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

Optimal. Leaf size=168 \[ \frac{b \sin ^3(c+d x)}{3 d \left (a^2+b^2\right )}+\frac{a^2 b \sin (c+d x)}{d \left (a^2+b^2\right )^2}+\frac{a \cos ^3(c+d x)}{3 d \left (a^2+b^2\right )}-\frac{a \cos (c+d x)}{d \left (a^2+b^2\right )}+\frac{a b^2 \cos (c+d x)}{d \left (a^2+b^2\right )^2}+\frac{a^3 b \tanh ^{-1}\left (\frac{b \cos (c+d x)-a \sin (c+d x)}{\sqrt{a^2+b^2}}\right )}{d \left (a^2+b^2\right )^{5/2}} \]

[Out]

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

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Rubi [A]  time = 0.22299, antiderivative size = 168, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 9, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.429, Rules used = {3518, 3109, 2564, 30, 2633, 3099, 3074, 206, 2638} \[ \frac{b \sin ^3(c+d x)}{3 d \left (a^2+b^2\right )}+\frac{a^2 b \sin (c+d x)}{d \left (a^2+b^2\right )^2}+\frac{a \cos ^3(c+d x)}{3 d \left (a^2+b^2\right )}-\frac{a \cos (c+d x)}{d \left (a^2+b^2\right )}+\frac{a b^2 \cos (c+d x)}{d \left (a^2+b^2\right )^2}+\frac{a^3 b \tanh ^{-1}\left (\frac{b \cos (c+d x)-a \sin (c+d x)}{\sqrt{a^2+b^2}}\right )}{d \left (a^2+b^2\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3518

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Int[(Sin[e + f*x]
^m*(a*Cos[e + f*x] + b*Sin[e + f*x])^n)/Cos[e + f*x]^n, x] /; FreeQ[{a, b, e, f}, x] && IntegerQ[(m - 1)/2] &&
ILtQ[n, 0] && ((LtQ[m, 5] && GtQ[n, -4]) || (EqQ[m, 5] && EqQ[n, -1]))

Rule 3109

Int[(cos[(c_.) + (d_.)*(x_)]^(m_.)*sin[(c_.) + (d_.)*(x_)]^(n_.))/(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(
c_.) + (d_.)*(x_)]), x_Symbol] :> Dist[b/(a^2 + b^2), Int[Cos[c + d*x]^m*Sin[c + d*x]^(n - 1), x], x] + (Dist[
a/(a^2 + b^2), Int[Cos[c + d*x]^(m - 1)*Sin[c + d*x]^n, x], x] - Dist[(a*b)/(a^2 + b^2), Int[(Cos[c + d*x]^(m
- 1)*Sin[c + d*x]^(n - 1))/(a*Cos[c + d*x] + b*Sin[c + d*x]), x], x]) /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 + b
^2, 0] && IGtQ[m, 0] && IGtQ[n, 0]

Rule 2564

Int[cos[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(a*f), Subst[Int[
x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Sin[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2] &&
!(IntegerQ[(m - 1)/2] && LtQ[0, m, n])

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2633

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]

Rule 3099

Int[sin[(c_.) + (d_.)*(x_)]^(m_)/(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)]), x_Symbol] :>
-Simp[(a*Sin[c + d*x]^(m - 1))/(d*(a^2 + b^2)*(m - 1)), x] + (Dist[a^2/(a^2 + b^2), Int[Sin[c + d*x]^(m - 2)/
(a*Cos[c + d*x] + b*Sin[c + d*x]), x], x] + Dist[b/(a^2 + b^2), Int[Sin[c + d*x]^(m - 1), x], x]) /; FreeQ[{a,
b, c, d}, x] && NeQ[a^2 + b^2, 0] && GtQ[m, 1]

Rule 3074

Int[(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> -Dist[d^(-1), Subst[Int
[1/(a^2 + b^2 - x^2), x], x, b*Cos[c + d*x] - a*Sin[c + d*x]], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 + b^2,
0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 2638

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

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

Mathematica [A]  time = 1.37492, size = 139, normalized size = 0.83 \[ \frac{\sqrt{a^2+b^2} \left (\left (3 a b^2-9 a^3\right ) \cos (c+d x)+a \left (a^2+b^2\right ) \cos (3 (c+d x))-2 b \sin (c+d x) \left (\left (a^2+b^2\right ) \cos (2 (c+d x))-7 a^2-b^2\right )\right )-24 a^3 b \tanh ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )-b}{\sqrt{a^2+b^2}}\right )}{12 d \left (a^2+b^2\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-24*a^3*b*ArcTanh[(-b + a*Tan[(c + d*x)/2])/Sqrt[a^2 + b^2]] + Sqrt[a^2 + b^2]*((-9*a^3 + 3*a*b^2)*Cos[c + d*
x] + a*(a^2 + b^2)*Cos[3*(c + d*x)] - 2*b*(-7*a^2 - b^2 + (a^2 + b^2)*Cos[2*(c + d*x)])*Sin[c + d*x]))/(12*(a^
2 + b^2)^(5/2)*d)

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Maple [A]  time = 0.065, size = 205, normalized size = 1.2 \begin{align*}{\frac{1}{d} \left ( -2\,{\frac{-b{a}^{2} \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{5}-a{b}^{2} \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}+ \left ( -10/3\,b{a}^{2}-4/3\,{b}^{3} \right ) \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}+2\,{a}^{3} \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-b{a}^{2}\tan \left ( 1/2\,dx+c/2 \right ) +2/3\,{a}^{3}-1/3\,a{b}^{2}}{ \left ({a}^{4}+2\,{a}^{2}{b}^{2}+{b}^{4} \right ) \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{3}}}-16\,{\frac{b{a}^{3}}{ \left ( 8\,{a}^{4}+16\,{a}^{2}{b}^{2}+8\,{b}^{4} \right ) \sqrt{{a}^{2}+{b}^{2}}}{\it Artanh} \left ( 1/2\,{\frac{2\,a\tan \left ( 1/2\,dx+c/2 \right ) -2\,b}{\sqrt{{a}^{2}+{b}^{2}}}} \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(-2/(a^4+2*a^2*b^2+b^4)*(-b*a^2*tan(1/2*d*x+1/2*c)^5-a*b^2*tan(1/2*d*x+1/2*c)^4+(-10/3*b*a^2-4/3*b^3)*tan(
1/2*d*x+1/2*c)^3+2*a^3*tan(1/2*d*x+1/2*c)^2-b*a^2*tan(1/2*d*x+1/2*c)+2/3*a^3-1/3*a*b^2)/(1+tan(1/2*d*x+1/2*c)^
2)^3-16*b*a^3/(8*a^4+16*a^2*b^2+8*b^4)/(a^2+b^2)^(1/2)*arctanh(1/2*(2*a*tan(1/2*d*x+1/2*c)-2*b)/(a^2+b^2)^(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)^3/(a+b*tan(d*x+c)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.27774, size = 597, normalized size = 3.55 \begin{align*} \frac{3 \, \sqrt{a^{2} + b^{2}} a^{3} b \log \left (\frac{2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) +{\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a^{2} - b^{2} - 2 \, \sqrt{a^{2} + b^{2}}{\left (b \cos \left (d x + c\right ) - a \sin \left (d x + c\right )\right )}}{2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) +{\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + b^{2}}\right ) + 2 \,{\left (a^{5} + 2 \, a^{3} b^{2} + a b^{4}\right )} \cos \left (d x + c\right )^{3} - 6 \,{\left (a^{5} + a^{3} b^{2}\right )} \cos \left (d x + c\right ) + 2 \,{\left (4 \, a^{4} b + 5 \, a^{2} b^{3} + b^{5} -{\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{6 \,{\left (a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}\right )} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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Giac [A]  time = 1.38606, size = 325, normalized size = 1.93 \begin{align*} \frac{\frac{3 \, a^{3} b \log \left (\frac{{\left | 2 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 2 \, b - 2 \, \sqrt{a^{2} + b^{2}} \right |}}{{\left | 2 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 2 \, b + 2 \, \sqrt{a^{2} + b^{2}} \right |}}\right )}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt{a^{2} + b^{2}}} + \frac{2 \,{\left (3 \, a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 3 \, a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 10 \, a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 4 \, b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 6 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 3 \, a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 2 \, a^{3} + a b^{2}\right )}}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )}{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )}^{3}}}{3 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(3*a^3*b*log(abs(2*a*tan(1/2*d*x + 1/2*c) - 2*b - 2*sqrt(a^2 + b^2))/abs(2*a*tan(1/2*d*x + 1/2*c) - 2*b +
2*sqrt(a^2 + b^2)))/((a^4 + 2*a^2*b^2 + b^4)*sqrt(a^2 + b^2)) + 2*(3*a^2*b*tan(1/2*d*x + 1/2*c)^5 + 3*a*b^2*ta
n(1/2*d*x + 1/2*c)^4 + 10*a^2*b*tan(1/2*d*x + 1/2*c)^3 + 4*b^3*tan(1/2*d*x + 1/2*c)^3 - 6*a^3*tan(1/2*d*x + 1/
2*c)^2 + 3*a^2*b*tan(1/2*d*x + 1/2*c) - 2*a^3 + a*b^2)/((a^4 + 2*a^2*b^2 + b^4)*(tan(1/2*d*x + 1/2*c)^2 + 1)^3
))/d