac{\sqrt{b}-i \sqrt{a} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{-a-b}}\right )+a \tan ^{-1}\left (\frac{\sqrt{b}+i \sqrt{a} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{-a-b}}\right )}{b^{3/2} d \sqrt{-a-b}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[c + d*x]^3/(a + b*Sin[c + d*x]^2),x]

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.069, size = 45, normalized size = 0.9 \begin{align*}{\frac{1}{d} \left ( -{\frac{\cos \left ( dx+c \right ) }{b}}+{\frac{a}{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 ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(d*x+c)^3/(a+sin(d*x+c)^2*b),x)

[Out]

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

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 1.79428, size = 381, normalized size = 7.33 \begin{align*} \left [\frac{\sqrt{a b + b^{2}} a \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 (a b + b^{2}\right )} \cos \left (d x + c\right )}{2 \,{\left (a b^{2} + b^{3}\right )} d}, -\frac{\sqrt{-a b - b^{2}} a \arctan \left (\frac{\sqrt{-a b - b^{2}} \cos \left (d x + c\right )}{a + b}\right ) +{\left (a b + b^{2}\right )} \cos \left (d x + c\right )}{{\left (a b^{2} + b^{3}\right )} d}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^3/(a+b*sin(d*x+c)^2),x, algorithm="fricas")

[Out]

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

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 1.16063, size = 77, normalized size = 1.48 \begin{align*} -\frac{a \arctan \left (\frac{b \cos \left (d x + c\right )}{\sqrt{-a b - b^{2}}}\right )}{\sqrt{-a b - b^{2}} b d} - \frac{\cos \left (d x + c\right )}{b d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^3/(a+b*sin(d*x+c)^2),x, algorithm="giac")

[Out]

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