Integrand size = 50, antiderivative size = 379 \[ \int \frac {a b B-a^2 C+b^2 B \sec (c+d x)+b^2 C \sec ^2(c+d x)}{(a+b \sec (c+d x))^{5/2}} \, dx=\frac {2 (b B-2 a C) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{a \sqrt {a+b} d}-\frac {2 (b B-2 a C) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{a \sqrt {a+b} d}-\frac {2 \sqrt {a+b} (b B-a C) \cot (c+d x) \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{a^2 d}+\frac {2 b^2 (b B-2 a C) \tan (c+d x)}{a \left (a^2-b^2\right ) d \sqrt {a+b \sec (c+d x)}} \]
2*(B*b-2*C*a)*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/d/(a+b)^(1/2)-2*(B*b-2*C*a)*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/d/(a+b)^(1/2)-2*(B*b-C*a)*cot(d*x+c)*Ellipti cPi((a+b*sec(d*x+c))^(1/2)/(a+b)^(1/2),(a+b)/a,((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)/a^2/d +2*b^2*(B*b-2*C*a)*tan(d*x+c)/a/(a^2-b^2)/d/(a+b*sec(d*x+c))^(1/2)
Leaf count is larger than twice the leaf count of optimal. \(1603\) vs. \(2(379)=758\).
Time = 15.51 (sec) , antiderivative size = 1603, normalized size of antiderivative = 4.23 \[ \int \frac {a b B-a^2 C+b^2 B \sec (c+d x)+b^2 C \sec ^2(c+d x)}{(a+b \sec (c+d x))^{5/2}} \, dx =\text {Too large to display} \]
Integrate[(a*b*B - a^2*C + b^2*B*Sec[c + d*x] + b^2*C*Sec[c + d*x]^2)/(a + b*Sec[c + d*x])^(5/2),x]
((b + a*Cos[c + d*x])^2*Sec[c + d*x]*(b*B - a*C + b*C*Sec[c + d*x])*((2*b* (b*B - 2*a*C)*Sin[c + d*x])/(a*(-a^2 + b^2)) - (2*(-(b^3*B*Sin[c + d*x]) + 2*a*b^2*C*Sin[c + d*x]))/(a*(a^2 - b^2)*(b + a*Cos[c + d*x]))))/(d*(b*C + b*B*Cos[c + d*x] - a*C*Cos[c + d*x])*(a + b*Sec[c + d*x])^(3/2)) - (2*(b + a*Cos[c + d*x])^(3/2)*Sqrt[Sec[c + d*x]]*(b*B - a*C + b*C*Sec[c + d*x])* Sqrt[(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(1 + Tan[(c + d *x)/2]^2)]*(-(a*b^2*B*Tan[(c + d*x)/2]) - b^3*B*Tan[(c + d*x)/2] + 2*a^2*b *C*Tan[(c + d*x)/2] + 2*a*b^2*C*Tan[(c + d*x)/2] + 2*a*b^2*B*Tan[(c + d*x) /2]^3 - 4*a^2*b*C*Tan[(c + d*x)/2]^3 - a*b^2*B*Tan[(c + d*x)/2]^5 + b^3*B* Tan[(c + d*x)/2]^5 + 2*a^2*b*C*Tan[(c + d*x)/2]^5 - 2*a*b^2*C*Tan[(c + d*x )/2]^5 - 2*a^2*b*B*EllipticPi[-1, ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b )]*Sqrt[1 - Tan[(c + d*x)/2]^2]*Sqrt[(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan [(c + d*x)/2]^2)/(a + b)] + 2*b^3*B*EllipticPi[-1, ArcSin[Tan[(c + d*x)/2] ], (a - b)/(a + b)]*Sqrt[1 - Tan[(c + d*x)/2]^2]*Sqrt[(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(a + b)] + 2*a^3*C*EllipticPi[-1, ArcSin [Tan[(c + d*x)/2]], (a - b)/(a + b)]*Sqrt[1 - Tan[(c + d*x)/2]^2]*Sqrt[(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(a + b)] - 2*a*b^2*C*El lipticPi[-1, ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)]*Sqrt[1 - Tan[(c + d*x)/2]^2]*Sqrt[(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(a + b)] - 2*a^2*b*B*EllipticPi[-1, ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + ...
Time = 1.53 (sec) , antiderivative size = 421, normalized size of antiderivative = 1.11, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {2014, 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.
| \(\displaystyle \int \frac {a^2 (-C)+a b B+b^2 B \sec (c+d x)+b^2 C \sec ^2(c+d x)}{(a+b \sec (c+d x))^{5/2}} \, dx\) |
| \(\Big \downarrow \) 2014 |
| \(\displaystyle \frac {\int \frac {C \sec (c+d x) b^3+(b B-a C) b^2}{(a+b \sec (c+d x))^{3/2}}dx}{b^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {C \csc \left (c+d x+\frac {\pi }{2}\right ) b^3+(b B-a C) b^2}{\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}}dx}{b^2}\) |
| \(\Big \downarrow \) 4411 |
| \(\displaystyle \frac {\frac {2 b^4 (b B-2 a C) \tan (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}-\frac {2 \int -\frac {-\left ((b B-2 a C) \sec ^2(c+d x) b^4\right )-a (b B-2 a C) \sec (c+d x) b^3+\left (a^2-b^2\right ) (b B-a C) b^2}{2 \sqrt {a+b \sec (c+d x)}}dx}{a \left (a^2-b^2\right )}}{b^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\int \frac {-\left ((b B-2 a C) \sec ^2(c+d x) b^4\right )-a (b B-2 a C) \sec (c+d x) b^3+\left (a^2-b^2\right ) (b B-a C) b^2}{\sqrt {a+b \sec (c+d x)}}dx}{a \left (a^2-b^2\right )}+\frac {2 b^4 (b B-2 a C) \tan (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{b^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\int \frac {-\left ((b B-2 a C) \csc \left (c+d x+\frac {\pi }{2}\right )^2 b^4\right )-a (b B-2 a C) \csc \left (c+d x+\frac {\pi }{2}\right ) b^3+\left (a^2-b^2\right ) (b B-a C) b^2}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{a \left (a^2-b^2\right )}+\frac {2 b^4 (b B-2 a C) \tan (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{b^2}\) |
| \(\Big \downarrow \) 4546 |
| \(\displaystyle \frac {\frac {\int \frac {\left (a^2-b^2\right ) (b B-a C) b^2+\left (b^4 (b B-2 a C)-a b^3 (b B-2 a C)\right ) \sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx-b^4 (b B-2 a C) \int \frac {\sec (c+d x) (\sec (c+d x)+1)}{\sqrt {a+b \sec (c+d x)}}dx}{a \left (a^2-b^2\right )}+\frac {2 b^4 (b B-2 a C) \tan (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{b^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\int \frac {\left (a^2-b^2\right ) (b B-a C) b^2+\left (b^4 (b B-2 a C)-a b^3 (b B-2 a C)\right ) \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-b^4 (b B-2 a C) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{a \left (a^2-b^2\right )}+\frac {2 b^4 (b B-2 a C) \tan (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{b^2}\) |
| \(\Big \downarrow \) 4409 |
| \(\displaystyle \frac {\frac {b^2 \left (a^2-b^2\right ) (b B-a C) \int \frac {1}{\sqrt {a+b \sec (c+d x)}}dx-\left (b^4 (b B-2 a C) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )-b^3 (a-b) (b B-2 a C) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx}{a \left (a^2-b^2\right )}+\frac {2 b^4 (b B-2 a C) \tan (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{b^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {b^2 \left (a^2-b^2\right ) (b B-a C) \int \frac {1}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\left (b^4 (b B-2 a C) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )-b^3 (a-b) (b B-2 a C) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{a \left (a^2-b^2\right )}+\frac {2 b^4 (b B-2 a C) \tan (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{b^2}\) |
| \(\Big \downarrow \) 4271 |
| \(\displaystyle \frac {\frac {-\left (b^4 (b B-2 a C) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )-b^3 (a-b) (b B-2 a C) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 b^2 \sqrt {a+b} \left (a^2-b^2\right ) (b B-a C) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{a d}}{a \left (a^2-b^2\right )}+\frac {2 b^4 (b B-2 a C) \tan (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{b^2}\) |
| \(\Big \downarrow \) 4319 |
| \(\displaystyle \frac {\frac {-\left (b^4 (b B-2 a C) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )-\frac {2 b^2 \sqrt {a+b} \left (a^2-b^2\right ) (b B-a C) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{a d}-\frac {2 b^2 (a-b) \sqrt {a+b} (b B-2 a C) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{d}}{a \left (a^2-b^2\right )}+\frac {2 b^4 (b B-2 a C) \tan (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{b^2}\) |
| \(\Big \downarrow \) 4492 |
| \(\displaystyle \frac {\frac {-\frac {2 b^2 \sqrt {a+b} \left (a^2-b^2\right ) (b B-a C) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{a d}-\frac {2 b^2 (a-b) \sqrt {a+b} (b B-2 a C) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{d}+\frac {2 b^2 (a-b) \sqrt {a+b} (b B-2 a C) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{d}}{a \left (a^2-b^2\right )}+\frac {2 b^4 (b B-2 a C) \tan (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{b^2}\) |
Int[(a*b*B - a^2*C + b^2*B*Sec[c + d*x] + b^2*C*Sec[c + d*x]^2)/(a + b*Sec [c + d*x])^(5/2),x]
(((2*(a - b)*b^2*Sqrt[a + b]*(b*B - 2*a*C)*Cot[c + d*x]*EllipticE[ArcSin[S qrt[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*(a - b)* b^2*Sqrt[a + b]*(b*B - 2*a*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))])/d - (2*b^2*Sqrt[a + b]*(a^2 - b^2)*(b*B - a*C)*Cot[c + d*x]*EllipticPi[(a + b)/a, ArcSin[Sqrt[a + b*Se c[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))])/(a*d))/(a*(a^2 - b^2)) + (2 *b^4*(b*B - 2*a*C)*Tan[c + d*x])/(a*(a^2 - b^2)*d*Sqrt[a + b*Sec[c + d*x]] ))/b^2
3.10.79.3.1 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[(u_.)*((a_) + (b_.)*(v_))^(m_)*((A_.) + (B_.)*(v_) + (C_.)*(v_)^2), x_S ymbol] :> Simp[1/b^2 Int[u*(a + b*v)^(m + 1)*Simp[b*B - a*C + b*C*v, x], x], x] /; FreeQ[{a, b, A, B, C}, x] && EqQ[A*b^2 - a*b*B + a^2*C, 0] && LeQ [m, -1]
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]
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_)]*(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]
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]
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[((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]
Leaf count of result is larger than twice the leaf count of optimal. \(3613\) vs. \(2(350)=700\).
Time = 17.15 (sec) , antiderivative size = 3614, normalized size of antiderivative = 9.54
| method | result | size |
| default | \(\text {Expression too large to display}\) | \(3614\) |
| parts | \(\text {Expression too large to display}\) | \(10182\) |
int((B*a*b-C*a^2+b^2*B*sec(d*x+c)+b^2*C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(5/ 2),x,method=_RETURNVERBOSE)
2/d/a/(a+b)/(a-b)*(-B*((a*(1-cos(d*x+c))^2*csc(d*x+c)^2-(1-cos(d*x+c))^2*b *csc(d*x+c)^2-a-b)*((1-cos(d*x+c))^2*csc(d*x+c)^2-1))^(1/2)*(-(1-cos(d*x+c ))^2*csc(d*x+c)^2+1)^(1/2)*(-(a*(1-cos(d*x+c))^2*csc(d*x+c)^2-(1-cos(d*x+c ))^2*b*csc(d*x+c)^2-a-b)/(a+b))^(1/2)*EllipticF(cot(d*x+c)-csc(d*x+c),((a- b)/(a+b))^(1/2))*a^2*b-B*((a*(1-cos(d*x+c))^2*csc(d*x+c)^2-(1-cos(d*x+c))^ 2*b*csc(d*x+c)^2-a-b)*((1-cos(d*x+c))^2*csc(d*x+c)^2-1))^(1/2)*(-(1-cos(d* x+c))^2*csc(d*x+c)^2+1)^(1/2)*(-(a*(1-cos(d*x+c))^2*csc(d*x+c)^2-(1-cos(d* x+c))^2*b*csc(d*x+c)^2-a-b)/(a+b))^(1/2)*EllipticF(cot(d*x+c)-csc(d*x+c),( (a-b)/(a+b))^(1/2))*a*b^2+B*((a*(1-cos(d*x+c))^2*csc(d*x+c)^2-(1-cos(d*x+c ))^2*b*csc(d*x+c)^2-a-b)*((1-cos(d*x+c))^2*csc(d*x+c)^2-1))^(1/2)*(-(1-cos (d*x+c))^2*csc(d*x+c)^2+1)^(1/2)*(-(a*(1-cos(d*x+c))^2*csc(d*x+c)^2-(1-cos (d*x+c))^2*b*csc(d*x+c)^2-a-b)/(a+b))^(1/2)*EllipticE(cot(d*x+c)-csc(d*x+c ),((a-b)/(a+b))^(1/2))*a*b^2+B*((a*(1-cos(d*x+c))^2*csc(d*x+c)^2-(1-cos(d* x+c))^2*b*csc(d*x+c)^2-a-b)*((1-cos(d*x+c))^2*csc(d*x+c)^2-1))^(1/2)*(-(1- cos(d*x+c))^2*csc(d*x+c)^2+1)^(1/2)*(-(a*(1-cos(d*x+c))^2*csc(d*x+c)^2-(1- cos(d*x+c))^2*b*csc(d*x+c)^2-a-b)/(a+b))^(1/2)*EllipticE(cot(d*x+c)-csc(d* x+c),((a-b)/(a+b))^(1/2))*b^3+2*B*((a*(1-cos(d*x+c))^2*csc(d*x+c)^2-(1-cos (d*x+c))^2*b*csc(d*x+c)^2-a-b)*((1-cos(d*x+c))^2*csc(d*x+c)^2-1))^(1/2)*(- (1-cos(d*x+c))^2*csc(d*x+c)^2+1)^(1/2)*(-(a*(1-cos(d*x+c))^2*csc(d*x+c)^2- (1-cos(d*x+c))^2*b*csc(d*x+c)^2-a-b)/(a+b))^(1/2)*EllipticPi(cot(d*x+c)...
\[ \int \frac {a b B-a^2 C+b^2 B \sec (c+d x)+b^2 C \sec ^2(c+d x)}{(a+b \sec (c+d x))^{5/2}} \, dx=\int { \frac {C b^{2} \sec \left (d x + c\right )^{2} + B b^{2} \sec \left (d x + c\right ) - C a^{2} + B a b}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]
integrate((B*a*b-C*a^2+b^2*B*sec(d*x+c)+b^2*C*sec(d*x+c)^2)/(a+b*sec(d*x+c ))^(5/2),x, algorithm="fricas")
integral((C*b*sec(d*x + c) - C*a + B*b)*sqrt(b*sec(d*x + c) + a)/(b^2*sec( d*x + c)^2 + 2*a*b*sec(d*x + c) + a^2), x)
\[ \int \frac {a b B-a^2 C+b^2 B \sec (c+d x)+b^2 C \sec ^2(c+d x)}{(a+b \sec (c+d x))^{5/2}} \, dx=- \int \left (- \frac {B b}{a \sqrt {a + b \sec {\left (c + d x \right )}} + b \sqrt {a + b \sec {\left (c + d x \right )}} \sec {\left (c + d x \right )}}\right )\, dx - \int \frac {C a}{a \sqrt {a + b \sec {\left (c + d x \right )}} + b \sqrt {a + b \sec {\left (c + d x \right )}} \sec {\left (c + d x \right )}}\, dx - \int \left (- \frac {C b \sec {\left (c + d x \right )}}{a \sqrt {a + b \sec {\left (c + d x \right )}} + b \sqrt {a + b \sec {\left (c + d x \right )}} \sec {\left (c + d x \right )}}\right )\, dx \]
-Integral(-B*b/(a*sqrt(a + b*sec(c + d*x)) + b*sqrt(a + b*sec(c + d*x))*se c(c + d*x)), x) - Integral(C*a/(a*sqrt(a + b*sec(c + d*x)) + b*sqrt(a + b* sec(c + d*x))*sec(c + d*x)), x) - Integral(-C*b*sec(c + d*x)/(a*sqrt(a + b *sec(c + d*x)) + b*sqrt(a + b*sec(c + d*x))*sec(c + d*x)), x)
Timed out. \[ \int \frac {a b B-a^2 C+b^2 B \sec (c+d x)+b^2 C \sec ^2(c+d x)}{(a+b \sec (c+d x))^{5/2}} \, dx=\text {Timed out} \]
integrate((B*a*b-C*a^2+b^2*B*sec(d*x+c)+b^2*C*sec(d*x+c)^2)/(a+b*sec(d*x+c ))^(5/2),x, algorithm="maxima")
\[ \int \frac {a b B-a^2 C+b^2 B \sec (c+d x)+b^2 C \sec ^2(c+d x)}{(a+b \sec (c+d x))^{5/2}} \, dx=\int { \frac {C b^{2} \sec \left (d x + c\right )^{2} + B b^{2} \sec \left (d x + c\right ) - C a^{2} + B a b}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]
integrate((B*a*b-C*a^2+b^2*B*sec(d*x+c)+b^2*C*sec(d*x+c)^2)/(a+b*sec(d*x+c ))^(5/2),x, algorithm="giac")
integrate((C*b^2*sec(d*x + c)^2 + B*b^2*sec(d*x + c) - C*a^2 + B*a*b)/(b*s ec(d*x + c) + a)^(5/2), x)
Timed out. \[ \int \frac {a b B-a^2 C+b^2 B \sec (c+d x)+b^2 C \sec ^2(c+d x)}{(a+b \sec (c+d x))^{5/2}} \, dx=\int \frac {\frac {B\,b^2}{\cos \left (c+d\,x\right )}-C\,a^2+\frac {C\,b^2}{{\cos \left (c+d\,x\right )}^2}+B\,a\,b}{{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \]
int(((B*b^2)/cos(c + d*x) - C*a^2 + (C*b^2)/cos(c + d*x)^2 + B*a*b)/(a + b /cos(c + d*x))^(5/2),x)