Integrand size = 25, antiderivative size = 105 \[ \int \frac {(a+a \sin (c+d x))^2}{\sqrt {e \cos (c+d x)}} \, dx=-\frac {10 a^2 \sqrt {e \cos (c+d x)}}{3 d e}+\frac {10 a^2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d \sqrt {e \cos (c+d x)}}-\frac {2 \sqrt {e \cos (c+d x)} \left (a^2+a^2 \sin (c+d x)\right )}{3 d e} \] Output:
-10/3*a^2*(e*cos(d*x+c))^(1/2)/d/e+10/3*a^2*cos(d*x+c)^(1/2)*InverseJacobi AM(1/2*d*x+1/2*c,2^(1/2))/d/(e*cos(d*x+c))^(1/2)-2/3*(e*cos(d*x+c))^(1/2)* (a^2+a^2*sin(d*x+c))/d/e
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 0.03 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.61 \[ \int \frac {(a+a \sin (c+d x))^2}{\sqrt {e \cos (c+d x)}} \, dx=-\frac {8 \sqrt [4]{2} a^2 \sqrt {e \cos (c+d x)} \operatorname {Hypergeometric2F1}\left (-\frac {5}{4},\frac {1}{4},\frac {5}{4},\frac {1}{2} (1-\sin (c+d x))\right )}{d e \sqrt [4]{1+\sin (c+d x)}} \] Input:
Integrate[(a + a*Sin[c + d*x])^2/Sqrt[e*Cos[c + d*x]],x]
Output:
(-8*2^(1/4)*a^2*Sqrt[e*Cos[c + d*x]]*Hypergeometric2F1[-5/4, 1/4, 5/4, (1 - Sin[c + d*x])/2])/(d*e*(1 + Sin[c + d*x])^(1/4))
Time = 0.48 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.98, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {3042, 3157, 3042, 3148, 3042, 3121, 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.
| \(\displaystyle \int \frac {(a \sin (c+d x)+a)^2}{\sqrt {e \cos (c+d x)}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a \sin (c+d x)+a)^2}{\sqrt {e \cos (c+d x)}}dx\) |
| \(\Big \downarrow \) 3157 |
| \(\displaystyle \frac {5}{3} a \int \frac {\sin (c+d x) a+a}{\sqrt {e \cos (c+d x)}}dx-\frac {2 \left (a^2 \sin (c+d x)+a^2\right ) \sqrt {e \cos (c+d x)}}{3 d e}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {5}{3} a \int \frac {\sin (c+d x) a+a}{\sqrt {e \cos (c+d x)}}dx-\frac {2 \left (a^2 \sin (c+d x)+a^2\right ) \sqrt {e \cos (c+d x)}}{3 d e}\) |
| \(\Big \downarrow \) 3148 |
| \(\displaystyle \frac {5}{3} a \left (a \int \frac {1}{\sqrt {e \cos (c+d x)}}dx-\frac {2 a \sqrt {e \cos (c+d x)}}{d e}\right )-\frac {2 \left (a^2 \sin (c+d x)+a^2\right ) \sqrt {e \cos (c+d x)}}{3 d e}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {5}{3} a \left (a \int \frac {1}{\sqrt {e \sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 a \sqrt {e \cos (c+d x)}}{d e}\right )-\frac {2 \left (a^2 \sin (c+d x)+a^2\right ) \sqrt {e \cos (c+d x)}}{3 d e}\) |
| \(\Big \downarrow \) 3121 |
| \(\displaystyle \frac {5}{3} a \left (\frac {a \sqrt {\cos (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx}{\sqrt {e \cos (c+d x)}}-\frac {2 a \sqrt {e \cos (c+d x)}}{d e}\right )-\frac {2 \left (a^2 \sin (c+d x)+a^2\right ) \sqrt {e \cos (c+d x)}}{3 d e}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {5}{3} a \left (\frac {a \sqrt {\cos (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{\sqrt {e \cos (c+d x)}}-\frac {2 a \sqrt {e \cos (c+d x)}}{d e}\right )-\frac {2 \left (a^2 \sin (c+d x)+a^2\right ) \sqrt {e \cos (c+d x)}}{3 d e}\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle \frac {5}{3} a \left (\frac {2 a \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d \sqrt {e \cos (c+d x)}}-\frac {2 a \sqrt {e \cos (c+d x)}}{d e}\right )-\frac {2 \left (a^2 \sin (c+d x)+a^2\right ) \sqrt {e \cos (c+d x)}}{3 d e}\) |
Input:
Int[(a + a*Sin[c + d*x])^2/Sqrt[e*Cos[c + d*x]],x]
Output:
(5*a*((-2*a*Sqrt[e*Cos[c + d*x]])/(d*e) + (2*a*Sqrt[Cos[c + d*x]]*Elliptic F[(c + d*x)/2, 2])/(d*Sqrt[e*Cos[c + d*x]])))/3 - (2*Sqrt[e*Cos[c + d*x]]* (a^2 + a^2*Sin[c + d*x]))/(3*d*e)
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[((b_)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*Sin[c + d*x]) ^n/Sin[c + d*x]^n Int[Sin[c + d*x]^n, x], x] /; FreeQ[{b, c, d}, x] && Lt Q[-1, n, 1] && IntegerQ[2*n]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)]), x_Symbol] :> Simp[(-b)*((g*Cos[e + f*x])^(p + 1)/(f*g*(p + 1))), x] + Simp[a Int[(g*Cos[e + f*x])^p, x], x] /; FreeQ[{a, b, e, f, g, p}, x] && (IntegerQ[2*p] || NeQ[a^2 - b^2, 0])
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)])^(m_), x_Symbol] :> Simp[(-b)*(g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + f*x])^(m - 1)/(f*g*(m + p))), x] + Simp[a*((2*m + p - 1)/(m + p)) Int[(g* Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m - 1), x], x] /; FreeQ[{a, b, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0] && GtQ[m, 0] && NeQ[m + p, 0] && Integers Q[2*m, 2*p]
Time = 0.81 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.45
| method | result | size |
| default | \(\frac {2 a^{2} \left (4 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-5 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+12 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}-6 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} e +e}\, d}\) | \(152\) |
| parts | \(\frac {2 a^{2} \sqrt {\cos \left (d x +c \right )}\, \operatorname {InverseJacobiAM}\left (\frac {d x}{2}+\frac {c}{2}, \sqrt {2}\right )}{d \sqrt {e \cos \left (d x +c \right )}}+\frac {4 a^{2} \sqrt {e \left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{3 \sqrt {-e \left (2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {e \left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )}\, d}-\frac {4 a^{2} \sqrt {e \cos \left (d x +c \right )}}{d e}\) | \(251\) |
Input:
int((a+a*sin(d*x+c))^2/(e*cos(d*x+c))^(1/2),x,method=_RETURNVERBOSE)
Output:
2/3/sin(1/2*d*x+1/2*c)/(-2*sin(1/2*d*x+1/2*c)^2*e+e)^(1/2)*a^2*(4*sin(1/2* d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)-2*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)- 5*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF( cos(1/2*d*x+1/2*c),2^(1/2))+12*sin(1/2*d*x+1/2*c)^3-6*sin(1/2*d*x+1/2*c))/ d
Result contains complex when optimal does not.
Time = 0.09 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.90 \[ \int \frac {(a+a \sin (c+d x))^2}{\sqrt {e \cos (c+d x)}} \, dx=-\frac {2 \, {\left (5 i \, \sqrt {\frac {1}{2}} a^{2} \sqrt {e} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - 5 i \, \sqrt {\frac {1}{2}} a^{2} \sqrt {e} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + {\left (a^{2} \sin \left (d x + c\right ) + 6 \, a^{2}\right )} \sqrt {e \cos \left (d x + c\right )}\right )}}{3 \, d e} \] Input:
integrate((a+a*sin(d*x+c))^2/(e*cos(d*x+c))^(1/2),x, algorithm="fricas")
Output:
-2/3*(5*I*sqrt(1/2)*a^2*sqrt(e)*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) - 5*I*sqrt(1/2)*a^2*sqrt(e)*weierstrassPInverse(-4, 0, cos (d*x + c) - I*sin(d*x + c)) + (a^2*sin(d*x + c) + 6*a^2)*sqrt(e*cos(d*x + c)))/(d*e)
Timed out. \[ \int \frac {(a+a \sin (c+d x))^2}{\sqrt {e \cos (c+d x)}} \, dx=\text {Timed out} \] Input:
integrate((a+a*sin(d*x+c))**2/(e*cos(d*x+c))**(1/2),x)
Output:
Timed out
\[ \int \frac {(a+a \sin (c+d x))^2}{\sqrt {e \cos (c+d x)}} \, dx=\int { \frac {{\left (a \sin \left (d x + c\right ) + a\right )}^{2}}{\sqrt {e \cos \left (d x + c\right )}} \,d x } \] Input:
integrate((a+a*sin(d*x+c))^2/(e*cos(d*x+c))^(1/2),x, algorithm="maxima")
Output:
integrate((a*sin(d*x + c) + a)^2/sqrt(e*cos(d*x + c)), x)
\[ \int \frac {(a+a \sin (c+d x))^2}{\sqrt {e \cos (c+d x)}} \, dx=\int { \frac {{\left (a \sin \left (d x + c\right ) + a\right )}^{2}}{\sqrt {e \cos \left (d x + c\right )}} \,d x } \] Input:
integrate((a+a*sin(d*x+c))^2/(e*cos(d*x+c))^(1/2),x, algorithm="giac")
Output:
integrate((a*sin(d*x + c) + a)^2/sqrt(e*cos(d*x + c)), x)
Timed out. \[ \int \frac {(a+a \sin (c+d x))^2}{\sqrt {e \cos (c+d x)}} \, dx=\int \frac {{\left (a+a\,\sin \left (c+d\,x\right )\right )}^2}{\sqrt {e\,\cos \left (c+d\,x\right )}} \,d x \] Input:
int((a + a*sin(c + d*x))^2/(e*cos(c + d*x))^(1/2),x)
Output:
int((a + a*sin(c + d*x))^2/(e*cos(c + d*x))^(1/2), x)
\[ \int \frac {(a+a \sin (c+d x))^2}{\sqrt {e \cos (c+d x)}} \, dx=\frac {\sqrt {e}\, a^{2} \left (-4 \sqrt {\cos \left (d x +c \right )}+\left (\int \frac {\sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )}d x \right ) d +\left (\int \frac {\sqrt {\cos \left (d x +c \right )}\, \sin \left (d x +c \right )^{2}}{\cos \left (d x +c \right )}d x \right ) d \right )}{d e} \] Input:
int((a+a*sin(d*x+c))^2/(e*cos(d*x+c))^(1/2),x)
Output:
(sqrt(e)*a**2*( - 4*sqrt(cos(c + d*x)) + int(sqrt(cos(c + d*x))/cos(c + d* x),x)*d + int((sqrt(cos(c + d*x))*sin(c + d*x)**2)/cos(c + d*x),x)*d))/(d* e)