3.87 \(\int \frac{\sin ^4(c+d x)}{a+b \sin ^2(c+d x)} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.114091, antiderivative size = 77, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used = {3187, 470, 522, 203, 205} \[ \frac{a^{3/2} \tan ^{-1}\left (\frac{\sqrt{a+b} \tan (c+d x)}{\sqrt{a}}\right )}{b^2 d \sqrt{a+b}}-\frac{x (2 a-b)}{2 b^2}-\frac{\sin (c+d x) \cos (c+d x)}{2 b d} \]

Antiderivative was successfully verified.

[In]

Int[Sin[c + d*x]^4/(a + b*Sin[c + d*x]^2),x]

[Out]

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

Rule 3187

Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.), x_Symbol] :> With[{ff = FreeF
actors[Tan[e + f*x], x]}, Dist[ff^(m + 1)/f, Subst[Int[(x^m*(a + (a + b)*ff^2*x^2)^p)/(1 + ff^2*x^2)^(m/2 + p
+ 1), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x] && IntegerQ[m/2] && IntegerQ[p]

Rule 470

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> -Simp[(a*e^(2
*n - 1)*(e*x)^(m - 2*n + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(b*n*(b*c - a*d)*(p + 1)), x] + Dist[e^(2
*n)/(b*n*(b*c - a*d)*(p + 1)), Int[(e*x)^(m - 2*n)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[a*c*(m - 2*n + 1) +
(a*d*(m - n + n*q + 1) + b*c*n*(p + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b*c - a*d, 0] &
& IGtQ[n, 0] && LtQ[p, -1] && GtQ[m - n + 1, n] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x]

Rule 522

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*((c_) + (d_.)*(x_)^(n_))), x_Symbol] :> Dist[(b*e - a*f
)/(b*c - a*d), Int[1/(a + b*x^n), x], x] - Dist[(d*e - c*f)/(b*c - a*d), Int[1/(c + d*x^n), x], x] /; FreeQ[{a
, b, c, d, e, f, n}, x]

Rule 203

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

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rubi steps

\begin{align*} \int \frac{\sin ^4(c+d x)}{a+b \sin ^2(c+d x)} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^4}{\left (1+x^2\right )^2 \left (a+(a+b) x^2\right )} \, dx,x,\tan (c+d x)\right )}{d}\\ &=-\frac{\cos (c+d x) \sin (c+d x)}{2 b d}+\frac{\operatorname{Subst}\left (\int \frac{a+(-a+b) x^2}{\left (1+x^2\right ) \left (a+(a+b) x^2\right )} \, dx,x,\tan (c+d x)\right )}{2 b d}\\ &=-\frac{\cos (c+d x) \sin (c+d x)}{2 b d}+\frac{a^2 \operatorname{Subst}\left (\int \frac{1}{a+(a+b) x^2} \, dx,x,\tan (c+d x)\right )}{b^2 d}-\frac{(2 a-b) \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\tan (c+d x)\right )}{2 b^2 d}\\ &=-\frac{(2 a-b) x}{2 b^2}+\frac{a^{3/2} \tan ^{-1}\left (\frac{\sqrt{a+b} \tan (c+d x)}{\sqrt{a}}\right )}{b^2 \sqrt{a+b} d}-\frac{\cos (c+d x) \sin (c+d x)}{2 b d}\\ \end{align*}

Mathematica [A]  time = 0.310148, size = 69, normalized size = 0.9 \[ -\frac{-\frac{4 a^{3/2} \tan ^{-1}\left (\frac{\sqrt{a+b} \tan (c+d x)}{\sqrt{a}}\right )}{\sqrt{a+b}}+2 (2 a-b) (c+d x)+b \sin (2 (c+d x))}{4 b^2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[c + d*x]^4/(a + b*Sin[c + d*x]^2),x]

[Out]

-(2*(2*a - b)*(c + d*x) - (4*a^(3/2)*ArcTan[(Sqrt[a + b]*Tan[c + d*x])/Sqrt[a]])/Sqrt[a + b] + b*Sin[2*(c + d*
x)])/(4*b^2*d)

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Maple [A]  time = 0.079, size = 94, normalized size = 1.2 \begin{align*} -{\frac{\tan \left ( dx+c \right ) }{2\,bd \left ( \left ( \tan \left ( dx+c \right ) \right ) ^{2}+1 \right ) }}+{\frac{\arctan \left ( \tan \left ( dx+c \right ) \right ) }{2\,bd}}-{\frac{\arctan \left ( \tan \left ( dx+c \right ) \right ) a}{{b}^{2}d}}+{\frac{{a}^{2}}{{b}^{2}d}\arctan \left ({ \left ( a+b \right ) \tan \left ( dx+c \right ){\frac{1}{\sqrt{a \left ( a+b \right ) }}}} \right ){\frac{1}{\sqrt{a \left ( a+b \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(d*x+c)^4/(a+sin(d*x+c)^2*b),x)

[Out]

-1/2/d/b*tan(d*x+c)/(tan(d*x+c)^2+1)+1/2/d/b*arctan(tan(d*x+c))-1/d/b^2*arctan(tan(d*x+c))*a+1/d*a^2/b^2/(a*(a
+b))^(1/2)*arctan((a+b)*tan(d*x+c)/(a*(a+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)^4/(a+b*sin(d*x+c)^2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.87633, size = 738, normalized size = 9.58 \begin{align*} \left [-\frac{2 \,{\left (2 \, a - b\right )} d x + 2 \, b \cos \left (d x + c\right ) \sin \left (d x + c\right ) - a \sqrt{-\frac{a}{a + b}} \log \left (\frac{{\left (8 \, a^{2} + 8 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{4} - 2 \,{\left (4 \, a^{2} + 5 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{2} - 4 \,{\left ({\left (2 \, a^{2} + 3 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{3} -{\left (a^{2} + 2 \, a b + b^{2}\right )} \cos \left (d x + c\right )\right )} \sqrt{-\frac{a}{a + b}} \sin \left (d x + c\right ) + a^{2} + 2 \, a b + b^{2}}{b^{2} \cos \left (d x + c\right )^{4} - 2 \,{\left (a b + b^{2}\right )} \cos \left (d x + c\right )^{2} + a^{2} + 2 \, a b + b^{2}}\right )}{4 \, b^{2} d}, -\frac{{\left (2 \, a - b\right )} d x + b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a \sqrt{\frac{a}{a + b}} \arctan \left (\frac{{\left ({\left (2 \, a + b\right )} \cos \left (d x + c\right )^{2} - a - b\right )} \sqrt{\frac{a}{a + b}}}{2 \, a \cos \left (d x + c\right ) \sin \left (d x + c\right )}\right )}{2 \, b^{2} d}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^4/(a+b*sin(d*x+c)^2),x, algorithm="fricas")

[Out]

[-1/4*(2*(2*a - b)*d*x + 2*b*cos(d*x + c)*sin(d*x + c) - a*sqrt(-a/(a + b))*log(((8*a^2 + 8*a*b + b^2)*cos(d*x
+ c)^4 - 2*(4*a^2 + 5*a*b + b^2)*cos(d*x + c)^2 - 4*((2*a^2 + 3*a*b + b^2)*cos(d*x + c)^3 - (a^2 + 2*a*b + b^
2)*cos(d*x + c))*sqrt(-a/(a + b))*sin(d*x + c) + a^2 + 2*a*b + b^2)/(b^2*cos(d*x + c)^4 - 2*(a*b + b^2)*cos(d*
x + c)^2 + a^2 + 2*a*b + b^2)))/(b^2*d), -1/2*((2*a - b)*d*x + b*cos(d*x + c)*sin(d*x + c) + a*sqrt(a/(a + b))
*arctan(1/2*((2*a + b)*cos(d*x + c)^2 - a - b)*sqrt(a/(a + b))/(a*cos(d*x + c)*sin(d*x + c))))/(b^2*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)**4/(a+b*sin(d*x+c)**2),x)

[Out]

Timed out

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Giac [A]  time = 1.19799, size = 154, normalized size = 2. \begin{align*} \frac{\frac{2 \,{\left (\pi \left \lfloor \frac{d x + c}{\pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (2 \, a + 2 \, b\right ) + \arctan \left (\frac{a \tan \left (d x + c\right ) + b \tan \left (d x + c\right )}{\sqrt{a^{2} + a b}}\right )\right )} a^{2}}{\sqrt{a^{2} + a b} b^{2}} - \frac{{\left (d x + c\right )}{\left (2 \, a - b\right )}}{b^{2}} - \frac{\tan \left (d x + c\right )}{{\left (\tan \left (d x + c\right )^{2} + 1\right )} b}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^4/(a+b*sin(d*x+c)^2),x, algorithm="giac")

[Out]

1/2*(2*(pi*floor((d*x + c)/pi + 1/2)*sgn(2*a + 2*b) + arctan((a*tan(d*x + c) + b*tan(d*x + c))/sqrt(a^2 + a*b)
))*a^2/(sqrt(a^2 + a*b)*b^2) - (d*x + c)*(2*a - b)/b^2 - tan(d*x + c)/((tan(d*x + c)^2 + 1)*b))/d