Integrand size = 25, antiderivative size = 204 \[ \int \frac {\cos ^4(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx=\frac {(a+b) \cos (e+f x) \sin (e+f x)}{a b f \sqrt {a+b \sin ^2(e+f x)}}+\frac {(2 a+b) \sqrt {\cos ^2(e+f x)} E\left (\arcsin (\sin (e+f x))\left |-\frac {b}{a}\right .\right ) \sec (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{a b^2 f \sqrt {\frac {a+b \sin ^2(e+f x)}{a}}}-\frac {2 (a+b) \sqrt {\cos ^2(e+f x)} \operatorname {EllipticF}\left (\arcsin (\sin (e+f x)),-\frac {b}{a}\right ) \sec (e+f x) \sqrt {\frac {a+b \sin ^2(e+f x)}{a}}}{b^2 f \sqrt {a+b \sin ^2(e+f x)}} \] Output:
(a+b)*cos(f*x+e)*sin(f*x+e)/a/b/f/(a+b*sin(f*x+e)^2)^(1/2)+(2*a+b)*(cos(f* x+e)^2)^(1/2)*EllipticE(sin(f*x+e),(-b/a)^(1/2))*sec(f*x+e)*(a+b*sin(f*x+e )^2)^(1/2)/a/b^2/f/((a+b*sin(f*x+e)^2)/a)^(1/2)-2*(a+b)*(cos(f*x+e)^2)^(1/ 2)*EllipticF(sin(f*x+e),(-b/a)^(1/2))*sec(f*x+e)*((a+b*sin(f*x+e)^2)/a)^(1 /2)/b^2/f/(a+b*sin(f*x+e)^2)^(1/2)
Time = 1.01 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.68 \[ \int \frac {\cos ^4(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx=\frac {2 a (2 a+b) \sqrt {\frac {2 a+b-b \cos (2 (e+f x))}{a}} E\left (e+f x\left |-\frac {b}{a}\right .\right )-(a+b) \left (4 a \sqrt {\frac {2 a+b-b \cos (2 (e+f x))}{a}} \operatorname {EllipticF}\left (e+f x,-\frac {b}{a}\right )-\sqrt {2} b \sin (2 (e+f x))\right )}{2 a b^2 f \sqrt {2 a+b-b \cos (2 (e+f x))}} \] Input:
Integrate[Cos[e + f*x]^4/(a + b*Sin[e + f*x]^2)^(3/2),x]
Output:
(2*a*(2*a + b)*Sqrt[(2*a + b - b*Cos[2*(e + f*x)])/a]*EllipticE[e + f*x, - (b/a)] - (a + b)*(4*a*Sqrt[(2*a + b - b*Cos[2*(e + f*x)])/a]*EllipticF[e + f*x, -(b/a)] - Sqrt[2]*b*Sin[2*(e + f*x)]))/(2*a*b^2*f*Sqrt[2*a + b - b*C os[2*(e + f*x)]])
Time = 0.40 (sec) , antiderivative size = 197, normalized size of antiderivative = 0.97, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {3042, 3671, 315, 25, 399, 323, 321, 330, 327}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {\cos ^4(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos (e+f x)^4}{\left (a+b \sin (e+f x)^2\right )^{3/2}}dx\) |
\(\Big \downarrow \) 3671 |
\(\displaystyle \frac {\sqrt {\cos ^2(e+f x)} \sec (e+f x) \int \frac {\left (1-\sin ^2(e+f x)\right )^{3/2}}{\left (b \sin ^2(e+f x)+a\right )^{3/2}}d\sin (e+f x)}{f}\) |
\(\Big \downarrow \) 315 |
\(\displaystyle \frac {\sqrt {\cos ^2(e+f x)} \sec (e+f x) \left (\frac {\int -\frac {a-(2 a+b) \sin ^2(e+f x)}{\sqrt {1-\sin ^2(e+f x)} \sqrt {b \sin ^2(e+f x)+a}}d\sin (e+f x)}{a b}+\frac {(a+b) \sqrt {1-\sin ^2(e+f x)} \sin (e+f x)}{a b \sqrt {a+b \sin ^2(e+f x)}}\right )}{f}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\sqrt {\cos ^2(e+f x)} \sec (e+f x) \left (\frac {(a+b) \sin (e+f x) \sqrt {1-\sin ^2(e+f x)}}{a b \sqrt {a+b \sin ^2(e+f x)}}-\frac {\int \frac {a-(2 a+b) \sin ^2(e+f x)}{\sqrt {1-\sin ^2(e+f x)} \sqrt {b \sin ^2(e+f x)+a}}d\sin (e+f x)}{a b}\right )}{f}\) |
\(\Big \downarrow \) 399 |
\(\displaystyle \frac {\sqrt {\cos ^2(e+f x)} \sec (e+f x) \left (\frac {(a+b) \sin (e+f x) \sqrt {1-\sin ^2(e+f x)}}{a b \sqrt {a+b \sin ^2(e+f x)}}-\frac {\frac {2 a (a+b) \int \frac {1}{\sqrt {1-\sin ^2(e+f x)} \sqrt {b \sin ^2(e+f x)+a}}d\sin (e+f x)}{b}-\frac {(2 a+b) \int \frac {\sqrt {b \sin ^2(e+f x)+a}}{\sqrt {1-\sin ^2(e+f x)}}d\sin (e+f x)}{b}}{a b}\right )}{f}\) |
\(\Big \downarrow \) 323 |
\(\displaystyle \frac {\sqrt {\cos ^2(e+f x)} \sec (e+f x) \left (\frac {(a+b) \sin (e+f x) \sqrt {1-\sin ^2(e+f x)}}{a b \sqrt {a+b \sin ^2(e+f x)}}-\frac {\frac {2 a (a+b) \sqrt {\frac {b \sin ^2(e+f x)}{a}+1} \int \frac {1}{\sqrt {1-\sin ^2(e+f x)} \sqrt {\frac {b \sin ^2(e+f x)}{a}+1}}d\sin (e+f x)}{b \sqrt {a+b \sin ^2(e+f x)}}-\frac {(2 a+b) \int \frac {\sqrt {b \sin ^2(e+f x)+a}}{\sqrt {1-\sin ^2(e+f x)}}d\sin (e+f x)}{b}}{a b}\right )}{f}\) |
\(\Big \downarrow \) 321 |
\(\displaystyle \frac {\sqrt {\cos ^2(e+f x)} \sec (e+f x) \left (\frac {(a+b) \sin (e+f x) \sqrt {1-\sin ^2(e+f x)}}{a b \sqrt {a+b \sin ^2(e+f x)}}-\frac {\frac {2 a (a+b) \sqrt {\frac {b \sin ^2(e+f x)}{a}+1} \operatorname {EllipticF}\left (\arcsin (\sin (e+f x)),-\frac {b}{a}\right )}{b \sqrt {a+b \sin ^2(e+f x)}}-\frac {(2 a+b) \int \frac {\sqrt {b \sin ^2(e+f x)+a}}{\sqrt {1-\sin ^2(e+f x)}}d\sin (e+f x)}{b}}{a b}\right )}{f}\) |
\(\Big \downarrow \) 330 |
\(\displaystyle \frac {\sqrt {\cos ^2(e+f x)} \sec (e+f x) \left (\frac {(a+b) \sin (e+f x) \sqrt {1-\sin ^2(e+f x)}}{a b \sqrt {a+b \sin ^2(e+f x)}}-\frac {\frac {2 a (a+b) \sqrt {\frac {b \sin ^2(e+f x)}{a}+1} \operatorname {EllipticF}\left (\arcsin (\sin (e+f x)),-\frac {b}{a}\right )}{b \sqrt {a+b \sin ^2(e+f x)}}-\frac {(2 a+b) \sqrt {a+b \sin ^2(e+f x)} \int \frac {\sqrt {\frac {b \sin ^2(e+f x)}{a}+1}}{\sqrt {1-\sin ^2(e+f x)}}d\sin (e+f x)}{b \sqrt {\frac {b \sin ^2(e+f x)}{a}+1}}}{a b}\right )}{f}\) |
\(\Big \downarrow \) 327 |
\(\displaystyle \frac {\sqrt {\cos ^2(e+f x)} \sec (e+f x) \left (\frac {(a+b) \sin (e+f x) \sqrt {1-\sin ^2(e+f x)}}{a b \sqrt {a+b \sin ^2(e+f x)}}-\frac {\frac {2 a (a+b) \sqrt {\frac {b \sin ^2(e+f x)}{a}+1} \operatorname {EllipticF}\left (\arcsin (\sin (e+f x)),-\frac {b}{a}\right )}{b \sqrt {a+b \sin ^2(e+f x)}}-\frac {(2 a+b) \sqrt {a+b \sin ^2(e+f x)} E\left (\arcsin (\sin (e+f x))\left |-\frac {b}{a}\right .\right )}{b \sqrt {\frac {b \sin ^2(e+f x)}{a}+1}}}{a b}\right )}{f}\) |
Input:
Int[Cos[e + f*x]^4/(a + b*Sin[e + f*x]^2)^(3/2),x]
Output:
(Sqrt[Cos[e + f*x]^2]*Sec[e + f*x]*(((a + b)*Sin[e + f*x]*Sqrt[1 - Sin[e + f*x]^2])/(a*b*Sqrt[a + b*Sin[e + f*x]^2]) - (-(((2*a + b)*EllipticE[ArcSi n[Sin[e + f*x]], -(b/a)]*Sqrt[a + b*Sin[e + f*x]^2])/(b*Sqrt[1 + (b*Sin[e + f*x]^2)/a])) + (2*a*(a + b)*EllipticF[ArcSin[Sin[e + f*x]], -(b/a)]*Sqrt [1 + (b*Sin[e + f*x]^2)/a])/(b*Sqrt[a + b*Sin[e + f*x]^2]))/(a*b)))/f
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim p[(a*d - c*b)*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q - 1)/(2*a*b*(p + 1))), x] - Simp[1/(2*a*b*(p + 1)) Int[(a + b*x^2)^(p + 1)*(c + d*x^2)^(q - 2)*S imp[c*(a*d - c*b*(2*p + 3)) + d*(a*d*(2*(q - 1) + 1) - b*c*(2*(p + q) + 1)) *x^2, x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, - 1] && GtQ[q, 1] && IntBinomialQ[a, b, c, d, 2, p, q, x]
Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> S imp[(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] && GtQ[a, 0] && !(NegQ[b/a] && SimplerSqrtQ[-b/a, -d/c])
Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> S imp[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]
Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Simp[ (Sqrt[a]/(Sqrt[c]*Rt[-d/c, 2]))*EllipticE[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]
Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Simp[ 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]
Int[((e_) + (f_.)*(x_)^2)/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_) ^2]), x_Symbol] :> Simp[f/b Int[Sqrt[a + b*x^2]/Sqrt[c + d*x^2], x], x] + Simp[(b*e - a*f)/b Int[1/(Sqrt[a + b*x^2]*Sqrt[c + d*x^2]), x], x] /; Fr eeQ[{a, b, c, d, e, f}, x] && !((PosQ[b/a] && PosQ[d/c]) || (NegQ[b/a] && (PosQ[d/c] || (GtQ[a, 0] && ( !GtQ[c, 0] || SimplerSqrtQ[-b/a, -d/c])))))
Int[cos[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^( p_.), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Simp[ff*(Sqrt[ Cos[e + f*x]^2]/(f*Cos[e + f*x])) Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*(a + b*ff^2*x^2)^p, x], x, Sin[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[m/2] && !IntegerQ[p]
Time = 3.72 (sec) , antiderivative size = 286, normalized size of antiderivative = 1.40
method | result | size |
default | \(-\frac {2 \sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {-\frac {b \cos \left (f x +e \right )^{2}}{a}+\frac {a +b}{a}}\, \operatorname {EllipticF}\left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) a^{2}+2 a b \sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {-\frac {b \cos \left (f x +e \right )^{2}}{a}+\frac {a +b}{a}}\, \operatorname {EllipticF}\left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right )-2 \sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {-\frac {b \cos \left (f x +e \right )^{2}}{a}+\frac {a +b}{a}}\, \operatorname {EllipticE}\left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) a^{2}-\sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {-\frac {b \cos \left (f x +e \right )^{2}}{a}+\frac {a +b}{a}}\, \operatorname {EllipticE}\left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) a b -b \cos \left (f x +e \right )^{2} \sin \left (f x +e \right ) a -b^{2} \cos \left (f x +e \right )^{2} \sin \left (f x +e \right )}{b^{2} a \cos \left (f x +e \right ) \sqrt {a +b \sin \left (f x +e \right )^{2}}\, f}\) | \(286\) |
Input:
int(cos(f*x+e)^4/(a+b*sin(f*x+e)^2)^(3/2),x,method=_RETURNVERBOSE)
Output:
-(2*(cos(f*x+e)^2)^(1/2)*(-b/a*cos(f*x+e)^2+(a+b)/a)^(1/2)*EllipticF(sin(f *x+e),(-b/a)^(1/2))*a^2+2*a*b*(cos(f*x+e)^2)^(1/2)*(-b/a*cos(f*x+e)^2+(a+b )/a)^(1/2)*EllipticF(sin(f*x+e),(-b/a)^(1/2))-2*(cos(f*x+e)^2)^(1/2)*(-b/a *cos(f*x+e)^2+(a+b)/a)^(1/2)*EllipticE(sin(f*x+e),(-b/a)^(1/2))*a^2-(cos(f *x+e)^2)^(1/2)*(-b/a*cos(f*x+e)^2+(a+b)/a)^(1/2)*EllipticE(sin(f*x+e),(-b/ a)^(1/2))*a*b-b*cos(f*x+e)^2*sin(f*x+e)*a-b^2*cos(f*x+e)^2*sin(f*x+e))/b^2 /a/cos(f*x+e)/(a+b*sin(f*x+e)^2)^(1/2)/f
\[ \int \frac {\cos ^4(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx=\int { \frac {\cos \left (f x + e\right )^{4}}{{\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}}} \,d x } \] Input:
integrate(cos(f*x+e)^4/(a+b*sin(f*x+e)^2)^(3/2),x, algorithm="fricas")
Output:
integral(sqrt(-b*cos(f*x + e)^2 + a + b)*cos(f*x + e)^4/(b^2*cos(f*x + e)^ 4 - 2*(a*b + b^2)*cos(f*x + e)^2 + a^2 + 2*a*b + b^2), x)
Timed out. \[ \int \frac {\cos ^4(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx=\text {Timed out} \] Input:
integrate(cos(f*x+e)**4/(a+b*sin(f*x+e)**2)**(3/2),x)
Output:
Timed out
\[ \int \frac {\cos ^4(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx=\int { \frac {\cos \left (f x + e\right )^{4}}{{\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}}} \,d x } \] Input:
integrate(cos(f*x+e)^4/(a+b*sin(f*x+e)^2)^(3/2),x, algorithm="maxima")
Output:
integrate(cos(f*x + e)^4/(b*sin(f*x + e)^2 + a)^(3/2), x)
\[ \int \frac {\cos ^4(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx=\int { \frac {\cos \left (f x + e\right )^{4}}{{\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}}} \,d x } \] Input:
integrate(cos(f*x+e)^4/(a+b*sin(f*x+e)^2)^(3/2),x, algorithm="giac")
Output:
integrate(cos(f*x + e)^4/(b*sin(f*x + e)^2 + a)^(3/2), x)
Timed out. \[ \int \frac {\cos ^4(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx=\int \frac {{\cos \left (e+f\,x\right )}^4}{{\left (b\,{\sin \left (e+f\,x\right )}^2+a\right )}^{3/2}} \,d x \] Input:
int(cos(e + f*x)^4/(a + b*sin(e + f*x)^2)^(3/2),x)
Output:
int(cos(e + f*x)^4/(a + b*sin(e + f*x)^2)^(3/2), x)
\[ \int \frac {\cos ^4(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx=\int \frac {\sqrt {\sin \left (f x +e \right )^{2} b +a}\, \cos \left (f x +e \right )^{4}}{\sin \left (f x +e \right )^{4} b^{2}+2 \sin \left (f x +e \right )^{2} a b +a^{2}}d x \] Input:
int(cos(f*x+e)^4/(a+b*sin(f*x+e)^2)^(3/2),x)
Output:
int((sqrt(sin(e + f*x)**2*b + a)*cos(e + f*x)**4)/(sin(e + f*x)**4*b**2 + 2*sin(e + f*x)**2*a*b + a**2),x)