Integrand size = 33, antiderivative size = 275 \[ \int \frac {\sec ^2(c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^{3/2}} \, dx=\frac {2 \left (a A b-2 a^2 B+b^2 B\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{b^3 \sqrt {a+b} d}+\frac {2 (A b-(2 a+b) B) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{b^2 \sqrt {a+b} d}+\frac {2 a (A b-a B) \tan (c+d x)}{b \left (a^2-b^2\right ) d \sqrt {a+b \sec (c+d x)}} \]
2*(A*a*b-2*B*a^2+B*b^2)*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)/b^3/d/(a+b)^(1/2)+2*(A*b-(2*a+b)*B)*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)/b^2/d/(a+b)^(1/2)+2*a*(A*b-B*a )*tan(d*x+c)/b/(a^2-b^2)/d/(a+b*sec(d*x+c))^(1/2)
Time = 20.31 (sec) , antiderivative size = 467, normalized size of antiderivative = 1.70 \[ \int \frac {\sec ^2(c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^{3/2}} \, dx=\frac {(b+a \cos (c+d x))^2 \sec ^2(c+d x) \left (\frac {2 \left (a A b-2 a^2 B+b^2 B\right ) \sin (c+d x)}{b^2 \left (-a^2+b^2\right )}-\frac {2 \left (a A b \sin (c+d x)-a^2 B \sin (c+d x)\right )}{b \left (-a^2+b^2\right ) (b+a \cos (c+d x))}\right )}{d (a+b \sec (c+d x))^{3/2}}+\frac {2 (b+a \cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x) \sqrt {\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)} \left (2 (a+b) \left (-a A b+2 a^2 B-b^2 B\right ) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {b+a \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} E\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {a-b}{a+b}\right )+2 b (a+b) (A b+(-2 a+b) B) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {b+a \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} \operatorname {EllipticF}\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {a-b}{a+b}\right )+\left (-a A b+2 a^2 B-b^2 B\right ) \cos (c+d x) (b+a \cos (c+d x)) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{b^2 \left (-a^2+b^2\right ) d \sqrt {\sec ^2\left (\frac {1}{2} (c+d x)\right )} (a+b \sec (c+d x))^{3/2}} \]
((b + a*Cos[c + d*x])^2*Sec[c + d*x]^2*((2*(a*A*b - 2*a^2*B + b^2*B)*Sin[c + d*x])/(b^2*(-a^2 + b^2)) - (2*(a*A*b*Sin[c + d*x] - a^2*B*Sin[c + d*x]) )/(b*(-a^2 + b^2)*(b + a*Cos[c + d*x]))))/(d*(a + b*Sec[c + d*x])^(3/2)) + (2*(b + a*Cos[c + d*x])*Sec[c + d*x]^(3/2)*Sqrt[Cos[(c + d*x)/2]^2*Sec[c + d*x]]*(2*(a + b)*(-(a*A*b) + 2*a^2*B - b^2*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]))]*Ellipt icE[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)] + 2*b*(a + b)*(A*b + (-2*a + b)*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]))]*EllipticF[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)] + (-(a*A*b) + 2*a^2*B - b^2*B)*Cos[c + d*x]*(b + a*Cos[c + d*x])*Se c[(c + d*x)/2]^2*Tan[(c + d*x)/2]))/(b^2*(-a^2 + b^2)*d*Sqrt[Sec[(c + d*x) /2]^2]*(a + b*Sec[c + d*x])^(3/2))
Time = 1.04 (sec) , antiderivative size = 302, normalized size of antiderivative = 1.10, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.242, Rules used = {3042, 4497, 27, 3042, 4493, 3042, 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.
| \(\displaystyle \int \frac {\sec ^2(c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^{3/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^2 \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}}dx\) |
| \(\Big \downarrow \) 4497 |
| \(\displaystyle \frac {2 \int -\frac {\sec (c+d x) \left (b (A b-a B)+\left (-2 B a^2+A b a+b^2 B\right ) \sec (c+d x)\right )}{2 \sqrt {a+b \sec (c+d x)}}dx}{b \left (a^2-b^2\right )}+\frac {2 a (A b-a B) \tan (c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {2 a (A b-a B) \tan (c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}-\frac {\int \frac {\sec (c+d x) \left (b (A b-a B)+\left (-2 B a^2+A b a+b^2 B\right ) \sec (c+d x)\right )}{\sqrt {a+b \sec (c+d x)}}dx}{b \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 a (A b-a B) \tan (c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}-\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (b (A b-a B)+\left (-2 B a^2+A b a+b^2 B\right ) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{b \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 4493 |
| \(\displaystyle \frac {2 a (A b-a B) \tan (c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}-\frac {\left (-2 a^2 B+a A b+b^2 B\right ) \int \frac {\sec (c+d x) (\sec (c+d x)+1)}{\sqrt {a+b \sec (c+d x)}}dx-(a-b) (A b-B (2 a+b)) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx}{b \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 a (A b-a B) \tan (c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}-\frac {\left (-2 a^2 B+a A b+b^2 B\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-(a-b) (A b-B (2 a+b)) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{b \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 4319 |
| \(\displaystyle \frac {2 a (A b-a B) \tan (c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}-\frac {\left (-2 a^2 B+a A b+b^2 B\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 (a-b) \sqrt {a+b} (A b-B (2 a+b)) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}}{b \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 4492 |
| \(\displaystyle \frac {2 a (A b-a B) \tan (c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}-\frac {-\frac {2 (a-b) \sqrt {a+b} \left (-2 a^2 B+a A b+b^2 B\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{b^2 d}-\frac {2 (a-b) \sqrt {a+b} (A b-B (2 a+b)) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}}{b \left (a^2-b^2\right )}\) |
-(((-2*(a - b)*Sqrt[a + b]*(a*A*b - 2*a^2*B + b^2*B)*Cot[c + d*x]*Elliptic E[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))])/(b^2* d) - (2*(a - b)*Sqrt[a + b]*(A*b - (2*a + b)*B)*Cot[c + d*x]*EllipticF[Arc Sin[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - S ec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])/(b*d))/(b* (a^2 - b^2))) + (2*a*(A*b - a*B)*Tan[c + d*x])/(b*(a^2 - b^2)*d*Sqrt[a + b *Sec[c + d*x]])
3.4.79.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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]
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]
Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/Sqrt[c sc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[(A - B) Int[Csc[e + f*x]/Sqrt[a + b*Csc[e + f*x]], x], x] + Simp[B 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} , x] && NeQ[a^2 - b^2, 0] && NeQ[A^2 - B^2, 0]
Int[csc[(e_.) + (f_.)*(x_)]^2*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*( csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[a*(A*b - a*B)*Cot[ e + f*x]*((a + b*Csc[e + f*x])^(m + 1)/(b*f*(m + 1)*(a^2 - b^2))), x] - Sim p[1/(b*(m + 1)*(a^2 - b^2)) Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m + 1) *Simp[b*(A*b - a*B)*(m + 1) - (a*A*b*(m + 2) - B*(a^2 + b^2*(m + 1)))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && NeQ[a^2 - b^2, 0] && LtQ[m, -1]
Leaf count of result is larger than twice the leaf count of optimal. \(3047\) vs. \(2(255)=510\).
Time = 23.72 (sec) , antiderivative size = 3048, normalized size of antiderivative = 11.08
| method | result | size |
| default | \(\text {Expression too large to display}\) | \(3048\) |
| parts | \(\text {Expression too large to display}\) | \(3048\) |
2/d/b^2/(a+b)/(a-b)*(-B*sin(d*x+c)*b^3-A*a^2*b*cos(d*x+c)*sin(d*x+c)+A*a*b ^2*cos(d*x+c)*sin(d*x+c)-B*a*b^2*cos(d*x+c)*sin(d*x+c)-B*a^2*b*cos(d*x+c)* sin(d*x+c)+2*B*EllipticF(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*(1/(a+ b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2 )*b^3*cos(d*x+c)+4*B*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(cos( d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticE(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b)) ^(1/2))*a^3*cos(d*x+c)-2*B*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2) *(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticE(cot(d*x+c)-csc(d*x+c),((a-b)/ (a+b))^(1/2))*b^3*cos(d*x+c)+2*A*EllipticF(cot(d*x+c)-csc(d*x+c),((a-b)/(a +b))^(1/2))*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(c os(d*x+c)+1))^(1/2)*b^3*cos(d*x+c)+A*EllipticF(cot(d*x+c)-csc(d*x+c),((a-b )/(a+b))^(1/2))*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c )/(cos(d*x+c)+1))^(1/2)*b^3*cos(d*x+c)^2+B*EllipticF(cot(d*x+c)-csc(d*x+c) ,((a-b)/(a+b))^(1/2))*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(cos (d*x+c)/(cos(d*x+c)+1))^(1/2)*b^3*cos(d*x+c)^2+2*B*(1/(a+b)*(b+a*cos(d*x+c ))/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticE(cot(d *x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*a^3*cos(d*x+c)^2-B*(1/(a+b)*(b+a*cos (d*x+c))/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticE (cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*b^3*cos(d*x+c)^2+A*(1/(a+b)*(b +a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticF(cot(d*x+c)-csc(d*x+c),((...
\[ \int \frac {\sec ^2(c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^{3/2}} \, dx=\int { \frac {{\left (B \sec \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{2}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x } \]
integral((B*sec(d*x + c)^3 + A*sec(d*x + c)^2)*sqrt(b*sec(d*x + c) + a)/(b ^2*sec(d*x + c)^2 + 2*a*b*sec(d*x + c) + a^2), x)
\[ \int \frac {\sec ^2(c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^{3/2}} \, dx=\int \frac {\left (A + B \sec {\left (c + d x \right )}\right ) \sec ^{2}{\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx \]
Timed out. \[ \int \frac {\sec ^2(c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^{3/2}} \, dx=\text {Timed out} \]
\[ \int \frac {\sec ^2(c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^{3/2}} \, dx=\int { \frac {{\left (B \sec \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{2}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {\sec ^2(c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^{3/2}} \, dx=\int \frac {A+\frac {B}{\cos \left (c+d\,x\right )}}{{\cos \left (c+d\,x\right )}^2\,{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{3/2}} \,d x \]