Integrand size = 41, antiderivative size = 185 \[ \int (a+b \sin (d+e x)) \sqrt {b^2+2 a b \sin (d+e x)+a^2 \sin ^2(d+e x)} \, dx=-\frac {\left (a^2+b^2\right ) \cos (d+e x) \sqrt {b^2+2 a b \sin (d+e x)+a^2 \sin ^2(d+e x)}}{e (b+a \sin (d+e x))}+\frac {3 a^2 b x \sqrt {b^2+2 a b \sin (d+e x)+a^2 \sin ^2(d+e x)}}{2 \left (a b+a^2 \sin (d+e x)\right )}-\frac {a^2 b \cos (d+e x) \sin (d+e x) \sqrt {b^2+2 a b \sin (d+e x)+a^2 \sin ^2(d+e x)}}{2 e \left (a b+a^2 \sin (d+e x)\right )} \]
-(a^2+b^2)*cos(e*x+d)*(b^2+2*a*b*sin(e*x+d)+a^2*sin(e*x+d)^2)^(1/2)/e/(b+a *sin(e*x+d))+3/2*a^2*b*x*(b^2+2*a*b*sin(e*x+d)+a^2*sin(e*x+d)^2)^(1/2)/(a* b+a^2*sin(e*x+d))-1/2*a^2*b*cos(e*x+d)*sin(e*x+d)*(b^2+2*a*b*sin(e*x+d)+a^ 2*sin(e*x+d)^2)^(1/2)/e/(a*b+a^2*sin(e*x+d))
Time = 0.39 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.38 \[ \int (a+b \sin (d+e x)) \sqrt {b^2+2 a b \sin (d+e x)+a^2 \sin ^2(d+e x)} \, dx=-\frac {\sqrt {(b+a \sin (d+e x))^2} \left (4 \left (a^2+b^2\right ) \cos (d+e x)+a b (-6 (d+e x)+\sin (2 (d+e x)))\right )}{4 e (b+a \sin (d+e x))} \]
-1/4*(Sqrt[(b + a*Sin[d + e*x])^2]*(4*(a^2 + b^2)*Cos[d + e*x] + a*b*(-6*( d + e*x) + Sin[2*(d + e*x)])))/(e*(b + a*Sin[d + e*x]))
Time = 0.35 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.54, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.122, Rules used = {3042, 3771, 27, 3042, 3213}
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.
\(\displaystyle \int (a+b \sin (d+e x)) \sqrt {a^2 \sin ^2(d+e x)+2 a b \sin (d+e x)+b^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int (a+b \sin (d+e x)) \sqrt {a^2 \sin (d+e x)^2+2 a b \sin (d+e x)+b^2}dx\) |
\(\Big \downarrow \) 3771 |
\(\displaystyle \frac {\sqrt {a^2 \sin ^2(d+e x)+2 a b \sin (d+e x)+b^2} \int 2 \left (\sin (d+e x) a^2+b a\right ) (a+b \sin (d+e x))dx}{2 \left (a^2 \sin (d+e x)+a b\right )}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\sqrt {a^2 \sin ^2(d+e x)+2 a b \sin (d+e x)+b^2} \int \left (\sin (d+e x) a^2+b a\right ) (a+b \sin (d+e x))dx}{a^2 \sin (d+e x)+a b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\sqrt {a^2 \sin ^2(d+e x)+2 a b \sin (d+e x)+b^2} \int \left (\sin (d+e x) a^2+b a\right ) (a+b \sin (d+e x))dx}{a^2 \sin (d+e x)+a b}\) |
\(\Big \downarrow \) 3213 |
\(\displaystyle \frac {\sqrt {a^2 \sin ^2(d+e x)+2 a b \sin (d+e x)+b^2} \left (-\frac {a \left (a^2+b^2\right ) \cos (d+e x)}{e}-\frac {a^2 b \sin (d+e x) \cos (d+e x)}{2 e}+\frac {3}{2} a^2 b x\right )}{a^2 \sin (d+e x)+a b}\) |
(((3*a^2*b*x)/2 - (a*(a^2 + b^2)*Cos[d + e*x])/e - (a^2*b*Cos[d + e*x]*Sin [d + e*x])/(2*e))*Sqrt[b^2 + 2*a*b*Sin[d + e*x] + a^2*Sin[d + e*x]^2])/(a* b + a^2*Sin[d + e*x])
3.6.5.3.1 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[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.) *(x_)]), x_Symbol] :> Simp[(2*a*c + b*d)*(x/2), x] + (-Simp[(b*c + a*d)*(Co s[e + f*x]/f), x] - Simp[b*d*Cos[e + f*x]*(Sin[e + f*x]/(2*f)), x]) /; Free Q[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]
Int[((A_) + (B_.)*sin[(d_.) + (e_.)*(x_)])*((a_) + (b_.)*sin[(d_.) + (e_.)* (x_)] + (c_.)*sin[(d_.) + (e_.)*(x_)]^2)^(n_), x_Symbol] :> Simp[(a + b*Sin [d + e*x] + c*Sin[d + e*x]^2)^n/(b + 2*c*Sin[d + e*x])^(2*n) Int[(A + B*S in[d + e*x])*(b + 2*c*Sin[d + e*x])^(2*n), x], x] /; FreeQ[{a, b, c, d, e, A, B}, x] && EqQ[b^2 - 4*a*c, 0] && !IntegerQ[n]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 1.81 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.39
method | result | size |
default | \(\frac {\operatorname {csgn}\left (b +a \sin \left (e x +d \right )\right ) \left (a b \left (-\frac {\cos \left (e x +d \right ) \sin \left (e x +d \right )}{2}+\frac {e x}{2}+\frac {d}{2}\right )-a^{2} \cos \left (e x +d \right )-b^{2} \cos \left (e x +d \right )+a b \left (e x +d \right )\right )}{e}\) | \(72\) |
parts | \(-\frac {a \left (a \cos \left (e x +d \right )-\left (e x +d \right ) b +a \right ) \operatorname {csgn}\left (b +a \sin \left (e x +d \right )\right )}{e}-\frac {b \,\operatorname {csgn}\left (b +a \sin \left (e x +d \right )\right ) \left (\cos \left (e x +d \right ) \sin \left (e x +d \right ) a +2 b \cos \left (e x +d \right )-\left (e x +d \right ) a +2 b \right )}{2 e}\) | \(89\) |
csgn(b+a*sin(e*x+d))/e*(a*b*(-1/2*cos(e*x+d)*sin(e*x+d)+1/2*e*x+1/2*d)-a^2 *cos(e*x+d)-b^2*cos(e*x+d)+a*b*(e*x+d))
Time = 0.27 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.23 \[ \int (a+b \sin (d+e x)) \sqrt {b^2+2 a b \sin (d+e x)+a^2 \sin ^2(d+e x)} \, dx=\frac {3 \, a b e x - a b \cos \left (e x + d\right ) \sin \left (e x + d\right ) - 2 \, {\left (a^{2} + b^{2}\right )} \cos \left (e x + d\right )}{2 \, e} \]
integrate((a+b*sin(e*x+d))*(b^2+2*a*b*sin(e*x+d)+a^2*sin(e*x+d)^2)^(1/2),x , algorithm="fricas")
\[ \int (a+b \sin (d+e x)) \sqrt {b^2+2 a b \sin (d+e x)+a^2 \sin ^2(d+e x)} \, dx=\int \left (a + b \sin {\left (d + e x \right )}\right ) \sqrt {\left (a \sin {\left (d + e x \right )} + b\right )^{2}}\, dx \]
Time = 0.37 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.01 \[ \int (a+b \sin (d+e x)) \sqrt {b^2+2 a b \sin (d+e x)+a^2 \sin ^2(d+e x)} \, dx=\frac {2 \, {\left (b \arctan \left (\frac {\sin \left (e x + d\right )}{\cos \left (e x + d\right ) + 1}\right ) - \frac {a}{\frac {\sin \left (e x + d\right )^{2}}{{\left (\cos \left (e x + d\right ) + 1\right )}^{2}} + 1}\right )} a + {\left (a \arctan \left (\frac {\sin \left (e x + d\right )}{\cos \left (e x + d\right ) + 1}\right ) - \frac {2 \, b + \frac {a \sin \left (e x + d\right )}{\cos \left (e x + d\right ) + 1} + \frac {2 \, b \sin \left (e x + d\right )^{2}}{{\left (\cos \left (e x + d\right ) + 1\right )}^{2}} - \frac {a \sin \left (e x + d\right )^{3}}{{\left (\cos \left (e x + d\right ) + 1\right )}^{3}}}{\frac {2 \, \sin \left (e x + d\right )^{2}}{{\left (\cos \left (e x + d\right ) + 1\right )}^{2}} + \frac {\sin \left (e x + d\right )^{4}}{{\left (\cos \left (e x + d\right ) + 1\right )}^{4}} + 1}\right )} b}{e} \]
integrate((a+b*sin(e*x+d))*(b^2+2*a*b*sin(e*x+d)+a^2*sin(e*x+d)^2)^(1/2),x , algorithm="maxima")
(2*(b*arctan(sin(e*x + d)/(cos(e*x + d) + 1)) - a/(sin(e*x + d)^2/(cos(e*x + d) + 1)^2 + 1))*a + (a*arctan(sin(e*x + d)/(cos(e*x + d) + 1)) - (2*b + a*sin(e*x + d)/(cos(e*x + d) + 1) + 2*b*sin(e*x + d)^2/(cos(e*x + d) + 1) ^2 - a*sin(e*x + d)^3/(cos(e*x + d) + 1)^3)/(2*sin(e*x + d)^2/(cos(e*x + d ) + 1)^2 + sin(e*x + d)^4/(cos(e*x + d) + 1)^4 + 1))*b)/e
Time = 0.27 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.51 \[ \int (a+b \sin (d+e x)) \sqrt {b^2+2 a b \sin (d+e x)+a^2 \sin ^2(d+e x)} \, dx=\frac {3}{2} \, a b x \mathrm {sgn}\left (a \sin \left (e x + d\right ) + b\right ) - \frac {a^{2} \cos \left (e x + d\right ) \mathrm {sgn}\left (a \sin \left (e x + d\right ) + b\right )}{e} - \frac {b^{2} \cos \left (e x + d\right ) \mathrm {sgn}\left (a \sin \left (e x + d\right ) + b\right )}{e} - \frac {a b \mathrm {sgn}\left (a \sin \left (e x + d\right ) + b\right ) \sin \left (2 \, e x + 2 \, d\right )}{4 \, e} \]
integrate((a+b*sin(e*x+d))*(b^2+2*a*b*sin(e*x+d)+a^2*sin(e*x+d)^2)^(1/2),x , algorithm="giac")
3/2*a*b*x*sgn(a*sin(e*x + d) + b) - a^2*cos(e*x + d)*sgn(a*sin(e*x + d) + b)/e - b^2*cos(e*x + d)*sgn(a*sin(e*x + d) + b)/e - 1/4*a*b*sgn(a*sin(e*x + d) + b)*sin(2*e*x + 2*d)/e
Timed out. \[ \int (a+b \sin (d+e x)) \sqrt {b^2+2 a b \sin (d+e x)+a^2 \sin ^2(d+e x)} \, dx=\int \left (a+b\,\sin \left (d+e\,x\right )\right )\,\sqrt {a^2\,{\sin \left (d+e\,x\right )}^2+2\,a\,b\,\sin \left (d+e\,x\right )+b^2} \,d x \]