\(\int \frac {a^2-b^2 \sec ^2(c+d x)}{(a+b \sec (c+d x))^{5/2}} \, dx\) [757]

Optimal result

Integrand size = 32, antiderivative size = 338 \[ \int \frac {a^2-b^2 \sec ^2(c+d x)}{(a+b \sec (c+d x))^{5/2}} \, dx=\frac {4 \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{\sqrt {a+b} d}-\frac {4 \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{\sqrt {a+b} d}-\frac {2 \sqrt {a+b} \cot (c+d x) \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{a d}+\frac {4 b^2 \tan (c+d x)}{\left (a^2-b^2\right ) d \sqrt {a+b \sec (c+d x)}} \] Output:

4*cot(d*x+c)*EllipticE((a+b*sec(d*x+c))^(1/2)/(a+b)^(1/2),((a+b)/(a-b))^(1 
/2))*(b*(1-sec(d*x+c))/(a+b))^(1/2)*(-b*(1+sec(d*x+c))/(a-b))^(1/2)/(a+b)^ 
(1/2)/d-4*cot(d*x+c)*EllipticF((a+b*sec(d*x+c))^(1/2)/(a+b)^(1/2),((a+b)/( 
a-b))^(1/2))*(b*(1-sec(d*x+c))/(a+b))^(1/2)*(-b*(1+sec(d*x+c))/(a-b))^(1/2 
)/(a+b)^(1/2)/d-2*(a+b)^(1/2)*cot(d*x+c)*EllipticPi((a+b*sec(d*x+c))^(1/2) 
/(a+b)^(1/2),(a+b)/a,((a+b)/(a-b))^(1/2))*(b*(1-sec(d*x+c))/(a+b))^(1/2)*( 
-b*(1+sec(d*x+c))/(a-b))^(1/2)/a/d+4*b^2*tan(d*x+c)/(a^2-b^2)/d/(a+b*sec(d 
*x+c))^(1/2)
 

Mathematica [A] (verified)

Time = 11.65 (sec) , antiderivative size = 498, normalized size of antiderivative = 1.47 \[ \int \frac {a^2-b^2 \sec ^2(c+d x)}{(a+b \sec (c+d x))^{5/2}} \, dx=\frac {(b+a \cos (c+d x))^2 \sec (c+d x) (a-b \sec (c+d x)) \left (\frac {4 b \sin (c+d x)}{-a^2+b^2}-\frac {4 b^2 \sin (c+d x)}{\left (-a^2+b^2\right ) (b+a \cos (c+d x))}\right )}{d (-b+a \cos (c+d x)) (a+b \sec (c+d x))^{3/2}}+\frac {4 (b+a \cos (c+d x)) \sec ^2\left (\frac {1}{2} (c+d x)\right ) (a-b \sec (c+d x)) \left (-2 b (a+b) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {b+a \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} E\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {a-b}{a+b}\right )+(a+b)^2 \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {b+a \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} \operatorname {EllipticF}\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {a-b}{a+b}\right )-2 \left (a^2-b^2\right ) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {b+a \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} \operatorname {EllipticPi}\left (-1,\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {a-b}{a+b}\right )-b \cos (c+d x) (b+a \cos (c+d x)) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{\left (a^2-b^2\right ) d (-b+a \cos (c+d x)) (a+b \sec (c+d x))^{3/2} \left (-1+\tan ^4\left (\frac {1}{2} (c+d x)\right )\right )} \] Input:

Integrate[(a^2 - b^2*Sec[c + d*x]^2)/(a + b*Sec[c + d*x])^(5/2),x]
 

Output:

((b + a*Cos[c + d*x])^2*Sec[c + d*x]*(a - b*Sec[c + d*x])*((4*b*Sin[c + d* 
x])/(-a^2 + b^2) - (4*b^2*Sin[c + d*x])/((-a^2 + b^2)*(b + a*Cos[c + d*x]) 
)))/(d*(-b + a*Cos[c + d*x])*(a + b*Sec[c + d*x])^(3/2)) + (4*(b + a*Cos[c 
 + d*x])*Sec[(c + d*x)/2]^2*(a - b*Sec[c + d*x])*(-2*b*(a + b)*Sqrt[Cos[c 
+ d*x]/(1 + Cos[c + d*x])]*Sqrt[(b + a*Cos[c + d*x])/((a + b)*(1 + Cos[c + 
 d*x]))]*EllipticE[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)] + (a + b)^2* 
Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[(b + a*Cos[c + d*x])/((a + b)*( 
1 + Cos[c + d*x]))]*EllipticF[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)] - 
 2*(a^2 - b^2)*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[(b + a*Cos[c + d 
*x])/((a + b)*(1 + Cos[c + d*x]))]*EllipticPi[-1, ArcSin[Tan[(c + d*x)/2]] 
, (a - b)/(a + b)] - b*Cos[c + d*x]*(b + a*Cos[c + d*x])*Sec[(c + d*x)/2]^ 
2*Tan[(c + d*x)/2]))/((a^2 - b^2)*d*(-b + a*Cos[c + d*x])*(a + b*Sec[c + d 
*x])^(3/2)*(-1 + Tan[(c + d*x)/2]^4))
 

Rubi [A] (verified)

Time = 1.39 (sec) , antiderivative size = 372, normalized size of antiderivative = 1.10, number of steps used = 14, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.438, Rules used = {3042, 4530, 25, 3042, 4411, 27, 3042, 4546, 3042, 4409, 3042, 4271, 4319, 4492}

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.

\(\displaystyle \int \frac {a^2-b^2 \sec ^2(c+d x)}{(a+b \sec (c+d x))^{5/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {a^2-b^2 \csc \left (c+d x+\frac {\pi }{2}\right )^2}{\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2}}dx\)

\(\Big \downarrow \) 4530

\(\displaystyle -\int -\frac {a-b \sec (c+d x)}{(a+b \sec (c+d x))^{3/2}}dx\)

\(\Big \downarrow \) 25

\(\displaystyle \int \frac {a-b \sec (c+d x)}{(a+b \sec (c+d x))^{3/2}}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {a-b \csc \left (c+d x+\frac {\pi }{2}\right )}{\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}}dx\)

\(\Big \downarrow \) 4411

\(\displaystyle \frac {4 b^2 \tan (c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}-\frac {2 \int -\frac {-2 b \sec (c+d x) a^2-2 b^2 \sec ^2(c+d x) a+\left (a^2-b^2\right ) a}{2 \sqrt {a+b \sec (c+d x)}}dx}{a \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {-2 b \sec (c+d x) a^2-2 b^2 \sec ^2(c+d x) a+\left (a^2-b^2\right ) a}{\sqrt {a+b \sec (c+d x)}}dx}{a \left (a^2-b^2\right )}+\frac {4 b^2 \tan (c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {-2 b \csc \left (c+d x+\frac {\pi }{2}\right ) a^2-2 b^2 \csc \left (c+d x+\frac {\pi }{2}\right )^2 a+\left (a^2-b^2\right ) a}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{a \left (a^2-b^2\right )}+\frac {4 b^2 \tan (c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}\)

\(\Big \downarrow \) 4546

\(\displaystyle \frac {\int \frac {a \left (a^2-b^2\right )+\left (2 a b^2-2 a^2 b\right ) \sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx-2 a b^2 \int \frac {\sec (c+d x) (\sec (c+d x)+1)}{\sqrt {a+b \sec (c+d x)}}dx}{a \left (a^2-b^2\right )}+\frac {4 b^2 \tan (c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {a \left (a^2-b^2\right )+\left (2 a b^2-2 a^2 b\right ) \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-2 a b^2 \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{a \left (a^2-b^2\right )}+\frac {4 b^2 \tan (c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}\)

\(\Big \downarrow \) 4409

\(\displaystyle \frac {a \left (a^2-b^2\right ) \int \frac {1}{\sqrt {a+b \sec (c+d x)}}dx-2 a b^2 \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-2 a b (a-b) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx}{a \left (a^2-b^2\right )}+\frac {4 b^2 \tan (c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {a \left (a^2-b^2\right ) \int \frac {1}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-2 a b^2 \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-2 a b (a-b) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{a \left (a^2-b^2\right )}+\frac {4 b^2 \tan (c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}\)

\(\Big \downarrow \) 4271

\(\displaystyle \frac {-2 a b^2 \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-2 a b (a-b) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 \sqrt {a+b} \left (a^2-b^2\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{d}}{a \left (a^2-b^2\right )}+\frac {4 b^2 \tan (c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}\)

\(\Big \downarrow \) 4319

\(\displaystyle \frac {-2 a b^2 \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 \sqrt {a+b} \left (a^2-b^2\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{d}-\frac {4 a (a-b) \sqrt {a+b} \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{d}}{a \left (a^2-b^2\right )}+\frac {4 b^2 \tan (c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}\)

\(\Big \downarrow \) 4492

\(\displaystyle \frac {-\frac {2 \sqrt {a+b} \left (a^2-b^2\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{d}-\frac {4 a (a-b) \sqrt {a+b} \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{d}+\frac {4 a (a-b) \sqrt {a+b} \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{d}}{a \left (a^2-b^2\right )}+\frac {4 b^2 \tan (c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}\)

Input:

Int[(a^2 - b^2*Sec[c + d*x]^2)/(a + b*Sec[c + d*x])^(5/2),x]
 

Output:

((4*a*(a - b)*Sqrt[a + b]*Cot[c + d*x]*EllipticE[ArcSin[Sqrt[a + b*Sec[c + 
 d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + b)] 
*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])/d - (4*a*(a - b)*Sqrt[a + b]*Cot 
[c + d*x]*EllipticF[ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]], (a + b)/ 
(a - b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]) 
)/(a - b))])/d - (2*Sqrt[a + b]*(a^2 - b^2)*Cot[c + d*x]*EllipticPi[(a + b 
)/a, ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[( 
b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])/d) 
/(a*(a^2 - b^2)) + (4*b^2*Tan[c + d*x])/((a^2 - b^2)*d*Sqrt[a + b*Sec[c + 
d*x]])
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 3042

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

rule 4271

Int[1/Sqrt[csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[2*(Rt[a 
 + b, 2]/(a*d*Cot[c + d*x]))*Sqrt[b*((1 - Csc[c + d*x])/(a + b))]*Sqrt[(-b) 
*((1 + Csc[c + d*x])/(a - b))]*EllipticPi[(a + b)/a, ArcSin[Sqrt[a + b*Csc[ 
c + d*x]]/Rt[a + b, 2]], (a + b)/(a - b)], x] /; FreeQ[{a, b, c, d}, x] && 
NeQ[a^2 - b^2, 0]
 

rule 4319

Int[csc[(e_.) + (f_.)*(x_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_S 
ymbol] :> Simp[-2*(Rt[a + b, 2]/(b*f*Cot[e + f*x]))*Sqrt[(b*(1 - Csc[e + f* 
x]))/(a + b)]*Sqrt[(-b)*((1 + Csc[e + f*x])/(a - b))]*EllipticF[ArcSin[Sqrt 
[a + b*Csc[e + f*x]]/Rt[a + b, 2]], (a + b)/(a - b)], x] /; FreeQ[{a, b, e, 
 f}, x] && NeQ[a^2 - b^2, 0]
 

rule 4409

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_ 
.) + (a_)], x_Symbol] :> Simp[c   Int[1/Sqrt[a + b*Csc[e + f*x]], x], x] + 
Simp[d   Int[Csc[e + f*x]/Sqrt[a + b*Csc[e + f*x]], x], x] /; FreeQ[{a, b, 
c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0]
 

rule 4411

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(d 
_.) + (c_)), x_Symbol] :> Simp[b*(b*c - a*d)*Cot[e + f*x]*((a + b*Csc[e + f 
*x])^(m + 1)/(a*f*(m + 1)*(a^2 - b^2))), x] + Simp[1/(a*(m + 1)*(a^2 - b^2) 
)   Int[(a + b*Csc[e + f*x])^(m + 1)*Simp[c*(a^2 - b^2)*(m + 1) - (a*(b*c - 
 a*d)*(m + 1))*Csc[e + f*x] + b*(b*c - a*d)*(m + 2)*Csc[e + f*x]^2, x], x], 
 x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] && N 
eQ[a^2 - b^2, 0] && IntegerQ[2*m]
 

rule 4492

Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/Sqrt[c 
sc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[-2*(A*b - a*B)*Rt[a 
 + b*(B/A), 2]*Sqrt[b*((1 - Csc[e + f*x])/(a + b))]*(Sqrt[(-b)*((1 + Csc[e 
+ f*x])/(a - b))]/(b^2*f*Cot[e + f*x]))*EllipticE[ArcSin[Sqrt[a + b*Csc[e + 
 f*x]]/Rt[a + b*(B/A), 2]], (a*A + b*B)/(a*A - b*B)], x] /; FreeQ[{a, b, e, 
 f, A, B}, x] && NeQ[a^2 - b^2, 0] && EqQ[A^2 - B^2, 0]
 

rule 4530

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(b_. 
) + (a_))^(m_.), x_Symbol] :> Simp[C/b^2   Int[(a + b*Csc[e + f*x])^(m + 1) 
*Simp[-a + b*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, C, m}, x] && 
 EqQ[A*b^2 + a^2*C, 0]
 

rule 4546

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. 
))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Int[(A + (B - C 
)*Csc[e + f*x])/Sqrt[a + b*Csc[e + f*x]], x] + Simp[C   Int[Csc[e + f*x]*(( 
1 + Csc[e + f*x])/Sqrt[a + b*Csc[e + f*x]]), x], x] /; FreeQ[{a, b, e, f, A 
, B, C}, x] && NeQ[a^2 - b^2, 0]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(731\) vs. \(2(309)=618\).

Time = 25.37 (sec) , antiderivative size = 732, normalized size of antiderivative = 2.17

method	result	size
default	\(\frac {4 \left (\left (-\left (1-\cos \left (d x +c \right )\right )^{3} \csc \left (d x +c \right )^{3}+\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ) a b +\left (\left (1-\cos \left (d x +c \right )\right )^{3} \csc \left (d x +c \right )^{3}-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right ) b^{2}-\sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {b +a \cos \left (d x +c \right )}{\left (a +b \right ) \left (\cos \left (d x +c \right )+1\right )}}\, \operatorname {EllipticF}\left (-\csc \left (d x +c \right )+\cot \left (d x +c \right ), \sqrt {\frac {a -b}{a +b}}\right ) a^{2}-2 \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {b +a \cos \left (d x +c \right )}{\left (a +b \right ) \left (\cos \left (d x +c \right )+1\right )}}\, \operatorname {EllipticF}\left (-\csc \left (d x +c \right )+\cot \left (d x +c \right ), \sqrt {\frac {a -b}{a +b}}\right ) a b -\sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {b +a \cos \left (d x +c \right )}{\left (a +b \right ) \left (\cos \left (d x +c \right )+1\right )}}\, \operatorname {EllipticF}\left (-\csc \left (d x +c \right )+\cot \left (d x +c \right ), \sqrt {\frac {a -b}{a +b}}\right ) b^{2}+2 \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {b +a \cos \left (d x +c \right )}{\left (a +b \right ) \left (\cos \left (d x +c \right )+1\right )}}\, \operatorname {EllipticE}\left (-\csc \left (d x +c \right )+\cot \left (d x +c \right ), \sqrt {\frac {a -b}{a +b}}\right ) a b +2 \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {b +a \cos \left (d x +c \right )}{\left (a +b \right ) \left (\cos \left (d x +c \right )+1\right )}}\, \operatorname {EllipticE}\left (-\csc \left (d x +c \right )+\cot \left (d x +c \right ), \sqrt {\frac {a -b}{a +b}}\right ) b^{2}+2 \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {b +a \cos \left (d x +c \right )}{\left (a +b \right ) \left (\cos \left (d x +c \right )+1\right )}}\, \operatorname {EllipticPi}\left (-\csc \left (d x +c \right )+\cot \left (d x +c \right ), -1, \sqrt {\frac {a -b}{a +b}}\right ) a^{2}-2 \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {b +a \cos \left (d x +c \right )}{\left (a +b \right ) \left (\cos \left (d x +c \right )+1\right )}}\, \operatorname {EllipticPi}\left (-\csc \left (d x +c \right )+\cot \left (d x +c \right ), -1, \sqrt {\frac {a -b}{a +b}}\right ) b^{2}\right ) \sqrt {a +b \sec \left (d x +c \right )}}{d \left (a -b \right ) \left (a +b \right ) \left (\left (1-\cos \left (d x +c \right )\right )^{2} a \csc \left (d x +c \right )^{2}-b \left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-a -b \right )}\)	\(732\)
parts	\(\text {Expression too large to display}\)	\(3101\)

Input:

int((a^2-b^2*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(5/2),x,method=_RETURNVERBOSE)
 

Output:

4/d/(a-b)/(a+b)*((-(1-cos(d*x+c))^3*csc(d*x+c)^3+csc(d*x+c)-cot(d*x+c))*a* 
b+((1-cos(d*x+c))^3*csc(d*x+c)^3-csc(d*x+c)+cot(d*x+c))*b^2-(cos(d*x+c)/(c 
os(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*Ellipt 
icF(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))*a^2-2*(cos(d*x+c)/(cos(d*x 
+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticF(-c 
sc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))*a*b-(cos(d*x+c)/(cos(d*x+c)+1))^ 
(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticF(-csc(d*x+c 
)+cot(d*x+c),((a-b)/(a+b))^(1/2))*b^2+2*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)* 
(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticE(-csc(d*x+c)+cot( 
d*x+c),((a-b)/(a+b))^(1/2))*a*b+2*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+ 
b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticE(-csc(d*x+c)+cot(d*x+c) 
,((a-b)/(a+b))^(1/2))*b^2+2*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+ 
a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticPi(-csc(d*x+c)+cot(d*x+c),-1,( 
(a-b)/(a+b))^(1/2))*a^2-2*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a* 
cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticPi(-csc(d*x+c)+cot(d*x+c),-1,((a 
-b)/(a+b))^(1/2))*b^2)*(a+b*sec(d*x+c))^(1/2)/((1-cos(d*x+c))^2*a*csc(d*x+ 
c)^2-b*(1-cos(d*x+c))^2*csc(d*x+c)^2-a-b)
 

Fricas [F]

\[ \int \frac {a^2-b^2 \sec ^2(c+d x)}{(a+b \sec (c+d x))^{5/2}} \, dx=\int { -\frac {b^{2} \sec \left (d x + c\right )^{2} - a^{2}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \] Input:

integrate((a^2-b^2*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(5/2),x, algorithm="fric 
as")
 

Output:

integral(-sqrt(b*sec(d*x + c) + a)*(b*sec(d*x + c) - a)/(b^2*sec(d*x + c)^ 
2 + 2*a*b*sec(d*x + c) + a^2), x)
 

Sympy [F]

\[ \int \frac {a^2-b^2 \sec ^2(c+d x)}{(a+b \sec (c+d x))^{5/2}} \, dx=\int \frac {a - b \sec {\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx \] Input:

integrate((a**2-b**2*sec(d*x+c)**2)/(a+b*sec(d*x+c))**(5/2),x)
 

Output:

Integral((a - b*sec(c + d*x))/(a + b*sec(c + d*x))**(3/2), x)
 

Maxima [F(-1)]

Timed out. \[ \int \frac {a^2-b^2 \sec ^2(c+d x)}{(a+b \sec (c+d x))^{5/2}} \, dx=\text {Timed out} \] Input:

integrate((a^2-b^2*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(5/2),x, algorithm="maxi 
ma")
 

Output:

Timed out
 

Giac [F]

\[ \int \frac {a^2-b^2 \sec ^2(c+d x)}{(a+b \sec (c+d x))^{5/2}} \, dx=\int { -\frac {b^{2} \sec \left (d x + c\right )^{2} - a^{2}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \] Input:

integrate((a^2-b^2*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(5/2),x, algorithm="giac 
")
 

Output:

integrate(-(b^2*sec(d*x + c)^2 - a^2)/(b*sec(d*x + c) + a)^(5/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {a^2-b^2 \sec ^2(c+d x)}{(a+b \sec (c+d x))^{5/2}} \, dx=-\int -\frac {a^2-\frac {b^2}{{\cos \left (c+d\,x\right )}^2}}{{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \] Input:

int((a^2 - b^2/cos(c + d*x)^2)/(a + b/cos(c + d*x))^(5/2),x)
 

Output:

-int(-(a^2 - b^2/cos(c + d*x)^2)/(a + b/cos(c + d*x))^(5/2), x)
 

Reduce [F]

\[ \int \frac {a^2-b^2 \sec ^2(c+d x)}{(a+b \sec (c+d x))^{5/2}} \, dx=\left (\int \frac {\sqrt {\sec \left (d x +c \right ) b +a}}{\sec \left (d x +c \right )^{2} b^{2}+2 \sec \left (d x +c \right ) a b +a^{2}}d x \right ) a -\left (\int \frac {\sqrt {\sec \left (d x +c \right ) b +a}\, \sec \left (d x +c \right )}{\sec \left (d x +c \right )^{2} b^{2}+2 \sec \left (d x +c \right ) a b +a^{2}}d x \right ) b \] Input:

int((a^2-b^2*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(5/2),x)
 

Output:

int(sqrt(sec(c + d*x)*b + a)/(sec(c + d*x)**2*b**2 + 2*sec(c + d*x)*a*b + 
a**2),x)*a - int((sqrt(sec(c + d*x)*b + a)*sec(c + d*x))/(sec(c + d*x)**2* 
b**2 + 2*sec(c + d*x)*a*b + a**2),x)*b