Integrand size = 25, antiderivative size = 198 \[ \int \sec ^4(e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx=-\frac {(2 a+b) E\left (e+f x\left |-\frac {b}{a}\right .\right ) \sqrt {a+b \sin ^2(e+f x)}}{3 (a+b) f \sqrt {\frac {a+b \sin ^2(e+f x)}{a}}}+\frac {2 a \operatorname {EllipticF}\left (e+f x,-\frac {b}{a}\right ) \sqrt {\frac {a+b \sin ^2(e+f x)}{a}}}{3 f \sqrt {a+b \sin ^2(e+f x)}}+\frac {(2 a+b) \sqrt {a+b \sin ^2(e+f x)} \tan (e+f x)}{3 (a+b) f}+\frac {\sec ^2(e+f x) \sqrt {a+b \sin ^2(e+f x)} \tan (e+f x)}{3 f} \] Output:
-1/3*(2*a+b)*EllipticE(sin(f*x+e),(-b/a)^(1/2))*(a+b*sin(f*x+e)^2)^(1/2)/( a+b)/f/((a+b*sin(f*x+e)^2)/a)^(1/2)+2/3*a*InverseJacobiAM(f*x+e,(-b/a)^(1/ 2))*((a+b*sin(f*x+e)^2)/a)^(1/2)/f/(a+b*sin(f*x+e)^2)^(1/2)+1/3*(2*a+b)*(a +b*sin(f*x+e)^2)^(1/2)*tan(f*x+e)/(a+b)/f+1/3*sec(f*x+e)^2*(a+b*sin(f*x+e) ^2)^(1/2)*tan(f*x+e)/f
Time = 2.30 (sec) , antiderivative size = 187, normalized size of antiderivative = 0.94 \[ \int \sec ^4(e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx=\frac {-2 a (2 a+b) \sqrt {\frac {2 a+b-b \cos (2 (e+f x))}{a}} E\left (e+f x\left |-\frac {b}{a}\right .\right )+4 a (a+b) \sqrt {\frac {2 a+b-b \cos (2 (e+f x))}{a}} \operatorname {EllipticF}\left (e+f x,-\frac {b}{a}\right )+\frac {\left (\left (8 a^2-4 b^2\right ) \cos (2 (e+f x))+(2 a+b) (8 a+5 b-b \cos (4 (e+f x)))\right ) \sec ^2(e+f x) \tan (e+f x)}{2 \sqrt {2}}}{6 (a+b) f \sqrt {2 a+b-b \cos (2 (e+f x))}} \] Input:
Integrate[Sec[e + f*x]^4*Sqrt[a + b*Sin[e + f*x]^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)] + 4*a*(a + b)*Sqrt[(2*a + b - b*Cos[2*(e + f*x)])/a]*EllipticF[e + f*x, -(b/a)] + (((8*a^2 - 4*b^2)*Cos[2*(e + f*x)] + (2*a + b)*(8*a + 5*b - b*Cos[4*(e + f*x)]))*Sec[e + f*x]^2*Tan[e + f*x])/(2*Sqrt[2]))/(6*(a + b )*f*Sqrt[2*a + b - b*Cos[2*(e + f*x)]])
Time = 0.44 (sec) , antiderivative size = 244, normalized size of antiderivative = 1.23, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.440, Rules used = {3042, 3671, 314, 25, 402, 27, 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.
\(\displaystyle \int \sec ^4(e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sqrt {a+b \sin (e+f x)^2}}{\cos (e+f x)^4}dx\) |
\(\Big \downarrow \) 3671 |
\(\displaystyle \frac {\sqrt {\cos ^2(e+f x)} \sec (e+f x) \int \frac {\sqrt {b \sin ^2(e+f x)+a}}{\left (1-\sin ^2(e+f x)\right )^{5/2}}d\sin (e+f x)}{f}\) |
\(\Big \downarrow \) 314 |
\(\displaystyle \frac {\sqrt {\cos ^2(e+f x)} \sec (e+f x) \left (\frac {\sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{3 \left (1-\sin ^2(e+f x)\right )^{3/2}}-\frac {1}{3} \int -\frac {b \sin ^2(e+f x)+2 a}{\left (1-\sin ^2(e+f x)\right )^{3/2} \sqrt {b \sin ^2(e+f x)+a}}d\sin (e+f x)\right )}{f}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\sqrt {\cos ^2(e+f x)} \sec (e+f x) \left (\frac {1}{3} \int \frac {b \sin ^2(e+f x)+2 a}{\left (1-\sin ^2(e+f x)\right )^{3/2} \sqrt {b \sin ^2(e+f x)+a}}d\sin (e+f x)+\frac {\sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{3 \left (1-\sin ^2(e+f x)\right )^{3/2}}\right )}{f}\) |
\(\Big \downarrow \) 402 |
\(\displaystyle \frac {\sqrt {\cos ^2(e+f x)} \sec (e+f x) \left (\frac {1}{3} \left (\frac {\int \frac {b \left (a-(2 a+b) \sin ^2(e+f x)\right )}{\sqrt {1-\sin ^2(e+f x)} \sqrt {b \sin ^2(e+f x)+a}}d\sin (e+f x)}{a+b}+\frac {(2 a+b) \sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{(a+b) \sqrt {1-\sin ^2(e+f x)}}\right )+\frac {\sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{3 \left (1-\sin ^2(e+f x)\right )^{3/2}}\right )}{f}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\sqrt {\cos ^2(e+f x)} \sec (e+f x) \left (\frac {1}{3} \left (\frac {b \int \frac {a-(2 a+b) \sin ^2(e+f x)}{\sqrt {1-\sin ^2(e+f x)} \sqrt {b \sin ^2(e+f x)+a}}d\sin (e+f x)}{a+b}+\frac {(2 a+b) \sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{(a+b) \sqrt {1-\sin ^2(e+f x)}}\right )+\frac {\sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{3 \left (1-\sin ^2(e+f x)\right )^{3/2}}\right )}{f}\) |
\(\Big \downarrow \) 399 |
\(\displaystyle \frac {\sqrt {\cos ^2(e+f x)} \sec (e+f x) \left (\frac {1}{3} \left (\frac {b \left (\frac {2 a (a+b) \int \frac {1}{\sqrt {1-\sin ^2(e+f x)} \sqrt {b \sin ^2(e+f x)+a}}d\sin (e+f x)}{b}-\frac {(2 a+b) \int \frac {\sqrt {b \sin ^2(e+f x)+a}}{\sqrt {1-\sin ^2(e+f x)}}d\sin (e+f x)}{b}\right )}{a+b}+\frac {(2 a+b) \sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{(a+b) \sqrt {1-\sin ^2(e+f x)}}\right )+\frac {\sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{3 \left (1-\sin ^2(e+f x)\right )^{3/2}}\right )}{f}\) |
\(\Big \downarrow \) 323 |
\(\displaystyle \frac {\sqrt {\cos ^2(e+f x)} \sec (e+f x) \left (\frac {1}{3} \left (\frac {b \left (\frac {2 a (a+b) \sqrt {\frac {b \sin ^2(e+f x)}{a}+1} \int \frac {1}{\sqrt {1-\sin ^2(e+f x)} \sqrt {\frac {b \sin ^2(e+f x)}{a}+1}}d\sin (e+f x)}{b \sqrt {a+b \sin ^2(e+f x)}}-\frac {(2 a+b) \int \frac {\sqrt {b \sin ^2(e+f x)+a}}{\sqrt {1-\sin ^2(e+f x)}}d\sin (e+f x)}{b}\right )}{a+b}+\frac {(2 a+b) \sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{(a+b) \sqrt {1-\sin ^2(e+f x)}}\right )+\frac {\sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{3 \left (1-\sin ^2(e+f x)\right )^{3/2}}\right )}{f}\) |
\(\Big \downarrow \) 321 |
\(\displaystyle \frac {\sqrt {\cos ^2(e+f x)} \sec (e+f x) \left (\frac {1}{3} \left (\frac {b \left (\frac {2 a (a+b) \sqrt {\frac {b \sin ^2(e+f x)}{a}+1} \operatorname {EllipticF}\left (\arcsin (\sin (e+f x)),-\frac {b}{a}\right )}{b \sqrt {a+b \sin ^2(e+f x)}}-\frac {(2 a+b) \int \frac {\sqrt {b \sin ^2(e+f x)+a}}{\sqrt {1-\sin ^2(e+f x)}}d\sin (e+f x)}{b}\right )}{a+b}+\frac {(2 a+b) \sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{(a+b) \sqrt {1-\sin ^2(e+f x)}}\right )+\frac {\sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{3 \left (1-\sin ^2(e+f x)\right )^{3/2}}\right )}{f}\) |
\(\Big \downarrow \) 330 |
\(\displaystyle \frac {\sqrt {\cos ^2(e+f x)} \sec (e+f x) \left (\frac {1}{3} \left (\frac {b \left (\frac {2 a (a+b) \sqrt {\frac {b \sin ^2(e+f x)}{a}+1} \operatorname {EllipticF}\left (\arcsin (\sin (e+f x)),-\frac {b}{a}\right )}{b \sqrt {a+b \sin ^2(e+f x)}}-\frac {(2 a+b) \sqrt {a+b \sin ^2(e+f x)} \int \frac {\sqrt {\frac {b \sin ^2(e+f x)}{a}+1}}{\sqrt {1-\sin ^2(e+f x)}}d\sin (e+f x)}{b \sqrt {\frac {b \sin ^2(e+f x)}{a}+1}}\right )}{a+b}+\frac {(2 a+b) \sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{(a+b) \sqrt {1-\sin ^2(e+f x)}}\right )+\frac {\sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{3 \left (1-\sin ^2(e+f x)\right )^{3/2}}\right )}{f}\) |
\(\Big \downarrow \) 327 |
\(\displaystyle \frac {\sqrt {\cos ^2(e+f x)} \sec (e+f x) \left (\frac {1}{3} \left (\frac {b \left (\frac {2 a (a+b) \sqrt {\frac {b \sin ^2(e+f x)}{a}+1} \operatorname {EllipticF}\left (\arcsin (\sin (e+f x)),-\frac {b}{a}\right )}{b \sqrt {a+b \sin ^2(e+f x)}}-\frac {(2 a+b) \sqrt {a+b \sin ^2(e+f x)} E\left (\arcsin (\sin (e+f x))\left |-\frac {b}{a}\right .\right )}{b \sqrt {\frac {b \sin ^2(e+f x)}{a}+1}}\right )}{a+b}+\frac {(2 a+b) \sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{(a+b) \sqrt {1-\sin ^2(e+f x)}}\right )+\frac {\sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{3 \left (1-\sin ^2(e+f x)\right )^{3/2}}\right )}{f}\) |
Input:
Int[Sec[e + f*x]^4*Sqrt[a + b*Sin[e + f*x]^2],x]
Output:
(Sqrt[Cos[e + f*x]^2]*Sec[e + f*x]*((Sin[e + f*x]*Sqrt[a + b*Sin[e + f*x]^ 2])/(3*(1 - Sin[e + f*x]^2)^(3/2)) + (((2*a + b)*Sin[e + f*x]*Sqrt[a + b*S in[e + f*x]^2])/((a + b)*Sqrt[1 - Sin[e + f*x]^2]) + (b*(-(((2*a + b)*Elli pticE[ArcSin[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))/3))/f
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim p[(-x)*(a + b*x^2)^(p + 1)*((c + d*x^2)^q/(2*a*(p + 1))), x] + Simp[1/(2*a* (p + 1)) Int[(a + b*x^2)^(p + 1)*(c + d*x^2)^(q - 1)*Simp[c*(2*p + 3) + d *(2*(p + q + 1) + 1)*x^2, x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && LtQ[0, 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. \(367\) vs. \(2(183)=366\).
Time = 4.50 (sec) , antiderivative size = 368, normalized size of antiderivative = 1.86
method | result | size |
default | \(\frac {\sqrt {-b \cos \left (f x +e \right )^{4}+\left (a +b \right ) \cos \left (f x +e \right )^{2}}\, b \left (2 a +b \right ) \cos \left (f x +e \right )^{4} \sin \left (f x +e \right )-2 \sqrt {-b \cos \left (f x +e \right )^{4}+\left (a +b \right ) \cos \left (f x +e \right )^{2}}\, a \left (a +b \right ) \cos \left (f x +e \right )^{2} \sin \left (f x +e \right )-\sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {-\frac {b \cos \left (f x +e \right )^{2}}{a}+\frac {a +b}{a}}\, \sqrt {-b \cos \left (f x +e \right )^{4}+\left (a +b \right ) \cos \left (f x +e \right )^{2}}\, a \left (2 \operatorname {EllipticF}\left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) a +2 \operatorname {EllipticF}\left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) b -2 \operatorname {EllipticE}\left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) a -\operatorname {EllipticE}\left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) b \right ) \cos \left (f x +e \right )^{2}-\sqrt {-b \cos \left (f x +e \right )^{4}+\left (a +b \right ) \cos \left (f x +e \right )^{2}}\, \left (a^{2}+2 a b +b^{2}\right ) \sin \left (f x +e \right )}{3 \sqrt {-\left (a +b \sin \left (f x +e \right )^{2}\right ) \left (\sin \left (f x +e \right )-1\right ) \left (1+\sin \left (f x +e \right )\right )}\, \left (a +b \right ) \left (\sin \left (f x +e \right )-1\right ) \left (1+\sin \left (f x +e \right )\right ) \cos \left (f x +e \right ) \sqrt {a +b \sin \left (f x +e \right )^{2}}\, f}\) | \(368\) |
Input:
int(sec(f*x+e)^4*(a+b*sin(f*x+e)^2)^(1/2),x,method=_RETURNVERBOSE)
Output:
1/3*((-b*cos(f*x+e)^4+(a+b)*cos(f*x+e)^2)^(1/2)*b*(2*a+b)*cos(f*x+e)^4*sin (f*x+e)-2*(-b*cos(f*x+e)^4+(a+b)*cos(f*x+e)^2)^(1/2)*a*(a+b)*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)*(-b*cos( f*x+e)^4+(a+b)*cos(f*x+e)^2)^(1/2)*a*(2*EllipticF(sin(f*x+e),(-b/a)^(1/2)) *a+2*EllipticF(sin(f*x+e),(-b/a)^(1/2))*b-2*EllipticE(sin(f*x+e),(-b/a)^(1 /2))*a-EllipticE(sin(f*x+e),(-b/a)^(1/2))*b)*cos(f*x+e)^2-(-b*cos(f*x+e)^4 +(a+b)*cos(f*x+e)^2)^(1/2)*(a^2+2*a*b+b^2)*sin(f*x+e))/(-(a+b*sin(f*x+e)^2 )*(sin(f*x+e)-1)*(1+sin(f*x+e)))^(1/2)/(a+b)/(sin(f*x+e)-1)/(1+sin(f*x+e)) /cos(f*x+e)/(a+b*sin(f*x+e)^2)^(1/2)/f
Result contains complex when optimal does not.
Time = 0.13 (sec) , antiderivative size = 789, normalized size of antiderivative = 3.98 \[ \int \sec ^4(e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx =\text {Too large to display} \] Input:
integrate(sec(f*x+e)^4*(a+b*sin(f*x+e)^2)^(1/2),x, algorithm="fricas")
Output:
1/6*((2*(-2*I*a*b - I*b^2)*sqrt(-b)*sqrt((a^2 + a*b)/b^2)*cos(f*x + e)^3 - (4*I*a^2 + 4*I*a*b + I*b^2)*sqrt(-b)*cos(f*x + e)^3)*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*(2*I*a*b + I*b^2)*sqrt(-b) *sqrt((a^2 + a*b)/b^2)*cos(f*x + e)^3 - (-4*I*a^2 - 4*I*a*b - I*b^2)*sqrt( -b)*cos(f*x + e)^3)*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*(2*(-I*a*b - I*b^2)*sqrt(-b)*sqrt((a^2 + a*b)/b^2)*cos(f*x + e)^3 + (-2*I*a^2 - I*a*b)*sqrt(-b)*cos(f*x + e)^3)*sqrt((2*b*sqrt((a^2 + a *b)/b^2) + 2*a + b)/b)*elliptic_f(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*(2*(I*a*b + I*b^2)*sqrt(-b)*sq rt((a^2 + a*b)/b^2)*cos(f*x + e)^3 + (2*I*a^2 + I*a*b)*sqrt(-b)*cos(f*x + e)^3)*sqrt((2*b*sqrt((a^2 + a*b)/b^2) + 2*a + b)/b)*elliptic_f(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*( (2*a*b + b^2)*cos(f*x + e)^2 + a*b + b^2)*sqrt(-b*cos(f*x + e)^2 + a + b)* sin(f*x + e))/((a*b + b^2)*f*cos(f*x + e)^3)
\[ \int \sec ^4(e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx=\int \sqrt {a + b \sin ^{2}{\left (e + f x \right )}} \sec ^{4}{\left (e + f x \right )}\, dx \] Input:
integrate(sec(f*x+e)**4*(a+b*sin(f*x+e)**2)**(1/2),x)
Output:
Integral(sqrt(a + b*sin(e + f*x)**2)*sec(e + f*x)**4, x)
\[ \int \sec ^4(e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx=\int { \sqrt {b \sin \left (f x + e\right )^{2} + a} \sec \left (f x + e\right )^{4} \,d x } \] Input:
integrate(sec(f*x+e)^4*(a+b*sin(f*x+e)^2)^(1/2),x, algorithm="maxima")
Output:
integrate(sqrt(b*sin(f*x + e)^2 + a)*sec(f*x + e)^4, x)
\[ \int \sec ^4(e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx=\int { \sqrt {b \sin \left (f x + e\right )^{2} + a} \sec \left (f x + e\right )^{4} \,d x } \] Input:
integrate(sec(f*x+e)^4*(a+b*sin(f*x+e)^2)^(1/2),x, algorithm="giac")
Output:
integrate(sqrt(b*sin(f*x + e)^2 + a)*sec(f*x + e)^4, x)
Timed out. \[ \int \sec ^4(e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx=\int \frac {\sqrt {b\,{\sin \left (e+f\,x\right )}^2+a}}{{\cos \left (e+f\,x\right )}^4} \,d x \] Input:
int((a + b*sin(e + f*x)^2)^(1/2)/cos(e + f*x)^4,x)
Output:
int((a + b*sin(e + f*x)^2)^(1/2)/cos(e + f*x)^4, x)
\[ \int \sec ^4(e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx=\int \sqrt {\sin \left (f x +e \right )^{2} b +a}\, \sec \left (f x +e \right )^{4}d x \] Input:
int(sec(f*x+e)^4*(a+b*sin(f*x+e)^2)^(1/2),x)
Output:
int(sqrt(sin(e + f*x)**2*b + a)*sec(e + f*x)**4,x)