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

Optimal. Leaf size=83 \[ \frac {a \cos (c+d x)}{2 b d (a+b) \left (a-b \cos ^2(c+d x)+b\right )}-\frac {(a+2 b) \tanh ^{-1}\left (\frac {\sqrt {b} \cos (c+d x)}{\sqrt {a+b}}\right )}{2 b^{3/2} d (a+b)^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.09, antiderivative size = 83, 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, 385, 208} \[ \frac {a \cos (c+d x)}{2 b d (a+b) \left (a-b \cos ^2(c+d x)+b\right )}-\frac {(a+2 b) \tanh ^{-1}\left (\frac {\sqrt {b} \cos (c+d x)}{\sqrt {a+b}}\right )}{2 b^{3/2} d (a+b)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [C]  time = 0.48, size = 160, normalized size = 1.93 \[ \frac {\frac {2 a \sqrt {b} \cos (c+d x)}{2 a-b \cos (2 (c+d x))+b}+\frac {(a+2 b) \tan ^{-1}\left (\frac {\sqrt {b}-i \sqrt {a} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a-b}}\right )}{\sqrt {-a-b}}+\frac {(a+2 b) \tan ^{-1}\left (\frac {\sqrt {b}+i \sqrt {a} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a-b}}\right )}{\sqrt {-a-b}}}{2 b^{3/2} d (a+b)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.46, size = 327, normalized size = 3.94 \[ \left [\frac {{\left ({\left (a b + 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} - a^{2} - 3 \, a b - 2 \, b^{2}\right )} \sqrt {a b + b^{2}} \log \left (-\frac {b \cos \left (d x + c\right )^{2} - 2 \, \sqrt {a b + b^{2}} \cos \left (d x + c\right ) + a + b}{b \cos \left (d x + c\right )^{2} - a - b}\right ) - 2 \, {\left (a^{2} b + a b^{2}\right )} \cos \left (d x + c\right )}{4 \, {\left ({\left (a^{2} b^{3} + 2 \, a b^{4} + b^{5}\right )} d \cos \left (d x + c\right )^{2} - {\left (a^{3} b^{2} + 3 \, a^{2} b^{3} + 3 \, a b^{4} + b^{5}\right )} d\right )}}, \frac {{\left ({\left (a b + 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} - a^{2} - 3 \, a b - 2 \, b^{2}\right )} \sqrt {-a b - b^{2}} \arctan \left (\frac {\sqrt {-a b - b^{2}} \cos \left (d x + c\right )}{a + b}\right ) - {\left (a^{2} b + a b^{2}\right )} \cos \left (d x + c\right )}{2 \, {\left ({\left (a^{2} b^{3} + 2 \, a b^{4} + b^{5}\right )} d \cos \left (d x + c\right )^{2} - {\left (a^{3} b^{2} + 3 \, a^{2} b^{3} + 3 \, a b^{4} + b^{5}\right )} d\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.18, size = 93, normalized size = 1.12 \[ \frac {{\left (a + 2 \, b\right )} \arctan \left (\frac {b \cos \left (d x + c\right )}{\sqrt {-a b - b^{2}}}\right )}{2 \, {\left (a b + b^{2}\right )} \sqrt {-a b - b^{2}} d} - \frac {a \cos \left (d x + c\right )}{2 \, {\left (b \cos \left (d x + c\right )^{2} - a - b\right )} {\left (a b + b^{2}\right )} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.29, size = 80, normalized size = 0.96 \[ \frac {-\frac {a \cos \left (d x +c \right )}{2 \left (a +b \right ) b \left (b \left (\cos ^{2}\left (d x +c \right )\right )-a -b \right )}-\frac {\left (a +2 b \right ) \arctanh \left (\frac {\cos \left (d x +c \right ) b}{\sqrt {\left (a +b \right ) b}}\right )}{2 \left (a +b \right ) b \sqrt {\left (a +b \right ) b}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.42, size = 111, normalized size = 1.34 \[ \frac {\frac {2 \, a \cos \left (d x + c\right )}{a^{2} b + 2 \, a b^{2} + b^{3} - {\left (a b^{2} + b^{3}\right )} \cos \left (d x + c\right )^{2}} + \frac {{\left (a + 2 \, b\right )} \log \left (\frac {b \cos \left (d x + c\right ) - \sqrt {{\left (a + b\right )} b}}{b \cos \left (d x + c\right ) + \sqrt {{\left (a + b\right )} b}}\right )}{\sqrt {{\left (a + b\right )} b} {\left (a b + b^{2}\right )}}}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 13.49, size = 71, normalized size = 0.86 \[ \frac {a\,\cos \left (c+d\,x\right )}{2\,b\,d\,\left (a+b\right )\,\left (-b\,{\cos \left (c+d\,x\right )}^2+a+b\right )}-\frac {\mathrm {atanh}\left (\frac {\sqrt {b}\,\cos \left (c+d\,x\right )}{\sqrt {a+b}}\right )\,\left (a+2\,b\right )}{2\,b^{3/2}\,d\,{\left (a+b\right )}^{3/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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)**2,x)

[Out]

Timed out

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