\(\int \sin ^4(e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx\) [127]

Optimal result

Integrand size = 25, antiderivative size = 259 \[ \int \sin ^4(e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx=-\frac {(a+4 b) \cos (e+f x) \sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{15 b f}-\frac {\cos (e+f x) \sin ^3(e+f x) \sqrt {a+b \sin ^2(e+f x)}}{5 f}-\frac {\left (2 a^2-3 a b-8 b^2\right ) \sqrt {\cos ^2(e+f x)} E\left (\arcsin (\sin (e+f x))\left |-\frac {b}{a}\right .\right ) \sec (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{15 b^2 f \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}}+\frac {2 a (a-2 b) (a+b) \sqrt {\cos ^2(e+f x)} \operatorname {EllipticF}\left (\arcsin (\sin (e+f x)),-\frac {b}{a}\right ) \sec (e+f x) \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}}{15 b^2 f \sqrt {a+b \sin ^2(e+f x)}} \]

[Out]

-1/15*(a+4*b)*cos(f*x+e)*sin(f*x+e)*(a+b*sin(f*x+e)^2)^(1/2)/b/f-1/5*cos(f*x+e)*sin(f*x+e)^3*(a+b*sin(f*x+e)^2
)^(1/2)/f-1/15*(2*a^2-3*a*b-8*b^2)*EllipticE(sin(f*x+e),(-b/a)^(1/2))*sec(f*x+e)*(cos(f*x+e)^2)^(1/2)*(a+b*sin
(f*x+e)^2)^(1/2)/b^2/f/(1+b*sin(f*x+e)^2/a)^(1/2)+2/15*a*(a-2*b)*(a+b)*EllipticF(sin(f*x+e),(-b/a)^(1/2))*sec(
f*x+e)*(cos(f*x+e)^2)^(1/2)*(1+b*sin(f*x+e)^2/a)^(1/2)/b^2/f/(a+b*sin(f*x+e)^2)^(1/2)

Rubi [A] (verified)

Time = 0.22 (sec) , antiderivative size = 259, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {3267, 489, 596, 538, 437, 435, 432, 430} \[ \int \sin ^4(e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx=-\frac {\left (2 a^2-3 a b-8 b^2\right ) \sqrt {\cos ^2(e+f x)} \sec (e+f x) \sqrt {a+b \sin ^2(e+f x)} E\left (\arcsin (\sin (e+f x))\left |-\frac {b}{a}\right .\right )}{15 b^2 f \sqrt {\frac {b \sin ^2(e+f x)}{a}+1}}+\frac {2 a (a-2 b) (a+b) \sqrt {\cos ^2(e+f x)} \sec (e+f x) \sqrt {\frac {b \sin ^2(e+f x)}{a}+1} \operatorname {EllipticF}\left (\arcsin (\sin (e+f x)),-\frac {b}{a}\right )}{15 b^2 f \sqrt {a+b \sin ^2(e+f x)}}-\frac {(a+4 b) \sin (e+f x) \cos (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{15 b f}-\frac {\sin ^3(e+f x) \cos (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{5 f} \]

[In]

Int[Sin[e + f*x]^4*Sqrt[a + b*Sin[e + f*x]^2],x]

[Out]

-1/15*((a + 4*b)*Cos[e + f*x]*Sin[e + f*x]*Sqrt[a + b*Sin[e + f*x]^2])/(b*f) - (Cos[e + f*x]*Sin[e + f*x]^3*Sq
rt[a + b*Sin[e + f*x]^2])/(5*f) - ((2*a^2 - 3*a*b - 8*b^2)*Sqrt[Cos[e + f*x]^2]*EllipticE[ArcSin[Sin[e + f*x]]
, -(b/a)]*Sec[e + f*x]*Sqrt[a + b*Sin[e + f*x]^2])/(15*b^2*f*Sqrt[1 + (b*Sin[e + f*x]^2)/a]) + (2*a*(a - 2*b)*
(a + b)*Sqrt[Cos[e + f*x]^2]*EllipticF[ArcSin[Sin[e + f*x]], -(b/a)]*Sec[e + f*x]*Sqrt[1 + (b*Sin[e + f*x]^2)/
a])/(15*b^2*f*Sqrt[a + b*Sin[e + f*x]^2])

Rule 430

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Simp[(1/(Sqrt[a]*Sqrt[c]*Rt[-d/c, 2]
))*EllipticF[ArcSin[Rt[-d/c, 2]*x], b*(c/(a*d))], x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] && Gt
Q[a, 0] &&  !(NegQ[b/a] && SimplerSqrtQ[-b/a, -d/c])

Rule 432

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Dist[Sqrt[1 + (d/c)*x^2]/Sqrt[c + d*
x^2], Int[1/(Sqrt[a + b*x^2]*Sqrt[1 + (d/c)*x^2]), x], x] /; FreeQ[{a, b, c, d}, x] &&  !GtQ[c, 0]

Rule 435

Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Simp[(Sqrt[a]/(Sqrt[c]*Rt[-d/c, 2]))*Ell
ipticE[ArcSin[Rt[-d/c, 2]*x], b*(c/(a*d))], x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] && GtQ[a, 0
]

Rule 437

Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Dist[Sqrt[a + b*x^2]/Sqrt[1 + (b/a)*x^2]
, Int[Sqrt[1 + (b/a)*x^2]/Sqrt[c + d*x^2], x], x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] &&  !GtQ
[a, 0]

Rule 489

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

Rule 538

Int[((e_) + (f_.)*(x_)^(n_))/(Sqrt[(a_) + (b_.)*(x_)^(n_)]*Sqrt[(c_) + (d_.)*(x_)^(n_)]), x_Symbol] :> Dist[f/
b, Int[Sqrt[a + b*x^n]/Sqrt[c + d*x^n], x], x] + Dist[(b*e - a*f)/b, Int[1/(Sqrt[a + b*x^n]*Sqrt[c + d*x^n]),
x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] &&  !(EqQ[n, 2] && ((PosQ[b/a] && PosQ[d/c]) || (NegQ[b/a] && (PosQ[
d/c] || (GtQ[a, 0] && ( !GtQ[c, 0] || SimplerSqrtQ[-b/a, -d/c]))))))

Rule 596

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

Rule 3267

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

Rubi steps \begin{align*} \text {integral}& = \frac {\left (\sqrt {\cos ^2(e+f x)} \sec (e+f x)\right ) \text {Subst}\left (\int \frac {x^4 \sqrt {a+b x^2}}{\sqrt {1-x^2}} \, dx,x,\sin (e+f x)\right )}{f} \\ & = -\frac {\cos (e+f x) \sin ^3(e+f x) \sqrt {a+b \sin ^2(e+f x)}}{5 f}+\frac {\left (\sqrt {\cos ^2(e+f x)} \sec (e+f x)\right ) \text {Subst}\left (\int \frac {x^2 \left (3 a+(a+4 b) x^2\right )}{\sqrt {1-x^2} \sqrt {a+b x^2}} \, dx,x,\sin (e+f x)\right )}{5 f} \\ & = -\frac {(a+4 b) \cos (e+f x) \sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{15 b f}-\frac {\cos (e+f x) \sin ^3(e+f x) \sqrt {a+b \sin ^2(e+f x)}}{5 f}+\frac {\left (\sqrt {\cos ^2(e+f x)} \sec (e+f x)\right ) \text {Subst}\left (\int \frac {a (a+4 b)+\left (-2 a^2+3 a b+8 b^2\right ) x^2}{\sqrt {1-x^2} \sqrt {a+b x^2}} \, dx,x,\sin (e+f x)\right )}{15 b f} \\ & = -\frac {(a+4 b) \cos (e+f x) \sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{15 b f}-\frac {\cos (e+f x) \sin ^3(e+f x) \sqrt {a+b \sin ^2(e+f x)}}{5 f}+\frac {\left (2 a (a-2 b) (a+b) \sqrt {\cos ^2(e+f x)} \sec (e+f x)\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1-x^2} \sqrt {a+b x^2}} \, dx,x,\sin (e+f x)\right )}{15 b^2 f}+\frac {\left (\left (-2 a^2+3 a b+8 b^2\right ) \sqrt {\cos ^2(e+f x)} \sec (e+f x)\right ) \text {Subst}\left (\int \frac {\sqrt {a+b x^2}}{\sqrt {1-x^2}} \, dx,x,\sin (e+f x)\right )}{15 b^2 f} \\ & = -\frac {(a+4 b) \cos (e+f x) \sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{15 b f}-\frac {\cos (e+f x) \sin ^3(e+f x) \sqrt {a+b \sin ^2(e+f x)}}{5 f}+\frac {\left (\left (-2 a^2+3 a b+8 b^2\right ) \sqrt {\cos ^2(e+f x)} \sec (e+f x) \sqrt {a+b \sin ^2(e+f x)}\right ) \text {Subst}\left (\int \frac {\sqrt {1+\frac {b x^2}{a}}}{\sqrt {1-x^2}} \, dx,x,\sin (e+f x)\right )}{15 b^2 f \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}}+\frac {\left (2 a (a-2 b) (a+b) \sqrt {\cos ^2(e+f x)} \sec (e+f x) \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1-x^2} \sqrt {1+\frac {b x^2}{a}}} \, dx,x,\sin (e+f x)\right )}{15 b^2 f \sqrt {a+b \sin ^2(e+f x)}} \\ & = -\frac {(a+4 b) \cos (e+f x) \sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{15 b f}-\frac {\cos (e+f x) \sin ^3(e+f x) \sqrt {a+b \sin ^2(e+f x)}}{5 f}-\frac {\left (2 a^2-3 a b-8 b^2\right ) \sqrt {\cos ^2(e+f x)} E\left (\arcsin (\sin (e+f x))\left |-\frac {b}{a}\right .\right ) \sec (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{15 b^2 f \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}}+\frac {2 a (a-2 b) (a+b) \sqrt {\cos ^2(e+f x)} \operatorname {EllipticF}\left (\arcsin (\sin (e+f x)),-\frac {b}{a}\right ) \sec (e+f x) \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}}{15 b^2 f \sqrt {a+b \sin ^2(e+f x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.08 (sec) , antiderivative size = 199, normalized size of antiderivative = 0.77 \[ \int \sin ^4(e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx=\frac {-16 a \left (2 a^2-3 a b-8 b^2\right ) \sqrt {\frac {2 a+b-b \cos (2 (e+f x))}{a}} E\left (e+f x\left |-\frac {b}{a}\right .\right )+32 a \left (a^2-a b-2 b^2\right ) \sqrt {\frac {2 a+b-b \cos (2 (e+f x))}{a}} \operatorname {EllipticF}\left (e+f x,-\frac {b}{a}\right )-\sqrt {2} b \left (8 a^2+48 a b+25 b^2-4 b (4 a+7 b) \cos (2 (e+f x))+3 b^2 \cos (4 (e+f x))\right ) \sin (2 (e+f x))}{240 b^2 f \sqrt {2 a+b-b \cos (2 (e+f x))}} \]

[In]

Integrate[Sin[e + f*x]^4*Sqrt[a + b*Sin[e + f*x]^2],x]

[Out]

(-16*a*(2*a^2 - 3*a*b - 8*b^2)*Sqrt[(2*a + b - b*Cos[2*(e + f*x)])/a]*EllipticE[e + f*x, -(b/a)] + 32*a*(a^2 -
a*b - 2*b^2)*Sqrt[(2*a + b - b*Cos[2*(e + f*x)])/a]*EllipticF[e + f*x, -(b/a)] - Sqrt[2]*b*(8*a^2 + 48*a*b +
25*b^2 - 4*b*(4*a + 7*b)*Cos[2*(e + f*x)] + 3*b^2*Cos[4*(e + f*x)])*Sin[2*(e + f*x)])/(240*b^2*f*Sqrt[2*a + b
- b*Cos[2*(e + f*x)]])

Maple [A] (verified)

Time = 3.37 (sec) , antiderivative size = 413, normalized size of antiderivative = 1.59

method	result	size
default	\(\frac {3 b^{3} \left (\sin ^{7}\left (f x +e \right )\right )+4 a \,b^{2} \left (\sin ^{5}\left (f x +e \right )\right )+b^{3} \left (\sin ^{5}\left (f x +e \right )\right )+2 \sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {\frac {a +b \left (\sin ^{2}\left (f x +e \right )\right )}{a}}\, F\left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) a^{3}-2 a^{2} \sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {\frac {a +b \left (\sin ^{2}\left (f x +e \right )\right )}{a}}\, F\left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) b -4 a \sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {\frac {a +b \left (\sin ^{2}\left (f x +e \right )\right )}{a}}\, F\left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) b^{2}-2 \sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {\frac {a +b \left (\sin ^{2}\left (f x +e \right )\right )}{a}}\, E\left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) a^{3}+3 \sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {\frac {a +b \left (\sin ^{2}\left (f x +e \right )\right )}{a}}\, E\left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) a^{2} b +8 \sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {\frac {a +b \left (\sin ^{2}\left (f x +e \right )\right )}{a}}\, E\left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) a \,b^{2}+a^{2} b \left (\sin ^{3}\left (f x +e \right )\right )-4 b^{3} \left (\sin ^{3}\left (f x +e \right )\right )-a^{2} b \sin \left (f x +e \right )-4 a \,b^{2} \sin \left (f x +e \right )}{15 b^{2} \cos \left (f x +e \right ) \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}\, f}\)	\(413\)

[In]

int(sin(f*x+e)^4*(a+b*sin(f*x+e)^2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/15*(3*b^3*sin(f*x+e)^7+4*a*b^2*sin(f*x+e)^5+b^3*sin(f*x+e)^5+2*(cos(f*x+e)^2)^(1/2)*((a+b*sin(f*x+e)^2)/a)^(
1/2)*EllipticF(sin(f*x+e),(-1/a*b)^(1/2))*a^3-2*a^2*(cos(f*x+e)^2)^(1/2)*((a+b*sin(f*x+e)^2)/a)^(1/2)*Elliptic
F(sin(f*x+e),(-1/a*b)^(1/2))*b-4*a*(cos(f*x+e)^2)^(1/2)*((a+b*sin(f*x+e)^2)/a)^(1/2)*EllipticF(sin(f*x+e),(-1/
a*b)^(1/2))*b^2-2*(cos(f*x+e)^2)^(1/2)*((a+b*sin(f*x+e)^2)/a)^(1/2)*EllipticE(sin(f*x+e),(-1/a*b)^(1/2))*a^3+3
*(cos(f*x+e)^2)^(1/2)*((a+b*sin(f*x+e)^2)/a)^(1/2)*EllipticE(sin(f*x+e),(-1/a*b)^(1/2))*a^2*b+8*(cos(f*x+e)^2)
^(1/2)*((a+b*sin(f*x+e)^2)/a)^(1/2)*EllipticE(sin(f*x+e),(-1/a*b)^(1/2))*a*b^2+a^2*b*sin(f*x+e)^3-4*b^3*sin(f*
x+e)^3-a^2*b*sin(f*x+e)-4*a*b^2*sin(f*x+e))/b^2/cos(f*x+e)/(a+b*sin(f*x+e)^2)^(1/2)/f

Fricas [F]

\[ \int \sin ^4(e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx=\int { \sqrt {b \sin \left (f x + e\right )^{2} + a} \sin \left (f x + e\right )^{4} \,d x } \]

[In]

integrate(sin(f*x+e)^4*(a+b*sin(f*x+e)^2)^(1/2),x, algorithm="fricas")

[Out]

integral((cos(f*x + e)^4 - 2*cos(f*x + e)^2 + 1)*sqrt(-b*cos(f*x + e)^2 + a + b), x)

Sympy [F(-1)]

Timed out. \[ \int \sin ^4(e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx=\text {Timed out} \]

[In]

integrate(sin(f*x+e)**4*(a+b*sin(f*x+e)**2)**(1/2),x)

[Out]

Timed out

Maxima [F]

\[ \int \sin ^4(e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx=\int { \sqrt {b \sin \left (f x + e\right )^{2} + a} \sin \left (f x + e\right )^{4} \,d x } \]

[In]

integrate(sin(f*x+e)^4*(a+b*sin(f*x+e)^2)^(1/2),x, algorithm="maxima")

[Out]

integrate(sqrt(b*sin(f*x + e)^2 + a)*sin(f*x + e)^4, x)

Giac [F]

\[ \int \sin ^4(e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx=\int { \sqrt {b \sin \left (f x + e\right )^{2} + a} \sin \left (f x + e\right )^{4} \,d x } \]

[In]

integrate(sin(f*x+e)^4*(a+b*sin(f*x+e)^2)^(1/2),x, algorithm="giac")

[Out]

integrate(sqrt(b*sin(f*x + e)^2 + a)*sin(f*x + e)^4, x)

Mupad [F(-1)]

Timed out. \[ \int \sin ^4(e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx=\int {\sin \left (e+f\,x\right )}^4\,\sqrt {b\,{\sin \left (e+f\,x\right )}^2+a} \,d x \]

[In]

int(sin(e + f*x)^4*(a + b*sin(e + f*x)^2)^(1/2),x)

[Out]

int(sin(e + f*x)^4*(a + b*sin(e + f*x)^2)^(1/2), x)