3.107 \(\int \frac{\sin ^4(c+d x)}{(a+b \sin ^2(c+d x))^3} \, dx\)

Optimal. Leaf size=110 \[ -\frac{\tan ^3(c+d x)}{4 d (a+b) \left ((a+b) \tan ^2(c+d x)+a\right )^2}-\frac{3 \tan (c+d x)}{8 d (a+b)^2 \left ((a+b) \tan ^2(c+d x)+a\right )}+\frac{3 \tan ^{-1}\left (\frac{\sqrt{a+b} \tan (c+d x)}{\sqrt{a}}\right )}{8 \sqrt{a} d (a+b)^{5/2}} \]

[Out]

(3*ArcTan[(Sqrt[a + b]*Tan[c + d*x])/Sqrt[a]])/(8*Sqrt[a]*(a + b)^(5/2)*d) - Tan[c + d*x]^3/(4*(a + b)*d*(a +
(a + b)*Tan[c + d*x]^2)^2) - (3*Tan[c + d*x])/(8*(a + b)^2*d*(a + (a + b)*Tan[c + d*x]^2))

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Rubi [A]  time = 0.0956382, antiderivative size = 110, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {3187, 288, 205} \[ -\frac{\tan ^3(c+d x)}{4 d (a+b) \left ((a+b) \tan ^2(c+d x)+a\right )^2}-\frac{3 \tan (c+d x)}{8 d (a+b)^2 \left ((a+b) \tan ^2(c+d x)+a\right )}+\frac{3 \tan ^{-1}\left (\frac{\sqrt{a+b} \tan (c+d x)}{\sqrt{a}}\right )}{8 \sqrt{a} d (a+b)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[Sin[c + d*x]^4/(a + b*Sin[c + d*x]^2)^3,x]

[Out]

(3*ArcTan[(Sqrt[a + b]*Tan[c + d*x])/Sqrt[a]])/(8*Sqrt[a]*(a + b)^(5/2)*d) - Tan[c + d*x]^3/(4*(a + b)*d*(a +
(a + b)*Tan[c + d*x]^2)^2) - (3*Tan[c + d*x])/(8*(a + b)^2*d*(a + (a + b)*Tan[c + d*x]^2))

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 288

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

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)}{\left (a+b \sin ^2(c+d x)\right )^3} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^4}{\left (a+(a+b) x^2\right )^3} \, dx,x,\tan (c+d x)\right )}{d}\\ &=-\frac{\tan ^3(c+d x)}{4 (a+b) d \left (a+(a+b) \tan ^2(c+d x)\right )^2}+\frac{3 \operatorname{Subst}\left (\int \frac{x^2}{\left (a+(a+b) x^2\right )^2} \, dx,x,\tan (c+d x)\right )}{4 (a+b) d}\\ &=-\frac{\tan ^3(c+d x)}{4 (a+b) d \left (a+(a+b) \tan ^2(c+d x)\right )^2}-\frac{3 \tan (c+d x)}{8 (a+b)^2 d \left (a+(a+b) \tan ^2(c+d x)\right )}+\frac{3 \operatorname{Subst}\left (\int \frac{1}{a+(a+b) x^2} \, dx,x,\tan (c+d x)\right )}{8 (a+b)^2 d}\\ &=\frac{3 \tan ^{-1}\left (\frac{\sqrt{a+b} \tan (c+d x)}{\sqrt{a}}\right )}{8 \sqrt{a} (a+b)^{5/2} d}-\frac{\tan ^3(c+d x)}{4 (a+b) d \left (a+(a+b) \tan ^2(c+d x)\right )^2}-\frac{3 \tan (c+d x)}{8 (a+b)^2 d \left (a+(a+b) \tan ^2(c+d x)\right )}\\ \end{align*}

Mathematica [A]  time = 1.24913, size = 97, normalized size = 0.88 \[ \frac{\frac{3 \tan ^{-1}\left (\frac{\sqrt{a+b} \tan (c+d x)}{\sqrt{a}}\right )}{\sqrt{a} (a+b)^{5/2}}+\frac{\sin (2 (c+d x)) ((2 a+5 b) \cos (2 (c+d x))-8 a-5 b)}{(a+b)^2 (2 a-b \cos (2 (c+d x))+b)^2}}{8 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[c + d*x]^4/(a + b*Sin[c + d*x]^2)^3,x]

[Out]

((3*ArcTan[(Sqrt[a + b]*Tan[c + d*x])/Sqrt[a]])/(Sqrt[a]*(a + b)^(5/2)) + ((-8*a - 5*b + (2*a + 5*b)*Cos[2*(c
+ d*x)])*Sin[2*(c + d*x)])/((a + b)^2*(2*a + b - b*Cos[2*(c + d*x)])^2))/(8*d)

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Maple [A]  time = 0.089, size = 136, normalized size = 1.2 \begin{align*} -{\frac{5\, \left ( \tan \left ( dx+c \right ) \right ) ^{3}}{8\,d \left ( a \left ( \tan \left ( dx+c \right ) \right ) ^{2}+ \left ( \tan \left ( dx+c \right ) \right ) ^{2}b+a \right ) ^{2} \left ( a+b \right ) }}-{\frac{3\,a\tan \left ( dx+c \right ) }{8\,d \left ( a \left ( \tan \left ( dx+c \right ) \right ) ^{2}+ \left ( \tan \left ( dx+c \right ) \right ) ^{2}b+a \right ) ^{2} \left ({a}^{2}+2\,ab+{b}^{2} \right ) }}+{\frac{3}{8\,d \left ({a}^{2}+2\,ab+{b}^{2} \right ) }\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)^3,x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.97294, size = 1567, normalized size = 14.25 \begin{align*} \text{result too large to display} \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)^3,x, algorithm="fricas")

[Out]

[-1/32*(3*(b^2*cos(d*x + c)^4 - 2*(a*b + b^2)*cos(d*x + c)^2 + a^2 + 2*a*b + b^2)*sqrt(-a^2 - 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 + b)*cos(d*x + c)^3 - (a + b
)*cos(d*x + c))*sqrt(-a^2 - 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)) - 4*((2*a^3 + 7*a^2*b + 5*a*b^2)*cos(d*x + c)^3 - 5*(a^3 + 2*a^2*b + a*b^2)*cos(
d*x + c))*sin(d*x + c))/((a^4*b^2 + 3*a^3*b^3 + 3*a^2*b^4 + a*b^5)*d*cos(d*x + c)^4 - 2*(a^5*b + 4*a^4*b^2 + 6
*a^3*b^3 + 4*a^2*b^4 + a*b^5)*d*cos(d*x + c)^2 + (a^6 + 5*a^5*b + 10*a^4*b^2 + 10*a^3*b^3 + 5*a^2*b^4 + a*b^5)
*d), -1/16*(3*(b^2*cos(d*x + c)^4 - 2*(a*b + b^2)*cos(d*x + c)^2 + a^2 + 2*a*b + b^2)*sqrt(a^2 + a*b)*arctan(1
/2*((2*a + b)*cos(d*x + c)^2 - a - b)/(sqrt(a^2 + a*b)*cos(d*x + c)*sin(d*x + c))) - 2*((2*a^3 + 7*a^2*b + 5*a
*b^2)*cos(d*x + c)^3 - 5*(a^3 + 2*a^2*b + a*b^2)*cos(d*x + c))*sin(d*x + c))/((a^4*b^2 + 3*a^3*b^3 + 3*a^2*b^4
+ a*b^5)*d*cos(d*x + c)^4 - 2*(a^5*b + 4*a^4*b^2 + 6*a^3*b^3 + 4*a^2*b^4 + a*b^5)*d*cos(d*x + c)^2 + (a^6 + 5
*a^5*b + 10*a^4*b^2 + 10*a^3*b^3 + 5*a^2*b^4 + a*b^5)*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)**3,x)

[Out]

Timed out

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Giac [A]  time = 1.17635, size = 205, normalized size = 1.86 \begin{align*} \frac{\frac{3 \,{\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 )}}{{\left (a^{2} + 2 \, a b + b^{2}\right )} \sqrt{a^{2} + a b}} - \frac{5 \, a \tan \left (d x + c\right )^{3} + 5 \, b \tan \left (d x + c\right )^{3} + 3 \, a \tan \left (d x + c\right )}{{\left (a \tan \left (d x + c\right )^{2} + b \tan \left (d x + c\right )^{2} + a\right )}^{2}{\left (a^{2} + 2 \, a b + b^{2}\right )}}}{8 \, 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)^3,x, algorithm="giac")

[Out]

1/8*(3*(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 + 2*a*b + b^2)*sqrt(a^2 + a*b)) - (5*a*tan(d*x + c)^3 + 5*b*tan(d*x + c)^3 + 3*a*tan(d*x + c))/((a*ta
n(d*x + c)^2 + b*tan(d*x + c)^2 + a)^2*(a^2 + 2*a*b + b^2)))/d