Integrand size = 23, antiderivative size = 77 \[ \int \frac {1}{\cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))} \, dx=-\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a d}-\frac {2 b \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{a (a+b) d}+\frac {2 \sin (c+d x)}{a d \sqrt {\cos (c+d x)}} \]
-2*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(sin(1/2*d*x+1 /2*c),2^(1/2))/a/d-2*b*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*Ell ipticPi(sin(1/2*d*x+1/2*c),2*b/(a+b),2^(1/2))/a/(a+b)/d+2*sin(d*x+c)/a/d/c os(d*x+c)^(1/2)
Leaf count is larger than twice the leaf count of optimal. \(195\) vs. \(2(77)=154\).
Time = 1.96 (sec) , antiderivative size = 195, normalized size of antiderivative = 2.53 \[ \int \frac {1}{\cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))} \, dx=-\frac {\frac {6 b \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{a+b}+\frac {2 a \left (2 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )-\frac {2 a \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{a+b}\right )}{b}-\frac {4 \sin (c+d x)}{\sqrt {\cos (c+d x)}}+\frac {2 \left (-2 a b E\left (\left .\arcsin \left (\sqrt {\cos (c+d x)}\right )\right |-1\right )+2 a (a+b) \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )+\left (-2 a^2+b^2\right ) \operatorname {EllipticPi}\left (-\frac {b}{a},\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )\right ) \sin (c+d x)}{a b \sqrt {\sin ^2(c+d x)}}}{2 a d} \]
-1/2*((6*b*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2])/(a + b) + (2*a*(2*El lipticF[(c + d*x)/2, 2] - (2*a*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2])/ (a + b)))/b - (4*Sin[c + d*x])/Sqrt[Cos[c + d*x]] + (2*(-2*a*b*EllipticE[A rcSin[Sqrt[Cos[c + d*x]]], -1] + 2*a*(a + b)*EllipticF[ArcSin[Sqrt[Cos[c + d*x]]], -1] + (-2*a^2 + b^2)*EllipticPi[-(b/a), ArcSin[Sqrt[Cos[c + d*x]] ], -1])*Sin[c + d*x])/(a*b*Sqrt[Sin[c + d*x]^2]))/(a*d)
Time = 0.57 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.435, Rules used = {3042, 3281, 27, 3042, 3538, 25, 27, 3042, 3119, 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.
| \(\displaystyle \int \frac {1}{\cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx\) |
| \(\Big \downarrow \) 3281 |
| \(\displaystyle \frac {2 \int -\frac {b \cos ^2(c+d x)+a \cos (c+d x)+b}{2 \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{a}+\frac {2 \sin (c+d x)}{a d \sqrt {\cos (c+d x)}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {2 \sin (c+d x)}{a d \sqrt {\cos (c+d x)}}-\frac {\int \frac {b \cos ^2(c+d x)+a \cos (c+d x)+b}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 \sin (c+d x)}{a d \sqrt {\cos (c+d x)}}-\frac {\int \frac {b \sin \left (c+d x+\frac {\pi }{2}\right )^2+a \sin \left (c+d x+\frac {\pi }{2}\right )+b}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a}\) |
| \(\Big \downarrow \) 3538 |
| \(\displaystyle \frac {2 \sin (c+d x)}{a d \sqrt {\cos (c+d x)}}-\frac {\int \sqrt {\cos (c+d x)}dx-\frac {\int -\frac {b^2}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{b}}{a}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {2 \sin (c+d x)}{a d \sqrt {\cos (c+d x)}}-\frac {\frac {\int \frac {b^2}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{b}+\int \sqrt {\cos (c+d x)}dx}{a}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {2 \sin (c+d x)}{a d \sqrt {\cos (c+d x)}}-\frac {b \int \frac {1}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx+\int \sqrt {\cos (c+d x)}dx}{a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 \sin (c+d x)}{a d \sqrt {\cos (c+d x)}}-\frac {b \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx+\int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{a}\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle \frac {2 \sin (c+d x)}{a d \sqrt {\cos (c+d x)}}-\frac {b \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx+\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}}{a}\) |
| \(\Big \downarrow \) 3284 |
| \(\displaystyle \frac {2 \sin (c+d x)}{a d \sqrt {\cos (c+d x)}}-\frac {\frac {2 b \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{d (a+b)}+\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}}{a}\) |
-(((2*EllipticE[(c + d*x)/2, 2])/d + (2*b*EllipticPi[(2*b)/(a + b), (c + d *x)/2, 2])/((a + b)*d))/a) + (2*Sin[c + d*x])/(a*d*Sqrt[Cos[c + d*x]])
3.6.86.3.1 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[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)* (c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b^2)*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[a*(b*c - a*d)*(m + 1) + b^2*d*(m + n + 2) - (b^2*c + b*(b*c - a*d)*(m + 1))*Sin[e + f*x] - b^2*d*(m + n + 3)*Si n[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a* d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -1] && IntegersQ[ 2*m, 2*n] && ((EqQ[a, 0] && IntegerQ[m] && !IntegerQ[n]) || !(IntegerQ[2* n] && LtQ[n, -1] && ((IntegerQ[n] && !IntegerQ[m]) || EqQ[a, 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[((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]
Leaf count of result is larger than twice the leaf count of optimal. \(353\) vs. \(2(127)=254\).
Time = 3.90 (sec) , antiderivative size = 354, normalized size of antiderivative = 4.60
| method | result | size |
| default | \(-\frac {2 \left (-2 \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \left (a -b \right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) a -\sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) b -b \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \Pi \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), -\frac {2 b}{a -b}, \sqrt {2}\right ) \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\right )}{a \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \left (a -b \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d}\) | \(354\) |
-2*(-2*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(a-b)*cos(1/2* d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2+(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c )^2)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*E llipticE(cos(1/2*d*x+1/2*c),2^(1/2))*a-(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d* x+1/2*c)^2)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^ (1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*b-b*(sin(1/2*d*x+1/2*c)^2)^(1/ 2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),-2*b/(a- b),2^(1/2))*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2))/a/(-2*si n(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(a-b)/sin(1/2*d*x+1/2*c)/(2 *cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d
Timed out. \[ \int \frac {1}{\cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {1}{\cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))} \, dx=\text {Timed out} \]
\[ \int \frac {1}{\cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))} \, dx=\int { \frac {1}{{\left (b \cos \left (d x + c\right ) + a\right )} \cos \left (d x + c\right )^{\frac {3}{2}}} \,d x } \]
\[ \int \frac {1}{\cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))} \, dx=\int { \frac {1}{{\left (b \cos \left (d x + c\right ) + a\right )} \cos \left (d x + c\right )^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {1}{\cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))} \, dx=\int \frac {1}{{\cos \left (c+d\,x\right )}^{3/2}\,\left (a+b\,\cos \left (c+d\,x\right )\right )} \,d x \]