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]

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

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Rubi [A]  time = 0.07, antiderivative size = 52, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, 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 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 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 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]

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*}

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Mathematica [C]  time = 0.24, size = 125, normalized size = 2.40 \[ -\frac {\sqrt {b} \sqrt {-a-b} \cos (c+d x)+a \tan ^{-1}\left (\frac {\sqrt {b}-i \sqrt {a} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a-b}}\right )+a \tan ^{-1}\left (\frac {\sqrt {b}+i \sqrt {a} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a-b}}\right )}{b^{3/2} d \sqrt {-a-b}} \]

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))

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fricas [A]  time = 0.47, size = 165, normalized size = 3.17 \[ \left [\frac {\sqrt {a b + b^{2}} a \log \left (\frac {b \cos \left (d x + c\right )^{2} + 2 \, \sqrt {a b + b^{2}} \cos \left (d x + c\right ) + a + b}{b \cos \left (d x + c\right )^{2} - a - b}\right ) - 2 \, {\left (a b + b^{2}\right )} \cos \left (d x + c\right )}{2 \, {\left (a b^{2} + b^{3}\right )} d}, -\frac {\sqrt {-a b - b^{2}} a \arctan \left (\frac {\sqrt {-a b - b^{2}} \cos \left (d x + c\right )}{a + b}\right ) + {\left (a b + b^{2}\right )} \cos \left (d x + c\right )}{{\left (a b^{2} + b^{3}\right )} d}\right ] \]

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)]

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giac [A]  time = 0.17, size = 57, normalized size = 1.10 \[ -\frac {a \arctan \left (\frac {b \cos \left (d x + c\right )}{\sqrt {-a b - b^{2}}}\right )}{\sqrt {-a b - b^{2}} b d} - \frac {\cos \left (d x + c\right )}{b d} \]

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)

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maple [A]  time = 0.27, size = 45, normalized size = 0.87 \[ \frac {-\frac {\cos \left (d x +c \right )}{b}+\frac {a \arctanh \left (\frac {\cos \left (d x +c \right ) b}{\sqrt {\left (a +b \right ) b}}\right )}{b \sqrt {\left (a +b \right ) b}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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maxima [A]  time = 0.43, size = 67, normalized size = 1.29 \[ -\frac {\frac {a \log \left (\frac {b \cos \left (d x + c\right ) - \sqrt {{\left (a + b\right )} b}}{b \cos \left (d x + c\right ) + \sqrt {{\left (a + b\right )} b}}\right )}{\sqrt {{\left (a + b\right )} b} b} + \frac {2 \, \cos \left (d x + c\right )}{b}}{2 \, d} \]

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]

-1/2*(a*log((b*cos(d*x + c) - sqrt((a + b)*b))/(b*cos(d*x + c) + sqrt((a + b)*b)))/(sqrt((a + b)*b)*b) + 2*cos
(d*x + c)/b)/d

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mupad [B]  time = 0.10, size = 44, normalized size = 0.85 \[ \frac {a\,\mathrm {atanh}\left (\frac {\sqrt {b}\,\cos \left (c+d\,x\right )}{\sqrt {a+b}}\right )}{b^{3/2}\,d\,\sqrt {a+b}}-\frac {\cos \left (c+d\,x\right )}{b\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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