Integrand size = 28, antiderivative size = 115 \[ \int \frac {(a+a \sin (e+f x))^2}{(c-c \sin (e+f x))^{3/2}} \, dx=-\frac {3 \sqrt {2} a^2 \text {arctanh}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{c^{3/2} f}+\frac {a^2 c \cos ^3(e+f x)}{f (c-c \sin (e+f x))^{5/2}}+\frac {3 a^2 \cos (e+f x)}{c f \sqrt {c-c \sin (e+f x)}} \] Output:
-3*2^(1/2)*a^2*arctanh(1/2*c^(1/2)*cos(f*x+e)*2^(1/2)/(c-c*sin(f*x+e))^(1/ 2))/c^(3/2)/f+a^2*c*cos(f*x+e)^3/f/(c-c*sin(f*x+e))^(5/2)+3*a^2*cos(f*x+e) /c/f/(c-c*sin(f*x+e))^(1/2)
Result contains complex when optimal does not.
Time = 7.89 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.30 \[ \int \frac {(a+a \sin (e+f x))^2}{(c-c \sin (e+f x))^{3/2}} \, dx=-\frac {a^2 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (3 \cos \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {3}{2} (e+f x)\right )+3 \sin \left (\frac {1}{2} (e+f x)\right )-(6+6 i) \sqrt [4]{-1} \arctan \left (\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt [4]{-1} \left (1+\tan \left (\frac {1}{4} (e+f x)\right )\right )\right ) (-1+\sin (e+f x))-\sin \left (\frac {3}{2} (e+f x)\right )\right )}{c f (-1+\sin (e+f x)) \sqrt {c-c \sin (e+f x)}} \] Input:
Integrate[(a + a*Sin[e + f*x])^2/(c - c*Sin[e + f*x])^(3/2),x]
Output:
-((a^2*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(3*Cos[(e + f*x)/2] + Cos[(3* (e + f*x))/2] + 3*Sin[(e + f*x)/2] - (6 + 6*I)*(-1)^(1/4)*ArcTan[(1/2 + I/ 2)*(-1)^(1/4)*(1 + Tan[(e + f*x)/4])]*(-1 + Sin[e + f*x]) - Sin[(3*(e + f* x))/2]))/(c*f*(-1 + Sin[e + f*x])*Sqrt[c - c*Sin[e + f*x]]))
Time = 0.64 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.07, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.321, Rules used = {3042, 3215, 3042, 3159, 3042, 3158, 3042, 3128, 219}
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.
| \(\displaystyle \int \frac {(a \sin (e+f x)+a)^2}{(c-c \sin (e+f x))^{3/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a \sin (e+f x)+a)^2}{(c-c \sin (e+f x))^{3/2}}dx\) |
| \(\Big \downarrow \) 3215 |
| \(\displaystyle a^2 c^2 \int \frac {\cos ^4(e+f x)}{(c-c \sin (e+f x))^{7/2}}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a^2 c^2 \int \frac {\cos (e+f x)^4}{(c-c \sin (e+f x))^{7/2}}dx\) |
| \(\Big \downarrow \) 3159 |
| \(\displaystyle a^2 c^2 \left (\frac {\cos ^3(e+f x)}{c f (c-c \sin (e+f x))^{5/2}}-\frac {3 \int \frac {\cos ^2(e+f x)}{(c-c \sin (e+f x))^{3/2}}dx}{2 c^2}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a^2 c^2 \left (\frac {\cos ^3(e+f x)}{c f (c-c \sin (e+f x))^{5/2}}-\frac {3 \int \frac {\cos (e+f x)^2}{(c-c \sin (e+f x))^{3/2}}dx}{2 c^2}\right )\) |
| \(\Big \downarrow \) 3158 |
| \(\displaystyle a^2 c^2 \left (\frac {\cos ^3(e+f x)}{c f (c-c \sin (e+f x))^{5/2}}-\frac {3 \left (\frac {2 \int \frac {1}{\sqrt {c-c \sin (e+f x)}}dx}{c}-\frac {2 \cos (e+f x)}{c f \sqrt {c-c \sin (e+f x)}}\right )}{2 c^2}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a^2 c^2 \left (\frac {\cos ^3(e+f x)}{c f (c-c \sin (e+f x))^{5/2}}-\frac {3 \left (\frac {2 \int \frac {1}{\sqrt {c-c \sin (e+f x)}}dx}{c}-\frac {2 \cos (e+f x)}{c f \sqrt {c-c \sin (e+f x)}}\right )}{2 c^2}\right )\) |
| \(\Big \downarrow \) 3128 |
| \(\displaystyle a^2 c^2 \left (\frac {\cos ^3(e+f x)}{c f (c-c \sin (e+f x))^{5/2}}-\frac {3 \left (-\frac {4 \int \frac {1}{2 c-\frac {c^2 \cos ^2(e+f x)}{c-c \sin (e+f x)}}d\left (-\frac {c \cos (e+f x)}{\sqrt {c-c \sin (e+f x)}}\right )}{c f}-\frac {2 \cos (e+f x)}{c f \sqrt {c-c \sin (e+f x)}}\right )}{2 c^2}\right )\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle a^2 c^2 \left (\frac {\cos ^3(e+f x)}{c f (c-c \sin (e+f x))^{5/2}}-\frac {3 \left (\frac {2 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{c^{3/2} f}-\frac {2 \cos (e+f x)}{c f \sqrt {c-c \sin (e+f x)}}\right )}{2 c^2}\right )\) |
Input:
Int[(a + a*Sin[e + f*x])^2/(c - c*Sin[e + f*x])^(3/2),x]
Output:
a^2*c^2*(Cos[e + f*x]^3/(c*f*(c - c*Sin[e + f*x])^(5/2)) - (3*((2*Sqrt[2]* ArcTanh[(Sqrt[c]*Cos[e + f*x])/(Sqrt[2]*Sqrt[c - c*Sin[e + f*x]])])/(c^(3/ 2)*f) - (2*Cos[e + f*x])/(c*f*Sqrt[c - c*Sin[e + f*x]])))/(2*c^2))
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[-2/d Subst[Int[1/(2*a - x^2), x], x, b*(Cos[c + d*x]/Sqrt[a + b*Sin[c + d*x]])], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)])^(m_), x_Symbol] :> Simp[g*(g*Cos[e + f*x])^(p - 1)*((a + b*Sin[e + f*x ])^(m + 1)/(b*f*(m + p))), x] + Simp[g^2*((p - 1)/(a*(m + p))) Int[(g*Cos [e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{a, b, e, f, g}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -1] && GtQ[p, 1] && (GtQ[m, -2] || EqQ[2*m + p + 1, 0] || (EqQ[m, -2] && IntegerQ[p])) && NeQ[m + p, 0] && In tegersQ[2*m, 2*p]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)])^(m_), x_Symbol] :> Simp[2*g*(g*Cos[e + f*x])^(p - 1)*((a + b*Sin[e + f *x])^(m + 1)/(b*f*(2*m + p + 1))), x] + Simp[g^2*((p - 1)/(b^2*(2*m + p + 1 ))) Int[(g*Cos[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^(m + 2), x], x] /; FreeQ[{a, b, e, f, g}, x] && EqQ[a^2 - b^2, 0] && LeQ[m, -2] && GtQ[p, 1] & & NeQ[2*m + p + 1, 0] && !ILtQ[m + p + 1, 0] && IntegersQ[2*m, 2*p]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[a^m*c^m Int[Cos[e + f*x]^(2*m)*(c + d*Sin[e + f*x])^(n - m), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && EqQ[ b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IntegerQ[m] && !(IntegerQ[n] && ((Lt Q[m, 0] && GtQ[n, 0]) || LtQ[0, n, m] || LtQ[m, n, 0]))
Time = 1.13 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.27
| method | result | size |
| default | \(-\frac {a^{2} \left (-3 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c \sin \left (f x +e \right )+2 \sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {c}\, \sin \left (f x +e \right )+3 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c -4 \sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {c}\right ) \sqrt {c \left (1+\sin \left (f x +e \right )\right )}}{c^{\frac {5}{2}} \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}\) | \(146\) |
| parts | \(\frac {a^{2} \left (-\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{2} \sin \left (f x +e \right )+2 \sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, c^{\frac {3}{2}}+\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{2}\right ) \sqrt {c \left (1+\sin \left (f x +e \right )\right )}}{4 c^{\frac {7}{2}} \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}-\frac {a^{2} \left (-7 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c \sin \left (f x +e \right )+8 \sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {c}\, \sin \left (f x +e \right )+7 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c -10 \sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {c}\right ) \sqrt {c \left (1+\sin \left (f x +e \right )\right )}}{4 c^{\frac {5}{2}} \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}+\frac {a^{2} \left (3 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c \sin \left (f x +e \right )-3 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c +2 \sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {c}\right ) \sqrt {c \left (1+\sin \left (f x +e \right )\right )}}{2 c^{\frac {5}{2}} \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}\) | \(394\) |
Input:
int((a+sin(f*x+e)*a)^2/(c-c*sin(f*x+e))^(3/2),x,method=_RETURNVERBOSE)
Output:
-1/c^(5/2)*a^2*(-3*2^(1/2)*arctanh(1/2*(c*(1+sin(f*x+e)))^(1/2)*2^(1/2)/c^ (1/2))*c*sin(f*x+e)+2*(c*(1+sin(f*x+e)))^(1/2)*c^(1/2)*sin(f*x+e)+3*2^(1/2 )*arctanh(1/2*(c*(1+sin(f*x+e)))^(1/2)*2^(1/2)/c^(1/2))*c-4*(c*(1+sin(f*x+ e)))^(1/2)*c^(1/2))*(c*(1+sin(f*x+e)))^(1/2)/cos(f*x+e)/(c-c*sin(f*x+e))^( 1/2)/f
Leaf count of result is larger than twice the leaf count of optimal. 299 vs. \(2 (102) = 204\).
Time = 0.09 (sec) , antiderivative size = 299, normalized size of antiderivative = 2.60 \[ \int \frac {(a+a \sin (e+f x))^2}{(c-c \sin (e+f x))^{3/2}} \, dx=\frac {\frac {3 \, \sqrt {2} {\left (a^{2} c \cos \left (f x + e\right )^{2} - a^{2} c \cos \left (f x + e\right ) - 2 \, a^{2} c + {\left (a^{2} c \cos \left (f x + e\right ) + 2 \, a^{2} c\right )} \sin \left (f x + e\right )\right )} \log \left (-\frac {\cos \left (f x + e\right )^{2} + {\left (\cos \left (f x + e\right ) - 2\right )} \sin \left (f x + e\right ) - \frac {2 \, \sqrt {2} \sqrt {-c \sin \left (f x + e\right ) + c} {\left (\cos \left (f x + e\right ) + \sin \left (f x + e\right ) + 1\right )}}{\sqrt {c}} + 3 \, \cos \left (f x + e\right ) + 2}{\cos \left (f x + e\right )^{2} + {\left (\cos \left (f x + e\right ) + 2\right )} \sin \left (f x + e\right ) - \cos \left (f x + e\right ) - 2}\right )}{\sqrt {c}} - 4 \, {\left (a^{2} \cos \left (f x + e\right )^{2} + 2 \, a^{2} \cos \left (f x + e\right ) + a^{2} - {\left (a^{2} \cos \left (f x + e\right ) - a^{2}\right )} \sin \left (f x + e\right )\right )} \sqrt {-c \sin \left (f x + e\right ) + c}}{2 \, {\left (c^{2} f \cos \left (f x + e\right )^{2} - c^{2} f \cos \left (f x + e\right ) - 2 \, c^{2} f + {\left (c^{2} f \cos \left (f x + e\right ) + 2 \, c^{2} f\right )} \sin \left (f x + e\right )\right )}} \] Input:
integrate((a+a*sin(f*x+e))^2/(c-c*sin(f*x+e))^(3/2),x, algorithm="fricas")
Output:
1/2*(3*sqrt(2)*(a^2*c*cos(f*x + e)^2 - a^2*c*cos(f*x + e) - 2*a^2*c + (a^2 *c*cos(f*x + e) + 2*a^2*c)*sin(f*x + e))*log(-(cos(f*x + e)^2 + (cos(f*x + e) - 2)*sin(f*x + e) - 2*sqrt(2)*sqrt(-c*sin(f*x + e) + c)*(cos(f*x + e) + sin(f*x + e) + 1)/sqrt(c) + 3*cos(f*x + e) + 2)/(cos(f*x + e)^2 + (cos(f *x + e) + 2)*sin(f*x + e) - cos(f*x + e) - 2))/sqrt(c) - 4*(a^2*cos(f*x + e)^2 + 2*a^2*cos(f*x + e) + a^2 - (a^2*cos(f*x + e) - a^2)*sin(f*x + e))*s qrt(-c*sin(f*x + e) + c))/(c^2*f*cos(f*x + e)^2 - c^2*f*cos(f*x + e) - 2*c ^2*f + (c^2*f*cos(f*x + e) + 2*c^2*f)*sin(f*x + e))
\[ \int \frac {(a+a \sin (e+f x))^2}{(c-c \sin (e+f x))^{3/2}} \, dx=a^{2} \left (\int \frac {2 \sin {\left (e + f x \right )}}{- c \sqrt {- c \sin {\left (e + f x \right )} + c} \sin {\left (e + f x \right )} + c \sqrt {- c \sin {\left (e + f x \right )} + c}}\, dx + \int \frac {\sin ^{2}{\left (e + f x \right )}}{- c \sqrt {- c \sin {\left (e + f x \right )} + c} \sin {\left (e + f x \right )} + c \sqrt {- c \sin {\left (e + f x \right )} + c}}\, dx + \int \frac {1}{- c \sqrt {- c \sin {\left (e + f x \right )} + c} \sin {\left (e + f x \right )} + c \sqrt {- c \sin {\left (e + f x \right )} + c}}\, dx\right ) \] Input:
integrate((a+a*sin(f*x+e))**2/(c-c*sin(f*x+e))**(3/2),x)
Output:
a**2*(Integral(2*sin(e + f*x)/(-c*sqrt(-c*sin(e + f*x) + c)*sin(e + f*x) + c*sqrt(-c*sin(e + f*x) + c)), x) + Integral(sin(e + f*x)**2/(-c*sqrt(-c*s in(e + f*x) + c)*sin(e + f*x) + c*sqrt(-c*sin(e + f*x) + c)), x) + Integra l(1/(-c*sqrt(-c*sin(e + f*x) + c)*sin(e + f*x) + c*sqrt(-c*sin(e + f*x) + c)), x))
\[ \int \frac {(a+a \sin (e+f x))^2}{(c-c \sin (e+f x))^{3/2}} \, dx=\int { \frac {{\left (a \sin \left (f x + e\right ) + a\right )}^{2}}{{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {3}{2}}} \,d x } \] Input:
integrate((a+a*sin(f*x+e))^2/(c-c*sin(f*x+e))^(3/2),x, algorithm="maxima")
Output:
integrate((a*sin(f*x + e) + a)^2/(-c*sin(f*x + e) + c)^(3/2), x)
Exception generated. \[ \int \frac {(a+a \sin (e+f x))^2}{(c-c \sin (e+f x))^{3/2}} \, dx=\text {Exception raised: TypeError} \] Input:
integrate((a+a*sin(f*x+e))^2/(c-c*sin(f*x+e))^(3/2),x, algorithm="giac")
Output:
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value
Timed out. \[ \int \frac {(a+a \sin (e+f x))^2}{(c-c \sin (e+f x))^{3/2}} \, dx=\int \frac {{\left (a+a\,\sin \left (e+f\,x\right )\right )}^2}{{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{3/2}} \,d x \] Input:
int((a + a*sin(e + f*x))^2/(c - c*sin(e + f*x))^(3/2),x)
Output:
int((a + a*sin(e + f*x))^2/(c - c*sin(e + f*x))^(3/2), x)
\[ \int \frac {(a+a \sin (e+f x))^2}{(c-c \sin (e+f x))^{3/2}} \, dx=\frac {\sqrt {c}\, a^{2} \left (\int \frac {\sqrt {-\sin \left (f x +e \right )+1}}{\sin \left (f x +e \right )^{2}-2 \sin \left (f x +e \right )+1}d x +\int \frac {\sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{2}}{\sin \left (f x +e \right )^{2}-2 \sin \left (f x +e \right )+1}d x +2 \left (\int \frac {\sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )}{\sin \left (f x +e \right )^{2}-2 \sin \left (f x +e \right )+1}d x \right )\right )}{c^{2}} \] Input:
int((a+a*sin(f*x+e))^2/(c-c*sin(f*x+e))^(3/2),x)
Output:
(sqrt(c)*a**2*(int(sqrt( - sin(e + f*x) + 1)/(sin(e + f*x)**2 - 2*sin(e + f*x) + 1),x) + int((sqrt( - sin(e + f*x) + 1)*sin(e + f*x)**2)/(sin(e + f* x)**2 - 2*sin(e + f*x) + 1),x) + 2*int((sqrt( - sin(e + f*x) + 1)*sin(e + f*x))/(sin(e + f*x)**2 - 2*sin(e + f*x) + 1),x)))/c**2