Integrand size = 21, antiderivative size = 131 \[ \int (a \cos (c+d x)+b \sin (c+d x))^{3/2} \, dx=-\frac {2 (b \cos (c+d x)-a \sin (c+d x)) \sqrt {a \cos (c+d x)+b \sin (c+d x)}}{3 d}+\frac {2 \left (a^2+b^2\right ) \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\tan ^{-1}(a,b)\right ),2\right ) \sqrt {\frac {a \cos (c+d x)+b \sin (c+d x)}{\sqrt {a^2+b^2}}}}{3 d \sqrt {a \cos (c+d x)+b \sin (c+d x)}} \]
-2/3*(b*cos(d*x+c)-a*sin(d*x+c))*(a*cos(d*x+c)+b*sin(d*x+c))^(1/2)/d+2/3*( a^2+b^2)*(cos(1/2*c+1/2*d*x-1/2*arctan(a,b))^2)^(1/2)/cos(1/2*c+1/2*d*x-1/ 2*arctan(a,b))*EllipticF(sin(1/2*c+1/2*d*x-1/2*arctan(a,b)),2^(1/2))*((a*c os(d*x+c)+b*sin(d*x+c))/(a^2+b^2)^(1/2))^(1/2)/d/(a*cos(d*x+c)+b*sin(d*x+c ))^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 1.12 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.09 \[ \int (a \cos (c+d x)+b \sin (c+d x))^{3/2} \, dx=\frac {2 \left ((-b \cos (c+d x)+a \sin (c+d x)) \sqrt {a \cos (c+d x)+b \sin (c+d x)}+\frac {\left (a^2+b^2\right ) \sqrt {\cos ^2\left (c+d x+\arctan \left (\frac {a}{b}\right )\right )} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};\sin ^2\left (c+d x+\arctan \left (\frac {a}{b}\right )\right )\right ) \tan \left (c+d x+\arctan \left (\frac {a}{b}\right )\right )}{\sqrt {\sqrt {1+\frac {a^2}{b^2}} b \sin \left (c+d x+\arctan \left (\frac {a}{b}\right )\right )}}\right )}{3 d} \]
(2*((-(b*Cos[c + d*x]) + a*Sin[c + d*x])*Sqrt[a*Cos[c + d*x] + b*Sin[c + d *x]] + ((a^2 + b^2)*Sqrt[Cos[c + d*x + ArcTan[a/b]]^2]*HypergeometricPFQ[{ 1/4, 1/2}, {5/4}, Sin[c + d*x + ArcTan[a/b]]^2]*Tan[c + d*x + ArcTan[a/b]] )/Sqrt[Sqrt[1 + a^2/b^2]*b*Sin[c + d*x + ArcTan[a/b]]]))/(3*d)
Time = 0.42 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3042, 3552, 3042, 3557, 3042, 3120}
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.
\(\displaystyle \int (a \cos (c+d x)+b \sin (c+d x))^{3/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int (a \cos (c+d x)+b \sin (c+d x))^{3/2}dx\) |
\(\Big \downarrow \) 3552 |
\(\displaystyle \frac {1}{3} \left (a^2+b^2\right ) \int \frac {1}{\sqrt {a \cos (c+d x)+b \sin (c+d x)}}dx-\frac {2 (b \cos (c+d x)-a \sin (c+d x)) \sqrt {a \cos (c+d x)+b \sin (c+d x)}}{3 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{3} \left (a^2+b^2\right ) \int \frac {1}{\sqrt {a \cos (c+d x)+b \sin (c+d x)}}dx-\frac {2 (b \cos (c+d x)-a \sin (c+d x)) \sqrt {a \cos (c+d x)+b \sin (c+d x)}}{3 d}\) |
\(\Big \downarrow \) 3557 |
\(\displaystyle \frac {\left (a^2+b^2\right ) \sqrt {\frac {a \cos (c+d x)+b \sin (c+d x)}{\sqrt {a^2+b^2}}} \int \frac {1}{\sqrt {\cos \left (c+d x-\tan ^{-1}(a,b)\right )}}dx}{3 \sqrt {a \cos (c+d x)+b \sin (c+d x)}}-\frac {2 (b \cos (c+d x)-a \sin (c+d x)) \sqrt {a \cos (c+d x)+b \sin (c+d x)}}{3 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\left (a^2+b^2\right ) \sqrt {\frac {a \cos (c+d x)+b \sin (c+d x)}{\sqrt {a^2+b^2}}} \int \frac {1}{\sqrt {\sin \left (c+d x-\tan ^{-1}(a,b)+\frac {\pi }{2}\right )}}dx}{3 \sqrt {a \cos (c+d x)+b \sin (c+d x)}}-\frac {2 (b \cos (c+d x)-a \sin (c+d x)) \sqrt {a \cos (c+d x)+b \sin (c+d x)}}{3 d}\) |
\(\Big \downarrow \) 3120 |
\(\displaystyle \frac {2 \left (a^2+b^2\right ) \sqrt {\frac {a \cos (c+d x)+b \sin (c+d x)}{\sqrt {a^2+b^2}}} \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\tan ^{-1}(a,b)\right ),2\right )}{3 d \sqrt {a \cos (c+d x)+b \sin (c+d x)}}-\frac {2 (b \cos (c+d x)-a \sin (c+d x)) \sqrt {a \cos (c+d x)+b \sin (c+d x)}}{3 d}\) |
(-2*(b*Cos[c + d*x] - a*Sin[c + d*x])*Sqrt[a*Cos[c + d*x] + b*Sin[c + d*x] ])/(3*d) + (2*(a^2 + b^2)*EllipticF[(c + d*x - ArcTan[a, b])/2, 2]*Sqrt[(a *Cos[c + d*x] + b*Sin[c + d*x])/Sqrt[a^2 + b^2]])/(3*d*Sqrt[a*Cos[c + d*x] + b*Sin[c + d*x]])
3.3.34.3.1 Defintions of rubi rules used
Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2 )*(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
Int[(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x _Symbol] :> Simp[(-(b*Cos[c + d*x] - a*Sin[c + d*x]))*((a*Cos[c + d*x] + b* Sin[c + d*x])^(n - 1)/(d*n)), x] + Simp[(n - 1)*((a^2 + b^2)/n) Int[(a*Co s[c + d*x] + b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{a, b, c, d}, x] && N eQ[a^2 + b^2, 0] && !IntegerQ[(n - 1)/2] && GtQ[n, 1]
Int[(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x _Symbol] :> Simp[(a*Cos[c + d*x] + b*Sin[c + d*x])^n/((a*Cos[c + d*x] + b*S in[c + d*x])/Sqrt[a^2 + b^2])^n Int[Cos[c + d*x - ArcTan[a, b]]^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && !(GeQ[n, 1] || LeQ[n, -1]) && !(GtQ[a^2 + b^2, 0] || EqQ[a^2 + b^2, 0])
Time = 1.08 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.26
method | result | size |
default | \(-\frac {\left (a^{2}+b^{2}\right ) \left (\sqrt {-\sin \left (d x +c -\arctan \left (-a , b\right )\right )+1}\, \sqrt {2 \sin \left (d x +c -\arctan \left (-a , b\right )\right )+2}\, \sqrt {\sin \left (d x +c -\arctan \left (-a , b\right )\right )}\, \operatorname {EllipticF}\left (\sqrt {-\sin \left (d x +c -\arctan \left (-a , b\right )\right )+1}, \frac {\sqrt {2}}{2}\right )-2 \sin \left (d x +c -\arctan \left (-a , b\right )\right )^{3}+2 \sin \left (d x +c -\arctan \left (-a , b\right )\right )\right )}{3 \cos \left (d x +c -\arctan \left (-a , b\right )\right ) \sqrt {\sin \left (d x +c -\arctan \left (-a , b\right )\right ) \sqrt {a^{2}+b^{2}}}\, d}\) | \(165\) |
-1/3*(a^2+b^2)*((-sin(d*x+c-arctan(-a,b))+1)^(1/2)*(2*sin(d*x+c-arctan(-a, b))+2)^(1/2)*sin(d*x+c-arctan(-a,b))^(1/2)*EllipticF((-sin(d*x+c-arctan(-a ,b))+1)^(1/2),1/2*2^(1/2))-2*sin(d*x+c-arctan(-a,b))^3+2*sin(d*x+c-arctan( -a,b)))/cos(d*x+c-arctan(-a,b))/(sin(d*x+c-arctan(-a,b))*(a^2+b^2)^(1/2))^ (1/2)/d
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.09 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.21 \[ \int (a \cos (c+d x)+b \sin (c+d x))^{3/2} \, dx=\frac {\sqrt {2} \sqrt {a - i \, b} {\left (-i \, a + b\right )} {\rm weierstrassPInverse}\left (-\frac {4 \, {\left (a^{2} + 2 i \, a b - b^{2}\right )}}{a^{2} + b^{2}}, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + \sqrt {2} \sqrt {a + i \, b} {\left (i \, a + b\right )} {\rm weierstrassPInverse}\left (-\frac {4 \, {\left (a^{2} - 2 i \, a b - b^{2}\right )}}{a^{2} + b^{2}}, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 2 \, \sqrt {a \cos \left (d x + c\right ) + b \sin \left (d x + c\right )} {\left (b \cos \left (d x + c\right ) - a \sin \left (d x + c\right )\right )}}{3 \, d} \]
1/3*(sqrt(2)*sqrt(a - I*b)*(-I*a + b)*weierstrassPInverse(-4*(a^2 + 2*I*a* b - b^2)/(a^2 + b^2), 0, cos(d*x + c) + I*sin(d*x + c)) + sqrt(2)*sqrt(a + I*b)*(I*a + b)*weierstrassPInverse(-4*(a^2 - 2*I*a*b - b^2)/(a^2 + b^2), 0, cos(d*x + c) - I*sin(d*x + c)) - 2*sqrt(a*cos(d*x + c) + b*sin(d*x + c) )*(b*cos(d*x + c) - a*sin(d*x + c)))/d
\[ \int (a \cos (c+d x)+b \sin (c+d x))^{3/2} \, dx=\int \left (a \cos {\left (c + d x \right )} + b \sin {\left (c + d x \right )}\right )^{\frac {3}{2}}\, dx \]
\[ \int (a \cos (c+d x)+b \sin (c+d x))^{3/2} \, dx=\int { {\left (a \cos \left (d x + c\right ) + b \sin \left (d x + c\right )\right )}^{\frac {3}{2}} \,d x } \]
\[ \int (a \cos (c+d x)+b \sin (c+d x))^{3/2} \, dx=\int { {\left (a \cos \left (d x + c\right ) + b \sin \left (d x + c\right )\right )}^{\frac {3}{2}} \,d x } \]
Timed out. \[ \int (a \cos (c+d x)+b \sin (c+d x))^{3/2} \, dx=\int {\left (a\,\cos \left (c+d\,x\right )+b\,\sin \left (c+d\,x\right )\right )}^{3/2} \,d x \]