Integrand size = 33, antiderivative size = 435 \[ \int \frac {\cos ^2(c+d x) (A+B \sec (c+d x))}{\sqrt {a+b \sec (c+d x)}} \, dx=-\frac {(a-b) \sqrt {a+b} (3 A b-4 a B) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{4 a^2 b d}-\frac {\sqrt {a+b} (3 A b-2 a (A+2 B)) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{4 a^2 d}-\frac {\sqrt {a+b} \left (4 a^2 A+3 A b^2-4 a b B\right ) \cot (c+d x) \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{4 a^3 d}-\frac {(3 A b-4 a B) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{4 a^2 d}+\frac {A \cos (c+d x) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{2 a d} \] Output:
-1/4*(a-b)*(a+b)^(1/2)*(3*A*b-4*B*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^2/b/d-1/4*(a+b)^(1/2)*(3*A*b-2*a*(A+2*B))*c ot(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^2/d-1/4 *(a+b)^(1/2)*(4*A*a^2+3*A*b^2-4*B*a*b)*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^3/d-1/4*(3*A*b-4*B*a)*(a+b*sec(d *x+c))^(1/2)*sin(d*x+c)/a^2/d+1/2*A*cos(d*x+c)*(a+b*sec(d*x+c))^(1/2)*sin( d*x+c)/a/d
Leaf count is larger than twice the leaf count of optimal. \(1216\) vs. \(2(435)=870\).
Time = 13.51 (sec) , antiderivative size = 1216, normalized size of antiderivative = 2.80 \[ \int \frac {\cos ^2(c+d x) (A+B \sec (c+d x))}{\sqrt {a+b \sec (c+d x)}} \, dx =\text {Too large to display} \] Input:
Integrate[(Cos[c + d*x]^2*(A + B*Sec[c + d*x]))/Sqrt[a + b*Sec[c + d*x]],x ]
Output:
(A*(b + a*Cos[c + d*x])*Sec[c + d*x]*Sin[2*(c + d*x)])/(4*a*d*Sqrt[a + b*S ec[c + d*x]]) + (Sqrt[b + a*Cos[c + d*x]]*Sqrt[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)]*(- 3*a*A*b*Tan[(c + d*x)/2] - 3*A*b^2*Tan[(c + d*x)/2] + 4*a^2*B*Tan[(c + d*x )/2] + 4*a*b*B*Tan[(c + d*x)/2] + 6*a*A*b*Tan[(c + d*x)/2]^3 - 8*a^2*B*Tan [(c + d*x)/2]^3 - 3*a*A*b*Tan[(c + d*x)/2]^5 + 3*A*b^2*Tan[(c + d*x)/2]^5 + 4*a^2*B*Tan[(c + d*x)/2]^5 - 4*a*b*B*Tan[(c + d*x)/2]^5 + 8*a^2*A*Ellipt icPi[-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)] + 6*A*b^2*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)] - 8*a*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)] + 8*a^2*A*EllipticPi[-1, ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)]*Tan[(c + d*x)/2]^2*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)] + 6 *A*b^2*EllipticPi[-1, ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)]*Tan[(c + d*x)/2]^2*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)] - 8*a*b*B*EllipticPi[-1, ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)]*Tan[(c + d*x)/2]^2*Sqrt[1 - Tan[(c + d*x)/2]...
Time = 1.81 (sec) , antiderivative size = 436, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.424, Rules used = {3042, 4522, 27, 3042, 4592, 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.
\(\displaystyle \int \frac {\cos ^2(c+d x) (A+B \sec (c+d x))}{\sqrt {a+b \sec (c+d x)}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {A+B \csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^2 \sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx\) |
\(\Big \downarrow \) 4522 |
\(\displaystyle \frac {A \sin (c+d x) \cos (c+d x) \sqrt {a+b \sec (c+d x)}}{2 a d}-\frac {\int \frac {\cos (c+d x) \left (-A b \sec ^2(c+d x)-2 a A \sec (c+d x)+3 A b-4 a B\right )}{2 \sqrt {a+b \sec (c+d x)}}dx}{2 a}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {A \sin (c+d x) \cos (c+d x) \sqrt {a+b \sec (c+d x)}}{2 a d}-\frac {\int \frac {\cos (c+d x) \left (-A b \sec ^2(c+d x)-2 a A \sec (c+d x)+3 A b-4 a B\right )}{\sqrt {a+b \sec (c+d x)}}dx}{4 a}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {A \sin (c+d x) \cos (c+d x) \sqrt {a+b \sec (c+d x)}}{2 a d}-\frac {\int \frac {-A b \csc \left (c+d x+\frac {\pi }{2}\right )^2-2 a A \csc \left (c+d x+\frac {\pi }{2}\right )+3 A b-4 a B}{\csc \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{4 a}\) |
\(\Big \downarrow \) 4592 |
\(\displaystyle \frac {A \sin (c+d x) \cos (c+d x) \sqrt {a+b \sec (c+d x)}}{2 a d}-\frac {\frac {(3 A b-4 a B) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{a d}-\frac {\int \frac {4 A a^2-4 b B a+2 A b \sec (c+d x) a+3 A b^2+b (3 A b-4 a B) \sec ^2(c+d x)}{2 \sqrt {a+b \sec (c+d x)}}dx}{a}}{4 a}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {A \sin (c+d x) \cos (c+d x) \sqrt {a+b \sec (c+d x)}}{2 a d}-\frac {\frac {(3 A b-4 a B) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{a d}-\frac {\int \frac {4 A a^2-4 b B a+2 A b \sec (c+d x) a+3 A b^2+b (3 A b-4 a B) \sec ^2(c+d x)}{\sqrt {a+b \sec (c+d x)}}dx}{2 a}}{4 a}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {A \sin (c+d x) \cos (c+d x) \sqrt {a+b \sec (c+d x)}}{2 a d}-\frac {\frac {(3 A b-4 a B) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{a d}-\frac {\int \frac {4 A a^2-4 b B a+2 A b \csc \left (c+d x+\frac {\pi }{2}\right ) a+3 A b^2+b (3 A b-4 a B) \csc \left (c+d x+\frac {\pi }{2}\right )^2}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{2 a}}{4 a}\) |
\(\Big \downarrow \) 4546 |
\(\displaystyle \frac {A \sin (c+d x) \cos (c+d x) \sqrt {a+b \sec (c+d x)}}{2 a d}-\frac {\frac {(3 A b-4 a B) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{a d}-\frac {\int \frac {4 A a^2-4 b B a+3 A b^2+(2 a A b-b (3 A b-4 a B)) \sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx+b (3 A b-4 a B) \int \frac {\sec (c+d x) (\sec (c+d x)+1)}{\sqrt {a+b \sec (c+d x)}}dx}{2 a}}{4 a}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {A \sin (c+d x) \cos (c+d x) \sqrt {a+b \sec (c+d x)}}{2 a d}-\frac {\frac {(3 A b-4 a B) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{a d}-\frac {\int \frac {4 A a^2-4 b B a+3 A b^2+(2 a A b-b (3 A b-4 a B)) \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+b (3 A b-4 a B) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{2 a}}{4 a}\) |
\(\Big \downarrow \) 4409 |
\(\displaystyle \frac {A \sin (c+d x) \cos (c+d x) \sqrt {a+b \sec (c+d x)}}{2 a d}-\frac {\frac {(3 A b-4 a B) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{a d}-\frac {\left (4 a^2 A-4 a b B+3 A b^2\right ) \int \frac {1}{\sqrt {a+b \sec (c+d x)}}dx+b (3 A b-4 a B) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-b (3 A b-2 a (A+2 B)) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx}{2 a}}{4 a}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {A \sin (c+d x) \cos (c+d x) \sqrt {a+b \sec (c+d x)}}{2 a d}-\frac {\frac {(3 A b-4 a B) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{a d}-\frac {\left (4 a^2 A-4 a b B+3 A b^2\right ) \int \frac {1}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-b (3 A b-2 a (A+2 B)) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+b (3 A b-4 a B) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{2 a}}{4 a}\) |
\(\Big \downarrow \) 4271 |
\(\displaystyle \frac {A \sin (c+d x) \cos (c+d x) \sqrt {a+b \sec (c+d x)}}{2 a d}-\frac {\frac {(3 A b-4 a B) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{a d}-\frac {-b (3 A b-2 a (A+2 B)) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+b (3 A b-4 a B) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 \sqrt {a+b} \left (4 a^2 A-4 a b B+3 A b^2\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{a d}}{2 a}}{4 a}\) |
\(\Big \downarrow \) 4319 |
\(\displaystyle \frac {A \sin (c+d x) \cos (c+d x) \sqrt {a+b \sec (c+d x)}}{2 a d}-\frac {\frac {(3 A b-4 a B) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{a d}-\frac {b (3 A b-4 a B) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 \sqrt {a+b} \left (4 a^2 A-4 a b B+3 A b^2\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{a d}-\frac {2 \sqrt {a+b} (3 A b-2 a (A+2 B)) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{d}}{2 a}}{4 a}\) |
\(\Big \downarrow \) 4492 |
\(\displaystyle \frac {A \sin (c+d x) \cos (c+d x) \sqrt {a+b \sec (c+d x)}}{2 a d}-\frac {\frac {(3 A b-4 a B) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{a d}-\frac {-\frac {2 \sqrt {a+b} \left (4 a^2 A-4 a b B+3 A b^2\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{a d}-\frac {2 \sqrt {a+b} (3 A b-2 a (A+2 B)) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{d}-\frac {2 (a-b) \sqrt {a+b} (3 A b-4 a B) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{b d}}{2 a}}{4 a}\) |
Input:
Int[(Cos[c + d*x]^2*(A + B*Sec[c + d*x]))/Sqrt[a + b*Sec[c + d*x]],x]
Output:
(A*Cos[c + d*x]*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x])/(2*a*d) - (-1/2*((- 2*(a - b)*Sqrt[a + b]*(3*A*b - 4*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))])/(b*d) - (2*Sqrt[a + b]*(3*A*b - 2*a*(A + 2*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]*(4*a^2*A + 3 *A*b^2 - 4*a*b*B)*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))])/(a*d))/a + ((3*A*b - 4*a*B)* Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x])/(a*d))/(4*a)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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_)]*(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_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[A*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*((d*Csc[e + f*x])^n/(a*f*n)), x] + Sim p[1/(a*d*n) Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[a*B* n - A*b*(m + n + 1) + A*a*(n + 1)*Csc[e + f*x] + A*b*(m + n + 2)*Csc[e + f* x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, B, m}, x] && NeQ[A*b - a*B, 0] && NeQ[a^2 - b^2, 0] && LeQ[n, -1]
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]
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. ))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a _))^(m_), x_Symbol] :> Simp[A*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*((d *Csc[e + f*x])^n/(a*f*n)), x] + Simp[1/(a*d*n) Int[(a + b*Csc[e + f*x])^m *(d*Csc[e + f*x])^(n + 1)*Simp[a*B*n - A*b*(m + n + 1) + a*(A + A*n + C*n)* Csc[e + f*x] + A*b*(m + n + 2)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d , e, f, A, B, C, m}, x] && NeQ[a^2 - b^2, 0] && LeQ[n, -1]
Leaf count of result is larger than twice the leaf count of optimal. \(1069\) vs. \(2(394)=788\).
Time = 10.34 (sec) , antiderivative size = 1070, normalized size of antiderivative = 2.46
Input:
int(cos(d*x+c)^2*(A+B*sec(d*x+c))/(a+b*sec(d*x+c))^(1/2),x,method=_RETURNV ERBOSE)
Output:
1/4/d/a^2*((-8*cos(d*x+c)^2-16*cos(d*x+c)-8)*A*(cos(d*x+c)/(1+cos(d*x+c))) ^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*a^2*EllipticPi(-csc (d*x+c)+cot(d*x+c),-1,((a-b)/(a+b))^(1/2))+(-6*cos(d*x+c)^2-12*cos(d*x+c)- 6)*A*(1/(a+b)*(b+a*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d* x+c)))^(1/2)*b^2*EllipticPi(-csc(d*x+c)+cot(d*x+c),-1,((a-b)/(a+b))^(1/2)) +(8*cos(d*x+c)^2+16*cos(d*x+c)+8)*B*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/( a+b)*(b+a*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*a*b*EllipticPi(-csc(d*x+c)+cot (d*x+c),-1,((a-b)/(a+b))^(1/2))+(3*cos(d*x+c)^2+6*cos(d*x+c)+3)*A*(1/(a+b) *(b+a*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)* a*b*EllipticE(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))+(3*cos(d*x+c)^2+ 6*cos(d*x+c)+3)*A*(1/(a+b)*(b+a*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*(cos(d*x +c)/(1+cos(d*x+c)))^(1/2)*b^2*EllipticE(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b ))^(1/2))+(-4*cos(d*x+c)^2-8*cos(d*x+c)-4)*B*(cos(d*x+c)/(1+cos(d*x+c)))^( 1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*a^2*EllipticE(-csc(d* x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))+(-4*cos(d*x+c)^2-8*cos(d*x+c)-4)*B*(c os(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(1+cos(d*x+c)))^ (1/2)*a*b*EllipticE(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))+(4*cos(d*x +c)^2+8*cos(d*x+c)+4)*A*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(b+a*co s(d*x+c))/(1+cos(d*x+c)))^(1/2)*a^2*EllipticF(-csc(d*x+c)+cot(d*x+c),((a-b )/(a+b))^(1/2))+(-2*cos(d*x+c)^2-4*cos(d*x+c)-2)*A*(cos(d*x+c)/(1+cos(d...
\[ \int \frac {\cos ^2(c+d x) (A+B \sec (c+d x))}{\sqrt {a+b \sec (c+d x)}} \, dx=\int { \frac {{\left (B \sec \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{2}}{\sqrt {b \sec \left (d x + c\right ) + a}} \,d x } \] Input:
integrate(cos(d*x+c)^2*(A+B*sec(d*x+c))/(a+b*sec(d*x+c))^(1/2),x, algorith m="fricas")
Output:
integral((B*cos(d*x + c)^2*sec(d*x + c) + A*cos(d*x + c)^2)/sqrt(b*sec(d*x + c) + a), x)
\[ \int \frac {\cos ^2(c+d x) (A+B \sec (c+d x))}{\sqrt {a+b \sec (c+d x)}} \, dx=\int \frac {\left (A + B \sec {\left (c + d x \right )}\right ) \cos ^{2}{\left (c + d x \right )}}{\sqrt {a + b \sec {\left (c + d x \right )}}}\, dx \] Input:
integrate(cos(d*x+c)**2*(A+B*sec(d*x+c))/(a+b*sec(d*x+c))**(1/2),x)
Output:
Integral((A + B*sec(c + d*x))*cos(c + d*x)**2/sqrt(a + b*sec(c + d*x)), x)
\[ \int \frac {\cos ^2(c+d x) (A+B \sec (c+d x))}{\sqrt {a+b \sec (c+d x)}} \, dx=\int { \frac {{\left (B \sec \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{2}}{\sqrt {b \sec \left (d x + c\right ) + a}} \,d x } \] Input:
integrate(cos(d*x+c)^2*(A+B*sec(d*x+c))/(a+b*sec(d*x+c))^(1/2),x, algorith m="maxima")
Output:
integrate((B*sec(d*x + c) + A)*cos(d*x + c)^2/sqrt(b*sec(d*x + c) + a), x)
\[ \int \frac {\cos ^2(c+d x) (A+B \sec (c+d x))}{\sqrt {a+b \sec (c+d x)}} \, dx=\int { \frac {{\left (B \sec \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{2}}{\sqrt {b \sec \left (d x + c\right ) + a}} \,d x } \] Input:
integrate(cos(d*x+c)^2*(A+B*sec(d*x+c))/(a+b*sec(d*x+c))^(1/2),x, algorith m="giac")
Output:
integrate((B*sec(d*x + c) + A)*cos(d*x + c)^2/sqrt(b*sec(d*x + c) + a), x)
Timed out. \[ \int \frac {\cos ^2(c+d x) (A+B \sec (c+d x))}{\sqrt {a+b \sec (c+d x)}} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^2\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}\right )}{\sqrt {a+\frac {b}{\cos \left (c+d\,x\right )}}} \,d x \] Input:
int((cos(c + d*x)^2*(A + B/cos(c + d*x)))/(a + b/cos(c + d*x))^(1/2),x)
Output:
int((cos(c + d*x)^2*(A + B/cos(c + d*x)))/(a + b/cos(c + d*x))^(1/2), x)
\[ \int \frac {\cos ^2(c+d x) (A+B \sec (c+d x))}{\sqrt {a+b \sec (c+d x)}} \, dx=\int \sqrt {\sec \left (d x +c \right ) b +a}\, \cos \left (d x +c \right )^{2}d x \] Input:
int(cos(d*x+c)^2*(A+B*sec(d*x+c))/(a+b*sec(d*x+c))^(1/2),x)
Output:
int(sqrt(sec(c + d*x)*b + a)*cos(c + d*x)**2,x)