Integrand size = 25, antiderivative size = 225 \[ \int \frac {\cos ^4(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx=\frac {(a+b) \cos (e+f x) \sin (e+f x)}{3 a b f \left (a+b \sin ^2(e+f x)\right )^{3/2}}-\frac {2 (a-b) \cos (e+f x) \sin (e+f x)}{3 a^2 b f \sqrt {a+b \sin ^2(e+f x)}}-\frac {2 (a-b) E\left (e+f x\left |-\frac {b}{a}\right .\right ) \sqrt {\frac {a+b \sin ^2(e+f x)}{a}}}{3 a b^2 f \sqrt {a+b \sin ^2(e+f x)}}+\frac {(2 a-b) \operatorname {EllipticF}\left (e+f x,-\frac {b}{a}\right ) \sqrt {\frac {a+b \sin ^2(e+f x)}{a}}}{3 a b^2 f \sqrt {a+b \sin ^2(e+f x)}} \] Output:
1/3*(a+b)*cos(f*x+e)*sin(f*x+e)/a/b/f/(a+b*sin(f*x+e)^2)^(3/2)-2/3*(a-b)*c os(f*x+e)*sin(f*x+e)/a^2/b/f/(a+b*sin(f*x+e)^2)^(1/2)-2/3*(a-b)*EllipticE( sin(f*x+e),(-b/a)^(1/2))*((a+b*sin(f*x+e)^2)/a)^(1/2)/a/b^2/f/(a+b*sin(f*x +e)^2)^(1/2)+1/3*(2*a-b)*InverseJacobiAM(f*x+e,(-b/a)^(1/2))*((a+b*sin(f*x +e)^2)/a)^(1/2)/a/b^2/f/(a+b*sin(f*x+e)^2)^(1/2)
Time = 2.04 (sec) , antiderivative size = 171, normalized size of antiderivative = 0.76 \[ \int \frac {\cos ^4(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx=\frac {-2 a^2 (a-b) \left (\frac {2 a+b-b \cos (2 (e+f x))}{a}\right )^{3/2} E\left (e+f x\left |-\frac {b}{a}\right .\right )+a^2 (2 a-b) \left (\frac {2 a+b-b \cos (2 (e+f x))}{a}\right )^{3/2} \operatorname {EllipticF}\left (e+f x,-\frac {b}{a}\right )-\sqrt {2} b \left (a^2-2 a b-b^2+b (-a+b) \cos (2 (e+f x))\right ) \sin (2 (e+f x))}{3 a^2 b^2 f (2 a+b-b \cos (2 (e+f x)))^{3/2}} \] Input:
Integrate[Cos[e + f*x]^4/(a + b*Sin[e + f*x]^2)^(5/2),x]
Output:
(-2*a^2*(a - b)*((2*a + b - b*Cos[2*(e + f*x)])/a)^(3/2)*EllipticE[e + f*x , -(b/a)] + a^2*(2*a - b)*((2*a + b - b*Cos[2*(e + f*x)])/a)^(3/2)*Ellipti cF[e + f*x, -(b/a)] - Sqrt[2]*b*(a^2 - 2*a*b - b^2 + b*(-a + b)*Cos[2*(e + f*x)])*Sin[2*(e + f*x)])/(3*a^2*b^2*f*(2*a + b - b*Cos[2*(e + f*x)])^(3/2 ))
Time = 0.48 (sec) , antiderivative size = 271, normalized size of antiderivative = 1.20, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 3671, 315, 25, 402, 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.
\(\displaystyle \int \frac {\cos ^4(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos (e+f x)^4}{\left (a+b \sin (e+f x)^2\right )^{5/2}}dx\) |
\(\Big \downarrow \) 3671 |
\(\displaystyle \frac {\sqrt {\cos ^2(e+f x)} \sec (e+f x) \int \frac {\left (1-\sin ^2(e+f x)\right )^{3/2}}{\left (b \sin ^2(e+f x)+a\right )^{5/2}}d\sin (e+f x)}{f}\) |
\(\Big \downarrow \) 315 |
\(\displaystyle \frac {\sqrt {\cos ^2(e+f x)} \sec (e+f x) \left (\frac {\int -\frac {-\left ((2 a-b) \sin ^2(e+f x)\right )+a-2 b}{\sqrt {1-\sin ^2(e+f x)} \left (b \sin ^2(e+f x)+a\right )^{3/2}}d\sin (e+f x)}{3 a b}+\frac {(a+b) \sqrt {1-\sin ^2(e+f x)} \sin (e+f x)}{3 a b \left (a+b \sin ^2(e+f x)\right )^{3/2}}\right )}{f}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\sqrt {\cos ^2(e+f x)} \sec (e+f x) \left (\frac {(a+b) \sin (e+f x) \sqrt {1-\sin ^2(e+f x)}}{3 a b \left (a+b \sin ^2(e+f x)\right )^{3/2}}-\frac {\int \frac {-\left ((2 a-b) \sin ^2(e+f x)\right )+a-2 b}{\sqrt {1-\sin ^2(e+f x)} \left (b \sin ^2(e+f x)+a\right )^{3/2}}d\sin (e+f x)}{3 a b}\right )}{f}\) |
\(\Big \downarrow \) 402 |
\(\displaystyle \frac {\sqrt {\cos ^2(e+f x)} \sec (e+f x) \left (\frac {(a+b) \sin (e+f x) \sqrt {1-\sin ^2(e+f x)}}{3 a b \left (a+b \sin ^2(e+f x)\right )^{3/2}}-\frac {\frac {2 (a-b) \sin (e+f x) \sqrt {1-\sin ^2(e+f x)}}{a \sqrt {a+b \sin ^2(e+f x)}}-\frac {\int \frac {a (a+b)-2 \left (a^2-b^2\right ) \sin ^2(e+f x)}{\sqrt {1-\sin ^2(e+f x)} \sqrt {b \sin ^2(e+f x)+a}}d\sin (e+f x)}{a (a+b)}}{3 a b}\right )}{f}\) |
\(\Big \downarrow \) 399 |
\(\displaystyle \frac {\sqrt {\cos ^2(e+f x)} \sec (e+f x) \left (\frac {(a+b) \sin (e+f x) \sqrt {1-\sin ^2(e+f x)}}{3 a b \left (a+b \sin ^2(e+f x)\right )^{3/2}}-\frac {\frac {2 (a-b) \sin (e+f x) \sqrt {1-\sin ^2(e+f x)}}{a \sqrt {a+b \sin ^2(e+f x)}}-\frac {\frac {a (2 a-b) (a+b) \int \frac {1}{\sqrt {1-\sin ^2(e+f x)} \sqrt {b \sin ^2(e+f x)+a}}d\sin (e+f x)}{b}-\frac {2 \left (a^2-b^2\right ) \int \frac {\sqrt {b \sin ^2(e+f x)+a}}{\sqrt {1-\sin ^2(e+f x)}}d\sin (e+f x)}{b}}{a (a+b)}}{3 a b}\right )}{f}\) |
\(\Big \downarrow \) 323 |
\(\displaystyle \frac {\sqrt {\cos ^2(e+f x)} \sec (e+f x) \left (\frac {(a+b) \sin (e+f x) \sqrt {1-\sin ^2(e+f x)}}{3 a b \left (a+b \sin ^2(e+f x)\right )^{3/2}}-\frac {\frac {2 (a-b) \sin (e+f x) \sqrt {1-\sin ^2(e+f x)}}{a \sqrt {a+b \sin ^2(e+f x)}}-\frac {\frac {a (2 a-b) (a+b) \sqrt {\frac {b \sin ^2(e+f x)}{a}+1} \int \frac {1}{\sqrt {1-\sin ^2(e+f x)} \sqrt {\frac {b \sin ^2(e+f x)}{a}+1}}d\sin (e+f x)}{b \sqrt {a+b \sin ^2(e+f x)}}-\frac {2 \left (a^2-b^2\right ) \int \frac {\sqrt {b \sin ^2(e+f x)+a}}{\sqrt {1-\sin ^2(e+f x)}}d\sin (e+f x)}{b}}{a (a+b)}}{3 a b}\right )}{f}\) |
\(\Big \downarrow \) 321 |
\(\displaystyle \frac {\sqrt {\cos ^2(e+f x)} \sec (e+f x) \left (\frac {(a+b) \sin (e+f x) \sqrt {1-\sin ^2(e+f x)}}{3 a b \left (a+b \sin ^2(e+f x)\right )^{3/2}}-\frac {\frac {2 (a-b) \sin (e+f x) \sqrt {1-\sin ^2(e+f x)}}{a \sqrt {a+b \sin ^2(e+f x)}}-\frac {\frac {a (2 a-b) (a+b) \sqrt {\frac {b \sin ^2(e+f x)}{a}+1} \operatorname {EllipticF}\left (\arcsin (\sin (e+f x)),-\frac {b}{a}\right )}{b \sqrt {a+b \sin ^2(e+f x)}}-\frac {2 \left (a^2-b^2\right ) \int \frac {\sqrt {b \sin ^2(e+f x)+a}}{\sqrt {1-\sin ^2(e+f x)}}d\sin (e+f x)}{b}}{a (a+b)}}{3 a b}\right )}{f}\) |
\(\Big \downarrow \) 330 |
\(\displaystyle \frac {\sqrt {\cos ^2(e+f x)} \sec (e+f x) \left (\frac {(a+b) \sin (e+f x) \sqrt {1-\sin ^2(e+f x)}}{3 a b \left (a+b \sin ^2(e+f x)\right )^{3/2}}-\frac {\frac {2 (a-b) \sin (e+f x) \sqrt {1-\sin ^2(e+f x)}}{a \sqrt {a+b \sin ^2(e+f x)}}-\frac {\frac {a (2 a-b) (a+b) \sqrt {\frac {b \sin ^2(e+f x)}{a}+1} \operatorname {EllipticF}\left (\arcsin (\sin (e+f x)),-\frac {b}{a}\right )}{b \sqrt {a+b \sin ^2(e+f x)}}-\frac {2 \left (a^2-b^2\right ) \sqrt {a+b \sin ^2(e+f x)} \int \frac {\sqrt {\frac {b \sin ^2(e+f x)}{a}+1}}{\sqrt {1-\sin ^2(e+f x)}}d\sin (e+f x)}{b \sqrt {\frac {b \sin ^2(e+f x)}{a}+1}}}{a (a+b)}}{3 a b}\right )}{f}\) |
\(\Big \downarrow \) 327 |
\(\displaystyle \frac {\sqrt {\cos ^2(e+f x)} \sec (e+f x) \left (\frac {(a+b) \sin (e+f x) \sqrt {1-\sin ^2(e+f x)}}{3 a b \left (a+b \sin ^2(e+f x)\right )^{3/2}}-\frac {\frac {2 (a-b) \sin (e+f x) \sqrt {1-\sin ^2(e+f x)}}{a \sqrt {a+b \sin ^2(e+f x)}}-\frac {\frac {a (2 a-b) (a+b) \sqrt {\frac {b \sin ^2(e+f x)}{a}+1} \operatorname {EllipticF}\left (\arcsin (\sin (e+f x)),-\frac {b}{a}\right )}{b \sqrt {a+b \sin ^2(e+f x)}}-\frac {2 \left (a^2-b^2\right ) \sqrt {a+b \sin ^2(e+f x)} E\left (\arcsin (\sin (e+f x))\left |-\frac {b}{a}\right .\right )}{b \sqrt {\frac {b \sin ^2(e+f x)}{a}+1}}}{a (a+b)}}{3 a b}\right )}{f}\) |
Input:
Int[Cos[e + f*x]^4/(a + b*Sin[e + f*x]^2)^(5/2),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])/(3*a*b*(a + b*Sin[e + f*x]^2)^(3/2)) - ((2*(a - b)*Sin[e + f*x]* Sqrt[1 - Sin[e + f*x]^2])/(a*Sqrt[a + b*Sin[e + f*x]^2]) - ((-2*(a^2 - b^2 )*EllipticE[ArcSin[Sin[e + f*x]], -(b/a)]*Sqrt[a + b*Sin[e + f*x]^2])/(b*S qrt[1 + (b*Sin[e + f*x]^2)/a]) + (a*(2*a - b)*(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*(a + b)))/(3*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[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_.)*((e_) + (f_.)*(x _)^2), x_Symbol] :> Simp[(-(b*e - a*f))*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^ (q + 1)/(a*2*(b*c - a*d)*(p + 1))), x] + Simp[1/(a*2*(b*c - a*d)*(p + 1)) Int[(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[c*(b*e - a*f) + e*2*(b*c - a*d) *(p + 1) + d*(b*e - a*f)*(2*(p + q + 2) + 1)*x^2, x], x], x] /; FreeQ[{a, b , c, d, e, f, q}, x] && LtQ[p, -1]
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]
Leaf count of result is larger than twice the leaf count of optimal. \(484\) vs. \(2(210)=420\).
Time = 4.27 (sec) , antiderivative size = 485, normalized size of antiderivative = 2.16
method | result | size |
default | \(\frac {\left (2 b^{2} a -2 b^{3}\right ) \cos \left (f x +e \right )^{4} \sin \left (f x +e \right )+\left (-a^{2} b +b^{2} a +2 b^{3}\right ) \cos \left (f x +e \right )^{2} \sin \left (f x +e \right )-\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}}\, a b \left (2 \operatorname {EllipticF}\left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) a -\operatorname {EllipticF}\left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) b -2 \operatorname {EllipticE}\left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) a +2 \operatorname {EllipticE}\left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) b \right ) \cos \left (f x +e \right )^{2}+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^{3}+\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} 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 ) a \,b^{2}-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^{3}+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^{2}}{3 a^{2} \left (a +b \sin \left (f x +e \right )^{2}\right )^{\frac {3}{2}} b^{2} \cos \left (f x +e \right ) f}\) | \(485\) |
Input:
int(cos(f*x+e)^4/(a+b*sin(f*x+e)^2)^(5/2),x,method=_RETURNVERBOSE)
Output:
1/3*((2*a*b^2-2*b^3)*cos(f*x+e)^4*sin(f*x+e)+(-a^2*b+a*b^2+2*b^3)*cos(f*x+ e)^2*sin(f*x+e)-(cos(f*x+e)^2)^(1/2)*(-b/a*cos(f*x+e)^2+(a+b)/a)^(1/2)*a*b *(2*EllipticF(sin(f*x+e),(-b/a)^(1/2))*a-EllipticF(sin(f*x+e),(-b/a)^(1/2) )*b-2*EllipticE(sin(f*x+e),(-b/a)^(1/2))*a+2*EllipticE(sin(f*x+e),(-b/a)^( 1/2))*b)*cos(f*x+e)^2+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^3+(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*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))*a*b^2-2*(cos(f*x+e)^2)^(1/2)*(-b/a*cos(f*x+e)^2+(a+b)/a)^(1/2)*Ell ipticE(sin(f*x+e),(-b/a)^(1/2))*a^3+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^2)/a^2/(a+b*sin( f*x+e)^2)^(3/2)/b^2/cos(f*x+e)/f
Result contains complex when optimal does not.
Time = 0.17 (sec) , antiderivative size = 1343, normalized size of antiderivative = 5.97 \[ \int \frac {\cos ^4(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx=\text {Too large to display} \] Input:
integrate(cos(f*x+e)^4/(a+b*sin(f*x+e)^2)^(5/2),x, algorithm="fricas")
Output:
1/3*((2*((-I*a*b^3 + I*b^4)*cos(f*x + e)^4 - I*a^3*b - I*a^2*b^2 + I*a*b^3 + I*b^4 - 2*(-I*a^2*b^2 + I*b^4)*cos(f*x + e)^2)*sqrt(-b)*sqrt((a^2 + a*b )/b^2) - ((2*I*a^2*b^2 - I*a*b^3 - I*b^4)*cos(f*x + e)^4 + 2*I*a^4 + 3*I*a ^3*b - I*a^2*b^2 - 3*I*a*b^3 - I*b^4 + 2*(-2*I*a^3*b - I*a^2*b^2 + 2*I*a*b ^3 + I*b^4)*cos(f*x + e)^2)*sqrt(-b))*sqrt((2*b*sqrt((a^2 + a*b)/b^2) + 2* a + b)/b)*elliptic_e(arcsin(sqrt((2*b*sqrt((a^2 + a*b)/b^2) + 2*a + b)/b)* (cos(f*x + e) + I*sin(f*x + e))), (8*a^2 + 8*a*b + b^2 - 4*(2*a*b + b^2)*s qrt((a^2 + a*b)/b^2))/b^2) + (2*((I*a*b^3 - I*b^4)*cos(f*x + e)^4 + I*a^3* b + I*a^2*b^2 - I*a*b^3 - I*b^4 - 2*(I*a^2*b^2 - I*b^4)*cos(f*x + e)^2)*sq rt(-b)*sqrt((a^2 + a*b)/b^2) - ((-2*I*a^2*b^2 + I*a*b^3 + I*b^4)*cos(f*x + e)^4 - 2*I*a^4 - 3*I*a^3*b + I*a^2*b^2 + 3*I*a*b^3 + I*b^4 + 2*(2*I*a^3*b + I*a^2*b^2 - 2*I*a*b^3 - I*b^4)*cos(f*x + e)^2)*sqrt(-b))*sqrt((2*b*sqrt ((a^2 + a*b)/b^2) + 2*a + b)/b)*elliptic_e(arcsin(sqrt((2*b*sqrt((a^2 + a* b)/b^2) + 2*a + b)/b)*(cos(f*x + e) - I*sin(f*x + e))), (8*a^2 + 8*a*b + b ^2 - 4*(2*a*b + b^2)*sqrt((a^2 + a*b)/b^2))/b^2) + (2*((I*a*b^3 - 2*I*b^4) *cos(f*x + e)^4 + I*a^3*b - 3*I*a*b^3 - 2*I*b^4 - 2*(I*a^2*b^2 - I*a*b^3 - 2*I*b^4)*cos(f*x + e)^2)*sqrt(-b)*sqrt((a^2 + a*b)/b^2) - ((-2*I*a^2*b^2 - I*a*b^3)*cos(f*x + e)^4 - 2*I*a^4 - 5*I*a^3*b - 4*I*a^2*b^2 - I*a*b^3 + 2*(2*I*a^3*b + 3*I*a^2*b^2 + I*a*b^3)*cos(f*x + e)^2)*sqrt(-b))*sqrt((2*b* sqrt((a^2 + a*b)/b^2) + 2*a + b)/b)*elliptic_f(arcsin(sqrt((2*b*sqrt((a...
Timed out. \[ \int \frac {\cos ^4(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx=\text {Timed out} \] Input:
integrate(cos(f*x+e)**4/(a+b*sin(f*x+e)**2)**(5/2),x)
Output:
Timed out
\[ \int \frac {\cos ^4(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx=\int { \frac {\cos \left (f x + e\right )^{4}}{{\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {5}{2}}} \,d x } \] Input:
integrate(cos(f*x+e)^4/(a+b*sin(f*x+e)^2)^(5/2),x, algorithm="maxima")
Output:
integrate(cos(f*x + e)^4/(b*sin(f*x + e)^2 + a)^(5/2), x)
\[ \int \frac {\cos ^4(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx=\int { \frac {\cos \left (f x + e\right )^{4}}{{\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {5}{2}}} \,d x } \] Input:
integrate(cos(f*x+e)^4/(a+b*sin(f*x+e)^2)^(5/2),x, algorithm="giac")
Output:
integrate(cos(f*x + e)^4/(b*sin(f*x + e)^2 + a)^(5/2), x)
Timed out. \[ \int \frac {\cos ^4(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx=\int \frac {{\cos \left (e+f\,x\right )}^4}{{\left (b\,{\sin \left (e+f\,x\right )}^2+a\right )}^{5/2}} \,d x \] Input:
int(cos(e + f*x)^4/(a + b*sin(e + f*x)^2)^(5/2),x)
Output:
int(cos(e + f*x)^4/(a + b*sin(e + f*x)^2)^(5/2), x)
\[ \int \frac {\cos ^4(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx=\int \frac {\sqrt {\sin \left (f x +e \right )^{2} b +a}\, \cos \left (f x +e \right )^{4}}{\sin \left (f x +e \right )^{6} b^{3}+3 \sin \left (f x +e \right )^{4} a \,b^{2}+3 \sin \left (f x +e \right )^{2} a^{2} b +a^{3}}d x \] Input:
int(cos(f*x+e)^4/(a+b*sin(f*x+e)^2)^(5/2),x)
Output:
int((sqrt(sin(e + f*x)**2*b + a)*cos(e + f*x)**4)/(sin(e + f*x)**6*b**3 + 3*sin(e + f*x)**4*a*b**2 + 3*sin(e + f*x)**2*a**2*b + a**3),x)