Integrand size = 28, antiderivative size = 192 \[ \int \frac {1}{(a+a \sin (e+f x))^2 (c-c \sin (e+f x))^{5/2}} \, dx=\frac {35 \text {arctanh}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{64 \sqrt {2} a^2 c^{5/2} f}+\frac {35 \cos (e+f x)}{64 a^2 c f (c-c \sin (e+f x))^{3/2}}+\frac {7 \sec (e+f x)}{24 a^2 c f (c-c \sin (e+f x))^{3/2}}-\frac {35 \sec (e+f x)}{48 a^2 c^2 f \sqrt {c-c \sin (e+f x)}}-\frac {\sec ^3(e+f x)}{3 a^2 c^2 f \sqrt {c-c \sin (e+f x)}} \] Output:
35/128*arctanh(1/2*c^(1/2)*cos(f*x+e)*2^(1/2)/(c-c*sin(f*x+e))^(1/2))*2^(1 /2)/a^2/c^(5/2)/f+35/64*cos(f*x+e)/a^2/c/f/(c-c*sin(f*x+e))^(3/2)+7/24*sec (f*x+e)/a^2/c/f/(c-c*sin(f*x+e))^(3/2)-35/48*sec(f*x+e)/a^2/c^2/f/(c-c*sin (f*x+e))^(1/2)-1/3*sec(f*x+e)^3/a^2/c^2/f/(c-c*sin(f*x+e))^(1/2)
Result contains complex when optimal does not.
Time = 6.46 (sec) , antiderivative size = 156, normalized size of antiderivative = 0.81 \[ \int \frac {1}{(a+a \sin (e+f x))^2 (c-c \sin (e+f x))^{5/2}} \, dx=-\frac {\left (\frac {1}{1536}+\frac {i}{1536}\right ) \sec ^3(e+f x) \left (840 \sqrt [4]{-1} \arctan \left (\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt [4]{-1} \left (1+\tan \left (\frac {1}{4} (e+f x)\right )\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^4 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^3+(1-i) (102+70 \cos (2 (e+f x))-329 \sin (e+f x)-105 \sin (3 (e+f x)))\right )}{a^2 c^2 f \sqrt {c-c \sin (e+f x)}} \] Input:
Integrate[1/((a + a*Sin[e + f*x])^2*(c - c*Sin[e + f*x])^(5/2)),x]
Output:
((-1/1536 - I/1536)*Sec[e + f*x]^3*(840*(-1)^(1/4)*ArcTan[(1/2 + I/2)*(-1) ^(1/4)*(1 + Tan[(e + f*x)/4])]*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^4*(Co s[(e + f*x)/2] + Sin[(e + f*x)/2])^3 + (1 - I)*(102 + 70*Cos[2*(e + f*x)] - 329*Sin[e + f*x] - 105*Sin[3*(e + f*x)])))/(a^2*c^2*f*Sqrt[c - c*Sin[e + f*x]])
Time = 0.90 (sec) , antiderivative size = 190, normalized size of antiderivative = 0.99, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.464, Rules used = {3042, 3215, 3042, 3166, 3042, 3160, 3042, 3166, 3042, 3129, 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.
| \(\displaystyle \int \frac {1}{(a \sin (e+f x)+a)^2 (c-c \sin (e+f x))^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{(a \sin (e+f x)+a)^2 (c-c \sin (e+f x))^{5/2}}dx\) |
| \(\Big \downarrow \) 3215 |
| \(\displaystyle \frac {\int \frac {\sec ^4(e+f x)}{\sqrt {c-c \sin (e+f x)}}dx}{a^2 c^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {1}{\cos (e+f x)^4 \sqrt {c-c \sin (e+f x)}}dx}{a^2 c^2}\) |
| \(\Big \downarrow \) 3166 |
| \(\displaystyle \frac {\frac {7}{6} c \int \frac {\sec ^2(e+f x)}{(c-c \sin (e+f x))^{3/2}}dx-\frac {\sec ^3(e+f x)}{3 f \sqrt {c-c \sin (e+f x)}}}{a^2 c^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {7}{6} c \int \frac {1}{\cos (e+f x)^2 (c-c \sin (e+f x))^{3/2}}dx-\frac {\sec ^3(e+f x)}{3 f \sqrt {c-c \sin (e+f x)}}}{a^2 c^2}\) |
| \(\Big \downarrow \) 3160 |
| \(\displaystyle \frac {\frac {7}{6} c \left (\frac {5 \int \frac {\sec ^2(e+f x)}{\sqrt {c-c \sin (e+f x)}}dx}{8 c}+\frac {\sec (e+f x)}{4 f (c-c \sin (e+f x))^{3/2}}\right )-\frac {\sec ^3(e+f x)}{3 f \sqrt {c-c \sin (e+f x)}}}{a^2 c^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {7}{6} c \left (\frac {5 \int \frac {1}{\cos (e+f x)^2 \sqrt {c-c \sin (e+f x)}}dx}{8 c}+\frac {\sec (e+f x)}{4 f (c-c \sin (e+f x))^{3/2}}\right )-\frac {\sec ^3(e+f x)}{3 f \sqrt {c-c \sin (e+f x)}}}{a^2 c^2}\) |
| \(\Big \downarrow \) 3166 |
| \(\displaystyle \frac {\frac {7}{6} c \left (\frac {5 \left (\frac {3}{2} c \int \frac {1}{(c-c \sin (e+f x))^{3/2}}dx-\frac {\sec (e+f x)}{f \sqrt {c-c \sin (e+f x)}}\right )}{8 c}+\frac {\sec (e+f x)}{4 f (c-c \sin (e+f x))^{3/2}}\right )-\frac {\sec ^3(e+f x)}{3 f \sqrt {c-c \sin (e+f x)}}}{a^2 c^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {7}{6} c \left (\frac {5 \left (\frac {3}{2} c \int \frac {1}{(c-c \sin (e+f x))^{3/2}}dx-\frac {\sec (e+f x)}{f \sqrt {c-c \sin (e+f x)}}\right )}{8 c}+\frac {\sec (e+f x)}{4 f (c-c \sin (e+f x))^{3/2}}\right )-\frac {\sec ^3(e+f x)}{3 f \sqrt {c-c \sin (e+f x)}}}{a^2 c^2}\) |
| \(\Big \downarrow \) 3129 |
| \(\displaystyle \frac {\frac {7}{6} c \left (\frac {5 \left (\frac {3}{2} c \left (\frac {\int \frac {1}{\sqrt {c-c \sin (e+f x)}}dx}{4 c}+\frac {\cos (e+f x)}{2 f (c-c \sin (e+f x))^{3/2}}\right )-\frac {\sec (e+f x)}{f \sqrt {c-c \sin (e+f x)}}\right )}{8 c}+\frac {\sec (e+f x)}{4 f (c-c \sin (e+f x))^{3/2}}\right )-\frac {\sec ^3(e+f x)}{3 f \sqrt {c-c \sin (e+f x)}}}{a^2 c^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {7}{6} c \left (\frac {5 \left (\frac {3}{2} c \left (\frac {\int \frac {1}{\sqrt {c-c \sin (e+f x)}}dx}{4 c}+\frac {\cos (e+f x)}{2 f (c-c \sin (e+f x))^{3/2}}\right )-\frac {\sec (e+f x)}{f \sqrt {c-c \sin (e+f x)}}\right )}{8 c}+\frac {\sec (e+f x)}{4 f (c-c \sin (e+f x))^{3/2}}\right )-\frac {\sec ^3(e+f x)}{3 f \sqrt {c-c \sin (e+f x)}}}{a^2 c^2}\) |
| \(\Big \downarrow \) 3128 |
| \(\displaystyle \frac {\frac {7}{6} c \left (\frac {5 \left (\frac {3}{2} c \left (\frac {\cos (e+f x)}{2 f (c-c \sin (e+f x))^{3/2}}-\frac {\int \frac {1}{2 c-\frac {c^2 \cos ^2(e+f x)}{c-c \sin (e+f x)}}d\left (-\frac {c \cos (e+f x)}{\sqrt {c-c \sin (e+f x)}}\right )}{2 c f}\right )-\frac {\sec (e+f x)}{f \sqrt {c-c \sin (e+f x)}}\right )}{8 c}+\frac {\sec (e+f x)}{4 f (c-c \sin (e+f x))^{3/2}}\right )-\frac {\sec ^3(e+f x)}{3 f \sqrt {c-c \sin (e+f x)}}}{a^2 c^2}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\frac {7}{6} c \left (\frac {5 \left (\frac {3}{2} c \left (\frac {\text {arctanh}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{2 \sqrt {2} c^{3/2} f}+\frac {\cos (e+f x)}{2 f (c-c \sin (e+f x))^{3/2}}\right )-\frac {\sec (e+f x)}{f \sqrt {c-c \sin (e+f x)}}\right )}{8 c}+\frac {\sec (e+f x)}{4 f (c-c \sin (e+f x))^{3/2}}\right )-\frac {\sec ^3(e+f x)}{3 f \sqrt {c-c \sin (e+f x)}}}{a^2 c^2}\) |
Input:
Int[1/((a + a*Sin[e + f*x])^2*(c - c*Sin[e + f*x])^(5/2)),x]
Output:
(-1/3*Sec[e + f*x]^3/(f*Sqrt[c - c*Sin[e + f*x]]) + (7*c*(Sec[e + f*x]/(4* f*(c - c*Sin[e + f*x])^(3/2)) + (5*(-(Sec[e + f*x]/(f*Sqrt[c - c*Sin[e + f *x]])) + (3*c*(ArcTanh[(Sqrt[c]*Cos[e + f*x])/(Sqrt[2]*Sqrt[c - c*Sin[e + f*x]])]/(2*Sqrt[2]*c^(3/2)*f) + Cos[e + f*x]/(2*f*(c - c*Sin[e + f*x])^(3/ 2))))/2))/(8*c)))/6)/(a^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[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b*Cos[c + d*x]*((a + b*Sin[c + d*x])^n/(a*d*(2*n + 1))), x] + Simp[(n + 1)/(a*(2*n + 1)) Int[(a + b*Sin[c + d*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0] && LtQ[n, -1] && IntegerQ[2*n]
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/(a*f*g*(2*m + p + 1))), x] + Simp[(m + p + 1)/(a*(2*m + p + 1)) 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] && LtQ[m, -1] && NeQ[2*m + p + 1, 0] & & IntegersQ[2*m, 2*p]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_. )*(x_)]], x_Symbol] :> Simp[(-b)*((g*Cos[e + f*x])^(p + 1)/(a*f*g*(p + 1)*S qrt[a + b*Sin[e + f*x]])), x] + Simp[a*((2*p + 1)/(2*g^2*(p + 1))) Int[(g *Cos[e + f*x])^(p + 2)/(a + b*Sin[e + f*x])^(3/2), x], x] /; FreeQ[{a, b, e , f, g}, x] && EqQ[a^2 - b^2, 0] && LtQ[p, -1] && IntegerQ[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 = 0.47 (sec) , antiderivative size = 233, normalized size of antiderivative = 1.21
| method | result | size |
| default | \(\frac {210 c^{\frac {7}{2}} \sin \left (f x +e \right )^{3}-70 c^{\frac {7}{2}} \sin \left (f x +e \right )^{2}-105 \sin \left (f x +e \right )^{2} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \left (c \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {3}{2}} c^{2}-322 c^{\frac {7}{2}} \sin \left (f x +e \right )+210 \sin \left (f x +e \right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \left (c \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {3}{2}} c^{2}+86 c^{\frac {7}{2}}-105 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{2} \left (c \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {3}{2}}}{384 c^{\frac {11}{2}} a^{2} \left (1+\sin \left (f x +e \right )\right ) \left (-1+\sin \left (f x +e \right )\right ) \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}\) | \(233\) |
Input:
int(1/(a+sin(f*x+e)*a)^2/(c-c*sin(f*x+e))^(5/2),x,method=_RETURNVERBOSE)
Output:
1/384*(210*c^(7/2)*sin(f*x+e)^3-70*c^(7/2)*sin(f*x+e)^2-105*sin(f*x+e)^2*2 ^(1/2)*arctanh(1/2*(c*(1+sin(f*x+e)))^(1/2)*2^(1/2)/c^(1/2))*(c*(1+sin(f*x +e)))^(3/2)*c^2-322*c^(7/2)*sin(f*x+e)+210*sin(f*x+e)*2^(1/2)*arctanh(1/2* (c*(1+sin(f*x+e)))^(1/2)*2^(1/2)/c^(1/2))*(c*(1+sin(f*x+e)))^(3/2)*c^2+86* c^(7/2)-105*2^(1/2)*arctanh(1/2*(c*(1+sin(f*x+e)))^(1/2)*2^(1/2)/c^(1/2))* c^2*(c*(1+sin(f*x+e)))^(3/2))/c^(11/2)/a^2/(1+sin(f*x+e))/(-1+sin(f*x+e))/ cos(f*x+e)/(c-c*sin(f*x+e))^(1/2)/f
Time = 0.10 (sec) , antiderivative size = 241, normalized size of antiderivative = 1.26 \[ \int \frac {1}{(a+a \sin (e+f x))^2 (c-c \sin (e+f x))^{5/2}} \, dx=\frac {105 \, \sqrt {2} {\left (\cos \left (f x + e\right )^{3} \sin \left (f x + e\right ) - \cos \left (f x + e\right )^{3}\right )} \sqrt {c} \log \left (-\frac {c \cos \left (f x + e\right )^{2} + 2 \, \sqrt {2} \sqrt {-c \sin \left (f x + e\right ) + c} \sqrt {c} {\left (\cos \left (f x + e\right ) + \sin \left (f x + e\right ) + 1\right )} + 3 \, c \cos \left (f x + e\right ) + {\left (c \cos \left (f x + e\right ) - 2 \, c\right )} \sin \left (f x + e\right ) + 2 \, c}{\cos \left (f x + e\right )^{2} + {\left (\cos \left (f x + e\right ) + 2\right )} \sin \left (f x + e\right ) - \cos \left (f x + e\right ) - 2}\right ) + 4 \, {\left (35 \, \cos \left (f x + e\right )^{2} - 7 \, {\left (15 \, \cos \left (f x + e\right )^{2} + 8\right )} \sin \left (f x + e\right ) + 8\right )} \sqrt {-c \sin \left (f x + e\right ) + c}}{768 \, {\left (a^{2} c^{3} f \cos \left (f x + e\right )^{3} \sin \left (f x + e\right ) - a^{2} c^{3} f \cos \left (f x + e\right )^{3}\right )}} \] Input:
integrate(1/(a+a*sin(f*x+e))^2/(c-c*sin(f*x+e))^(5/2),x, algorithm="fricas ")
Output:
1/768*(105*sqrt(2)*(cos(f*x + e)^3*sin(f*x + e) - cos(f*x + e)^3)*sqrt(c)* log(-(c*cos(f*x + e)^2 + 2*sqrt(2)*sqrt(-c*sin(f*x + e) + c)*sqrt(c)*(cos( f*x + e) + sin(f*x + e) + 1) + 3*c*cos(f*x + e) + (c*cos(f*x + e) - 2*c)*s in(f*x + e) + 2*c)/(cos(f*x + e)^2 + (cos(f*x + e) + 2)*sin(f*x + e) - cos (f*x + e) - 2)) + 4*(35*cos(f*x + e)^2 - 7*(15*cos(f*x + e)^2 + 8)*sin(f*x + e) + 8)*sqrt(-c*sin(f*x + e) + c))/(a^2*c^3*f*cos(f*x + e)^3*sin(f*x + e) - a^2*c^3*f*cos(f*x + e)^3)
\[ \int \frac {1}{(a+a \sin (e+f x))^2 (c-c \sin (e+f x))^{5/2}} \, dx=\frac {\int \frac {1}{c^{2} \sqrt {- c \sin {\left (e + f x \right )} + c} \sin ^{4}{\left (e + f x \right )} - 2 c^{2} \sqrt {- c \sin {\left (e + f x \right )} + c} \sin ^{2}{\left (e + f x \right )} + c^{2} \sqrt {- c \sin {\left (e + f x \right )} + c}}\, dx}{a^{2}} \] Input:
integrate(1/(a+a*sin(f*x+e))**2/(c-c*sin(f*x+e))**(5/2),x)
Output:
Integral(1/(c**2*sqrt(-c*sin(e + f*x) + c)*sin(e + f*x)**4 - 2*c**2*sqrt(- c*sin(e + f*x) + c)*sin(e + f*x)**2 + c**2*sqrt(-c*sin(e + f*x) + c)), x)/ a**2
Timed out. \[ \int \frac {1}{(a+a \sin (e+f x))^2 (c-c \sin (e+f x))^{5/2}} \, dx=\text {Timed out} \] Input:
integrate(1/(a+a*sin(f*x+e))^2/(c-c*sin(f*x+e))^(5/2),x, algorithm="maxima ")
Output:
Timed out
Exception generated. \[ \int \frac {1}{(a+a \sin (e+f x))^2 (c-c \sin (e+f x))^{5/2}} \, dx=\text {Exception raised: TypeError} \] Input:
integrate(1/(a+a*sin(f*x+e))^2/(c-c*sin(f*x+e))^(5/2),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. \[ \int \frac {1}{(a+a \sin (e+f x))^2 (c-c \sin (e+f x))^{5/2}} \, dx=\int \frac {1}{{\left (a+a\,\sin \left (e+f\,x\right )\right )}^2\,{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{5/2}} \,d x \] Input:
int(1/((a + a*sin(e + f*x))^2*(c - c*sin(e + f*x))^(5/2)),x)
Output:
int(1/((a + a*sin(e + f*x))^2*(c - c*sin(e + f*x))^(5/2)), x)
\[ \int \frac {1}{(a+a \sin (e+f x))^2 (c-c \sin (e+f x))^{5/2}} \, dx=-\frac {\sqrt {c}\, \left (\int \frac {\sqrt {-\sin \left (f x +e \right )+1}}{\sin \left (f x +e \right )^{5}-\sin \left (f x +e \right )^{4}-2 \sin \left (f x +e \right )^{3}+2 \sin \left (f x +e \right )^{2}+\sin \left (f x +e \right )-1}d x \right )}{a^{2} c^{3}} \] Input:
int(1/(a+a*sin(f*x+e))^2/(c-c*sin(f*x+e))^(5/2),x)
Output:
( - sqrt(c)*int(sqrt( - sin(e + f*x) + 1)/(sin(e + f*x)**5 - sin(e + f*x)* *4 - 2*sin(e + f*x)**3 + 2*sin(e + f*x)**2 + sin(e + f*x) - 1),x))/(a**2*c **3)