Integrand size = 41, antiderivative size = 271 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec (c+d x)}{(a+b \cos (c+d x))^{3/2}} \, dx=-\frac {2 \left (A b^2-a (b B-a C)\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{a b \left (a^2-b^2\right ) d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}+\frac {2 C \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{b d \sqrt {a+b \cos (c+d x)}}+\frac {2 A \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{a d \sqrt {a+b \cos (c+d x)}}+\frac {2 \left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{a \left (a^2-b^2\right ) d \sqrt {a+b \cos (c+d x)}} \]
2*(A*b^2-a*(B*b-C*a))*sin(d*x+c)/a/(a^2-b^2)/d/(a+b*cos(d*x+c))^(1/2)-2*(A *b^2-a*(B*b-C*a))*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*Elliptic E(sin(1/2*d*x+1/2*c),2^(1/2)*(b/(a+b))^(1/2))*(a+b*cos(d*x+c))^(1/2)/a/b/( a^2-b^2)/d/((a+b*cos(d*x+c))/(a+b))^(1/2)+2*C*(cos(1/2*d*x+1/2*c)^2)^(1/2) /cos(1/2*d*x+1/2*c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2)*(b/(a+b))^(1/2))* ((a+b*cos(d*x+c))/(a+b))^(1/2)/b/d/(a+b*cos(d*x+c))^(1/2)+2*A*(cos(1/2*d*x +1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticPi(sin(1/2*d*x+1/2*c),2,2^(1/2 )*(b/(a+b))^(1/2))*((a+b*cos(d*x+c))/(a+b))^(1/2)/a/d/(a+b*cos(d*x+c))^(1/ 2)
\[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec (c+d x)}{(a+b \cos (c+d x))^{3/2}} \, dx=\int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec (c+d x)}{(a+b \cos (c+d x))^{3/2}} \, dx \]
Integrate[((A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x])/(a + b*Co s[c + d*x])^(3/2),x]
Integrate[((A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x])/(a + b*Co s[c + d*x])^(3/2), x]
Time = 2.18 (sec) , antiderivative size = 292, normalized size of antiderivative = 1.08, number of steps used = 18, number of rules used = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.439, Rules used = {3042, 3534, 27, 3042, 3538, 25, 3042, 3134, 3042, 3132, 3481, 3042, 3142, 3042, 3140, 3286, 3042, 3284}
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.
\(\displaystyle \int \frac {\sec (c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^{3/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {A+B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right ) \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}}dx\) |
\(\Big \downarrow \) 3534 |
\(\displaystyle \frac {2 \int \frac {\left (-\left (\left (A b^2-a (b B-a C)\right ) \cos ^2(c+d x)\right )-a (A b+C b-a B) \cos (c+d x)+A \left (a^2-b^2\right )\right ) \sec (c+d x)}{2 \sqrt {a+b \cos (c+d x)}}dx}{a \left (a^2-b^2\right )}+\frac {2 \sin (c+d x) \left (A b^2-a (b B-a C)\right )}{a d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {\left (-\left (\left (A b^2-a (b B-a C)\right ) \cos ^2(c+d x)\right )-a (A b+C b-a B) \cos (c+d x)+A \left (a^2-b^2\right )\right ) \sec (c+d x)}{\sqrt {a+b \cos (c+d x)}}dx}{a \left (a^2-b^2\right )}+\frac {2 \sin (c+d x) \left (A b^2-a (b B-a C)\right )}{a d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {\left (a (b B-a C)-A b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2-a (A b+C b-a B) \sin \left (c+d x+\frac {\pi }{2}\right )+A \left (a^2-b^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{a \left (a^2-b^2\right )}+\frac {2 \sin (c+d x) \left (A b^2-a (b B-a C)\right )}{a d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}\) |
\(\Big \downarrow \) 3538 |
\(\displaystyle \frac {-\frac {\int -\frac {\left (A b \left (a^2-b^2\right )+a C \cos (c+d x) \left (a^2-b^2\right )\right ) \sec (c+d x)}{\sqrt {a+b \cos (c+d x)}}dx}{b}-\frac {\left (A b^2-a (b B-a C)\right ) \int \sqrt {a+b \cos (c+d x)}dx}{b}}{a \left (a^2-b^2\right )}+\frac {2 \sin (c+d x) \left (A b^2-a (b B-a C)\right )}{a d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\frac {\int \frac {\left (A b \left (a^2-b^2\right )+a C \cos (c+d x) \left (a^2-b^2\right )\right ) \sec (c+d x)}{\sqrt {a+b \cos (c+d x)}}dx}{b}-\frac {\left (A b^2-a (b B-a C)\right ) \int \sqrt {a+b \cos (c+d x)}dx}{b}}{a \left (a^2-b^2\right )}+\frac {2 \sin (c+d x) \left (A b^2-a (b B-a C)\right )}{a d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {\int \frac {A b \left (a^2-b^2\right )+a C \sin \left (c+d x+\frac {\pi }{2}\right ) \left (a^2-b^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}-\frac {\left (A b^2-a (b B-a C)\right ) \int \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{b}}{a \left (a^2-b^2\right )}+\frac {2 \sin (c+d x) \left (A b^2-a (b B-a C)\right )}{a d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}\) |
\(\Big \downarrow \) 3134 |
\(\displaystyle \frac {\frac {\int \frac {A b \left (a^2-b^2\right )+a C \sin \left (c+d x+\frac {\pi }{2}\right ) \left (a^2-b^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}-\frac {\left (A b^2-a (b B-a C)\right ) \sqrt {a+b \cos (c+d x)} \int \sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}}dx}{b \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{a \left (a^2-b^2\right )}+\frac {2 \sin (c+d x) \left (A b^2-a (b B-a C)\right )}{a d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {\int \frac {A b \left (a^2-b^2\right )+a C \sin \left (c+d x+\frac {\pi }{2}\right ) \left (a^2-b^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}-\frac {\left (A b^2-a (b B-a C)\right ) \sqrt {a+b \cos (c+d x)} \int \sqrt {\frac {a}{a+b}+\frac {b \sin \left (c+d x+\frac {\pi }{2}\right )}{a+b}}dx}{b \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{a \left (a^2-b^2\right )}+\frac {2 \sin (c+d x) \left (A b^2-a (b B-a C)\right )}{a d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}\) |
\(\Big \downarrow \) 3132 |
\(\displaystyle \frac {\frac {\int \frac {A b \left (a^2-b^2\right )+a C \sin \left (c+d x+\frac {\pi }{2}\right ) \left (a^2-b^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}-\frac {2 \left (A b^2-a (b B-a C)\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{b d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{a \left (a^2-b^2\right )}+\frac {2 \sin (c+d x) \left (A b^2-a (b B-a C)\right )}{a d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}\) |
\(\Big \downarrow \) 3481 |
\(\displaystyle \frac {\frac {A b \left (a^2-b^2\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \cos (c+d x)}}dx+a C \left (a^2-b^2\right ) \int \frac {1}{\sqrt {a+b \cos (c+d x)}}dx}{b}-\frac {2 \left (A b^2-a (b B-a C)\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{b d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{a \left (a^2-b^2\right )}+\frac {2 \sin (c+d x) \left (A b^2-a (b B-a C)\right )}{a d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {A b \left (a^2-b^2\right ) \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+a C \left (a^2-b^2\right ) \int \frac {1}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}-\frac {2 \left (A b^2-a (b B-a C)\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{b d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{a \left (a^2-b^2\right )}+\frac {2 \sin (c+d x) \left (A b^2-a (b B-a C)\right )}{a d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}\) |
\(\Big \downarrow \) 3142 |
\(\displaystyle \frac {\frac {A b \left (a^2-b^2\right ) \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {a C \left (a^2-b^2\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \int \frac {1}{\sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}}}dx}{\sqrt {a+b \cos (c+d x)}}}{b}-\frac {2 \left (A b^2-a (b B-a C)\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{b d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{a \left (a^2-b^2\right )}+\frac {2 \sin (c+d x) \left (A b^2-a (b B-a C)\right )}{a d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {A b \left (a^2-b^2\right ) \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {a C \left (a^2-b^2\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \int \frac {1}{\sqrt {\frac {a}{a+b}+\frac {b \sin \left (c+d x+\frac {\pi }{2}\right )}{a+b}}}dx}{\sqrt {a+b \cos (c+d x)}}}{b}-\frac {2 \left (A b^2-a (b B-a C)\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{b d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{a \left (a^2-b^2\right )}+\frac {2 \sin (c+d x) \left (A b^2-a (b B-a C)\right )}{a d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}\) |
\(\Big \downarrow \) 3140 |
\(\displaystyle \frac {\frac {A b \left (a^2-b^2\right ) \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 a C \left (a^2-b^2\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{d \sqrt {a+b \cos (c+d x)}}}{b}-\frac {2 \left (A b^2-a (b B-a C)\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{b d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{a \left (a^2-b^2\right )}+\frac {2 \sin (c+d x) \left (A b^2-a (b B-a C)\right )}{a d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}\) |
\(\Big \downarrow \) 3286 |
\(\displaystyle \frac {\frac {\frac {A b \left (a^2-b^2\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \int \frac {\sec (c+d x)}{\sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}}}dx}{\sqrt {a+b \cos (c+d x)}}+\frac {2 a C \left (a^2-b^2\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{d \sqrt {a+b \cos (c+d x)}}}{b}-\frac {2 \left (A b^2-a (b B-a C)\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{b d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{a \left (a^2-b^2\right )}+\frac {2 \sin (c+d x) \left (A b^2-a (b B-a C)\right )}{a d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {\frac {A b \left (a^2-b^2\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {\frac {a}{a+b}+\frac {b \sin \left (c+d x+\frac {\pi }{2}\right )}{a+b}}}dx}{\sqrt {a+b \cos (c+d x)}}+\frac {2 a C \left (a^2-b^2\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{d \sqrt {a+b \cos (c+d x)}}}{b}-\frac {2 \left (A b^2-a (b B-a C)\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{b d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{a \left (a^2-b^2\right )}+\frac {2 \sin (c+d x) \left (A b^2-a (b B-a C)\right )}{a d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}\) |
\(\Big \downarrow \) 3284 |
\(\displaystyle \frac {2 \sin (c+d x) \left (A b^2-a (b B-a C)\right )}{a d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}+\frac {\frac {\frac {2 A b \left (a^2-b^2\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{d \sqrt {a+b \cos (c+d x)}}+\frac {2 a C \left (a^2-b^2\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{d \sqrt {a+b \cos (c+d x)}}}{b}-\frac {2 \left (A b^2-a (b B-a C)\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{b d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{a \left (a^2-b^2\right )}\) |
((-2*(A*b^2 - a*(b*B - a*C))*Sqrt[a + b*Cos[c + d*x]]*EllipticE[(c + d*x)/ 2, (2*b)/(a + b)])/(b*d*Sqrt[(a + b*Cos[c + d*x])/(a + b)]) + ((2*a*(a^2 - b^2)*C*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*b)/(a + b)])/(d*Sqrt[a + b*Cos[c + d*x]]) + (2*A*b*(a^2 - b^2)*Sqrt[(a + b*Cos[ c + d*x])/(a + b)]*EllipticPi[2, (c + d*x)/2, (2*b)/(a + b)])/(d*Sqrt[a + b*Cos[c + d*x]]))/b)/(a*(a^2 - b^2)) + (2*(A*b^2 - a*(b*B - a*C))*Sin[c + d*x])/(a*(a^2 - b^2)*d*Sqrt[a + b*Cos[c + d*x]])
3.11.50.3.1 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[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[2*(Sqrt[a + b]/d)*EllipticE[(1/2)*(c - Pi/2 + d*x), 2*(b/(a + b))], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]
Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[Sqrt[a + b*Sin[c + d*x]]/Sqrt[(a + b*Sin[c + d*x])/(a + b)] Int[Sqrt[a/(a + b) + ( b/(a + b))*Sin[c + d*x]], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2 , 0] && !GtQ[a + b, 0]
Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/(d*S qrt[a + b]))*EllipticF[(1/2)*(c - Pi/2 + d*x), 2*(b/(a + b))], x] /; FreeQ[ {a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]
Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[Sqrt[(a + b*Sin[c + d*x])/(a + b)]/Sqrt[a + b*Sin[c + d*x]] Int[1/Sqrt[a/(a + b) + (b/(a + b))*Sin[c + d*x]], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && !GtQ[a + b, 0]
Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp[(2/(f*(a + b)*Sqrt[c + d]))*EllipticPi[ 2*(b/(a + b)), (1/2)*(e - Pi/2 + f*x), 2*(d/(c + d))], x] /; FreeQ[{a, b, c , d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[c + d, 0]
Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp[Sqrt[(c + d*Sin[e + f*x])/(c + d)]/Sqrt [c + d*Sin[e + f*x]] Int[1/((a + b*Sin[e + f*x])*Sqrt[c/(c + d) + (d/(c + d))*Sin[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] && NeQ[c^2 - d^2, 0] && !GtQ[c + d, 0]
Int[(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)]))/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[ B/d Int[(a + b*Sin[e + f*x])^m, x], x] - Simp[(B*c - A*d)/d Int[(a + b* Sin[e + f*x])^m/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(A*b^2 - a*b*B + a^2*C))*Cos[e + f*x ]*(a + b*Sin[e + f*x])^(m + 1)*((c + d*Sin[e + f*x])^(n + 1)/(f*(m + 1)*(b* c - a*d)*(a^2 - b^2))), x] + Simp[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)) Int [(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[(m + 1)*(b*c - a* d)*(a*A - b*B + a*C) + d*(A*b^2 - a*b*B + a^2*C)*(m + n + 2) - (c*(A*b^2 - a*b*B + a^2*C) + (m + 1)*(b*c - a*d)*(A*b - a*B + b*C))*Sin[e + f*x] - d*(A *b^2 - a*b*B + a^2*C)*(m + n + 3)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b , c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -1] && ((EqQ[a, 0] && IntegerQ[m] && !IntegerQ [n]) || !(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] && !IntegerQ[m]) | | EqQ[a, 0])))
Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^ 2)/(Sqrt[(a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> Simp[C/(b*d) Int[Sqrt[a + b*Sin[e + f*x]], x] , x] - Simp[1/(b*d) Int[Simp[a*c*C - A*b*d + (b*c*C - b*B*d + a*C*d)*Sin[ e + f*x], x]/(Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])), x], x] /; Fre eQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0 ] && NeQ[c^2 - d^2, 0]
Time = 5.80 (sec) , antiderivative size = 547, normalized size of antiderivative = 2.02
method | result | size |
default | \(-\frac {\sqrt {-\left (-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b -a +b \right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (\frac {2 C \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {\frac {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +a -b}{a -b}}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {-\frac {2 b}{a -b}}\right )}{b \sqrt {-2 b \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (a +b \right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}}-\frac {2 A \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {\frac {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +a -b}{a -b}}\, \Pi \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), 2, \sqrt {-\frac {2 b}{a -b}}\right )}{a \sqrt {-2 b \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (a +b \right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}}-\frac {2 \left (-A \,b^{2}+B a b -a^{2} C \right ) \sqrt {-2 b \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (a +b \right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (2 b \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-\frac {2 b \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a -b}+\frac {a +b}{a -b}}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {-\frac {2 b}{a -b}}\right ) a -\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-\frac {2 b \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a -b}+\frac {a +b}{a -b}}\, b E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {-\frac {2 b}{a -b}}\right )\right )}{a b \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \left (2 b \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-a -b \right ) \left (a^{2}-b^{2}\right )}\right )}{\sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 b \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+a +b}\, d}\) | \(547\) |
parts | \(\text {Expression too large to display}\) | \(968\) |
int((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)/(a+b*cos(d*x+c))^(3/2),x,me thod=_RETURNVERBOSE)
-(-(-2*cos(1/2*d*x+1/2*c)^2*b-a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*C/b*(sin (1/2*d*x+1/2*c)^2)^(1/2)*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)/(a-b))^(1/2)/(-2* b*sin(1/2*d*x+1/2*c)^4+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2 *d*x+1/2*c),(-2*b/(a-b))^(1/2))-2*A/a*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*cos (1/2*d*x+1/2*c)^2*b+a-b)/(a-b))^(1/2)/(-2*b*sin(1/2*d*x+1/2*c)^4+(a+b)*sin (1/2*d*x+1/2*c)^2)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),2,(-2*b/(a-b))^(1/2 ))-2*(-A*b^2+B*a*b-C*a^2)/a/b/sin(1/2*d*x+1/2*c)^2/(2*b*sin(1/2*d*x+1/2*c) ^2-a-b)/(a^2-b^2)*(-2*b*sin(1/2*d*x+1/2*c)^4+(a+b)*sin(1/2*d*x+1/2*c)^2)^( 1/2)*(2*b*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+(sin(1/2*d*x+1/2*c)^2)^( 1/2)*(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*EllipticE(cos(1/2 *d*x+1/2*c),(-2*b/(a-b))^(1/2))*a-(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b) *sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*b*EllipticE(cos(1/2*d*x+1/2*c),(- 2*b/(a-b))^(1/2))))/sin(1/2*d*x+1/2*c)/(-2*b*sin(1/2*d*x+1/2*c)^2+a+b)^(1/ 2)/d
Timed out. \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec (c+d x)}{(a+b \cos (c+d x))^{3/2}} \, dx=\text {Timed out} \]
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)/(a+b*cos(d*x+c))^(3/2 ),x, algorithm="fricas")
\[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec (c+d x)}{(a+b \cos (c+d x))^{3/2}} \, dx=\int \frac {\left (A + B \cos {\left (c + d x \right )} + C \cos ^{2}{\left (c + d x \right )}\right ) \sec {\left (c + d x \right )}}{\left (a + b \cos {\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx \]
Integral((A + B*cos(c + d*x) + C*cos(c + d*x)**2)*sec(c + d*x)/(a + b*cos( c + d*x))**(3/2), x)
\[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec (c+d x)}{(a+b \cos (c+d x))^{3/2}} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x } \]
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)/(a+b*cos(d*x+c))^(3/2 ),x, algorithm="maxima")
integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*sec(d*x + c)/(b*cos(d*x + c) + a)^(3/2), x)
\[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec (c+d x)}{(a+b \cos (c+d x))^{3/2}} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x } \]
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)/(a+b*cos(d*x+c))^(3/2 ),x, algorithm="giac")
integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*sec(d*x + c)/(b*cos(d*x + c) + a)^(3/2), x)
Timed out. \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec (c+d x)}{(a+b \cos (c+d x))^{3/2}} \, dx=\int \frac {C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A}{\cos \left (c+d\,x\right )\,{\left (a+b\,\cos \left (c+d\,x\right )\right )}^{3/2}} \,d x \]