Integrand size = 35, antiderivative size = 347 \[ \int \frac {\cos ^2(e+f x)}{\sqrt {d \sin (e+f x)} (a+b \sin (e+f x))^{5/2}} \, dx=\frac {2 \cos (e+f x) \sqrt {d \sin (e+f x)}}{3 a d f (a+b \sin (e+f x))^{3/2}}+\frac {4 b \cos (e+f x)}{3 a \left (a^2-b^2\right ) f \sqrt {d \sin (e+f x)} \sqrt {a+b \sin (e+f x)}}-\frac {4 b \sqrt {\frac {a (1-\csc (e+f x))}{a+b}} \sqrt {\frac {a (1+\csc (e+f x))}{a-b}} E\left (\arcsin \left (\frac {\sqrt {d} \sqrt {a+b \sin (e+f x)}}{\sqrt {a+b} \sqrt {d \sin (e+f x)}}\right )|-\frac {a+b}{a-b}\right ) \tan (e+f x)}{3 a^3 \sqrt {a+b} \sqrt {d} f}-\frac {4 \sqrt {\frac {a (1-\csc (e+f x))}{a+b}} \sqrt {\frac {a (1+\csc (e+f x))}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {d} \sqrt {a+b \sin (e+f x)}}{\sqrt {a+b} \sqrt {d \sin (e+f x)}}\right ),-\frac {a+b}{a-b}\right ) \tan (e+f x)}{3 a^2 \sqrt {a+b} \sqrt {d} f} \]
2/3*cos(f*x+e)*(d*sin(f*x+e))^(1/2)/a/d/f/(a+b*sin(f*x+e))^(3/2)+4/3*b*cos (f*x+e)/a/(a^2-b^2)/f/(d*sin(f*x+e))^(1/2)/(a+b*sin(f*x+e))^(1/2)-4/3*b*El lipticE(d^(1/2)*(a+b*sin(f*x+e))^(1/2)/(a+b)^(1/2)/(d*sin(f*x+e))^(1/2),(( -a-b)/(a-b))^(1/2))*(a*(1-csc(f*x+e))/(a+b))^(1/2)*(a*(1+csc(f*x+e))/(a-b) )^(1/2)*tan(f*x+e)/a^3/f/(a+b)^(1/2)/d^(1/2)-4/3*EllipticF(d^(1/2)*(a+b*si n(f*x+e))^(1/2)/(a+b)^(1/2)/(d*sin(f*x+e))^(1/2),((-a-b)/(a-b))^(1/2))*(a* (1-csc(f*x+e))/(a+b))^(1/2)*(a*(1+csc(f*x+e))/(a-b))^(1/2)*tan(f*x+e)/a^2/ f/(a+b)^(1/2)/d^(1/2)
Leaf count is larger than twice the leaf count of optimal. \(3348\) vs. \(2(347)=694\).
Time = 25.25 (sec) , antiderivative size = 3348, normalized size of antiderivative = 9.65 \[ \int \frac {\cos ^2(e+f x)}{\sqrt {d \sin (e+f x)} (a+b \sin (e+f x))^{5/2}} \, dx=\text {Result too large to show} \]
(Sin[e + f*x]*Sqrt[a + b*Sin[e + f*x]]*((2*Cos[e + f*x])/(3*a*(a + b*Sin[e + f*x])^2) - (4*b^2*Cos[e + f*x])/(3*a^2*(a^2 - b^2)*(a + b*Sin[e + f*x]) )))/(f*Sqrt[d*Sin[e + f*x]]) + (4*Sqrt[a + b*Sin[e + f*x]]*((2*Sqrt[a + b* Sin[e + f*x]])/(3*a*(a^2 - b^2)*Sqrt[Sin[e + f*x]]) - (4*b*Sqrt[Sin[e + f* x]]*Sqrt[a + b*Sin[e + f*x]])/(3*a^2*(a^2 - b^2)))*(-2*b*Sin[(e + f*x)/2]^ 2 - (2*a*(-(b*EllipticE[ArcSin[Sqrt[(-b + Sqrt[-a^2 + b^2] - a*Tan[(e + f* x)/2])/Sqrt[-a^2 + b^2]]/Sqrt[2]], (2*Sqrt[-a^2 + b^2])/(-b + Sqrt[-a^2 + b^2])]*Tan[(e + f*x)/2]) + a*EllipticF[ArcSin[Sqrt[(b + Sqrt[-a^2 + b^2] + a*Tan[(e + f*x)/2])/Sqrt[-a^2 + b^2]]/Sqrt[2]], (2*Sqrt[-a^2 + b^2])/(b + Sqrt[-a^2 + b^2])]*Sqrt[(a*Tan[(e + f*x)/2])/(-b + Sqrt[-a^2 + b^2])]*Sqr t[-((a*Tan[(e + f*x)/2])/(b + Sqrt[-a^2 + b^2]))]))/(Sqrt[-a^2 + b^2]*Sqrt [(a*Sec[(e + f*x)/2]^2*(a + b*Sin[e + f*x]))/(a^2 - b^2)]*Sqrt[(a*Tan[(e + f*x)/2])/(-b + Sqrt[-a^2 + b^2])])))/(3*a^2*(a^2 - b^2)*f*Sqrt[d*Sin[e + f*x]]*((2*b*Cos[e + f*x]*(-2*b*Sin[(e + f*x)/2]^2 - (2*a*(-(b*EllipticE[Ar cSin[Sqrt[(-b + Sqrt[-a^2 + b^2] - a*Tan[(e + f*x)/2])/Sqrt[-a^2 + b^2]]/S qrt[2]], (2*Sqrt[-a^2 + b^2])/(-b + Sqrt[-a^2 + b^2])]*Tan[(e + f*x)/2]) + a*EllipticF[ArcSin[Sqrt[(b + Sqrt[-a^2 + b^2] + a*Tan[(e + f*x)/2])/Sqrt[ -a^2 + b^2]]/Sqrt[2]], (2*Sqrt[-a^2 + b^2])/(b + Sqrt[-a^2 + b^2])]*Sqrt[( a*Tan[(e + f*x)/2])/(-b + Sqrt[-a^2 + b^2])]*Sqrt[-((a*Tan[(e + f*x)/2])/( b + Sqrt[-a^2 + b^2]))]))/(Sqrt[-a^2 + b^2]*Sqrt[(a*Sec[(e + f*x)/2]^2*...
Time = 1.31 (sec) , antiderivative size = 370, normalized size of antiderivative = 1.07, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.257, Rules used = {3042, 3366, 3042, 3279, 3042, 3477, 3042, 3295, 3473}
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.
\(\displaystyle \int \frac {\cos ^2(e+f x)}{\sqrt {d \sin (e+f x)} (a+b \sin (e+f x))^{5/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos (e+f x)^2}{\sqrt {d \sin (e+f x)} (a+b \sin (e+f x))^{5/2}}dx\) |
\(\Big \downarrow \) 3366 |
\(\displaystyle \frac {2 \int \frac {1}{\sqrt {d \sin (e+f x)} (a+b \sin (e+f x))^{3/2}}dx}{3 a}+\frac {2 \cos (e+f x) \sqrt {d \sin (e+f x)}}{3 a d f (a+b \sin (e+f x))^{3/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 \int \frac {1}{\sqrt {d \sin (e+f x)} (a+b \sin (e+f x))^{3/2}}dx}{3 a}+\frac {2 \cos (e+f x) \sqrt {d \sin (e+f x)}}{3 a d f (a+b \sin (e+f x))^{3/2}}\) |
\(\Big \downarrow \) 3279 |
\(\displaystyle \frac {2 \left (\frac {d \int \frac {b+a \sin (e+f x)}{(d \sin (e+f x))^{3/2} \sqrt {a+b \sin (e+f x)}}dx}{a^2-b^2}+\frac {2 b \cos (e+f x)}{f \left (a^2-b^2\right ) \sqrt {d \sin (e+f x)} \sqrt {a+b \sin (e+f x)}}\right )}{3 a}+\frac {2 \cos (e+f x) \sqrt {d \sin (e+f x)}}{3 a d f (a+b \sin (e+f x))^{3/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 \left (\frac {d \int \frac {b+a \sin (e+f x)}{(d \sin (e+f x))^{3/2} \sqrt {a+b \sin (e+f x)}}dx}{a^2-b^2}+\frac {2 b \cos (e+f x)}{f \left (a^2-b^2\right ) \sqrt {d \sin (e+f x)} \sqrt {a+b \sin (e+f x)}}\right )}{3 a}+\frac {2 \cos (e+f x) \sqrt {d \sin (e+f x)}}{3 a d f (a+b \sin (e+f x))^{3/2}}\) |
\(\Big \downarrow \) 3477 |
\(\displaystyle \frac {2 \left (\frac {d \left (\frac {(a-b) \int \frac {1}{\sqrt {d \sin (e+f x)} \sqrt {a+b \sin (e+f x)}}dx}{d}+b \int \frac {\sin (e+f x)+1}{(d \sin (e+f x))^{3/2} \sqrt {a+b \sin (e+f x)}}dx\right )}{a^2-b^2}+\frac {2 b \cos (e+f x)}{f \left (a^2-b^2\right ) \sqrt {d \sin (e+f x)} \sqrt {a+b \sin (e+f x)}}\right )}{3 a}+\frac {2 \cos (e+f x) \sqrt {d \sin (e+f x)}}{3 a d f (a+b \sin (e+f x))^{3/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 \left (\frac {d \left (\frac {(a-b) \int \frac {1}{\sqrt {d \sin (e+f x)} \sqrt {a+b \sin (e+f x)}}dx}{d}+b \int \frac {\sin (e+f x)+1}{(d \sin (e+f x))^{3/2} \sqrt {a+b \sin (e+f x)}}dx\right )}{a^2-b^2}+\frac {2 b \cos (e+f x)}{f \left (a^2-b^2\right ) \sqrt {d \sin (e+f x)} \sqrt {a+b \sin (e+f x)}}\right )}{3 a}+\frac {2 \cos (e+f x) \sqrt {d \sin (e+f x)}}{3 a d f (a+b \sin (e+f x))^{3/2}}\) |
\(\Big \downarrow \) 3295 |
\(\displaystyle \frac {2 \left (\frac {d \left (b \int \frac {\sin (e+f x)+1}{(d \sin (e+f x))^{3/2} \sqrt {a+b \sin (e+f x)}}dx-\frac {2 (a-b) \sqrt {a+b} \tan (e+f x) \sqrt {\frac {a (1-\csc (e+f x))}{a+b}} \sqrt {\frac {a (\csc (e+f x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {d} \sqrt {a+b \sin (e+f x)}}{\sqrt {a+b} \sqrt {d \sin (e+f x)}}\right ),-\frac {a+b}{a-b}\right )}{a d^{3/2} f}\right )}{a^2-b^2}+\frac {2 b \cos (e+f x)}{f \left (a^2-b^2\right ) \sqrt {d \sin (e+f x)} \sqrt {a+b \sin (e+f x)}}\right )}{3 a}+\frac {2 \cos (e+f x) \sqrt {d \sin (e+f x)}}{3 a d f (a+b \sin (e+f x))^{3/2}}\) |
\(\Big \downarrow \) 3473 |
\(\displaystyle \frac {2 \left (\frac {d \left (-\frac {2 b (a-b) \sqrt {a+b} \tan (e+f x) \sqrt {\frac {a (1-\csc (e+f x))}{a+b}} \sqrt {\frac {a (\csc (e+f x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {d} \sqrt {a+b \sin (e+f x)}}{\sqrt {a+b} \sqrt {d \sin (e+f x)}}\right )|-\frac {a+b}{a-b}\right )}{a^2 d^{3/2} f}-\frac {2 (a-b) \sqrt {a+b} \tan (e+f x) \sqrt {\frac {a (1-\csc (e+f x))}{a+b}} \sqrt {\frac {a (\csc (e+f x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {d} \sqrt {a+b \sin (e+f x)}}{\sqrt {a+b} \sqrt {d \sin (e+f x)}}\right ),-\frac {a+b}{a-b}\right )}{a d^{3/2} f}\right )}{a^2-b^2}+\frac {2 b \cos (e+f x)}{f \left (a^2-b^2\right ) \sqrt {d \sin (e+f x)} \sqrt {a+b \sin (e+f x)}}\right )}{3 a}+\frac {2 \cos (e+f x) \sqrt {d \sin (e+f x)}}{3 a d f (a+b \sin (e+f x))^{3/2}}\) |
(2*Cos[e + f*x]*Sqrt[d*Sin[e + f*x]])/(3*a*d*f*(a + b*Sin[e + f*x])^(3/2)) + (2*((2*b*Cos[e + f*x])/((a^2 - b^2)*f*Sqrt[d*Sin[e + f*x]]*Sqrt[a + b*S in[e + f*x]]) + (d*((-2*(a - b)*b*Sqrt[a + b]*Sqrt[(a*(1 - Csc[e + f*x]))/ (a + b)]*Sqrt[(a*(1 + Csc[e + f*x]))/(a - b)]*EllipticE[ArcSin[(Sqrt[d]*Sq rt[a + b*Sin[e + f*x]])/(Sqrt[a + b]*Sqrt[d*Sin[e + f*x]])], -((a + b)/(a - b))]*Tan[e + f*x])/(a^2*d^(3/2)*f) - (2*(a - b)*Sqrt[a + b]*Sqrt[(a*(1 - Csc[e + f*x]))/(a + b)]*Sqrt[(a*(1 + Csc[e + f*x]))/(a - b)]*EllipticF[Ar cSin[(Sqrt[d]*Sqrt[a + b*Sin[e + f*x]])/(Sqrt[a + b]*Sqrt[d*Sin[e + f*x]]) ], -((a + b)/(a - b))]*Tan[e + f*x])/(a*d^(3/2)*f)))/(a^2 - b^2)))/(3*a)
3.11.91.3.1 Defintions of rubi rules used
Int[1/(Sqrt[(d_.)*sin[(e_.) + (f_.)*(x_)]]*((a_) + (b_.)*sin[(e_.) + (f_.)* (x_)])^(3/2)), x_Symbol] :> Simp[2*b*(Cos[e + f*x]/(f*(a^2 - b^2)*Sqrt[a + b*Sin[e + f*x]]*Sqrt[d*Sin[e + f*x]])), x] + Simp[d/(a^2 - b^2) Int[(b + a*Sin[e + f*x])/(Sqrt[a + b*Sin[e + f*x]]*(d*Sin[e + f*x])^(3/2)), x], x] / ; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0]
Int[1/(Sqrt[(d_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(a_) + (b_.)*sin[(e_.) + (f _.)*(x_)]]), x_Symbol] :> Simp[-2*(Tan[e + f*x]/(a*f))*Rt[(a + b)/d, 2]*Sqr t[a*((1 - Csc[e + f*x])/(a + b))]*Sqrt[a*((1 + Csc[e + f*x])/(a - b))]*Elli pticF[ArcSin[Sqrt[a + b*Sin[e + f*x]]/Sqrt[d*Sin[e + f*x]]/Rt[(a + b)/d, 2] ], -(a + b)/(a - b)], x] /; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0] && PosQ[(a + b)/d]
Int[((cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*( x_)])^(m_))/Sqrt[(d_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Simp[(-g)*(g* Cos[e + f*x])^(p - 1)*Sqrt[d*Sin[e + f*x]]*((a + b*Sin[e + f*x])^(m + 1)/(a *d*f*(m + 1))), x] + Simp[g^2*((2*m + 3)/(2*a*(m + 1))) Int[(g*Cos[e + f* x])^(p - 2)*((a + b*Sin[e + f*x])^(m + 1)/Sqrt[d*Sin[e + f*x]]), x], x] /; FreeQ[{a, b, d, e, f, g}, x] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && EqQ[m + p + 1/2, 0]
Int[((A_) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((b_.)*sin[(e_.) + (f_.)*(x_)]) ^(3/2)*Sqrt[(c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp[-2*A* (c - d)*(Tan[e + f*x]/(f*b*c^2))*Rt[(c + d)/b, 2]*Sqrt[c*((1 + Csc[e + f*x] )/(c - d))]*Sqrt[c*((1 - Csc[e + f*x])/(c + d))]*EllipticE[ArcSin[Sqrt[c + d*Sin[e + f*x]]/Sqrt[b*Sin[e + f*x]]/Rt[(c + d)/b, 2]], -(c + d)/(c - d)], x] /; FreeQ[{b, c, d, e, f, A, B}, x] && NeQ[c^2 - d^2, 0] && EqQ[A, B] && PosQ[(c + d)/b]
Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((a_.) + (b_.)*sin[(e_.) + (f_ .)*(x_)])^(3/2)*Sqrt[(c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> S imp[(A - B)/(a - b) Int[1/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f* x]]), x], x] - Simp[(A*b - a*B)/(a - b) Int[(1 + Sin[e + f*x])/((a + b*Si n[e + f*x])^(3/2)*Sqrt[c + d*Sin[e + f*x]]), x], x] /; FreeQ[{a, b, c, d, e , f, A, B}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && NeQ[A, B]
Leaf count of result is larger than twice the leaf count of optimal. \(3196\) vs. \(2(307)=614\).
Time = 2.41 (sec) , antiderivative size = 3197, normalized size of antiderivative = 9.21
2/3/f*2^(1/2)/(a^2-b^2)/a^3/(d/((1-cos(f*x+e))^2*csc(f*x+e)^2+1)*(csc(f*x+ e)-cot(f*x+e)))^(1/2)*((a*(1-cos(f*x+e))^2*csc(f*x+e)^2+2*b*(csc(f*x+e)-co t(f*x+e))+a)/((1-cos(f*x+e))^2*csc(f*x+e)^2+1))^(1/2)*((-a^2+b^2)^(1/2)*2^ (1/2)*(1/(b+(-a^2+b^2)^(1/2))*(a*(csc(f*x+e)-cot(f*x+e))+(-a^2+b^2)^(1/2)+ b))^(1/2)*(1/(-a^2+b^2)^(1/2)*(-a*(csc(f*x+e)-cot(f*x+e))+(-a^2+b^2)^(1/2) -b))^(1/2)*(-a/(b+(-a^2+b^2)^(1/2))*(csc(f*x+e)-cot(f*x+e)))^(1/2)*Ellipti cF((1/(b+(-a^2+b^2)^(1/2))*(a*(csc(f*x+e)-cot(f*x+e))+(-a^2+b^2)^(1/2)+b)) ^(1/2),1/2*2^(1/2)*((b+(-a^2+b^2)^(1/2))/(-a^2+b^2)^(1/2))^(1/2))*a^3*(1-c os(f*x+e))^2*csc(f*x+e)^2-2*(-a^2+b^2)^(1/2)*2^(1/2)*(1/(b+(-a^2+b^2)^(1/2 ))*(a*(csc(f*x+e)-cot(f*x+e))+(-a^2+b^2)^(1/2)+b))^(1/2)*(1/(-a^2+b^2)^(1/ 2)*(-a*(csc(f*x+e)-cot(f*x+e))+(-a^2+b^2)^(1/2)-b))^(1/2)*(-a/(b+(-a^2+b^2 )^(1/2))*(csc(f*x+e)-cot(f*x+e)))^(1/2)*EllipticE((1/(b+(-a^2+b^2)^(1/2))* (a*(csc(f*x+e)-cot(f*x+e))+(-a^2+b^2)^(1/2)+b))^(1/2),1/2*2^(1/2)*((b+(-a^ 2+b^2)^(1/2))/(-a^2+b^2)^(1/2))^(1/2))*a*b^2*(1-cos(f*x+e))^2*csc(f*x+e)^2 +2*2^(1/2)*(1/(b+(-a^2+b^2)^(1/2))*(a*(csc(f*x+e)-cot(f*x+e))+(-a^2+b^2)^( 1/2)+b))^(1/2)*(1/(-a^2+b^2)^(1/2)*(-a*(csc(f*x+e)-cot(f*x+e))+(-a^2+b^2)^ (1/2)-b))^(1/2)*(-a/(b+(-a^2+b^2)^(1/2))*(csc(f*x+e)-cot(f*x+e)))^(1/2)*El lipticE((1/(b+(-a^2+b^2)^(1/2))*(a*(csc(f*x+e)-cot(f*x+e))+(-a^2+b^2)^(1/2 )+b))^(1/2),1/2*2^(1/2)*((b+(-a^2+b^2)^(1/2))/(-a^2+b^2)^(1/2))^(1/2))*a^3 *b*(1-cos(f*x+e))^2*csc(f*x+e)^2-2*2^(1/2)*(1/(b+(-a^2+b^2)^(1/2))*(a*(...
\[ \int \frac {\cos ^2(e+f x)}{\sqrt {d \sin (e+f x)} (a+b \sin (e+f x))^{5/2}} \, dx=\int { \frac {\cos \left (f x + e\right )^{2}}{{\left (b \sin \left (f x + e\right ) + a\right )}^{\frac {5}{2}} \sqrt {d \sin \left (f x + e\right )}} \,d x } \]
integral(sqrt(b*sin(f*x + e) + a)*sqrt(d*sin(f*x + e))*cos(f*x + e)^2/(b^3 *d*cos(f*x + e)^4 - (3*a^2*b + 2*b^3)*d*cos(f*x + e)^2 + (3*a^2*b + b^3)*d - (3*a*b^2*d*cos(f*x + e)^2 - (a^3 + 3*a*b^2)*d)*sin(f*x + e)), x)
Timed out. \[ \int \frac {\cos ^2(e+f x)}{\sqrt {d \sin (e+f x)} (a+b \sin (e+f x))^{5/2}} \, dx=\text {Timed out} \]
\[ \int \frac {\cos ^2(e+f x)}{\sqrt {d \sin (e+f x)} (a+b \sin (e+f x))^{5/2}} \, dx=\int { \frac {\cos \left (f x + e\right )^{2}}{{\left (b \sin \left (f x + e\right ) + a\right )}^{\frac {5}{2}} \sqrt {d \sin \left (f x + e\right )}} \,d x } \]
Timed out. \[ \int \frac {\cos ^2(e+f x)}{\sqrt {d \sin (e+f x)} (a+b \sin (e+f x))^{5/2}} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {\cos ^2(e+f x)}{\sqrt {d \sin (e+f x)} (a+b \sin (e+f x))^{5/2}} \, dx=\int \frac {{\cos \left (e+f\,x\right )}^2}{\sqrt {d\,\sin \left (e+f\,x\right )}\,{\left (a+b\,\sin \left (e+f\,x\right )\right )}^{5/2}} \,d x \]