Integrand size = 27, antiderivative size = 147 \[ \int \frac {(c+d \sin (e+f x))^2}{(a+a \sin (e+f x))^{5/2}} \, dx=-\frac {\left (3 c^2+10 c d+19 d^2\right ) \text {arctanh}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a+a \sin (e+f x)}}\right )}{16 \sqrt {2} a^{5/2} f}-\frac {3 (c-d) (c+3 d) \cos (e+f x)}{16 a f (a+a \sin (e+f x))^{3/2}}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))}{4 f (a+a \sin (e+f x))^{5/2}} \] Output:
-1/32*(3*c^2+10*c*d+19*d^2)*arctanh(1/2*a^(1/2)*cos(f*x+e)*2^(1/2)/(a+a*si n(f*x+e))^(1/2))*2^(1/2)/a^(5/2)/f-3/16*(c-d)*(c+3*d)*cos(f*x+e)/a/f/(a+a* sin(f*x+e))^(3/2)-1/4*(c-d)*cos(f*x+e)*(c+d*sin(f*x+e))/f/(a+a*sin(f*x+e)) ^(5/2)
Result contains complex when optimal does not.
Time = 1.59 (sec) , antiderivative size = 252, normalized size of antiderivative = 1.71 \[ \int \frac {(c+d \sin (e+f x))^2}{(a+a \sin (e+f x))^{5/2}} \, dx=\frac {\left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (8 (c-d)^2 \sin \left (\frac {1}{2} (e+f x)\right )-4 (c-d)^2 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )+2 \left (3 c^2+10 c d-13 d^2\right ) \sin \left (\frac {1}{2} (e+f x)\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^2-(c-d) (3 c+13 d) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^3+(1+i) (-1)^{3/4} \left (3 c^2+10 c d+19 d^2\right ) \text {arctanh}\left (\left (\frac {1}{2}+\frac {i}{2}\right ) (-1)^{3/4} \left (-1+\tan \left (\frac {1}{4} (e+f x)\right )\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^4\right )}{16 f (a (1+\sin (e+f x)))^{5/2}} \] Input:
Integrate[(c + d*Sin[e + f*x])^2/(a + a*Sin[e + f*x])^(5/2),x]
Output:
((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(8*(c - d)^2*Sin[(e + f*x)/2] - 4*( c - d)^2*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2]) + 2*(3*c^2 + 10*c*d - 13*d^ 2)*Sin[(e + f*x)/2]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^2 - (c - d)*(3*c + 13*d)*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3 + (1 + I)*(-1)^(3/4)*(3*c ^2 + 10*c*d + 19*d^2)*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 = 0.59 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.04, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.296, Rules used = {3042, 3239, 27, 3042, 3229, 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.
| \(\displaystyle \int \frac {(c+d \sin (e+f x))^2}{(a \sin (e+f x)+a)^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(c+d \sin (e+f x))^2}{(a \sin (e+f x)+a)^{5/2}}dx\) |
| \(\Big \downarrow \) 3239 |
| \(\displaystyle -\frac {\int -\frac {a \left (3 c^2+7 d c-2 d^2\right )+a d (c+7 d) \sin (e+f x)}{2 (\sin (e+f x) a+a)^{3/2}}dx}{4 a^2}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))}{4 f (a \sin (e+f x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {a \left (3 c^2+7 d c-2 d^2\right )+a d (c+7 d) \sin (e+f x)}{(\sin (e+f x) a+a)^{3/2}}dx}{8 a^2}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))}{4 f (a \sin (e+f x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {a \left (3 c^2+7 d c-2 d^2\right )+a d (c+7 d) \sin (e+f x)}{(\sin (e+f x) a+a)^{3/2}}dx}{8 a^2}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))}{4 f (a \sin (e+f x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 3229 |
| \(\displaystyle \frac {\frac {1}{4} \left (3 c^2+10 c d+19 d^2\right ) \int \frac {1}{\sqrt {\sin (e+f x) a+a}}dx-\frac {3 a (c-d) (c+3 d) \cos (e+f x)}{2 f (a \sin (e+f x)+a)^{3/2}}}{8 a^2}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))}{4 f (a \sin (e+f x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{4} \left (3 c^2+10 c d+19 d^2\right ) \int \frac {1}{\sqrt {\sin (e+f x) a+a}}dx-\frac {3 a (c-d) (c+3 d) \cos (e+f x)}{2 f (a \sin (e+f x)+a)^{3/2}}}{8 a^2}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))}{4 f (a \sin (e+f x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 3128 |
| \(\displaystyle \frac {-\frac {\left (3 c^2+10 c d+19 d^2\right ) \int \frac {1}{2 a-\frac {a^2 \cos ^2(e+f x)}{\sin (e+f x) a+a}}d\frac {a \cos (e+f x)}{\sqrt {\sin (e+f x) a+a}}}{2 f}-\frac {3 a (c-d) (c+3 d) \cos (e+f x)}{2 f (a \sin (e+f x)+a)^{3/2}}}{8 a^2}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))}{4 f (a \sin (e+f x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {-\frac {\left (3 c^2+10 c d+19 d^2\right ) \text {arctanh}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a \sin (e+f x)+a}}\right )}{2 \sqrt {2} \sqrt {a} f}-\frac {3 a (c-d) (c+3 d) \cos (e+f x)}{2 f (a \sin (e+f x)+a)^{3/2}}}{8 a^2}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))}{4 f (a \sin (e+f x)+a)^{5/2}}\) |
Input:
Int[(c + d*Sin[e + f*x])^2/(a + a*Sin[e + f*x])^(5/2),x]
Output:
-1/4*((c - d)*Cos[e + f*x]*(c + d*Sin[e + f*x]))/(f*(a + a*Sin[e + f*x])^( 5/2)) + (-1/2*((3*c^2 + 10*c*d + 19*d^2)*ArcTanh[(Sqrt[a]*Cos[e + f*x])/(S qrt[2]*Sqrt[a + a*Sin[e + f*x]])])/(Sqrt[2]*Sqrt[a]*f) - (3*a*(c - d)*(c + 3*d)*Cos[e + f*x])/(2*f*(a + a*Sin[e + f*x])^(3/2)))/(8*a^2)
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_.)*(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[(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)]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + ( f_.)*(x_)])^2, x_Symbol] :> Simp[(b*c - a*d)*Cos[e + f*x]*(a + b*Sin[e + f* x])^m*((c + d*Sin[e + f*x])/(a*f*(2*m + 1))), x] + Simp[1/(a*b*(2*m + 1)) Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[a*c*d*(m - 1) + b*(d^2 + c^2*(m + 1) ) + d*(a*d*(m - 1) + b*c*(m + 2))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && LtQ[m, -1]
Leaf count of result is larger than twice the leaf count of optimal. \(378\) vs. \(2(128)=256\).
Time = 2.03 (sec) , antiderivative size = 379, normalized size of antiderivative = 2.58
| method | result | size |
| default | \(-\frac {\left (-\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -\sin \left (f x +e \right ) a}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{2} \left (3 c^{2}+10 c d +19 d^{2}\right ) \cos \left (f x +e \right )^{2}+2 \sin \left (f x +e \right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -\sin \left (f x +e \right ) a}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{2} \left (3 c^{2}+10 c d +19 d^{2}\right )+20 \sqrt {a -\sin \left (f x +e \right ) a}\, a^{\frac {3}{2}} c^{2}+24 \sqrt {a -\sin \left (f x +e \right ) a}\, a^{\frac {3}{2}} c d -44 \sqrt {a -\sin \left (f x +e \right ) a}\, a^{\frac {3}{2}} d^{2}-6 \left (a -\sin \left (f x +e \right ) a \right )^{\frac {3}{2}} \sqrt {a}\, c^{2}-20 \left (a -\sin \left (f x +e \right ) a \right )^{\frac {3}{2}} \sqrt {a}\, c d +26 \left (a -\sin \left (f x +e \right ) a \right )^{\frac {3}{2}} \sqrt {a}\, d^{2}+6 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -\sin \left (f x +e \right ) a}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{2} c^{2}+20 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -\sin \left (f x +e \right ) a}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{2} c d +38 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -\sin \left (f x +e \right ) a}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{2} d^{2}\right ) \sqrt {-a \left (-1+\sin \left (f x +e \right )\right )}}{32 a^{\frac {9}{2}} \left (1+\sin \left (f x +e \right )\right ) \cos \left (f x +e \right ) \sqrt {a +\sin \left (f x +e \right ) a}\, f}\) | \(379\) |
| parts | \(-\frac {c^{2} \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^{2} \left (19 \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}-38 \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 )+44 \sqrt {a -\sin \left (f x +e \right ) a}\, a^{\frac {3}{2}}-26 \left (a -\sin \left (f x +e \right ) a \right )^{\frac {3}{2}} \sqrt {a}-38 \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 {c 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 )}}{16 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}\) | \(592\) |
Input:
int((c+d*sin(f*x+e))^2/(a+sin(f*x+e)*a)^(5/2),x,method=_RETURNVERBOSE)
Output:
-1/32*(-2^(1/2)*arctanh(1/2*(a-sin(f*x+e)*a)^(1/2)*2^(1/2)/a^(1/2))*a^2*(3 *c^2+10*c*d+19*d^2)*cos(f*x+e)^2+2*sin(f*x+e)*2^(1/2)*arctanh(1/2*(a-sin(f *x+e)*a)^(1/2)*2^(1/2)/a^(1/2))*a^2*(3*c^2+10*c*d+19*d^2)+20*(a-sin(f*x+e) *a)^(1/2)*a^(3/2)*c^2+24*(a-sin(f*x+e)*a)^(1/2)*a^(3/2)*c*d-44*(a-sin(f*x+ e)*a)^(1/2)*a^(3/2)*d^2-6*(a-sin(f*x+e)*a)^(3/2)*a^(1/2)*c^2-20*(a-sin(f*x +e)*a)^(3/2)*a^(1/2)*c*d+26*(a-sin(f*x+e)*a)^(3/2)*a^(1/2)*d^2+6*2^(1/2)*a rctanh(1/2*(a-sin(f*x+e)*a)^(1/2)*2^(1/2)/a^(1/2))*a^2*c^2+20*2^(1/2)*arct anh(1/2*(a-sin(f*x+e)*a)^(1/2)*2^(1/2)/a^(1/2))*a^2*c*d+38*2^(1/2)*arctanh (1/2*(a-sin(f*x+e)*a)^(1/2)*2^(1/2)/a^(1/2))*a^2*d^2)*(-a*(-1+sin(f*x+e))) ^(1/2)/a^(9/2)/(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. 492 vs. \(2 (128) = 256\).
Time = 0.11 (sec) , antiderivative size = 492, normalized size of antiderivative = 3.35 \[ \int \frac {(c+d \sin (e+f x))^2}{(a+a \sin (e+f x))^{5/2}} \, dx=\frac {\sqrt {2} {\left ({\left (3 \, c^{2} + 10 \, c d + 19 \, d^{2}\right )} \cos \left (f x + e\right )^{3} + 3 \, {\left (3 \, c^{2} + 10 \, c d + 19 \, d^{2}\right )} \cos \left (f x + e\right )^{2} - 12 \, c^{2} - 40 \, c d - 76 \, d^{2} - 2 \, {\left (3 \, c^{2} + 10 \, c d + 19 \, d^{2}\right )} \cos \left (f x + e\right ) + {\left ({\left (3 \, c^{2} + 10 \, c d + 19 \, d^{2}\right )} \cos \left (f x + e\right )^{2} - 12 \, c^{2} - 40 \, c d - 76 \, d^{2} - 2 \, {\left (3 \, c^{2} + 10 \, c d + 19 \, d^{2}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )\right )} \sqrt {a} \log \left (-\frac {a \cos \left (f x + e\right )^{2} - 2 \, \sqrt {2} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {a} {\left (\cos \left (f x + e\right ) - \sin \left (f x + e\right ) + 1\right )} + 3 \, a \cos \left (f x + e\right ) - {\left (a \cos \left (f x + e\right ) - 2 \, a\right )} \sin \left (f x + e\right ) + 2 \, a}{\cos \left (f x + e\right )^{2} - {\left (\cos \left (f x + e\right ) + 2\right )} \sin \left (f x + e\right ) - \cos \left (f x + e\right ) - 2}\right ) + 4 \, {\left ({\left (3 \, c^{2} + 10 \, c d - 13 \, d^{2}\right )} \cos \left (f x + e\right )^{2} + 4 \, c^{2} - 8 \, c d + 4 \, d^{2} + {\left (7 \, c^{2} + 2 \, c d - 9 \, d^{2}\right )} \cos \left (f x + e\right ) - {\left (4 \, c^{2} - 8 \, c d + 4 \, d^{2} - {\left (3 \, c^{2} + 10 \, c d - 13 \, d^{2}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )\right )} \sqrt {a \sin \left (f x + e\right ) + a}}{64 \, {\left (a^{3} f \cos \left (f x + e\right )^{3} + 3 \, a^{3} f \cos \left (f x + e\right )^{2} - 2 \, a^{3} f \cos \left (f x + e\right ) - 4 \, a^{3} f + {\left (a^{3} f \cos \left (f x + e\right )^{2} - 2 \, a^{3} f \cos \left (f x + e\right ) - 4 \, a^{3} f\right )} \sin \left (f x + e\right )\right )}} \] Input:
integrate((c+d*sin(f*x+e))^2/(a+a*sin(f*x+e))^(5/2),x, algorithm="fricas")
Output:
1/64*(sqrt(2)*((3*c^2 + 10*c*d + 19*d^2)*cos(f*x + e)^3 + 3*(3*c^2 + 10*c* d + 19*d^2)*cos(f*x + e)^2 - 12*c^2 - 40*c*d - 76*d^2 - 2*(3*c^2 + 10*c*d + 19*d^2)*cos(f*x + e) + ((3*c^2 + 10*c*d + 19*d^2)*cos(f*x + e)^2 - 12*c^ 2 - 40*c*d - 76*d^2 - 2*(3*c^2 + 10*c*d + 19*d^2)*cos(f*x + e))*sin(f*x + e))*sqrt(a)*log(-(a*cos(f*x + e)^2 - 2*sqrt(2)*sqrt(a*sin(f*x + e) + a)*sq rt(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^2 + 10*c*d - 13*d^2)*cos(f*x + e)^2 + 4*c^2 - 8*c*d + 4*d^2 + (7*c^2 + 2*c*d - 9*d^2)*cos(f*x + e) - (4*c^2 - 8 *c*d + 4*d^2 - (3*c^2 + 10*c*d - 13*d^2)*cos(f*x + e))*sin(f*x + e))*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))
\[ \int \frac {(c+d \sin (e+f x))^2}{(a+a \sin (e+f x))^{5/2}} \, dx=\int \frac {\left (c + d \sin {\left (e + f x \right )}\right )^{2}}{\left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{\frac {5}{2}}}\, dx \] Input:
integrate((c+d*sin(f*x+e))**2/(a+a*sin(f*x+e))**(5/2),x)
Output:
Integral((c + d*sin(e + f*x))**2/(a*(sin(e + f*x) + 1))**(5/2), x)
\[ \int \frac {(c+d \sin (e+f x))^2}{(a+a \sin (e+f x))^{5/2}} \, dx=\int { \frac {{\left (d \sin \left (f x + e\right ) + c\right )}^{2}}{{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {5}{2}}} \,d x } \] Input:
integrate((c+d*sin(f*x+e))^2/(a+a*sin(f*x+e))^(5/2),x, algorithm="maxima")
Output:
integrate((d*sin(f*x + e) + c)^2/(a*sin(f*x + e) + a)^(5/2), x)
Exception generated. \[ \int \frac {(c+d \sin (e+f x))^2}{(a+a \sin (e+f x))^{5/2}} \, dx=\text {Exception raised: TypeError} \] Input:
integrate((c+d*sin(f*x+e))^2/(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. \[ \int \frac {(c+d \sin (e+f x))^2}{(a+a \sin (e+f x))^{5/2}} \, dx=\int \frac {{\left (c+d\,\sin \left (e+f\,x\right )\right )}^2}{{\left (a+a\,\sin \left (e+f\,x\right )\right )}^{5/2}} \,d x \] Input:
int((c + d*sin(e + f*x))^2/(a + a*sin(e + f*x))^(5/2),x)
Output:
int((c + d*sin(e + f*x))^2/(a + a*sin(e + f*x))^(5/2), x)
\[ \int \frac {(c+d \sin (e+f x))^2}{(a+a \sin (e+f x))^{5/2}} \, dx=\frac {\sqrt {a}\, \left (\left (\int \frac {\sqrt {\sin \left (f x +e \right )+1}}{\sin \left (f x +e \right )^{3}+3 \sin \left (f x +e \right )^{2}+3 \sin \left (f x +e \right )+1}d x \right ) c^{2}+\left (\int \frac {\sqrt {\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{2}}{\sin \left (f x +e \right )^{3}+3 \sin \left (f x +e \right )^{2}+3 \sin \left (f x +e \right )+1}d x \right ) d^{2}+2 \left (\int \frac {\sqrt {\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )}{\sin \left (f x +e \right )^{3}+3 \sin \left (f x +e \right )^{2}+3 \sin \left (f x +e \right )+1}d x \right ) c d \right )}{a^{3}} \] Input:
int((c+d*sin(f*x+e))^2/(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**2 + int((sqrt(sin(e + f*x) + 1)*sin(e + f*x)** 2)/(sin(e + f*x)**3 + 3*sin(e + f*x)**2 + 3*sin(e + f*x) + 1),x)*d**2 + 2* int((sqrt(sin(e + f*x) + 1)*sin(e + f*x))/(sin(e + f*x)**3 + 3*sin(e + f*x )**2 + 3*sin(e + f*x) + 1),x)*c*d))/a**3