Integrand size = 22, antiderivative size = 186 \[ \int \frac {1}{(a+b \cos (d+e x)+c \sin (d+e x))^{3/2}} \, dx=\frac {2 (c \cos (d+e x)-b \sin (d+e x))}{\left (a^2-b^2-c^2\right ) e \sqrt {a+b \cos (d+e x)+c \sin (d+e x)}}+\frac {2 E\left (\frac {1}{2} \left (d+e x-\tan ^{-1}(b,c)\right )|\frac {2 \sqrt {b^2+c^2}}{a+\sqrt {b^2+c^2}}\right ) \sqrt {a+b \cos (d+e x)+c \sin (d+e x)}}{\left (a^2-b^2-c^2\right ) e \sqrt {\frac {a+b \cos (d+e x)+c \sin (d+e x)}{a+\sqrt {b^2+c^2}}}} \]
2*(c*cos(e*x+d)-b*sin(e*x+d))/(a^2-b^2-c^2)/e/(a+b*cos(e*x+d)+c*sin(e*x+d) )^(1/2)+2*(cos(1/2*d+1/2*e*x-1/2*arctan(b,c))^2)^(1/2)/cos(1/2*d+1/2*e*x-1 /2*arctan(b,c))*EllipticE(sin(1/2*d+1/2*e*x-1/2*arctan(b,c)),2^(1/2)*((b^2 +c^2)^(1/2)/(a+(b^2+c^2)^(1/2)))^(1/2))*(a+b*cos(e*x+d)+c*sin(e*x+d))^(1/2 )/(a^2-b^2-c^2)/e/((a+b*cos(e*x+d)+c*sin(e*x+d))/(a+(b^2+c^2)^(1/2)))^(1/2 )
Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.
Time = 6.44 (sec) , antiderivative size = 1540, normalized size of antiderivative = 8.28 \[ \int \frac {1}{(a+b \cos (d+e x)+c \sin (d+e x))^{3/2}} \, dx =\text {Too large to display} \]
(Sqrt[a + b*Cos[d + e*x] + c*Sin[d + e*x]]*((-2*(b^2 + c^2))/(b*c*(-a^2 + b^2 + c^2)) + (2*(a*c + b^2*Sin[d + e*x] + c^2*Sin[d + e*x]))/(b*(-a^2 + b ^2 + c^2)*(a + b*Cos[d + e*x] + c*Sin[d + e*x]))))/e - (2*a*AppellF1[1/2, 1/2, 1/2, 3/2, -((a + Sqrt[1 + b^2/c^2]*c*Sin[d + e*x + ArcTan[b/c]])/(Sqr t[1 + b^2/c^2]*(1 - a/(Sqrt[1 + b^2/c^2]*c))*c)), -((a + Sqrt[1 + b^2/c^2] *c*Sin[d + e*x + ArcTan[b/c]])/(Sqrt[1 + b^2/c^2]*(-1 - a/(Sqrt[1 + b^2/c^ 2]*c))*c))]*Sec[d + e*x + ArcTan[b/c]]*Sqrt[(c*Sqrt[(b^2 + c^2)/c^2] - c*S qrt[(b^2 + c^2)/c^2]*Sin[d + e*x + ArcTan[b/c]])/(a + c*Sqrt[(b^2 + c^2)/c ^2])]*Sqrt[a + c*Sqrt[(b^2 + c^2)/c^2]*Sin[d + e*x + ArcTan[b/c]]]*Sqrt[(c *Sqrt[(b^2 + c^2)/c^2] + c*Sqrt[(b^2 + c^2)/c^2]*Sin[d + e*x + ArcTan[b/c] ])/(-a + c*Sqrt[(b^2 + c^2)/c^2])])/(Sqrt[1 + b^2/c^2]*c*(-a^2 + b^2 + c^2 )*e) - (b^2*(-((c*AppellF1[-1/2, -1/2, -1/2, 1/2, -((a + b*Sqrt[1 + c^2/b^ 2]*Cos[d + e*x - ArcTan[c/b]])/(b*Sqrt[1 + c^2/b^2]*(1 - a/(b*Sqrt[1 + c^2 /b^2])))), -((a + b*Sqrt[1 + c^2/b^2]*Cos[d + e*x - ArcTan[c/b]])/(b*Sqrt[ 1 + c^2/b^2]*(-1 - a/(b*Sqrt[1 + c^2/b^2]))))]*Sin[d + e*x - ArcTan[c/b]]) /(b*Sqrt[1 + c^2/b^2]*Sqrt[(b*Sqrt[(b^2 + c^2)/b^2] - b*Sqrt[(b^2 + c^2)/b ^2]*Cos[d + e*x - ArcTan[c/b]])/(a + b*Sqrt[(b^2 + c^2)/b^2])]*Sqrt[a + b* Sqrt[(b^2 + c^2)/b^2]*Cos[d + e*x - ArcTan[c/b]]]*Sqrt[(b*Sqrt[(b^2 + c^2) /b^2] + b*Sqrt[(b^2 + c^2)/b^2]*Cos[d + e*x - ArcTan[c/b]])/(-a + b*Sqrt[( b^2 + c^2)/b^2])])) - ((2*b*(a + b*Sqrt[1 + c^2/b^2]*Cos[d + e*x - ArcT...
Time = 0.55 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {3042, 3607, 3042, 3598, 3042, 3132}
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.
\(\displaystyle \int \frac {1}{(a+b \cos (d+e x)+c \sin (d+e x))^{3/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{(a+b \cos (d+e x)+c \sin (d+e x))^{3/2}}dx\) |
\(\Big \downarrow \) 3607 |
\(\displaystyle \frac {\int \sqrt {a+b \cos (d+e x)+c \sin (d+e x)}dx}{a^2-b^2-c^2}+\frac {2 (c \cos (d+e x)-b \sin (d+e x))}{e \left (a^2-b^2-c^2\right ) \sqrt {a+b \cos (d+e x)+c \sin (d+e x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \sqrt {a+b \cos (d+e x)+c \sin (d+e x)}dx}{a^2-b^2-c^2}+\frac {2 (c \cos (d+e x)-b \sin (d+e x))}{e \left (a^2-b^2-c^2\right ) \sqrt {a+b \cos (d+e x)+c \sin (d+e x)}}\) |
\(\Big \downarrow \) 3598 |
\(\displaystyle \frac {\sqrt {a+b \cos (d+e x)+c \sin (d+e x)} \int \sqrt {\frac {a}{a+\sqrt {b^2+c^2}}+\frac {\sqrt {b^2+c^2} \cos \left (d+e x-\tan ^{-1}(b,c)\right )}{a+\sqrt {b^2+c^2}}}dx}{\left (a^2-b^2-c^2\right ) \sqrt {\frac {a+b \cos (d+e x)+c \sin (d+e x)}{a+\sqrt {b^2+c^2}}}}+\frac {2 (c \cos (d+e x)-b \sin (d+e x))}{e \left (a^2-b^2-c^2\right ) \sqrt {a+b \cos (d+e x)+c \sin (d+e x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\sqrt {a+b \cos (d+e x)+c \sin (d+e x)} \int \sqrt {\frac {a}{a+\sqrt {b^2+c^2}}+\frac {\sqrt {b^2+c^2} \sin \left (d+e x-\tan ^{-1}(b,c)+\frac {\pi }{2}\right )}{a+\sqrt {b^2+c^2}}}dx}{\left (a^2-b^2-c^2\right ) \sqrt {\frac {a+b \cos (d+e x)+c \sin (d+e x)}{a+\sqrt {b^2+c^2}}}}+\frac {2 (c \cos (d+e x)-b \sin (d+e x))}{e \left (a^2-b^2-c^2\right ) \sqrt {a+b \cos (d+e x)+c \sin (d+e x)}}\) |
\(\Big \downarrow \) 3132 |
\(\displaystyle \frac {2 \sqrt {a+b \cos (d+e x)+c \sin (d+e x)} E\left (\frac {1}{2} \left (d+e x-\tan ^{-1}(b,c)\right )|\frac {2 \sqrt {b^2+c^2}}{a+\sqrt {b^2+c^2}}\right )}{e \left (a^2-b^2-c^2\right ) \sqrt {\frac {a+b \cos (d+e x)+c \sin (d+e x)}{a+\sqrt {b^2+c^2}}}}+\frac {2 (c \cos (d+e x)-b \sin (d+e x))}{e \left (a^2-b^2-c^2\right ) \sqrt {a+b \cos (d+e x)+c \sin (d+e x)}}\) |
(2*(c*Cos[d + e*x] - b*Sin[d + e*x]))/((a^2 - b^2 - c^2)*e*Sqrt[a + b*Cos[ d + e*x] + c*Sin[d + e*x]]) + (2*EllipticE[(d + e*x - ArcTan[b, c])/2, (2* Sqrt[b^2 + c^2])/(a + Sqrt[b^2 + c^2])]*Sqrt[a + b*Cos[d + e*x] + c*Sin[d + e*x]])/((a^2 - b^2 - c^2)*e*Sqrt[(a + b*Cos[d + e*x] + c*Sin[d + e*x])/( a + Sqrt[b^2 + c^2])])
3.5.14.3.1 Defintions of rubi rules used
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[cos[(d_.) + (e_.)*(x_)]*(b_.) + (a_) + (c_.)*sin[(d_.) + (e_.)*(x_ )]], x_Symbol] :> Simp[Sqrt[a + b*Cos[d + e*x] + c*Sin[d + e*x]]/Sqrt[(a + b*Cos[d + e*x] + c*Sin[d + e*x])/(a + Sqrt[b^2 + c^2])] Int[Sqrt[a/(a + S qrt[b^2 + c^2]) + (Sqrt[b^2 + c^2]/(a + Sqrt[b^2 + c^2]))*Cos[d + e*x - Arc Tan[b, c]]], x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[a^2 - b^2 - c^2, 0] && NeQ[b^2 + c^2, 0] && !GtQ[a + Sqrt[b^2 + c^2], 0]
Int[(cos[(d_.) + (e_.)*(x_)]*(b_.) + (a_) + (c_.)*sin[(d_.) + (e_.)*(x_)])^ (-3/2), x_Symbol] :> Simp[2*((c*Cos[d + e*x] - b*Sin[d + e*x])/(e*(a^2 - b^ 2 - c^2)*Sqrt[a + b*Cos[d + e*x] + c*Sin[d + e*x]])), x] + Simp[1/(a^2 - b^ 2 - c^2) Int[Sqrt[a + b*Cos[d + e*x] + c*Sin[d + e*x]], x], x] /; FreeQ[{ a, b, c, d, e}, x] && NeQ[a^2 - b^2 - c^2, 0]
Leaf count of result is larger than twice the leaf count of optimal. \(2644\) vs. \(2(213)=426\).
Time = 2.44 (sec) , antiderivative size = 2645, normalized size of antiderivative = 14.22
-(-(-b^2*sin(e*x+d-arctan(-b,c))-c^2*sin(e*x+d-arctan(-b,c))-a*(b^2+c^2)^( 1/2))*cos(e*x+d-arctan(-b,c))^2/(b^2+c^2)^(1/2))^(1/2)/(b^2+c^2)^(3/2)*(b^ 2*sin(e*x+d-arctan(-b,c))+c^2*sin(e*x+d-arctan(-b,c))+a*(b^2+c^2)^(1/2))*( cos(e*x+d-arctan(-b,c))^2*((b^2+c^2)^(1/2)*sin(e*x+d-arctan(-b,c))+a)*(b^2 +c^2))^(1/2)*(sin(e*x+d-arctan(-b,c))^2*b^2+sin(e*x+d-arctan(-b,c))^2*c^2- a^2)/(sin(e*x+d-arctan(-b,c))^2*b^2+sin(e*x+d-arctan(-b,c))^2*c^2+2*sin(e* x+d-arctan(-b,c))*a*(b^2+c^2)^(1/2)+a^2)/(a*(((b^2+c^2)^(1/2)*sin(e*x+d-ar ctan(-b,c))+a)*cos(e*x+d-arctan(-b,c))^2)^(1/2)-sin(e*x+d-arctan(-b,c))*(c os(e*x+d-arctan(-b,c))^2*((b^2+c^2)^(1/2)*sin(e*x+d-arctan(-b,c))+a)*(b^2+ c^2))^(1/2))*(-(b^2+c^2)^(1/2)*(-b^2-c^2)*cos(e*x+d-arctan(-b,c))^2/(a^2-b ^2-c^2)/(cos(e*x+d-arctan(-b,c))^2*((b^2+c^2)^(1/2)*sin(e*x+d-arctan(-b,c) )+a)*(b^2+c^2))^(1/2)+a*(b^2+c^2)/(a^2-b^2-c^2)*(1/(b^2+c^2)^(1/2)*a+1)*(( (b^2+c^2)^(1/2)*sin(e*x+d-arctan(-b,c))+a)/(a+(b^2+c^2)^(1/2)))^(1/2)*((si n(e*x+d-arctan(-b,c))+1)*(b^2+c^2)^(1/2)/(-a+(b^2+c^2)^(1/2)))^(1/2)*((-si n(e*x+d-arctan(-b,c))+1)*(b^2+c^2)^(1/2)/(a+(b^2+c^2)^(1/2)))^(1/2)/(cos(e *x+d-arctan(-b,c))^2*((b^2+c^2)^(1/2)*sin(e*x+d-arctan(-b,c))+a)*(b^2+c^2) )^(1/2)*EllipticF((((b^2+c^2)^(1/2)*sin(e*x+d-arctan(-b,c))+a)/(a+(b^2+c^2 )^(1/2)))^(1/2),((-a-(b^2+c^2)^(1/2))/(-a+(b^2+c^2)^(1/2)))^(1/2))+2*(-(b^ 2+c^2)^(3/2)+2*(b^2+c^2)^(1/2)*b^2+2*(b^2+c^2)^(1/2)*c^2)/(2*a^2-2*b^2-2*c ^2)*(1/(b^2+c^2)^(1/2)*a+1)*(((b^2+c^2)^(1/2)*sin(e*x+d-arctan(-b,c))+a...
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.15 (sec) , antiderivative size = 1737, normalized size of antiderivative = 9.34 \[ \int \frac {1}{(a+b \cos (d+e x)+c \sin (d+e x))^{3/2}} \, dx=\text {Too large to display} \]
-1/3*((sqrt(2)*(-I*a*b^2 - a*b*c)*cos(e*x + d) + sqrt(2)*(-I*a*b*c - a*c^2 )*sin(e*x + d) + sqrt(2)*(-I*a^2*b - a^2*c))*sqrt(b + I*c)*weierstrassPInv erse(4/3*(4*a^2*b^2 - 3*b^4 - 4*a^2*c^2 + 6*I*b*c^3 + 3*c^4 - 2*I*(4*a^2*b - 3*b^3)*c)/(b^4 + 2*b^2*c^2 + c^4), -8/27*(8*a^3*b^3 - 9*a*b^5 + 27*a*b* c^4 - 9*I*a*c^5 + 2*I*(4*a^3 + 9*a*b^2)*c^3 - 6*(4*a^3*b - 3*a*b^3)*c^2 - 3*I*(8*a^3*b^2 - 9*a*b^4)*c)/(b^6 + 3*b^4*c^2 + 3*b^2*c^4 + c^6), 1/3*(2*a *b - 2*I*a*c + 3*(b^2 + c^2)*cos(e*x + d) - 3*(I*b^2 + I*c^2)*sin(e*x + d) )/(b^2 + c^2)) + (sqrt(2)*(I*a*b^2 - a*b*c)*cos(e*x + d) + sqrt(2)*(I*a*b* c - a*c^2)*sin(e*x + d) + sqrt(2)*(I*a^2*b - a^2*c))*sqrt(b - I*c)*weierst rassPInverse(4/3*(4*a^2*b^2 - 3*b^4 - 4*a^2*c^2 - 6*I*b*c^3 + 3*c^4 + 2*I* (4*a^2*b - 3*b^3)*c)/(b^4 + 2*b^2*c^2 + c^4), -8/27*(8*a^3*b^3 - 9*a*b^5 + 27*a*b*c^4 + 9*I*a*c^5 - 2*I*(4*a^3 + 9*a*b^2)*c^3 - 6*(4*a^3*b - 3*a*b^3 )*c^2 + 3*I*(8*a^3*b^2 - 9*a*b^4)*c)/(b^6 + 3*b^4*c^2 + 3*b^2*c^4 + c^6), 1/3*(2*a*b + 2*I*a*c + 3*(b^2 + c^2)*cos(e*x + d) - 3*(-I*b^2 - I*c^2)*sin (e*x + d))/(b^2 + c^2)) - 3*(sqrt(2)*(-I*b^3 - I*b*c^2)*cos(e*x + d) + sqr t(2)*(-I*b^2*c - I*c^3)*sin(e*x + d) + sqrt(2)*(-I*a*b^2 - I*a*c^2))*sqrt( b + I*c)*weierstrassZeta(4/3*(4*a^2*b^2 - 3*b^4 - 4*a^2*c^2 + 6*I*b*c^3 + 3*c^4 - 2*I*(4*a^2*b - 3*b^3)*c)/(b^4 + 2*b^2*c^2 + c^4), -8/27*(8*a^3*b^3 - 9*a*b^5 + 27*a*b*c^4 - 9*I*a*c^5 + 2*I*(4*a^3 + 9*a*b^2)*c^3 - 6*(4*a^3 *b - 3*a*b^3)*c^2 - 3*I*(8*a^3*b^2 - 9*a*b^4)*c)/(b^6 + 3*b^4*c^2 + 3*b...
\[ \int \frac {1}{(a+b \cos (d+e x)+c \sin (d+e x))^{3/2}} \, dx=\int \frac {1}{\left (a + b \cos {\left (d + e x \right )} + c \sin {\left (d + e x \right )}\right )^{\frac {3}{2}}}\, dx \]
\[ \int \frac {1}{(a+b \cos (d+e x)+c \sin (d+e x))^{3/2}} \, dx=\int { \frac {1}{{\left (b \cos \left (e x + d\right ) + c \sin \left (e x + d\right ) + a\right )}^{\frac {3}{2}}} \,d x } \]
\[ \int \frac {1}{(a+b \cos (d+e x)+c \sin (d+e x))^{3/2}} \, dx=\int { \frac {1}{{\left (b \cos \left (e x + d\right ) + c \sin \left (e x + d\right ) + a\right )}^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {1}{(a+b \cos (d+e x)+c \sin (d+e x))^{3/2}} \, dx=\int \frac {1}{{\left (a+b\,\cos \left (d+e\,x\right )+c\,\sin \left (d+e\,x\right )\right )}^{3/2}} \,d x \]