\(\int \sec ^2(e+f x) (a+b \sin ^2(e+f x))^{3/2} \, dx\) [277]

Optimal result

Integrand size = 25, antiderivative size = 212 \[ \int \sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx=-\frac {(a+2 b) \sqrt {\cos ^2(e+f x)} E\left (\arcsin (\sin (e+f x))\left |-\frac {b}{a}\right .\right ) \sec (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{f \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}}+\frac {a (a+b) \sqrt {\cos ^2(e+f x)} \operatorname {EllipticF}\left (\arcsin (\sin (e+f x)),-\frac {b}{a}\right ) \sec (e+f x) \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}}{f \sqrt {a+b \sin ^2(e+f x)}}-\frac {b \sqrt {a+b \sin ^2(e+f x)} \tan (e+f x)}{f}+\frac {(a+2 b) \sqrt {a+b \sin ^2(e+f x)} \tan (e+f x)}{f} \] Output:

-(a+2*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)/f/(1+b*sin(f*x+e)^2/a)^(1/2)+a*(a+b)*(cos(f*x+e 
)^2)^(1/2)*EllipticF(sin(f*x+e),(-b/a)^(1/2))*sec(f*x+e)*(1+b*sin(f*x+e)^2 
/a)^(1/2)/f/(a+b*sin(f*x+e)^2)^(1/2)-b*(a+b*sin(f*x+e)^2)^(1/2)*tan(f*x+e) 
/f+(a+2*b)*(a+b*sin(f*x+e)^2)^(1/2)*tan(f*x+e)/f
 

Mathematica [A] (verified)

Time = 1.41 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.68 \[ \int \sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx=\frac {-2 a (a+2 b) \sqrt {\frac {2 a+b-b \cos (2 (e+f x))}{a}} E\left (e+f x\left |-\frac {b}{a}\right .\right )+(a+b) \left (2 a \sqrt {\frac {2 a+b-b \cos (2 (e+f x))}{a}} \operatorname {EllipticF}\left (e+f x,-\frac {b}{a}\right )+\sqrt {2} (2 a+b-b \cos (2 (e+f x))) \tan (e+f x)\right )}{2 f \sqrt {2 a+b-b \cos (2 (e+f x))}} \] Input:

Integrate[Sec[e + f*x]^2*(a + b*Sin[e + f*x]^2)^(3/2),x]
 

Output:

(-2*a*(a + 2*b)*Sqrt[(2*a + b - b*Cos[2*(e + f*x)])/a]*EllipticE[e + f*x, 
-(b/a)] + (a + b)*(2*a*Sqrt[(2*a + b - b*Cos[2*(e + f*x)])/a]*EllipticF[e 
+ f*x, -(b/a)] + Sqrt[2]*(2*a + b - b*Cos[2*(e + f*x)])*Tan[e + f*x]))/(2* 
f*Sqrt[2*a + b - b*Cos[2*(e + f*x)]])
 

Rubi [A] (verified)

Time = 0.39 (sec) , antiderivative size = 185, normalized size of antiderivative = 0.87, 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, 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.

\(\displaystyle \int \sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\left (a+b \sin (e+f x)^2\right )^{3/2}}{\cos (e+f x)^2}dx\)

\(\Big \downarrow \) 3671

\(\displaystyle \frac {\sqrt {\cos ^2(e+f x)} \sec (e+f x) \int \frac {\left (b \sin ^2(e+f x)+a\right )^{3/2}}{\left (1-\sin ^2(e+f x)\right )^{3/2}}d\sin (e+f x)}{f}\)

\(\Big \downarrow \) 315

\(\displaystyle \frac {\sqrt {\cos ^2(e+f x)} \sec (e+f x) \left (\frac {(a+b) \sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{\sqrt {1-\sin ^2(e+f x)}}-\int \frac {b \left ((a+2 b) \sin ^2(e+f x)+a\right )}{\sqrt {1-\sin ^2(e+f x)} \sqrt {b \sin ^2(e+f x)+a}}d\sin (e+f x)\right )}{f}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\sqrt {\cos ^2(e+f x)} \sec (e+f x) \left (\frac {(a+b) \sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{\sqrt {1-\sin ^2(e+f x)}}-b \int \frac {(a+2 b) \sin ^2(e+f x)+a}{\sqrt {1-\sin ^2(e+f x)} \sqrt {b \sin ^2(e+f x)+a}}d\sin (e+f x)\right )}{f}\)

\(\Big \downarrow \) 399

\(\displaystyle \frac {\sqrt {\cos ^2(e+f x)} \sec (e+f x) \left (\frac {(a+b) \sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{\sqrt {1-\sin ^2(e+f x)}}-b \left (\frac {(a+2 b) \int \frac {\sqrt {b \sin ^2(e+f x)+a}}{\sqrt {1-\sin ^2(e+f x)}}d\sin (e+f x)}{b}-\frac {a (a+b) \int \frac {1}{\sqrt {1-\sin ^2(e+f x)} \sqrt {b \sin ^2(e+f x)+a}}d\sin (e+f x)}{b}\right )\right )}{f}\)

\(\Big \downarrow \) 323

\(\displaystyle \frac {\sqrt {\cos ^2(e+f x)} \sec (e+f x) \left (\frac {(a+b) \sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{\sqrt {1-\sin ^2(e+f x)}}-b \left (\frac {(a+2 b) \int \frac {\sqrt {b \sin ^2(e+f x)+a}}{\sqrt {1-\sin ^2(e+f x)}}d\sin (e+f x)}{b}-\frac {a (a+b) \sqrt {\frac {b \sin ^2(e+f x)}{a}+1} \int \frac {1}{\sqrt {1-\sin ^2(e+f x)} \sqrt {\frac {b \sin ^2(e+f x)}{a}+1}}d\sin (e+f x)}{b \sqrt {a+b \sin ^2(e+f x)}}\right )\right )}{f}\)

\(\Big \downarrow \) 321

\(\displaystyle \frac {\sqrt {\cos ^2(e+f x)} \sec (e+f x) \left (\frac {(a+b) \sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{\sqrt {1-\sin ^2(e+f x)}}-b \left (\frac {(a+2 b) \int \frac {\sqrt {b \sin ^2(e+f x)+a}}{\sqrt {1-\sin ^2(e+f x)}}d\sin (e+f x)}{b}-\frac {a (a+b) \sqrt {\frac {b \sin ^2(e+f x)}{a}+1} \operatorname {EllipticF}\left (\arcsin (\sin (e+f x)),-\frac {b}{a}\right )}{b \sqrt {a+b \sin ^2(e+f x)}}\right )\right )}{f}\)

\(\Big \downarrow \) 330

\(\displaystyle \frac {\sqrt {\cos ^2(e+f x)} \sec (e+f x) \left (\frac {(a+b) \sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{\sqrt {1-\sin ^2(e+f x)}}-b \left (\frac {(a+2 b) \sqrt {a+b \sin ^2(e+f x)} \int \frac {\sqrt {\frac {b \sin ^2(e+f x)}{a}+1}}{\sqrt {1-\sin ^2(e+f x)}}d\sin (e+f x)}{b \sqrt {\frac {b \sin ^2(e+f x)}{a}+1}}-\frac {a (a+b) \sqrt {\frac {b \sin ^2(e+f x)}{a}+1} \operatorname {EllipticF}\left (\arcsin (\sin (e+f x)),-\frac {b}{a}\right )}{b \sqrt {a+b \sin ^2(e+f x)}}\right )\right )}{f}\)

\(\Big \downarrow \) 327

\(\displaystyle \frac {\sqrt {\cos ^2(e+f x)} \sec (e+f x) \left (\frac {(a+b) \sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{\sqrt {1-\sin ^2(e+f x)}}-b \left (\frac {(a+2 b) \sqrt {a+b \sin ^2(e+f x)} E\left (\arcsin (\sin (e+f x))\left |-\frac {b}{a}\right .\right )}{b \sqrt {\frac {b \sin ^2(e+f x)}{a}+1}}-\frac {a (a+b) \sqrt {\frac {b \sin ^2(e+f x)}{a}+1} \operatorname {EllipticF}\left (\arcsin (\sin (e+f x)),-\frac {b}{a}\right )}{b \sqrt {a+b \sin ^2(e+f x)}}\right )\right )}{f}\)

Input:

Int[Sec[e + f*x]^2*(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[a + b*Sin[e 
 + f*x]^2])/Sqrt[1 - Sin[e + f*x]^2] - b*(((a + 2*b)*EllipticE[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]) - (a*(a + b)*EllipticF[ArcSin[Sin[e + f*x]], -(b/a)]*Sqrt[1 + (b*S 
in[e + f*x]^2)/a])/(b*Sqrt[a + b*Sin[e + f*x]^2]))))/f
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 315

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]
 

rule 321

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])
 

rule 323

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]
 

rule 327

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]
 

rule 330

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]
 

rule 399

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])))))
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3671

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]
 

Maple [A] (verified)

Time = 4.26 (sec) , antiderivative size = 365, normalized size of antiderivative = 1.72

method	result	size
default	\(\frac {\sqrt {-b \cos \left (f x +e \right )^{4}+\left (a +b \right ) \cos \left (f x +e \right )^{2}}\, \left (\sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {-\frac {b \cos \left (f x +e \right )^{2}}{a}+\frac {a +b}{a}}\, \operatorname {EllipticF}\left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) a^{2}+a b \sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {-\frac {b \cos \left (f x +e \right )^{2}}{a}+\frac {a +b}{a}}\, \operatorname {EllipticF}\left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right )-\sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {-\frac {b \cos \left (f x +e \right )^{2}}{a}+\frac {a +b}{a}}\, \operatorname {EllipticE}\left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) a^{2}-2 \sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {-\frac {b \cos \left (f x +e \right )^{2}}{a}+\frac {a +b}{a}}\, \operatorname {EllipticE}\left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) a b -b \cos \left (f x +e \right )^{2} \sin \left (f x +e \right ) a -b^{2} \cos \left (f x +e \right )^{2} \sin \left (f x +e \right )+a^{2} \sin \left (f x +e \right )+2 a b \sin \left (f x +e \right )+b^{2} \sin \left (f x +e \right )\right )}{\sqrt {-\left (a +b \sin \left (f x +e \right )^{2}\right ) \left (\sin \left (f x +e \right )-1\right ) \left (1+\sin \left (f x +e \right )\right )}\, \cos \left (f x +e \right ) \sqrt {a +b \sin \left (f x +e \right )^{2}}\, f}\)	\(365\)

Input:

int(sec(f*x+e)^2*(a+b*sin(f*x+e)^2)^(3/2),x,method=_RETURNVERBOSE)
 

Output:

(-b*cos(f*x+e)^4+(a+b)*cos(f*x+e)^2)^(1/2)*((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+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))-(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^2-2*(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)+a^2*sin(f*x+e)+2*a*b*sin(f*x+e)+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)/cos(f*x+ 
e)/(a+b*sin(f*x+e)^2)^(1/2)/f
                                                                                    
                                                                                    
 

Fricas [F]

\[ \int \sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx=\int { {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} \sec \left (f x + e\right )^{2} \,d x } \] Input:

integrate(sec(f*x+e)^2*(a+b*sin(f*x+e)^2)^(3/2),x, algorithm="fricas")
 

Output:

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

Sympy [F(-1)]

Timed out. \[ \int \sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx=\text {Timed out} \] Input:

integrate(sec(f*x+e)**2*(a+b*sin(f*x+e)**2)**(3/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx=\int { {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} \sec \left (f x + e\right )^{2} \,d x } \] Input:

integrate(sec(f*x+e)^2*(a+b*sin(f*x+e)^2)^(3/2),x, algorithm="maxima")
 

Output:

integrate((b*sin(f*x + e)^2 + a)^(3/2)*sec(f*x + e)^2, x)
 

Giac [F(-1)]

Timed out. \[ \int \sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx=\text {Timed out} \] Input:

integrate(sec(f*x+e)^2*(a+b*sin(f*x+e)^2)^(3/2),x, algorithm="giac")
 

Output:

Timed out
 

Mupad [F(-1)]

Timed out. \[ \int \sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx=\int \frac {{\left (b\,{\sin \left (e+f\,x\right )}^2+a\right )}^{3/2}}{{\cos \left (e+f\,x\right )}^2} \,d x \] Input:

int((a + b*sin(e + f*x)^2)^(3/2)/cos(e + f*x)^2,x)
 

Output:

int((a + b*sin(e + f*x)^2)^(3/2)/cos(e + f*x)^2, x)
 

Reduce [F]

\[ \int \sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx=\left (\int \sqrt {\sin \left (f x +e \right )^{2} b +a}\, \sec \left (f x +e \right )^{2} \sin \left (f x +e \right )^{2}d x \right ) b +\left (\int \sqrt {\sin \left (f x +e \right )^{2} b +a}\, \sec \left (f x +e \right )^{2}d x \right ) a \] Input:

int(sec(f*x+e)^2*(a+b*sin(f*x+e)^2)^(3/2),x)
 

Output:

int(sqrt(sin(e + f*x)**2*b + a)*sec(e + f*x)**2*sin(e + f*x)**2,x)*b + int 
(sqrt(sin(e + f*x)**2*b + a)*sec(e + f*x)**2,x)*a