Integrand size = 42, antiderivative size = 397 \[ \int \sec ^2(c+d x) \sqrt {a+b \sec (c+d x)} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {2 (a-b) \sqrt {a+b} \left (14 a^2 b B-63 b^3 B-8 a^3 C-19 a b^2 C\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{105 b^4 d}+\frac {2 (a-b) \sqrt {a+b} \left (b^2 (63 B-25 C)+2 a b (7 B-3 C)-8 a^2 C\right ) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{105 b^3 d}+\frac {2 \left (7 a b B-4 a^2 C+25 b^2 C\right ) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{105 b^2 d}+\frac {2 (7 b B+a C) \sec (c+d x) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{35 b d}+\frac {2 C \sec ^2(c+d x) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{7 d} \]
2/105*(a-b)*(14*B*a^2*b-63*B*b^3-8*C*a^3-19*C*a*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))*(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^4/d+2/105*(a-b) *(b^2*(63*B-25*C)+2*a*b*(7*B-3*C)-8*C*a^2)*cot(d*x+c)*EllipticF((a+b*sec(d *x+c))^(1/2)/(a+b)^(1/2),((a+b)/(a-b))^(1/2))*(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+2/105*(7*B*a*b-4*C*a^ 2+25*C*b^2)*(a+b*sec(d*x+c))^(1/2)*tan(d*x+c)/b^2/d+2/35*(7*B*b+C*a)*sec(d *x+c)*(a+b*sec(d*x+c))^(1/2)*tan(d*x+c)/b/d+2/7*C*sec(d*x+c)^2*(a+b*sec(d* x+c))^(1/2)*tan(d*x+c)/d
Leaf count is larger than twice the leaf count of optimal. \(3330\) vs. \(2(397)=794\).
Time = 22.56 (sec) , antiderivative size = 3330, normalized size of antiderivative = 8.39 \[ \int \sec ^2(c+d x) \sqrt {a+b \sec (c+d x)} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Result too large to show} \]
(Sqrt[a + b*Sec[c + d*x]]*((2*(-14*a^2*b*B + 63*b^3*B + 8*a^3*C + 19*a*b^2 *C)*Sin[c + d*x])/(105*b^3) + (2*Sec[c + d*x]^2*(7*b*B*Sin[c + d*x] + a*C* Sin[c + d*x]))/(35*b) + (2*Sec[c + d*x]*(7*a*b*B*Sin[c + d*x] - 4*a^2*C*Si n[c + d*x] + 25*b^2*C*Sin[c + d*x]))/(105*b^2) + (2*C*Sec[c + d*x]^2*Tan[c + d*x])/7))/d - (2*((2*a^2*B)/(15*b*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) - (3*b*B)/(5*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) - (19*a* C)/(105*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) - (8*a^3*C)/(105*b^2* Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) - (2*a*B*Sqrt[Sec[c + d*x]])/ (15*Sqrt[b + a*Cos[c + d*x]]) + (2*a^3*B*Sqrt[Sec[c + d*x]])/(15*b^2*Sqrt[ b + a*Cos[c + d*x]]) - (8*a^4*C*Sqrt[Sec[c + d*x]])/(105*b^3*Sqrt[b + a*Co s[c + d*x]]) - (17*a^2*C*Sqrt[Sec[c + d*x]])/(105*b*Sqrt[b + a*Cos[c + d*x ]]) + (5*b*C*Sqrt[Sec[c + d*x]])/(21*Sqrt[b + a*Cos[c + d*x]]) - (3*a*B*Co s[2*(c + d*x)]*Sqrt[Sec[c + d*x]])/(5*Sqrt[b + a*Cos[c + d*x]]) + (2*a^3*B *Cos[2*(c + d*x)]*Sqrt[Sec[c + d*x]])/(15*b^2*Sqrt[b + a*Cos[c + d*x]]) - (8*a^4*C*Cos[2*(c + d*x)]*Sqrt[Sec[c + d*x]])/(105*b^3*Sqrt[b + a*Cos[c + d*x]]) - (19*a^2*C*Cos[2*(c + d*x)]*Sqrt[Sec[c + d*x]])/(105*b*Sqrt[b + a* Cos[c + d*x]]))*Sqrt[Cos[(c + d*x)/2]^2*Sec[c + d*x]]*Sqrt[a + b*Sec[c + d *x]]*(2*(a + b)*(-14*a^2*b*B + 63*b^3*B + 8*a^3*C + 19*a*b^2*C)*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)] - 2*b*(a...
Time = 1.88 (sec) , antiderivative size = 413, normalized size of antiderivative = 1.04, number of steps used = 16, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {3042, 4560, 3042, 4519, 27, 3042, 4580, 27, 3042, 4570, 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.
| \(\displaystyle \int \sec ^2(c+d x) \sqrt {a+b \sec (c+d x)} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right )^2 \sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )} \left (B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )dx\) |
| \(\Big \downarrow \) 4560 |
| \(\displaystyle \int \sec ^3(c+d x) \sqrt {a+b \sec (c+d x)} (B+C \sec (c+d x))dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right )^3 \sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )} \left (B+C \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx\) |
| \(\Big \downarrow \) 4519 |
| \(\displaystyle \frac {2}{7} \int \frac {\sec ^2(c+d x) \left ((7 b B+a C) \sec ^2(c+d x)+(7 a B+5 b C) \sec (c+d x)+4 a C\right )}{2 \sqrt {a+b \sec (c+d x)}}dx+\frac {2 C \tan (c+d x) \sec ^2(c+d x) \sqrt {a+b \sec (c+d x)}}{7 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{7} \int \frac {\sec ^2(c+d x) \left ((7 b B+a C) \sec ^2(c+d x)+(7 a B+5 b C) \sec (c+d x)+4 a C\right )}{\sqrt {a+b \sec (c+d x)}}dx+\frac {2 C \tan (c+d x) \sec ^2(c+d x) \sqrt {a+b \sec (c+d x)}}{7 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{7} \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^2 \left ((7 b B+a C) \csc \left (c+d x+\frac {\pi }{2}\right )^2+(7 a B+5 b C) \csc \left (c+d x+\frac {\pi }{2}\right )+4 a C\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 C \tan (c+d x) \sec ^2(c+d x) \sqrt {a+b \sec (c+d x)}}{7 d}\) |
| \(\Big \downarrow \) 4580 |
| \(\displaystyle \frac {1}{7} \left (\frac {2 \int \frac {\sec (c+d x) \left (\left (-4 C a^2+7 b B a+25 b^2 C\right ) \sec ^2(c+d x)+b (21 b B+23 a C) \sec (c+d x)+2 a (7 b B+a C)\right )}{2 \sqrt {a+b \sec (c+d x)}}dx}{5 b}+\frac {2 (a C+7 b B) \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}\right )+\frac {2 C \tan (c+d x) \sec ^2(c+d x) \sqrt {a+b \sec (c+d x)}}{7 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{7} \left (\frac {\int \frac {\sec (c+d x) \left (\left (-4 C a^2+7 b B a+25 b^2 C\right ) \sec ^2(c+d x)+b (21 b B+23 a C) \sec (c+d x)+2 a (7 b B+a C)\right )}{\sqrt {a+b \sec (c+d x)}}dx}{5 b}+\frac {2 (a C+7 b B) \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}\right )+\frac {2 C \tan (c+d x) \sec ^2(c+d x) \sqrt {a+b \sec (c+d x)}}{7 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{7} \left (\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\left (-4 C a^2+7 b B a+25 b^2 C\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2+b (21 b B+23 a C) \csc \left (c+d x+\frac {\pi }{2}\right )+2 a (7 b B+a C)\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{5 b}+\frac {2 (a C+7 b B) \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}\right )+\frac {2 C \tan (c+d x) \sec ^2(c+d x) \sqrt {a+b \sec (c+d x)}}{7 d}\) |
| \(\Big \downarrow \) 4570 |
| \(\displaystyle \frac {1}{7} \left (\frac {\frac {2 \int \frac {\sec (c+d x) \left (b \left (2 C a^2+49 b B a+25 b^2 C\right )-\left (-8 C a^3+14 b B a^2-19 b^2 C a-63 b^3 B\right ) \sec (c+d x)\right )}{2 \sqrt {a+b \sec (c+d x)}}dx}{3 b}+\frac {2 \left (-4 a^2 C+7 a b B+25 b^2 C\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 b d}}{5 b}+\frac {2 (a C+7 b B) \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}\right )+\frac {2 C \tan (c+d x) \sec ^2(c+d x) \sqrt {a+b \sec (c+d x)}}{7 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{7} \left (\frac {\frac {\int \frac {\sec (c+d x) \left (b \left (2 C a^2+49 b B a+25 b^2 C\right )-\left (-8 C a^3+14 b B a^2-19 b^2 C a-63 b^3 B\right ) \sec (c+d x)\right )}{\sqrt {a+b \sec (c+d x)}}dx}{3 b}+\frac {2 \left (-4 a^2 C+7 a b B+25 b^2 C\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 b d}}{5 b}+\frac {2 (a C+7 b B) \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}\right )+\frac {2 C \tan (c+d x) \sec ^2(c+d x) \sqrt {a+b \sec (c+d x)}}{7 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{7} \left (\frac {\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (b \left (2 C a^2+49 b B a+25 b^2 C\right )+\left (8 C a^3-14 b B a^2+19 b^2 C a+63 b^3 B\right ) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 b}+\frac {2 \left (-4 a^2 C+7 a b B+25 b^2 C\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 b d}}{5 b}+\frac {2 (a C+7 b B) \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}\right )+\frac {2 C \tan (c+d x) \sec ^2(c+d x) \sqrt {a+b \sec (c+d x)}}{7 d}\) |
| \(\Big \downarrow \) 4493 |
| \(\displaystyle \frac {1}{7} \left (\frac {\frac {(a-b) \left (-8 a^2 C+a b (14 B-6 C)+b^2 (63 B-25 C)\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx-\left (-8 a^3 C+14 a^2 b B-19 a b^2 C-63 b^3 B\right ) \int \frac {\sec (c+d x) (\sec (c+d x)+1)}{\sqrt {a+b \sec (c+d x)}}dx}{3 b}+\frac {2 \left (-4 a^2 C+7 a b B+25 b^2 C\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 b d}}{5 b}+\frac {2 (a C+7 b B) \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}\right )+\frac {2 C \tan (c+d x) \sec ^2(c+d x) \sqrt {a+b \sec (c+d x)}}{7 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{7} \left (\frac {\frac {(a-b) \left (-8 a^2 C+a b (14 B-6 C)+b^2 (63 B-25 C)\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\left (-8 a^3 C+14 a^2 b B-19 a b^2 C-63 b^3 B\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 b}+\frac {2 \left (-4 a^2 C+7 a b B+25 b^2 C\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 b d}}{5 b}+\frac {2 (a C+7 b B) \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}\right )+\frac {2 C \tan (c+d x) \sec ^2(c+d x) \sqrt {a+b \sec (c+d x)}}{7 d}\) |
| \(\Big \downarrow \) 4319 |
| \(\displaystyle \frac {1}{7} \left (\frac {\frac {\frac {2 (a-b) \sqrt {a+b} \left (-8 a^2 C+a b (14 B-6 C)+b^2 (63 B-25 C)\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}-\left (-8 a^3 C+14 a^2 b B-19 a b^2 C-63 b^3 B\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 b}+\frac {2 \left (-4 a^2 C+7 a b B+25 b^2 C\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 b d}}{5 b}+\frac {2 (a C+7 b B) \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}\right )+\frac {2 C \tan (c+d x) \sec ^2(c+d x) \sqrt {a+b \sec (c+d x)}}{7 d}\) |
| \(\Big \downarrow \) 4492 |
| \(\displaystyle \frac {1}{7} \left (\frac {\frac {2 \left (-4 a^2 C+7 a b B+25 b^2 C\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 b d}+\frac {\frac {2 (a-b) \sqrt {a+b} \left (-8 a^2 C+a b (14 B-6 C)+b^2 (63 B-25 C)\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}+\frac {2 (a-b) \sqrt {a+b} \left (-8 a^3 C+14 a^2 b B-19 a b^2 C-63 b^3 B\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{b^2 d}}{3 b}}{5 b}+\frac {2 (a C+7 b B) \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}\right )+\frac {2 C \tan (c+d x) \sec ^2(c+d x) \sqrt {a+b \sec (c+d x)}}{7 d}\) |
(2*C*Sec[c + d*x]^2*Sqrt[a + b*Sec[c + d*x]]*Tan[c + d*x])/(7*d) + ((2*(7* b*B + a*C)*Sec[c + d*x]*Sqrt[a + b*Sec[c + d*x]]*Tan[c + d*x])/(5*b*d) + ( ((2*(a - b)*Sqrt[a + b]*(14*a^2*b*B - 63*b^3*B - 8*a^3*C - 19*a*b^2*C)*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))])/(b^2*d) + (2*(a - b)*Sqrt[a + b]*(b^2*(63*B - 25*C) + a*b*(14 *B - 6*C) - 8*a^2*C)*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))])/(b*d))/(3*b) + (2*(7*a*b*B - 4*a^2*C + 25*b^2*C)*Sqrt[a + b*Sec[c + d*x]]*Tan[c + d*x])/(3*b*d))/(5*b))/7
3.9.15.3.1 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_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-B)*d* Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^(n - 1)/(f*(m + n))), x] + Simp[d/(m + n) Int[(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^(n - 1)*Simp[a*B*(n - 1) + (b*B*(m + n - 1) + a*A*(m + n))*Csc[e + f*x] + (a* B*m + A*b*(m + n))*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, B }, x] && NeQ[A*b - a*B, 0] && NeQ[a^2 - b^2, 0] && LtQ[0, m, 1] && GtQ[n, 0 ]
Int[((a_.) + csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_. )*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*((c_.) + csc[(e_.) + (f_.) *(x_)]*(d_.))^(n_.), x_Symbol] :> Simp[1/b^2 Int[(a + b*Csc[e + f*x])^(m + 1)*(c + d*Csc[e + f*x])^n*(b*B - a*C + b*C*Csc[e + f*x]), x], x] /; FreeQ [{a, b, c, d, e, f, A, B, C, m, n}, x] && EqQ[A*b^2 - a*b*B + a^2*C, 0]
Int[csc[(e_.) + (f_.)*(x_)]*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e _.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_S ymbol] :> Simp[(-C)*Cot[e + f*x]*((a + b*Csc[e + f*x])^(m + 1)/(b*f*(m + 2) )), x] + Simp[1/(b*(m + 2)) Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*Simp[ b*A*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && !LtQ[m, -1]
Int[csc[(e_.) + (f_.)*(x_)]^2*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[ (e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x _Symbol] :> Simp[(-C)*Csc[e + f*x]*Cot[e + f*x]*((a + b*Csc[e + f*x])^(m + 1)/(b*f*(m + 3))), x] + Simp[1/(b*(m + 3)) Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*Simp[a*C + b*(C*(m + 2) + A*(m + 3))*Csc[e + f*x] - (2*a*C - b*B* (m + 3))*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] & & NeQ[a^2 - b^2, 0] && !LtQ[m, -1]
Leaf count of result is larger than twice the leaf count of optimal. \(4326\) vs. \(2(363)=726\).
Time = 26.61 (sec) , antiderivative size = 4327, normalized size of antiderivative = 10.90
| method | result | size |
| parts | \(\text {Expression too large to display}\) | \(4327\) |
| default | \(\text {Expression too large to display}\) | \(4370\) |
int(sec(d*x+c)^2*(B*sec(d*x+c)+C*sec(d*x+c)^2)*(a+b*sec(d*x+c))^(1/2),x,me thod=_RETURNVERBOSE)
2/15*B/d/b^2*(a+b*sec(d*x+c))^(1/2)/(b+a*cos(d*x+c))/(cos(d*x+c)+1)*(4*a*b ^2*tan(d*x+c)+9*sin(d*x+c)*cos(d*x+c)*a*b^2-4*EllipticE(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)*a^2*b*cos(d*x+c)+9*sin(d*x+c)*b^3+4*a*b^2 *sin(d*x+c)+3*b^3*tan(d*x+c)+a^2*b*cos(d*x+c)*sin(d*x+c)-9*(1/(a+b)*(b+a*c os(d*x+c))/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*Ellipti cF(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*b^3-2*(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)*EllipticE(cot( d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*a^3+9*(1/(a+b)*(b+a*cos(d*x+c))/(co s(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+18*EllipticE(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)*a*b^2*cos(d*x+c)+4*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)*(c os(d*x+c)/(cos(d*x+c)+1))^(1/2)*a^2*b*cos(d*x+c)-14*EllipticF(cot(d*x+c)-c sc(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)*a*b^2*cos(d*x+c)-2*EllipticE(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)*a^2*b*cos(d*x+c)^2+9*Elliptic E(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*(1/(a+b)*(b+a*cos(d*x+c))/...
\[ \int \sec ^2(c+d x) \sqrt {a+b \sec (c+d x)} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right )\right )} \sqrt {b \sec \left (d x + c\right ) + a} \sec \left (d x + c\right )^{2} \,d x } \]
integrate(sec(d*x+c)^2*(B*sec(d*x+c)+C*sec(d*x+c)^2)*(a+b*sec(d*x+c))^(1/2 ),x, algorithm="fricas")
\[ \int \sec ^2(c+d x) \sqrt {a+b \sec (c+d x)} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int \left (B + C \sec {\left (c + d x \right )}\right ) \sqrt {a + b \sec {\left (c + d x \right )}} \sec ^{3}{\left (c + d x \right )}\, dx \]
\[ \int \sec ^2(c+d x) \sqrt {a+b \sec (c+d x)} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right )\right )} \sqrt {b \sec \left (d x + c\right ) + a} \sec \left (d x + c\right )^{2} \,d x } \]
integrate(sec(d*x+c)^2*(B*sec(d*x+c)+C*sec(d*x+c)^2)*(a+b*sec(d*x+c))^(1/2 ),x, algorithm="maxima")
\[ \int \sec ^2(c+d x) \sqrt {a+b \sec (c+d x)} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right )\right )} \sqrt {b \sec \left (d x + c\right ) + a} \sec \left (d x + c\right )^{2} \,d x } \]
integrate(sec(d*x+c)^2*(B*sec(d*x+c)+C*sec(d*x+c)^2)*(a+b*sec(d*x+c))^(1/2 ),x, algorithm="giac")
Timed out. \[ \int \sec ^2(c+d x) \sqrt {a+b \sec (c+d x)} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int \frac {\left (\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )\,\sqrt {a+\frac {b}{\cos \left (c+d\,x\right )}}}{{\cos \left (c+d\,x\right )}^2} \,d x \]