Integrand size = 23, antiderivative size = 183 \[ \int \frac {\sin ^4(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx=\frac {15 \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a+a \sin (c+d x)}}\right )}{2 \sqrt {2} a^{3/2} d}+\frac {\cos (c+d x) \sin ^3(c+d x)}{2 d (a+a \sin (c+d x))^{3/2}}-\frac {31 \cos (c+d x)}{5 a d \sqrt {a+a \sin (c+d x)}}-\frac {9 \cos (c+d x) \sin ^2(c+d x)}{10 a d \sqrt {a+a \sin (c+d x)}}+\frac {13 \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{10 a^2 d} \] Output:
15/4*arctanh(1/2*a^(1/2)*cos(d*x+c)*2^(1/2)/(a+a*sin(d*x+c))^(1/2))*2^(1/2 )/a^(3/2)/d+1/2*cos(d*x+c)*sin(d*x+c)^3/d/(a+a*sin(d*x+c))^(3/2)-31/5*cos( d*x+c)/a/d/(a+a*sin(d*x+c))^(1/2)-9/10*cos(d*x+c)*sin(d*x+c)^2/a/d/(a+a*si n(d*x+c))^(1/2)+13/10*cos(d*x+c)*(a+a*sin(d*x+c))^(1/2)/a^2/d
Result contains complex when optimal does not.
Time = 0.95 (sec) , antiderivative size = 178, normalized size of antiderivative = 0.97 \[ \int \frac {\sin ^4(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx=\frac {\left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right ) \left (-55 \cos \left (\frac {1}{2} (c+d x)\right )-41 \cos \left (\frac {3}{2} (c+d x)\right )-3 \cos \left (\frac {5}{2} (c+d x)\right )+\cos \left (\frac {7}{2} (c+d x)\right )+55 \sin \left (\frac {1}{2} (c+d x)\right )-(150+150 i) (-1)^{3/4} \text {arctanh}\left (\left (\frac {1}{2}+\frac {i}{2}\right ) (-1)^{3/4} \left (-1+\tan \left (\frac {1}{4} (c+d x)\right )\right )\right ) (1+\sin (c+d x))-41 \sin \left (\frac {3}{2} (c+d x)\right )+3 \sin \left (\frac {5}{2} (c+d x)\right )+\sin \left (\frac {7}{2} (c+d x)\right )\right )}{20 d (a (1+\sin (c+d x)))^{3/2}} \] Input:
Integrate[Sin[c + d*x]^4/(a + a*Sin[c + d*x])^(3/2),x]
Output:
((Cos[(c + d*x)/2] + Sin[(c + d*x)/2])*(-55*Cos[(c + d*x)/2] - 41*Cos[(3*( c + d*x))/2] - 3*Cos[(5*(c + d*x))/2] + Cos[(7*(c + d*x))/2] + 55*Sin[(c + d*x)/2] - (150 + 150*I)*(-1)^(3/4)*ArcTanh[(1/2 + I/2)*(-1)^(3/4)*(-1 + T an[(c + d*x)/4])]*(1 + Sin[c + d*x]) - 41*Sin[(3*(c + d*x))/2] + 3*Sin[(5* (c + d*x))/2] + Sin[(7*(c + d*x))/2]))/(20*d*(a*(1 + Sin[c + d*x]))^(3/2))
Time = 1.09 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.09, number of steps used = 17, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.696, Rules used = {3042, 3244, 27, 3042, 3462, 27, 3042, 3447, 3042, 3502, 27, 3042, 3230, 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.
| \(\displaystyle \int \frac {\sin ^4(c+d x)}{(a \sin (c+d x)+a)^{3/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (c+d x)^4}{(a \sin (c+d x)+a)^{3/2}}dx\) |
| \(\Big \downarrow \) 3244 |
| \(\displaystyle \frac {\sin ^3(c+d x) \cos (c+d x)}{2 d (a \sin (c+d x)+a)^{3/2}}-\frac {\int \frac {3 \sin ^2(c+d x) (2 a-3 a \sin (c+d x))}{2 \sqrt {\sin (c+d x) a+a}}dx}{2 a^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\sin ^3(c+d x) \cos (c+d x)}{2 d (a \sin (c+d x)+a)^{3/2}}-\frac {3 \int \frac {\sin ^2(c+d x) (2 a-3 a \sin (c+d x))}{\sqrt {\sin (c+d x) a+a}}dx}{4 a^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\sin ^3(c+d x) \cos (c+d x)}{2 d (a \sin (c+d x)+a)^{3/2}}-\frac {3 \int \frac {\sin (c+d x)^2 (2 a-3 a \sin (c+d x))}{\sqrt {\sin (c+d x) a+a}}dx}{4 a^2}\) |
| \(\Big \downarrow \) 3462 |
| \(\displaystyle \frac {\sin ^3(c+d x) \cos (c+d x)}{2 d (a \sin (c+d x)+a)^{3/2}}-\frac {3 \left (\frac {2 \int -\frac {\sin (c+d x) \left (12 a^2-13 a^2 \sin (c+d x)\right )}{2 \sqrt {\sin (c+d x) a+a}}dx}{5 a}+\frac {6 a \sin ^2(c+d x) \cos (c+d x)}{5 d \sqrt {a \sin (c+d x)+a}}\right )}{4 a^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\sin ^3(c+d x) \cos (c+d x)}{2 d (a \sin (c+d x)+a)^{3/2}}-\frac {3 \left (\frac {6 a \sin ^2(c+d x) \cos (c+d x)}{5 d \sqrt {a \sin (c+d x)+a}}-\frac {\int \frac {\sin (c+d x) \left (12 a^2-13 a^2 \sin (c+d x)\right )}{\sqrt {\sin (c+d x) a+a}}dx}{5 a}\right )}{4 a^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\sin ^3(c+d x) \cos (c+d x)}{2 d (a \sin (c+d x)+a)^{3/2}}-\frac {3 \left (\frac {6 a \sin ^2(c+d x) \cos (c+d x)}{5 d \sqrt {a \sin (c+d x)+a}}-\frac {\int \frac {\sin (c+d x) \left (12 a^2-13 a^2 \sin (c+d x)\right )}{\sqrt {\sin (c+d x) a+a}}dx}{5 a}\right )}{4 a^2}\) |
| \(\Big \downarrow \) 3447 |
| \(\displaystyle \frac {\sin ^3(c+d x) \cos (c+d x)}{2 d (a \sin (c+d x)+a)^{3/2}}-\frac {3 \left (\frac {6 a \sin ^2(c+d x) \cos (c+d x)}{5 d \sqrt {a \sin (c+d x)+a}}-\frac {\int \frac {12 a^2 \sin (c+d x)-13 a^2 \sin ^2(c+d x)}{\sqrt {\sin (c+d x) a+a}}dx}{5 a}\right )}{4 a^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\sin ^3(c+d x) \cos (c+d x)}{2 d (a \sin (c+d x)+a)^{3/2}}-\frac {3 \left (\frac {6 a \sin ^2(c+d x) \cos (c+d x)}{5 d \sqrt {a \sin (c+d x)+a}}-\frac {\int \frac {12 a^2 \sin (c+d x)-13 a^2 \sin (c+d x)^2}{\sqrt {\sin (c+d x) a+a}}dx}{5 a}\right )}{4 a^2}\) |
| \(\Big \downarrow \) 3502 |
| \(\displaystyle \frac {\sin ^3(c+d x) \cos (c+d x)}{2 d (a \sin (c+d x)+a)^{3/2}}-\frac {3 \left (\frac {6 a \sin ^2(c+d x) \cos (c+d x)}{5 d \sqrt {a \sin (c+d x)+a}}-\frac {\frac {2 \int -\frac {13 a^3-62 a^3 \sin (c+d x)}{2 \sqrt {\sin (c+d x) a+a}}dx}{3 a}+\frac {26 a \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{3 d}}{5 a}\right )}{4 a^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\sin ^3(c+d x) \cos (c+d x)}{2 d (a \sin (c+d x)+a)^{3/2}}-\frac {3 \left (\frac {6 a \sin ^2(c+d x) \cos (c+d x)}{5 d \sqrt {a \sin (c+d x)+a}}-\frac {\frac {26 a \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{3 d}-\frac {\int \frac {13 a^3-62 a^3 \sin (c+d x)}{\sqrt {\sin (c+d x) a+a}}dx}{3 a}}{5 a}\right )}{4 a^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\sin ^3(c+d x) \cos (c+d x)}{2 d (a \sin (c+d x)+a)^{3/2}}-\frac {3 \left (\frac {6 a \sin ^2(c+d x) \cos (c+d x)}{5 d \sqrt {a \sin (c+d x)+a}}-\frac {\frac {26 a \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{3 d}-\frac {\int \frac {13 a^3-62 a^3 \sin (c+d x)}{\sqrt {\sin (c+d x) a+a}}dx}{3 a}}{5 a}\right )}{4 a^2}\) |
| \(\Big \downarrow \) 3230 |
| \(\displaystyle \frac {\sin ^3(c+d x) \cos (c+d x)}{2 d (a \sin (c+d x)+a)^{3/2}}-\frac {3 \left (\frac {6 a \sin ^2(c+d x) \cos (c+d x)}{5 d \sqrt {a \sin (c+d x)+a}}-\frac {\frac {26 a \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{3 d}-\frac {75 a^3 \int \frac {1}{\sqrt {\sin (c+d x) a+a}}dx+\frac {124 a^3 \cos (c+d x)}{d \sqrt {a \sin (c+d x)+a}}}{3 a}}{5 a}\right )}{4 a^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\sin ^3(c+d x) \cos (c+d x)}{2 d (a \sin (c+d x)+a)^{3/2}}-\frac {3 \left (\frac {6 a \sin ^2(c+d x) \cos (c+d x)}{5 d \sqrt {a \sin (c+d x)+a}}-\frac {\frac {26 a \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{3 d}-\frac {75 a^3 \int \frac {1}{\sqrt {\sin (c+d x) a+a}}dx+\frac {124 a^3 \cos (c+d x)}{d \sqrt {a \sin (c+d x)+a}}}{3 a}}{5 a}\right )}{4 a^2}\) |
| \(\Big \downarrow \) 3128 |
| \(\displaystyle \frac {\sin ^3(c+d x) \cos (c+d x)}{2 d (a \sin (c+d x)+a)^{3/2}}-\frac {3 \left (\frac {6 a \sin ^2(c+d x) \cos (c+d x)}{5 d \sqrt {a \sin (c+d x)+a}}-\frac {\frac {26 a \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{3 d}-\frac {\frac {124 a^3 \cos (c+d x)}{d \sqrt {a \sin (c+d x)+a}}-\frac {150 a^3 \int \frac {1}{2 a-\frac {a^2 \cos ^2(c+d x)}{\sin (c+d x) a+a}}d\frac {a \cos (c+d x)}{\sqrt {\sin (c+d x) a+a}}}{d}}{3 a}}{5 a}\right )}{4 a^2}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\sin ^3(c+d x) \cos (c+d x)}{2 d (a \sin (c+d x)+a)^{3/2}}-\frac {3 \left (\frac {6 a \sin ^2(c+d x) \cos (c+d x)}{5 d \sqrt {a \sin (c+d x)+a}}-\frac {\frac {26 a \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{3 d}-\frac {\frac {124 a^3 \cos (c+d x)}{d \sqrt {a \sin (c+d x)+a}}-\frac {75 \sqrt {2} a^{5/2} \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a \sin (c+d x)+a}}\right )}{d}}{3 a}}{5 a}\right )}{4 a^2}\) |
Input:
Int[Sin[c + d*x]^4/(a + a*Sin[c + d*x])^(3/2),x]
Output:
(Cos[c + d*x]*Sin[c + d*x]^3)/(2*d*(a + a*Sin[c + d*x])^(3/2)) - (3*((6*a* Cos[c + d*x]*Sin[c + d*x]^2)/(5*d*Sqrt[a + a*Sin[c + d*x]]) - ((26*a*Cos[c + d*x]*Sqrt[a + a*Sin[c + d*x]])/(3*d) - ((-75*Sqrt[2]*a^(5/2)*ArcTanh[(S qrt[a]*Cos[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sin[c + d*x]])])/d + (124*a^3*Cos [c + d*x])/(d*Sqrt[a + a*Sin[c + d*x]]))/(3*a))/(5*a)))/(4*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[(-d)*Cos[e + f*x]*((a + b*Sin[e + f*x])^m/( f*(m + 1))), x] + Simp[(a*d*m + b*c*(m + 1))/(b*(m + 1)) Int[(a + b*Sin[e + f*x])^m, x], x] /; FreeQ[{a, b, c, d, e, f, m}, 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_)])^(n_), x_Symbol] :> Simp[(b*c - a*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n - 1)/(a*f*(2*m + 1))), x] + Simp[1/(a*b* (2*m + 1)) Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n - 2)* Simp[b*(c^2*(m + 1) + d^2*(n - 1)) + a*c*d*(m - n + 1) + d*(a*d*(m - n + 1) + b*c*(m + n))*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] && NeQ[c^2 - d^2, 0] && LtQ[m, -1] && GtQ[n, 1] && (IntegersQ[2*m, 2*n] || (IntegerQ[m] && EqQ[c, 0]))
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Int[(a + b*Sin[e + f*x])^m*(A*c + (B*c + A*d)*Sin[e + f*x] + B*d*Sin[e + f*x]^2), x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim p[(-B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^n/(f*(m + n + 1))), x] + Simp[1/(b*(m + n + 1)) Int[(a + b*Sin[e + f*x])^m*(c + d*S in[e + f*x])^(n - 1)*Simp[A*b*c*(m + n + 1) + B*(a*c*m + b*d*n) + (A*b*d*(m + n + 1) + B*(a*d*m + b*c*n))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[n, 0] && (IntegerQ[n] || EqQ[m + 1/2, 0])
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Co s[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Simp[1/(b*(m + 2)) Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && !LtQ[m, -1]
Time = 0.35 (sec) , antiderivative size = 193, normalized size of antiderivative = 1.05
| method | result | size |
| default | \(\frac {\left (-8 \left (a -a \sin \left (d x +c \right )\right )^{\frac {5}{2}} \sqrt {a}\, \sin \left (d x +c \right )+75 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{3} \sin \left (d x +c \right )-8 \left (a -a \sin \left (d x +c \right )\right )^{\frac {5}{2}} \sqrt {a}+75 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{3}-80 \sqrt {a -a \sin \left (d x +c \right )}\, a^{\frac {5}{2}} \sin \left (d x +c \right )-90 \sqrt {a -a \sin \left (d x +c \right )}\, a^{\frac {5}{2}}\right ) \sqrt {-a \left (\sin \left (d x +c \right )-1\right )}}{20 a^{\frac {9}{2}} \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d}\) | \(193\) |
Input:
int(1/(a+a*sin(d*x+c))^(3/2)*sin(d*x+c)^4,x,method=_RETURNVERBOSE)
Output:
1/20/a^(9/2)*(-8*(a-a*sin(d*x+c))^(5/2)*a^(1/2)*sin(d*x+c)+75*2^(1/2)*arct anh(1/2*(a-a*sin(d*x+c))^(1/2)*2^(1/2)/a^(1/2))*a^3*sin(d*x+c)-8*(a-a*sin( d*x+c))^(5/2)*a^(1/2)+75*2^(1/2)*arctanh(1/2*(a-a*sin(d*x+c))^(1/2)*2^(1/2 )/a^(1/2))*a^3-80*(a-a*sin(d*x+c))^(1/2)*a^(5/2)*sin(d*x+c)-90*(a-a*sin(d* x+c))^(1/2)*a^(5/2))*(-a*(sin(d*x+c)-1))^(1/2)/cos(d*x+c)/(a+a*sin(d*x+c)) ^(1/2)/d
Leaf count of result is larger than twice the leaf count of optimal. 314 vs. \(2 (156) = 312\).
Time = 0.10 (sec) , antiderivative size = 314, normalized size of antiderivative = 1.72 \[ \int \frac {\sin ^4(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx=\frac {75 \, \sqrt {2} {\left (\cos \left (d x + c\right )^{2} - {\left (\cos \left (d x + c\right ) + 2\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 2\right )} \sqrt {a} \log \left (-\frac {a \cos \left (d x + c\right )^{2} + 2 \, \sqrt {2} \sqrt {a \sin \left (d x + c\right ) + a} \sqrt {a} {\left (\cos \left (d x + c\right ) - \sin \left (d x + c\right ) + 1\right )} + 3 \, a \cos \left (d x + c\right ) - {\left (a \cos \left (d x + c\right ) - 2 \, a\right )} \sin \left (d x + c\right ) + 2 \, a}{\cos \left (d x + c\right )^{2} - {\left (\cos \left (d x + c\right ) + 2\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 2}\right ) - 4 \, {\left (4 \, \cos \left (d x + c\right )^{4} - 4 \, \cos \left (d x + c\right )^{3} - 48 \, \cos \left (d x + c\right )^{2} + {\left (4 \, \cos \left (d x + c\right )^{3} + 8 \, \cos \left (d x + c\right )^{2} - 40 \, \cos \left (d x + c\right ) + 5\right )} \sin \left (d x + c\right ) - 45 \, \cos \left (d x + c\right ) - 5\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{40 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} - a^{2} d \cos \left (d x + c\right ) - 2 \, a^{2} d - {\left (a^{2} d \cos \left (d x + c\right ) + 2 \, a^{2} d\right )} \sin \left (d x + c\right )\right )}} \] Input:
integrate(sin(d*x+c)^4/(a+a*sin(d*x+c))^(3/2),x, algorithm="fricas")
Output:
1/40*(75*sqrt(2)*(cos(d*x + c)^2 - (cos(d*x + c) + 2)*sin(d*x + c) - cos(d *x + c) - 2)*sqrt(a)*log(-(a*cos(d*x + c)^2 + 2*sqrt(2)*sqrt(a*sin(d*x + c ) + a)*sqrt(a)*(cos(d*x + c) - sin(d*x + c) + 1) + 3*a*cos(d*x + c) - (a*c os(d*x + c) - 2*a)*sin(d*x + c) + 2*a)/(cos(d*x + c)^2 - (cos(d*x + c) + 2 )*sin(d*x + c) - cos(d*x + c) - 2)) - 4*(4*cos(d*x + c)^4 - 4*cos(d*x + c) ^3 - 48*cos(d*x + c)^2 + (4*cos(d*x + c)^3 + 8*cos(d*x + c)^2 - 40*cos(d*x + c) + 5)*sin(d*x + c) - 45*cos(d*x + c) - 5)*sqrt(a*sin(d*x + c) + a))/( a^2*d*cos(d*x + c)^2 - a^2*d*cos(d*x + c) - 2*a^2*d - (a^2*d*cos(d*x + c) + 2*a^2*d)*sin(d*x + c))
\[ \int \frac {\sin ^4(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx=\int \frac {\sin ^{4}{\left (c + d x \right )}}{\left (a \left (\sin {\left (c + d x \right )} + 1\right )\right )^{\frac {3}{2}}}\, dx \] Input:
integrate(sin(d*x+c)**4/(a+a*sin(d*x+c))**(3/2),x)
Output:
Integral(sin(c + d*x)**4/(a*(sin(c + d*x) + 1))**(3/2), x)
\[ \int \frac {\sin ^4(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx=\int { \frac {\sin \left (d x + c\right )^{4}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x } \] Input:
integrate(sin(d*x+c)^4/(a+a*sin(d*x+c))^(3/2),x, algorithm="maxima")
Output:
integrate(sin(d*x + c)^4/(a*sin(d*x + c) + a)^(3/2), x)
Time = 0.16 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.08 \[ \int \frac {\sin ^4(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx=-\frac {\frac {75 \, \sqrt {2} \log \left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{a^{\frac {3}{2}} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {75 \, \sqrt {2} \log \left (-\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{a^{\frac {3}{2}} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} + \frac {10 \, \sqrt {2} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )} a^{\frac {3}{2}} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {32 \, \sqrt {2} {\left (2 \, a^{\frac {17}{2}} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 5 \, a^{\frac {17}{2}} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{a^{10} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{40 \, d} \] Input:
integrate(sin(d*x+c)^4/(a+a*sin(d*x+c))^(3/2),x, algorithm="giac")
Output:
-1/40*(75*sqrt(2)*log(sin(-1/4*pi + 1/2*d*x + 1/2*c) + 1)/(a^(3/2)*sgn(cos (-1/4*pi + 1/2*d*x + 1/2*c))) - 75*sqrt(2)*log(-sin(-1/4*pi + 1/2*d*x + 1/ 2*c) + 1)/(a^(3/2)*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))) + 10*sqrt(2)*sin(- 1/4*pi + 1/2*d*x + 1/2*c)/((sin(-1/4*pi + 1/2*d*x + 1/2*c)^2 - 1)*a^(3/2)* sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))) - 32*sqrt(2)*(2*a^(17/2)*sin(-1/4*pi + 1/2*d*x + 1/2*c)^5 + 5*a^(17/2)*sin(-1/4*pi + 1/2*d*x + 1/2*c))/(a^10*sg n(cos(-1/4*pi + 1/2*d*x + 1/2*c))))/d
Timed out. \[ \int \frac {\sin ^4(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx=\int \frac {{\sin \left (c+d\,x\right )}^4}{{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{3/2}} \,d x \] Input:
int(sin(c + d*x)^4/(a + a*sin(c + d*x))^(3/2),x)
Output:
int(sin(c + d*x)^4/(a + a*sin(c + d*x))^(3/2), x)
\[ \int \frac {\sin ^4(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx=\frac {\sqrt {a}\, \left (\int \frac {\sqrt {\sin \left (d x +c \right )+1}\, \sin \left (d x +c \right )^{4}}{\sin \left (d x +c \right )^{2}+2 \sin \left (d x +c \right )+1}d x \right )}{a^{2}} \] Input:
int(sin(d*x+c)^4/(a+a*sin(d*x+c))^(3/2),x)
Output:
(sqrt(a)*int((sqrt(sin(c + d*x) + 1)*sin(c + d*x)**4)/(sin(c + d*x)**2 + 2 *sin(c + d*x) + 1),x))/a**2