3.2.7 \(\int \frac {\sin ^4(c+d x)}{(a+b \sin ^2(c+d x))^3} \, dx\) [107]

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

[Out]

3/8*arctan((a+b)^(1/2)*tan(d*x+c)/a^(1/2))/(a+b)^(5/2)/d/a^(1/2)-1/4*tan(d*x+c)^3/(a+b)/d/(a+(a+b)*tan(d*x+c)^
2)^2-3/8*tan(d*x+c)/(a+b)^2/d/(a+(a+b)*tan(d*x+c)^2)

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Rubi [A]
time = 0.07, antiderivative size = 110, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {3266, 294, 211} \begin {gather*} \frac {3 \text {ArcTan}\left (\frac {\sqrt {a+b} \tan (c+d x)}{\sqrt {a}}\right )}{8 \sqrt {a} d (a+b)^{5/2}}-\frac {3 \tan (c+d x)}{8 d (a+b)^2 \left ((a+b) \tan ^2(c+d x)+a\right )}-\frac {\tan ^3(c+d x)}{4 d (a+b) \left ((a+b) \tan ^2(c+d x)+a\right )^2} \end {gather*}

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 211

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

Rule 294

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 3266

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]

Rubi steps

\begin {align*} \int \frac {\sin ^4(c+d x)}{\left (a+b \sin ^2(c+d x)\right )^3} \, dx &=\frac {\text {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 \text {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 \text {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*}

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Mathematica [A]
time = 0.89, size = 97, normalized size = 0.88 \begin {gather*} \frac {\frac {3 \tan ^{-1}\left (\frac {\sqrt {a+b} \tan (c+d x)}{\sqrt {a}}\right )}{\sqrt {a} (a+b)^{5/2}}+\frac {(-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} \end {gather*}

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 = 0.24, size = 109, normalized size = 0.99

method	result	size
derivativedivides	\(\frac {\frac {-\frac {5 \left (\tan ^{3}\left (d x +c \right )\right )}{8 \left (a +b \right )}-\frac {3 a \tan \left (d x +c \right )}{8 \left (a^{2}+2 a b +b^{2}\right )}}{\left (a \left (\tan ^{2}\left (d x +c \right )\right )+b \left (\tan ^{2}\left (d x +c \right )\right )+a \right )^{2}}+\frac {3 \arctan \left (\frac {\left (a +b \right ) \tan \left (d x +c \right )}{\sqrt {a \left (a +b \right )}}\right )}{8 \left (a^{2}+2 a b +b^{2}\right ) \sqrt {a \left (a +b \right )}}}{d}\)	\(109\)
default	\(\frac {\frac {-\frac {5 \left (\tan ^{3}\left (d x +c \right )\right )}{8 \left (a +b \right )}-\frac {3 a \tan \left (d x +c \right )}{8 \left (a^{2}+2 a b +b^{2}\right )}}{\left (a \left (\tan ^{2}\left (d x +c \right )\right )+b \left (\tan ^{2}\left (d x +c \right )\right )+a \right )^{2}}+\frac {3 \arctan \left (\frac {\left (a +b \right ) \tan \left (d x +c \right )}{\sqrt {a \left (a +b \right )}}\right )}{8 \left (a^{2}+2 a b +b^{2}\right ) \sqrt {a \left (a +b \right )}}}{d}\)	\(109\)
risch	\(\frac {i \left (-8 b \,{\mathrm e}^{6 i \left (d x +c \right )} a^{2}-16 a \,b^{2} {\mathrm e}^{6 i \left (d x +c \right )}-5 b^{3} {\mathrm e}^{6 i \left (d x +c \right )}+16 a^{3} {\mathrm e}^{4 i \left (d x +c \right )}+56 b \,{\mathrm e}^{4 i \left (d x +c \right )} a^{2}+46 a \,b^{2} {\mathrm e}^{4 i \left (d x +c \right )}+15 b^{3} {\mathrm e}^{4 i \left (d x +c \right )}-8 b \,{\mathrm e}^{2 i \left (d x +c \right )} a^{2}-32 a \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-15 b^{3} {\mathrm e}^{2 i \left (d x +c \right )}+2 a \,b^{2}+5 b^{3}\right )}{4 b^{2} \left (a +b \right )^{2} d \left (-b \,{\mathrm e}^{4 i \left (d x +c \right )}+4 a \,{\mathrm e}^{2 i \left (d x +c \right )}+2 b \,{\mathrm e}^{2 i \left (d x +c \right )}-b \right )^{2}}-\frac {3 \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {2 i a^{2}+2 i a b +2 a \sqrt {-a^{2}-a b}+b \sqrt {-a^{2}-a b}}{b \sqrt {-a^{2}-a b}}\right )}{16 \sqrt {-a^{2}-a b}\, \left (a +b \right )^{2} d}+\frac {3 \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {-2 i a^{2}-2 i a b +2 a \sqrt {-a^{2}-a b}+b \sqrt {-a^{2}-a b}}{b \sqrt {-a^{2}-a b}}\right )}{16 \sqrt {-a^{2}-a b}\, \left (a +b \right )^{2} d}\)	\(398\)

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,method=_RETURNVERBOSE)

[Out]

1/d*((-5/8/(a+b)*tan(d*x+c)^3-3/8*a/(a^2+2*a*b+b^2)*tan(d*x+c))/(a*tan(d*x+c)^2+b*tan(d*x+c)^2+a)^2+3/8/(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 [A]
time = 0.53, size = 158, normalized size = 1.44 \begin {gather*} -\frac {\frac {5 \, {\left (a + b\right )} \tan \left (d x + c\right )^{3} + 3 \, a \tan \left (d x + c\right )}{{\left (a^{4} + 4 \, a^{3} b + 6 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4}\right )} \tan \left (d x + c\right )^{4} + a^{4} + 2 \, a^{3} b + a^{2} b^{2} + 2 \, {\left (a^{4} + 3 \, a^{3} b + 3 \, a^{2} b^{2} + a b^{3}\right )} \tan \left (d x + c\right )^{2}} - \frac {3 \, \arctan \left (\frac {{\left (a + b\right )} \tan \left (d x + c\right )}{\sqrt {{\left (a + b\right )} a}}\right )}{\sqrt {{\left (a + b\right )} a} {\left (a^{2} + 2 \, a b + b^{2}\right )}}}{8 \, d} \end {gather*}

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]

-1/8*((5*(a + b)*tan(d*x + c)^3 + 3*a*tan(d*x + c))/((a^4 + 4*a^3*b + 6*a^2*b^2 + 4*a*b^3 + b^4)*tan(d*x + c)^
4 + a^4 + 2*a^3*b + a^2*b^2 + 2*(a^4 + 3*a^3*b + 3*a^2*b^2 + a*b^3)*tan(d*x + c)^2) - 3*arctan((a + b)*tan(d*x
+ c)/sqrt((a + b)*a))/(sqrt((a + b)*a)*(a^2 + 2*a*b + b^2)))/d

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 291 vs. \(2 (96) = 192\).
time = 0.43, size = 683, normalized size = 6.21 \begin {gather*} \left [-\frac {3 \, {\left (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 )} \sqrt {-a^{2} - 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 + b\right )} \cos \left (d x + c\right )^{3} - {\left (a + b\right )} \cos \left (d x + c\right )\right )} \sqrt {-a^{2} - 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 \, {\left ({\left (2 \, a^{3} + 7 \, a^{2} b + 5 \, a b^{2}\right )} \cos \left (d x + c\right )^{3} - 5 \, {\left (a^{3} + 2 \, a^{2} b + a b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{32 \, {\left ({\left (a^{4} b^{2} + 3 \, a^{3} b^{3} + 3 \, a^{2} b^{4} + a b^{5}\right )} d \cos \left (d x + c\right )^{4} - 2 \, {\left (a^{5} b + 4 \, a^{4} b^{2} + 6 \, a^{3} b^{3} + 4 \, a^{2} b^{4} + a b^{5}\right )} d \cos \left (d x + c\right )^{2} + {\left (a^{6} + 5 \, a^{5} b + 10 \, a^{4} b^{2} + 10 \, a^{3} b^{3} + 5 \, a^{2} b^{4} + a b^{5}\right )} d\right )}}, -\frac {3 \, {\left (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 )} \sqrt {a^{2} + a b} \arctan \left (\frac {{\left (2 \, a + b\right )} \cos \left (d x + c\right )^{2} - a - b}{2 \, \sqrt {a^{2} + a b} \cos \left (d x + c\right ) \sin \left (d x + c\right )}\right ) - 2 \, {\left ({\left (2 \, a^{3} + 7 \, a^{2} b + 5 \, a b^{2}\right )} \cos \left (d x + c\right )^{3} - 5 \, {\left (a^{3} + 2 \, a^{2} b + a b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{16 \, {\left ({\left (a^{4} b^{2} + 3 \, a^{3} b^{3} + 3 \, a^{2} b^{4} + a b^{5}\right )} d \cos \left (d x + c\right )^{4} - 2 \, {\left (a^{5} b + 4 \, a^{4} b^{2} + 6 \, a^{3} b^{3} + 4 \, a^{2} b^{4} + a b^{5}\right )} d \cos \left (d x + c\right )^{2} + {\left (a^{6} + 5 \, a^{5} b + 10 \, a^{4} b^{2} + 10 \, a^{3} b^{3} + 5 \, a^{2} b^{4} + a b^{5}\right )} d\right )}}\right ] \end {gather*}

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)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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 = 0.48, size = 152, normalized size = 1.38 \begin {gather*} \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 {gather*}

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

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Mupad [B]
time = 13.69, size = 149, normalized size = 1.35 \begin {gather*} \frac {3\,\mathrm {atan}\left (\frac {3\,\mathrm {tan}\left (c+d\,x\right )\,\left (2\,a+2\,b\right )\,\left (\frac {8\,a^2}{3}+\frac {16\,a\,b}{3}+\frac {8\,b^2}{3}\right )}{16\,\sqrt {a}\,{\left (a+b\right )}^{5/2}}\right )}{8\,\sqrt {a}\,d\,{\left (a+b\right )}^{5/2}}-\frac {\frac {5\,{\mathrm {tan}\left (c+d\,x\right )}^3}{8\,\left (a+b\right )}+\frac {3\,a\,\mathrm {tan}\left (c+d\,x\right )}{8\,\left (a^2+2\,a\,b+b^2\right )}}{d\,\left ({\mathrm {tan}\left (c+d\,x\right )}^4\,\left (a^2+2\,a\,b+b^2\right )+a^2+{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (2\,a^2+2\,b\,a\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(c + d*x)^4/(a + b*sin(c + d*x)^2)^3,x)

[Out]

(3*atan((3*tan(c + d*x)*(2*a + 2*b)*((16*a*b)/3 + (8*a^2)/3 + (8*b^2)/3))/(16*a^(1/2)*(a + b)^(5/2))))/(8*a^(1
/2)*d*(a + b)^(5/2)) - ((5*tan(c + d*x)^3)/(8*(a + b)) + (3*a*tan(c + d*x))/(8*(2*a*b + a^2 + b^2)))/(d*(tan(c
+ d*x)^4*(2*a*b + a^2 + b^2) + a^2 + tan(c + d*x)^2*(2*a*b + 2*a^2)))

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