Integrand size = 25, antiderivative size = 126 \[ \int \frac {c+d \sin (e+f x)}{(a+a \sin (e+f x))^{5/2}} \, dx=-\frac {(3 c+5 d) \text {arctanh}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a+a \sin (e+f x)}}\right )}{16 \sqrt {2} a^{5/2} f}-\frac {(c-d) \cos (e+f x)}{4 f (a+a \sin (e+f x))^{5/2}}-\frac {(3 c+5 d) \cos (e+f x)}{16 a f (a+a \sin (e+f x))^{3/2}} \] Output:
-1/32*(3*c+5*d)*arctanh(1/2*a^(1/2)*cos(f*x+e)*2^(1/2)/(a+a*sin(f*x+e))^(1 /2))*2^(1/2)/a^(5/2)/f-1/4*(c-d)*cos(f*x+e)/f/(a+a*sin(f*x+e))^(5/2)-1/16* (3*c+5*d)*cos(f*x+e)/a/f/(a+a*sin(f*x+e))^(3/2)
Result contains complex when optimal does not.
Time = 0.71 (sec) , antiderivative size = 227, normalized size of antiderivative = 1.80 \[ \int \frac {c+d \sin (e+f x)}{(a+a \sin (e+f x))^{5/2}} \, dx=\frac {\left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (8 (c-d) \sin \left (\frac {1}{2} (e+f x)\right )+4 (-c+d) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )+2 (3 c+5 d) \sin \left (\frac {1}{2} (e+f x)\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^2-(3 c+5 d) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^3+(1+i) (-1)^{3/4} (3 c+5 d) \text {arctanh}\left (\left (\frac {1}{2}+\frac {i}{2}\right ) (-1)^{3/4} \left (-1+\tan \left (\frac {1}{4} (e+f x)\right )\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^4\right )}{16 f (a (1+\sin (e+f x)))^{5/2}} \] Input:
Integrate[(c + d*Sin[e + f*x])/(a + a*Sin[e + f*x])^(5/2),x]
Output:
((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(8*(c - d)*Sin[(e + f*x)/2] + 4*(-c + d)*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2]) + 2*(3*c + 5*d)*Sin[(e + f*x)/ 2]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^2 - (3*c + 5*d)*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3 + (1 + I)*(-1)^(3/4)*(3*c + 5*d)*ArcTanh[(1/2 + I/2 )*(-1)^(3/4)*(-1 + Tan[(e + f*x)/4])]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2] )^4))/(16*f*(a*(1 + Sin[e + f*x]))^(5/2))
Time = 0.44 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.98, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {3042, 3229, 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.
\(\displaystyle \int \frac {c+d \sin (e+f x)}{(a \sin (e+f x)+a)^{5/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {c+d \sin (e+f x)}{(a \sin (e+f x)+a)^{5/2}}dx\) |
\(\Big \downarrow \) 3229 |
\(\displaystyle \frac {(3 c+5 d) \int \frac {1}{(\sin (e+f x) a+a)^{3/2}}dx}{8 a}-\frac {(c-d) \cos (e+f x)}{4 f (a \sin (e+f x)+a)^{5/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {(3 c+5 d) \int \frac {1}{(\sin (e+f x) a+a)^{3/2}}dx}{8 a}-\frac {(c-d) \cos (e+f x)}{4 f (a \sin (e+f x)+a)^{5/2}}\) |
\(\Big \downarrow \) 3129 |
\(\displaystyle \frac {(3 c+5 d) \left (\frac {\int \frac {1}{\sqrt {\sin (e+f x) a+a}}dx}{4 a}-\frac {\cos (e+f x)}{2 f (a \sin (e+f x)+a)^{3/2}}\right )}{8 a}-\frac {(c-d) \cos (e+f x)}{4 f (a \sin (e+f x)+a)^{5/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {(3 c+5 d) \left (\frac {\int \frac {1}{\sqrt {\sin (e+f x) a+a}}dx}{4 a}-\frac {\cos (e+f x)}{2 f (a \sin (e+f x)+a)^{3/2}}\right )}{8 a}-\frac {(c-d) \cos (e+f x)}{4 f (a \sin (e+f x)+a)^{5/2}}\) |
\(\Big \downarrow \) 3128 |
\(\displaystyle \frac {(3 c+5 d) \left (-\frac {\int \frac {1}{2 a-\frac {a^2 \cos ^2(e+f x)}{\sin (e+f x) a+a}}d\frac {a \cos (e+f x)}{\sqrt {\sin (e+f x) a+a}}}{2 a f}-\frac {\cos (e+f x)}{2 f (a \sin (e+f x)+a)^{3/2}}\right )}{8 a}-\frac {(c-d) \cos (e+f x)}{4 f (a \sin (e+f x)+a)^{5/2}}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {(3 c+5 d) \left (-\frac {\text {arctanh}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a \sin (e+f x)+a}}\right )}{2 \sqrt {2} a^{3/2} f}-\frac {\cos (e+f x)}{2 f (a \sin (e+f x)+a)^{3/2}}\right )}{8 a}-\frac {(c-d) \cos (e+f x)}{4 f (a \sin (e+f x)+a)^{5/2}}\) |
Input:
Int[(c + d*Sin[e + f*x])/(a + a*Sin[e + f*x])^(5/2),x]
Output:
-1/4*((c - d)*Cos[e + f*x])/(f*(a + a*Sin[e + f*x])^(5/2)) + ((3*c + 5*d)* (-1/2*ArcTanh[(Sqrt[a]*Cos[e + f*x])/(Sqrt[2]*Sqrt[a + a*Sin[e + f*x]])]/( Sqrt[2]*a^(3/2)*f) - Cos[e + f*x]/(2*f*(a + a*Sin[e + f*x])^(3/2))))/(8*a)
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[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*c - a*d)*Cos[e + f*x]*((a + b*Sin[e + f* x])^m/(a*f*(2*m + 1))), x] + Simp[(a*d*m + b*c*(m + 1))/(a*b*(2*m + 1)) I nt[(a + b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)]
Leaf count of result is larger than twice the leaf count of optimal. \(278\) vs. \(2(107)=214\).
Time = 1.61 (sec) , antiderivative size = 279, normalized size of antiderivative = 2.21
method | result | size |
default | \(-\frac {\left (-\operatorname {arctanh}\left (\frac {\sqrt {a -\sin \left (f x +e \right ) a}\, \sqrt {2}}{2 \sqrt {a}}\right ) \sqrt {2}\, a^{3} \left (3 c +5 d \right ) \cos \left (f x +e \right )^{2}+2 \sin \left (f x +e \right ) \operatorname {arctanh}\left (\frac {\sqrt {a -\sin \left (f x +e \right ) a}\, \sqrt {2}}{2 \sqrt {a}}\right ) \sqrt {2}\, a^{3} \left (3 c +5 d \right )+20 \sqrt {a -\sin \left (f x +e \right ) a}\, a^{\frac {5}{2}} c +12 \sqrt {a -\sin \left (f x +e \right ) a}\, a^{\frac {5}{2}} d -6 \left (a -\sin \left (f x +e \right ) a \right )^{\frac {3}{2}} a^{\frac {3}{2}} c -10 \left (a -\sin \left (f x +e \right ) a \right )^{\frac {3}{2}} a^{\frac {3}{2}} d +6 \,\operatorname {arctanh}\left (\frac {\sqrt {a -\sin \left (f x +e \right ) a}\, \sqrt {2}}{2 \sqrt {a}}\right ) \sqrt {2}\, a^{3} c +10 \,\operatorname {arctanh}\left (\frac {\sqrt {a -\sin \left (f x +e \right ) a}\, \sqrt {2}}{2 \sqrt {a}}\right ) \sqrt {2}\, a^{3} d \right ) \sqrt {-a \left (-1+\sin \left (f x +e \right )\right )}}{32 a^{\frac {11}{2}} \left (1+\sin \left (f x +e \right )\right ) \cos \left (f x +e \right ) \sqrt {a +\sin \left (f x +e \right ) a}\, f}\) | \(279\) |
parts | \(-\frac {c \left (-3 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -\sin \left (f x +e \right ) a}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{2} \cos \left (f x +e \right )^{2}+6 \sqrt {a -\sin \left (f x +e \right ) a}\, \sin \left (f x +e \right ) a^{\frac {3}{2}}+6 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -\sin \left (f x +e \right ) a}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{2} \sin \left (f x +e \right )+14 \sqrt {a -\sin \left (f x +e \right ) a}\, a^{\frac {3}{2}}+6 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -\sin \left (f x +e \right ) a}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{2}\right ) \sqrt {-a \left (-1+\sin \left (f x +e \right )\right )}}{32 a^{\frac {9}{2}} \left (1+\sin \left (f x +e \right )\right ) \cos \left (f x +e \right ) \sqrt {a +\sin \left (f x +e \right ) a}\, f}-\frac {d \left (-5 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -\sin \left (f x +e \right ) a}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{3} \cos \left (f x +e \right )^{2}+12 \sqrt {a -\sin \left (f x +e \right ) a}\, a^{\frac {5}{2}}-10 a^{\frac {3}{2}} \left (a -\sin \left (f x +e \right ) a \right )^{\frac {3}{2}}+10 \,\operatorname {arctanh}\left (\frac {\sqrt {a -\sin \left (f x +e \right ) a}\, \sqrt {2}}{2 \sqrt {a}}\right ) \sqrt {2}\, \sin \left (f x +e \right ) a^{3}+10 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -\sin \left (f x +e \right ) a}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{3}\right ) \sqrt {-a \left (-1+\sin \left (f x +e \right )\right )}}{32 a^{\frac {11}{2}} \left (1+\sin \left (f x +e \right )\right ) \cos \left (f x +e \right ) \sqrt {a +\sin \left (f x +e \right ) a}\, f}\) | \(394\) |
Input:
int((c+d*sin(f*x+e))/(a+sin(f*x+e)*a)^(5/2),x,method=_RETURNVERBOSE)
Output:
-1/32*(-arctanh(1/2*(a-sin(f*x+e)*a)^(1/2)*2^(1/2)/a^(1/2))*2^(1/2)*a^3*(3 *c+5*d)*cos(f*x+e)^2+2*sin(f*x+e)*arctanh(1/2*(a-sin(f*x+e)*a)^(1/2)*2^(1/ 2)/a^(1/2))*2^(1/2)*a^3*(3*c+5*d)+20*(a-sin(f*x+e)*a)^(1/2)*a^(5/2)*c+12*( a-sin(f*x+e)*a)^(1/2)*a^(5/2)*d-6*(a-sin(f*x+e)*a)^(3/2)*a^(3/2)*c-10*(a-s in(f*x+e)*a)^(3/2)*a^(3/2)*d+6*arctanh(1/2*(a-sin(f*x+e)*a)^(1/2)*2^(1/2)/ a^(1/2))*2^(1/2)*a^3*c+10*arctanh(1/2*(a-sin(f*x+e)*a)^(1/2)*2^(1/2)/a^(1/ 2))*2^(1/2)*a^3*d)*(-a*(-1+sin(f*x+e)))^(1/2)/a^(11/2)/(1+sin(f*x+e))/cos( f*x+e)/(a+sin(f*x+e)*a)^(1/2)/f
Leaf count of result is larger than twice the leaf count of optimal. 392 vs. \(2 (107) = 214\).
Time = 0.11 (sec) , antiderivative size = 392, normalized size of antiderivative = 3.11 \[ \int \frac {c+d \sin (e+f x)}{(a+a \sin (e+f x))^{5/2}} \, dx=\frac {\sqrt {2} {\left ({\left (3 \, c + 5 \, d\right )} \cos \left (f x + e\right )^{3} + 3 \, {\left (3 \, c + 5 \, d\right )} \cos \left (f x + e\right )^{2} - 2 \, {\left (3 \, c + 5 \, d\right )} \cos \left (f x + e\right ) + {\left ({\left (3 \, c + 5 \, d\right )} \cos \left (f x + e\right )^{2} - 2 \, {\left (3 \, c + 5 \, d\right )} \cos \left (f x + e\right ) - 12 \, c - 20 \, d\right )} \sin \left (f x + e\right ) - 12 \, c - 20 \, d\right )} \sqrt {a} \log \left (-\frac {a \cos \left (f x + e\right )^{2} - 2 \, \sqrt {2} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {a} {\left (\cos \left (f x + e\right ) - \sin \left (f x + e\right ) + 1\right )} + 3 \, a \cos \left (f x + e\right ) - {\left (a \cos \left (f x + e\right ) - 2 \, a\right )} \sin \left (f x + e\right ) + 2 \, a}{\cos \left (f x + e\right )^{2} - {\left (\cos \left (f x + e\right ) + 2\right )} \sin \left (f x + e\right ) - \cos \left (f x + e\right ) - 2}\right ) + 4 \, {\left ({\left (3 \, c + 5 \, d\right )} \cos \left (f x + e\right )^{2} + {\left (7 \, c + d\right )} \cos \left (f x + e\right ) + {\left ({\left (3 \, c + 5 \, d\right )} \cos \left (f x + e\right ) - 4 \, c + 4 \, d\right )} \sin \left (f x + e\right ) + 4 \, c - 4 \, d\right )} \sqrt {a \sin \left (f x + e\right ) + a}}{64 \, {\left (a^{3} f \cos \left (f x + e\right )^{3} + 3 \, a^{3} f \cos \left (f x + e\right )^{2} - 2 \, a^{3} f \cos \left (f x + e\right ) - 4 \, a^{3} f + {\left (a^{3} f \cos \left (f x + e\right )^{2} - 2 \, a^{3} f \cos \left (f x + e\right ) - 4 \, a^{3} f\right )} \sin \left (f x + e\right )\right )}} \] Input:
integrate((c+d*sin(f*x+e))/(a+a*sin(f*x+e))^(5/2),x, algorithm="fricas")
Output:
1/64*(sqrt(2)*((3*c + 5*d)*cos(f*x + e)^3 + 3*(3*c + 5*d)*cos(f*x + e)^2 - 2*(3*c + 5*d)*cos(f*x + e) + ((3*c + 5*d)*cos(f*x + e)^2 - 2*(3*c + 5*d)* cos(f*x + e) - 12*c - 20*d)*sin(f*x + e) - 12*c - 20*d)*sqrt(a)*log(-(a*co s(f*x + e)^2 - 2*sqrt(2)*sqrt(a*sin(f*x + e) + a)*sqrt(a)*(cos(f*x + e) - sin(f*x + e) + 1) + 3*a*cos(f*x + e) - (a*cos(f*x + e) - 2*a)*sin(f*x + e) + 2*a)/(cos(f*x + e)^2 - (cos(f*x + e) + 2)*sin(f*x + e) - cos(f*x + e) - 2)) + 4*((3*c + 5*d)*cos(f*x + e)^2 + (7*c + d)*cos(f*x + e) + ((3*c + 5* d)*cos(f*x + e) - 4*c + 4*d)*sin(f*x + e) + 4*c - 4*d)*sqrt(a*sin(f*x + e) + a))/(a^3*f*cos(f*x + e)^3 + 3*a^3*f*cos(f*x + e)^2 - 2*a^3*f*cos(f*x + e) - 4*a^3*f + (a^3*f*cos(f*x + e)^2 - 2*a^3*f*cos(f*x + e) - 4*a^3*f)*sin (f*x + e))
\[ \int \frac {c+d \sin (e+f x)}{(a+a \sin (e+f x))^{5/2}} \, dx=\int \frac {c + d \sin {\left (e + f x \right )}}{\left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{\frac {5}{2}}}\, dx \] Input:
integrate((c+d*sin(f*x+e))/(a+a*sin(f*x+e))**(5/2),x)
Output:
Integral((c + d*sin(e + f*x))/(a*(sin(e + f*x) + 1))**(5/2), x)
\[ \int \frac {c+d \sin (e+f x)}{(a+a \sin (e+f x))^{5/2}} \, dx=\int { \frac {d \sin \left (f x + e\right ) + c}{{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {5}{2}}} \,d x } \] Input:
integrate((c+d*sin(f*x+e))/(a+a*sin(f*x+e))^(5/2),x, algorithm="maxima")
Output:
integrate((d*sin(f*x + e) + c)/(a*sin(f*x + e) + a)^(5/2), x)
Exception generated. \[ \int \frac {c+d \sin (e+f x)}{(a+a \sin (e+f x))^{5/2}} \, dx=\text {Exception raised: TypeError} \] Input:
integrate((c+d*sin(f*x+e))/(a+a*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. \[ \int \frac {c+d \sin (e+f x)}{(a+a \sin (e+f x))^{5/2}} \, dx=\int \frac {c+d\,\sin \left (e+f\,x\right )}{{\left (a+a\,\sin \left (e+f\,x\right )\right )}^{5/2}} \,d x \] Input:
int((c + d*sin(e + f*x))/(a + a*sin(e + f*x))^(5/2),x)
Output:
int((c + d*sin(e + f*x))/(a + a*sin(e + f*x))^(5/2), x)
\[ \int \frac {c+d \sin (e+f x)}{(a+a \sin (e+f x))^{5/2}} \, dx=\frac {\sqrt {a}\, \left (\left (\int \frac {\sqrt {\sin \left (f x +e \right )+1}}{\sin \left (f x +e \right )^{3}+3 \sin \left (f x +e \right )^{2}+3 \sin \left (f x +e \right )+1}d x \right ) c +\left (\int \frac {\sqrt {\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )}{\sin \left (f x +e \right )^{3}+3 \sin \left (f x +e \right )^{2}+3 \sin \left (f x +e \right )+1}d x \right ) d \right )}{a^{3}} \] Input:
int((c+d*sin(f*x+e))/(a+a*sin(f*x+e))^(5/2),x)
Output:
(sqrt(a)*(int(sqrt(sin(e + f*x) + 1)/(sin(e + f*x)**3 + 3*sin(e + f*x)**2 + 3*sin(e + f*x) + 1),x)*c + int((sqrt(sin(e + f*x) + 1)*sin(e + f*x))/(si n(e + f*x)**3 + 3*sin(e + f*x)**2 + 3*sin(e + f*x) + 1),x)*d))/a**3