Integrand size = 29, antiderivative size = 185 \[ \int \frac {1}{\sqrt {3+3 \sin (e+f x)} (c+d \sin (e+f x))^{5/2}} \, dx=-\frac {\sqrt {\frac {2}{3}} \text {arctanh}\left (\frac {\sqrt {\frac {3}{2}} \sqrt {c-d} \cos (e+f x)}{\sqrt {3+3 \sin (e+f x)} \sqrt {c+d \sin (e+f x)}}\right )}{(c-d)^{5/2} f}+\frac {2 d \cos (e+f x)}{3 \left (c^2-d^2\right ) f \sqrt {3+3 \sin (e+f x)} (c+d \sin (e+f x))^{3/2}}+\frac {2 d (5 c+d) \cos (e+f x)}{3 \left (c^2-d^2\right )^2 f \sqrt {3+3 \sin (e+f x)} \sqrt {c+d \sin (e+f x)}} \]
-arctanh(1/2*cos(f*x+e)*a^(1/2)*(c-d)^(1/2)*2^(1/2)/(a+a*sin(f*x+e))^(1/2) /(c+d*sin(f*x+e))^(1/2))*2^(1/2)/(c-d)^(5/2)/f/a^(1/2)+2/3*d*cos(f*x+e)/(c ^2-d^2)/f/(c+d*sin(f*x+e))^(3/2)/(a+a*sin(f*x+e))^(1/2)+2/3*d*(5*c+d)*cos( f*x+e)/(c^2-d^2)^2/f/(a+a*sin(f*x+e))^(1/2)/(c+d*sin(f*x+e))^(1/2)
Leaf count is larger than twice the leaf count of optimal. \(390\) vs. \(2(185)=370\).
Time = 3.10 (sec) , antiderivative size = 390, normalized size of antiderivative = 2.11 \[ \int \frac {1}{\sqrt {3+3 \sin (e+f x)} (c+d \sin (e+f x))^{5/2}} \, dx=\frac {\frac {2 d \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (6 c^2+c d-d^2+d (5 c+d) \sin (e+f x)\right )}{(c+d)^2 (c+d \sin (e+f x))}+\frac {3 \left (\log \left (1+\tan \left (\frac {1}{2} (e+f x)\right )\right )-\log \left (c-d+2 \sqrt {c-d} \sqrt {\frac {1}{1+\cos (e+f x)}} \sqrt {c+d \sin (e+f x)}+(-c+d) \tan \left (\frac {1}{2} (e+f x)\right )\right )\right )}{\frac {\sec ^2\left (\frac {1}{2} (e+f x)\right )}{2+2 \tan \left (\frac {1}{2} (e+f x)\right )}-\frac {-\frac {1}{2} (c-d) \sec ^2\left (\frac {1}{2} (e+f x)\right )+\frac {\sqrt {c-d} \left (\frac {1}{1+\cos (e+f x)}\right )^{3/2} (d+d \cos (e+f x)+c \sin (e+f x))}{\sqrt {c+d \sin (e+f x)}}}{c-d+2 \sqrt {c-d} \sqrt {\frac {1}{1+\cos (e+f x)}} \sqrt {c+d \sin (e+f x)}+(-c+d) \tan \left (\frac {1}{2} (e+f x)\right )}}}{3 \sqrt {3} (c-d)^2 f \sqrt {1+\sin (e+f x)} \sqrt {c+d \sin (e+f x)}} \]
((2*d*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(Cos[(e + f*x)/2] + Sin[(e + f *x)/2])*(6*c^2 + c*d - d^2 + d*(5*c + d)*Sin[e + f*x]))/((c + d)^2*(c + d* Sin[e + f*x])) + (3*(Log[1 + Tan[(e + f*x)/2]] - Log[c - d + 2*Sqrt[c - d] *Sqrt[(1 + Cos[e + f*x])^(-1)]*Sqrt[c + d*Sin[e + f*x]] + (-c + d)*Tan[(e + f*x)/2]]))/(Sec[(e + f*x)/2]^2/(2 + 2*Tan[(e + f*x)/2]) - (-1/2*((c - d) *Sec[(e + f*x)/2]^2) + (Sqrt[c - d]*((1 + Cos[e + f*x])^(-1))^(3/2)*(d + d *Cos[e + f*x] + c*Sin[e + f*x]))/Sqrt[c + d*Sin[e + f*x]])/(c - d + 2*Sqrt [c - d]*Sqrt[(1 + Cos[e + f*x])^(-1)]*Sqrt[c + d*Sin[e + f*x]] + (-c + d)* Tan[(e + f*x)/2])))/(3*Sqrt[3]*(c - d)^2*f*Sqrt[1 + Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]])
Time = 0.80 (sec) , antiderivative size = 225, normalized size of antiderivative = 1.22, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.276, Rules used = {3042, 3258, 3042, 3463, 27, 3042, 3261, 221}
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.
| \(\displaystyle \int \frac {1}{\sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))^{5/2}}dx\) |
| \(\Big \downarrow \) 3258 |
| \(\displaystyle \frac {\int \frac {a (3 c+d)-2 a d \sin (e+f x)}{\sqrt {\sin (e+f x) a+a} (c+d \sin (e+f x))^{3/2}}dx}{3 a \left (c^2-d^2\right )}+\frac {2 d \cos (e+f x)}{3 f \left (c^2-d^2\right ) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {a (3 c+d)-2 a d \sin (e+f x)}{\sqrt {\sin (e+f x) a+a} (c+d \sin (e+f x))^{3/2}}dx}{3 a \left (c^2-d^2\right )}+\frac {2 d \cos (e+f x)}{3 f \left (c^2-d^2\right ) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 3463 |
| \(\displaystyle \frac {\frac {2 a d (5 c+d) \cos (e+f x)}{f \left (c^2-d^2\right ) \sqrt {a \sin (e+f x)+a} \sqrt {c+d \sin (e+f x)}}-\frac {2 \int -\frac {3 a^2 (c+d)^2}{2 \sqrt {\sin (e+f x) a+a} \sqrt {c+d \sin (e+f x)}}dx}{a \left (c^2-d^2\right )}}{3 a \left (c^2-d^2\right )}+\frac {2 d \cos (e+f x)}{3 f \left (c^2-d^2\right ) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {3 a (c+d)^2 \int \frac {1}{\sqrt {\sin (e+f x) a+a} \sqrt {c+d \sin (e+f x)}}dx}{c^2-d^2}+\frac {2 a d (5 c+d) \cos (e+f x)}{f \left (c^2-d^2\right ) \sqrt {a \sin (e+f x)+a} \sqrt {c+d \sin (e+f x)}}}{3 a \left (c^2-d^2\right )}+\frac {2 d \cos (e+f x)}{3 f \left (c^2-d^2\right ) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {3 a (c+d)^2 \int \frac {1}{\sqrt {\sin (e+f x) a+a} \sqrt {c+d \sin (e+f x)}}dx}{c^2-d^2}+\frac {2 a d (5 c+d) \cos (e+f x)}{f \left (c^2-d^2\right ) \sqrt {a \sin (e+f x)+a} \sqrt {c+d \sin (e+f x)}}}{3 a \left (c^2-d^2\right )}+\frac {2 d \cos (e+f x)}{3 f \left (c^2-d^2\right ) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 3261 |
| \(\displaystyle \frac {\frac {2 a d (5 c+d) \cos (e+f x)}{f \left (c^2-d^2\right ) \sqrt {a \sin (e+f x)+a} \sqrt {c+d \sin (e+f x)}}-\frac {6 a^2 (c+d)^2 \int \frac {1}{2 a^2-\frac {a^3 (c-d) \cos ^2(e+f x)}{(\sin (e+f x) a+a) (c+d \sin (e+f x))}}d\frac {a \cos (e+f x)}{\sqrt {\sin (e+f x) a+a} \sqrt {c+d \sin (e+f x)}}}{f \left (c^2-d^2\right )}}{3 a \left (c^2-d^2\right )}+\frac {2 d \cos (e+f x)}{3 f \left (c^2-d^2\right ) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {\frac {2 a d (5 c+d) \cos (e+f x)}{f \left (c^2-d^2\right ) \sqrt {a \sin (e+f x)+a} \sqrt {c+d \sin (e+f x)}}-\frac {3 \sqrt {2} \sqrt {a} (c+d)^2 \text {arctanh}\left (\frac {\sqrt {a} \sqrt {c-d} \cos (e+f x)}{\sqrt {2} \sqrt {a \sin (e+f x)+a} \sqrt {c+d \sin (e+f x)}}\right )}{f \sqrt {c-d} \left (c^2-d^2\right )}}{3 a \left (c^2-d^2\right )}+\frac {2 d \cos (e+f x)}{3 f \left (c^2-d^2\right ) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))^{3/2}}\) |
(2*d*Cos[e + f*x])/(3*(c^2 - d^2)*f*Sqrt[a + a*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(3/2)) + ((-3*Sqrt[2]*Sqrt[a]*(c + d)^2*ArcTanh[(Sqrt[a]*Sqrt[c - d]*Cos[e + f*x])/(Sqrt[2]*Sqrt[a + a*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x] ])])/(Sqrt[c - d]*(c^2 - d^2)*f) + (2*a*d*(5*c + d)*Cos[e + f*x])/((c^2 - d^2)*f*Sqrt[a + a*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]]))/(3*a*(c^2 - d^2 ))
3.6.93.3.1 Defintions of rubi rules used
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[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)/Sqrt[(a_) + (b_.)*sin[(e_. ) + (f_.)*(x_)]], x_Symbol] :> Simp[(-d)*Cos[e + f*x]*((c + d*Sin[e + f*x]) ^(n + 1)/(f*(n + 1)*(c^2 - d^2)*Sqrt[a + b*Sin[e + f*x]])), x] - Simp[1/(2* b*(n + 1)*(c^2 - d^2)) Int[(c + d*Sin[e + f*x])^(n + 1)*(Simp[a*d - 2*b*c *(n + 1) + b*d*(2*n + 3)*Sin[e + f*x], x]/Sqrt[a + b*Sin[e + f*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[n, -1] && IntegerQ[2*n]
Int[1/(Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(c_.) + (d_.)*sin[(e _.) + (f_.)*(x_)]]), x_Symbol] :> Simp[-2*(a/f) Subst[Int[1/(2*b^2 - (a*c - b*d)*x^2), x], x, b*(Cos[e + f*x]/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*S in[e + f*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]
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*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n + 1)/(f*(n + 1)*(c^2 - d^2))), x] + Simp[1/(b*(n + 1)*(c^2 - d^2)) Int[(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n + 1)*Simp[A*(a*d*m + b*c*(n + 1)) - B*(a*c*m + b*d*(n + 1)) + b*(B*c - A*d)*(m + n + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && Eq Q[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[n, -1] && (IntegerQ[n] || EqQ[m + 1/2, 0])
Leaf count of result is larger than twice the leaf count of optimal. \(2406\) vs. \(2(164)=328\).
Time = 5.53 (sec) , antiderivative size = 2407, normalized size of antiderivative = 13.01
1/3/f*(3*2^(1/2)*((c+d*sin(f*x+e))/(cos(f*x+e)+1))^(1/2)*ln(-2*((2*c-2*d)^ (1/2)*2^(1/2)*((c+d*sin(f*x+e))/(cos(f*x+e)+1))^(1/2)*sin(f*x+e)+c*sin(f*x +e)-d*sin(f*x+e)+c*cos(f*x+e)-d*cos(f*x+e)-c+d)/(cos(f*x+e)-1-sin(f*x+e))) *c^2*cos(f*x+e)^2*d+6*2^(1/2)*((c+d*sin(f*x+e))/(cos(f*x+e)+1))^(1/2)*ln(- 2*((2*c-2*d)^(1/2)*2^(1/2)*((c+d*sin(f*x+e))/(cos(f*x+e)+1))^(1/2)*sin(f*x +e)+c*sin(f*x+e)-d*sin(f*x+e)+c*cos(f*x+e)-d*cos(f*x+e)-c+d)/(cos(f*x+e)-1 -sin(f*x+e)))*c*cos(f*x+e)^2*d^2-3*cos(f*x+e)*sin(f*x+e)*2^(1/2)*((c+d*sin (f*x+e))/(cos(f*x+e)+1))^(1/2)*ln(-2*((2*c-2*d)^(1/2)*2^(1/2)*((c+d*sin(f* x+e))/(cos(f*x+e)+1))^(1/2)*sin(f*x+e)+c*sin(f*x+e)-d*sin(f*x+e)+c*cos(f*x +e)-d*cos(f*x+e)-c+d)/(cos(f*x+e)-1-sin(f*x+e)))*c^2*d-6*cos(f*x+e)*sin(f* x+e)*2^(1/2)*((c+d*sin(f*x+e))/(cos(f*x+e)+1))^(1/2)*ln(-2*((2*c-2*d)^(1/2 )*2^(1/2)*((c+d*sin(f*x+e))/(cos(f*x+e)+1))^(1/2)*sin(f*x+e)+c*sin(f*x+e)- d*sin(f*x+e)+c*cos(f*x+e)-d*cos(f*x+e)-c+d)/(cos(f*x+e)-1-sin(f*x+e)))*c*d ^2-3*2^(1/2)*cos(f*x+e)*sin(f*x+e)*((c+d*sin(f*x+e))/(cos(f*x+e)+1))^(1/2) *ln(-2*((2*c-2*d)^(1/2)*2^(1/2)*((c+d*sin(f*x+e))/(cos(f*x+e)+1))^(1/2)*si n(f*x+e)+c*sin(f*x+e)-d*sin(f*x+e)+c*cos(f*x+e)-d*cos(f*x+e)-c+d)/(cos(f*x +e)-1-sin(f*x+e)))*d^3-3*ln(-2*((2*c-2*d)^(1/2)*2^(1/2)*((c+d*sin(f*x+e))/ (cos(f*x+e)+1))^(1/2)*sin(f*x+e)+c*sin(f*x+e)-d*sin(f*x+e)+c*cos(f*x+e)-d* cos(f*x+e)-c+d)/(cos(f*x+e)-1-sin(f*x+e)))*d^3*2^(1/2)*((c+d*sin(f*x+e))/( cos(f*x+e)+1))^(1/2)*sin(f*x+e)^2-3*2^(1/2)*((c+d*sin(f*x+e))/(cos(f*x+...
Leaf count of result is larger than twice the leaf count of optimal. 798 vs. \(2 (164) = 328\).
Time = 0.58 (sec) , antiderivative size = 1855, normalized size of antiderivative = 10.03 \[ \int \frac {1}{\sqrt {3+3 \sin (e+f x)} (c+d \sin (e+f x))^{5/2}} \, dx=\text {Too large to display} \]
[-1/12*(8*(6*c^2*d - 4*c*d^2 - 2*d^3 + (5*c*d^2 + d^3)*cos(f*x + e)^2 + (6 *c^2*d + c*d^2 - d^3)*cos(f*x + e) - (6*c^2*d - 4*c*d^2 - 2*d^3 - (5*c*d^2 + d^3)*cos(f*x + e))*sin(f*x + e))*sqrt(a*sin(f*x + e) + a)*sqrt(d*sin(f* x + e) + c) + 3*sqrt(2)*(a*c^4 + 4*a*c^3*d + 6*a*c^2*d^2 + 4*a*c*d^3 + a*d ^4 - (a*c^2*d^2 + 2*a*c*d^3 + a*d^4)*cos(f*x + e)^3 - (2*a*c^3*d + 5*a*c^2 *d^2 + 4*a*c*d^3 + a*d^4)*cos(f*x + e)^2 + (a*c^4 + 2*a*c^3*d + 2*a*c^2*d^ 2 + 2*a*c*d^3 + a*d^4)*cos(f*x + e) + (a*c^4 + 4*a*c^3*d + 6*a*c^2*d^2 + 4 *a*c*d^3 + a*d^4 - (a*c^2*d^2 + 2*a*c*d^3 + a*d^4)*cos(f*x + e)^2 + 2*(a*c ^3*d + 2*a*c^2*d^2 + a*c*d^3)*cos(f*x + e))*sin(f*x + e))*log(((c^2 - 14*c *d + 17*d^2)*cos(f*x + e)^3 - (13*c^2 - 22*c*d - 3*d^2)*cos(f*x + e)^2 - 4 *sqrt(2)*((c^2 - 4*c*d + 3*d^2)*cos(f*x + e)^2 - 4*c^2 + 8*c*d - 4*d^2 - ( 3*c^2 - 4*c*d + d^2)*cos(f*x + e) + (4*c^2 - 8*c*d + 4*d^2 + (c^2 - 4*c*d + 3*d^2)*cos(f*x + e))*sin(f*x + e))*sqrt(a*sin(f*x + e) + a)*sqrt(d*sin(f *x + e) + c)/sqrt(a*c - a*d) - 4*c^2 - 8*c*d - 4*d^2 - 2*(9*c^2 - 14*c*d + 9*d^2)*cos(f*x + e) + ((c^2 - 14*c*d + 17*d^2)*cos(f*x + e)^2 - 4*c^2 - 8 *c*d - 4*d^2 + 2*(7*c^2 - 18*c*d + 7*d^2)*cos(f*x + e))*sin(f*x + e))/(cos (f*x + e)^3 + 3*cos(f*x + e)^2 + (cos(f*x + e)^2 - 2*cos(f*x + e) - 4)*sin (f*x + e) - 2*cos(f*x + e) - 4))/sqrt(a*c - a*d))/((a*c^4*d^2 - 2*a*c^2*d^ 4 + a*d^6)*f*cos(f*x + e)^3 + (2*a*c^5*d + a*c^4*d^2 - 4*a*c^3*d^3 - 2*a*c ^2*d^4 + 2*a*c*d^5 + a*d^6)*f*cos(f*x + e)^2 - (a*c^6 - a*c^4*d^2 - a*c...
\[ \int \frac {1}{\sqrt {3+3 \sin (e+f x)} (c+d \sin (e+f x))^{5/2}} \, dx=\int \frac {1}{\sqrt {a \left (\sin {\left (e + f x \right )} + 1\right )} \left (c + d \sin {\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx \]
\[ \int \frac {1}{\sqrt {3+3 \sin (e+f x)} (c+d \sin (e+f x))^{5/2}} \, dx=\int { \frac {1}{\sqrt {a \sin \left (f x + e\right ) + a} {\left (d \sin \left (f x + e\right ) + c\right )}^{\frac {5}{2}}} \,d x } \]
\[ \int \frac {1}{\sqrt {3+3 \sin (e+f x)} (c+d \sin (e+f x))^{5/2}} \, dx=\int { \frac {1}{\sqrt {a \sin \left (f x + e\right ) + a} {\left (d \sin \left (f x + e\right ) + c\right )}^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {1}{\sqrt {3+3 \sin (e+f x)} (c+d \sin (e+f x))^{5/2}} \, dx=\int \frac {1}{\sqrt {a+a\,\sin \left (e+f\,x\right )}\,{\left (c+d\,\sin \left (e+f\,x\right )\right )}^{5/2}} \,d x \]