Integrand size = 27, antiderivative size = 210 \[ \int \frac {(a+a \sin (c+d x))^{3/2}}{(e \cos (c+d x))^{3/2}} \, dx=\frac {4 a \sqrt {a+a \sin (c+d x)}}{d e \sqrt {e \cos (c+d x)}}-\frac {2 a^2 \text {arcsinh}\left (\frac {\sqrt {e \cos (c+d x)}}{\sqrt {e}}\right ) \sqrt {1+\cos (c+d x)} \sqrt {a+a \sin (c+d x)}}{d e^{3/2} (a+a \cos (c+d x)+a \sin (c+d x))}-\frac {2 a^2 \arctan \left (\frac {\sqrt {e} \sin (c+d x)}{\sqrt {e \cos (c+d x)} \sqrt {1+\cos (c+d x)}}\right ) \sqrt {1+\cos (c+d x)} \sqrt {a+a \sin (c+d x)}}{d e^{3/2} (a+a \cos (c+d x)+a \sin (c+d x))} \]
4*a*(a+a*sin(d*x+c))^(1/2)/d/e/(e*cos(d*x+c))^(1/2)-2*a^2*arcsinh((e*cos(d *x+c))^(1/2)/e^(1/2))*(1+cos(d*x+c))^(1/2)*(a+a*sin(d*x+c))^(1/2)/d/e^(3/2 )/(a+a*cos(d*x+c)+a*sin(d*x+c))-2*a^2*arctan(sin(d*x+c)*e^(1/2)/(e*cos(d*x +c))^(1/2)/(1+cos(d*x+c))^(1/2))*(1+cos(d*x+c))^(1/2)*(a+a*sin(d*x+c))^(1/ 2)/d/e^(3/2)/(a+a*cos(d*x+c)+a*sin(d*x+c))
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.07 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.36 \[ \int \frac {(a+a \sin (c+d x))^{3/2}}{(e \cos (c+d x))^{3/2}} \, dx=\frac {4 \sqrt [4]{2} \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},-\frac {1}{4},\frac {3}{4},\frac {1}{2} (1-\sin (c+d x))\right ) (a (1+\sin (c+d x)))^{3/2}}{d e \sqrt {e \cos (c+d x)} (1+\sin (c+d x))^{5/4}} \]
(4*2^(1/4)*Hypergeometric2F1[-1/4, -1/4, 3/4, (1 - Sin[c + d*x])/2]*(a*(1 + Sin[c + d*x]))^(3/2))/(d*e*Sqrt[e*Cos[c + d*x]]*(1 + Sin[c + d*x])^(5/4) )
Time = 0.81 (sec) , antiderivative size = 213, normalized size of antiderivative = 1.01, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.407, Rules used = {3042, 3155, 3042, 3163, 3042, 25, 3254, 216, 3312, 63, 222}
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.
| \(\displaystyle \int \frac {(a \sin (c+d x)+a)^{3/2}}{(e \cos (c+d x))^{3/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a \sin (c+d x)+a)^{3/2}}{(e \cos (c+d x))^{3/2}}dx\) |
| \(\Big \downarrow \) 3155 |
| \(\displaystyle \frac {4 a \sqrt {a \sin (c+d x)+a}}{d e \sqrt {e \cos (c+d x)}}-\frac {a^2 \int \frac {\sqrt {e \cos (c+d x)}}{\sqrt {\sin (c+d x) a+a}}dx}{e^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {4 a \sqrt {a \sin (c+d x)+a}}{d e \sqrt {e \cos (c+d x)}}-\frac {a^2 \int \frac {\sqrt {e \cos (c+d x)}}{\sqrt {\sin (c+d x) a+a}}dx}{e^2}\) |
| \(\Big \downarrow \) 3163 |
| \(\displaystyle \frac {4 a \sqrt {a \sin (c+d x)+a}}{d e \sqrt {e \cos (c+d x)}}-\frac {a^2 \left (\frac {e \sqrt {\cos (c+d x)+1} \sqrt {a \sin (c+d x)+a} \int \frac {\sqrt {\cos (c+d x)+1}}{\sqrt {e \cos (c+d x)}}dx}{a \sin (c+d x)+a \cos (c+d x)+a}-\frac {e \sqrt {\cos (c+d x)+1} \sqrt {a \sin (c+d x)+a} \int \frac {\sin (c+d x)}{\sqrt {e \cos (c+d x)} \sqrt {\cos (c+d x)+1}}dx}{a \sin (c+d x)+a \cos (c+d x)+a}\right )}{e^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {4 a \sqrt {a \sin (c+d x)+a}}{d e \sqrt {e \cos (c+d x)}}-\frac {a^2 \left (\frac {e \sqrt {\cos (c+d x)+1} \sqrt {a \sin (c+d x)+a} \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )+1}}{\sqrt {e \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{a \sin (c+d x)+a \cos (c+d x)+a}-\frac {e \sqrt {\cos (c+d x)+1} \sqrt {a \sin (c+d x)+a} \int -\frac {\cos \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {e \sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )+1}}dx}{a \sin (c+d x)+a \cos (c+d x)+a}\right )}{e^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {4 a \sqrt {a \sin (c+d x)+a}}{d e \sqrt {e \cos (c+d x)}}-\frac {a^2 \left (\frac {e \sqrt {\cos (c+d x)+1} \sqrt {a \sin (c+d x)+a} \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )+1}}{\sqrt {e \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{a \sin (c+d x)+a \cos (c+d x)+a}+\frac {e \sqrt {\cos (c+d x)+1} \sqrt {a \sin (c+d x)+a} \int \frac {\cos \left (\frac {1}{2} (2 c+\pi )+d x\right )}{\sqrt {e \sin \left (\frac {1}{2} (2 c+\pi )+d x\right )} \sqrt {\sin \left (\frac {1}{2} (2 c+\pi )+d x\right )+1}}dx}{a \sin (c+d x)+a \cos (c+d x)+a}\right )}{e^2}\) |
| \(\Big \downarrow \) 3254 |
| \(\displaystyle \frac {4 a \sqrt {a \sin (c+d x)+a}}{d e \sqrt {e \cos (c+d x)}}-\frac {a^2 \left (\frac {e \sqrt {\cos (c+d x)+1} \sqrt {a \sin (c+d x)+a} \int \frac {\cos \left (\frac {1}{2} (2 c+\pi )+d x\right )}{\sqrt {e \sin \left (\frac {1}{2} (2 c+\pi )+d x\right )} \sqrt {\sin \left (\frac {1}{2} (2 c+\pi )+d x\right )+1}}dx}{a \sin (c+d x)+a \cos (c+d x)+a}-\frac {2 e \sqrt {\cos (c+d x)+1} \sqrt {a \sin (c+d x)+a} \int \frac {1}{\frac {\sin (c+d x) \tan (c+d x)}{\cos (c+d x)+1}+1}d\left (-\frac {\sin (c+d x)}{\sqrt {e \cos (c+d x)} \sqrt {\cos (c+d x)+1}}\right )}{d (a \sin (c+d x)+a \cos (c+d x)+a)}\right )}{e^2}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {4 a \sqrt {a \sin (c+d x)+a}}{d e \sqrt {e \cos (c+d x)}}-\frac {a^2 \left (\frac {e \sqrt {\cos (c+d x)+1} \sqrt {a \sin (c+d x)+a} \int \frac {\cos \left (\frac {1}{2} (2 c+\pi )+d x\right )}{\sqrt {e \sin \left (\frac {1}{2} (2 c+\pi )+d x\right )} \sqrt {\sin \left (\frac {1}{2} (2 c+\pi )+d x\right )+1}}dx}{a \sin (c+d x)+a \cos (c+d x)+a}+\frac {2 \sqrt {e} \sqrt {\cos (c+d x)+1} \sqrt {a \sin (c+d x)+a} \arctan \left (\frac {\sqrt {e} \sin (c+d x)}{\sqrt {\cos (c+d x)+1} \sqrt {e \cos (c+d x)}}\right )}{d (a \sin (c+d x)+a \cos (c+d x)+a)}\right )}{e^2}\) |
| \(\Big \downarrow \) 3312 |
| \(\displaystyle \frac {4 a \sqrt {a \sin (c+d x)+a}}{d e \sqrt {e \cos (c+d x)}}-\frac {a^2 \left (\frac {e \sqrt {\cos (c+d x)+1} \sqrt {a \sin (c+d x)+a} \int \frac {1}{\sqrt {e \cos (c+d x)} \sqrt {\cos (c+d x)+1}}d\cos (c+d x)}{d (a \sin (c+d x)+a \cos (c+d x)+a)}+\frac {2 \sqrt {e} \sqrt {\cos (c+d x)+1} \sqrt {a \sin (c+d x)+a} \arctan \left (\frac {\sqrt {e} \sin (c+d x)}{\sqrt {\cos (c+d x)+1} \sqrt {e \cos (c+d x)}}\right )}{d (a \sin (c+d x)+a \cos (c+d x)+a)}\right )}{e^2}\) |
| \(\Big \downarrow \) 63 |
| \(\displaystyle \frac {4 a \sqrt {a \sin (c+d x)+a}}{d e \sqrt {e \cos (c+d x)}}-\frac {a^2 \left (\frac {2 \sqrt {\cos (c+d x)+1} \sqrt {a \sin (c+d x)+a} \int \frac {1}{\sqrt {\cos (c+d x)+1}}d\sqrt {e \cos (c+d x)}}{d (a \sin (c+d x)+a \cos (c+d x)+a)}+\frac {2 \sqrt {e} \sqrt {\cos (c+d x)+1} \sqrt {a \sin (c+d x)+a} \arctan \left (\frac {\sqrt {e} \sin (c+d x)}{\sqrt {\cos (c+d x)+1} \sqrt {e \cos (c+d x)}}\right )}{d (a \sin (c+d x)+a \cos (c+d x)+a)}\right )}{e^2}\) |
| \(\Big \downarrow \) 222 |
| \(\displaystyle \frac {4 a \sqrt {a \sin (c+d x)+a}}{d e \sqrt {e \cos (c+d x)}}-\frac {a^2 \left (\frac {2 \sqrt {e} \sqrt {\cos (c+d x)+1} \sqrt {a \sin (c+d x)+a} \text {arcsinh}\left (\frac {\sqrt {e \cos (c+d x)}}{\sqrt {e}}\right )}{d (a \sin (c+d x)+a \cos (c+d x)+a)}+\frac {2 \sqrt {e} \sqrt {\cos (c+d x)+1} \sqrt {a \sin (c+d x)+a} \arctan \left (\frac {\sqrt {e} \sin (c+d x)}{\sqrt {\cos (c+d x)+1} \sqrt {e \cos (c+d x)}}\right )}{d (a \sin (c+d x)+a \cos (c+d x)+a)}\right )}{e^2}\) |
(4*a*Sqrt[a + a*Sin[c + d*x]])/(d*e*Sqrt[e*Cos[c + d*x]]) - (a^2*((2*Sqrt[ e]*ArcSinh[Sqrt[e*Cos[c + d*x]]/Sqrt[e]]*Sqrt[1 + Cos[c + d*x]]*Sqrt[a + a *Sin[c + d*x]])/(d*(a + a*Cos[c + d*x] + a*Sin[c + d*x])) + (2*Sqrt[e]*Arc Tan[(Sqrt[e]*Sin[c + d*x])/(Sqrt[e*Cos[c + d*x]]*Sqrt[1 + Cos[c + d*x]])]* Sqrt[1 + Cos[c + d*x]]*Sqrt[a + a*Sin[c + d*x]])/(d*(a + a*Cos[c + d*x] + a*Sin[c + d*x]))))/e^2
3.3.84.3.1 Defintions of rubi rules used
Int[1/(Sqrt[(b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Simp[2/b S ubst[Int[1/Sqrt[c + d*(x^2/b)], x], x, Sqrt[b*x]], x] /; FreeQ[{b, c, d}, x ] && GtQ[c, 0]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[Rt[b, 2]*(x/Sqrt [a])]/Rt[b, 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)])^(m_), x_Symbol] :> Simp[-2*b*(g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + f*x])^(m - 1)/(f*g*(p + 1))), x] + Simp[b^2*((2*m + p - 1)/(g^2*(p + 1))) Int[(g*Cos[e + f*x])^(p + 2)*(a + b*Sin[e + f*x])^(m - 2), x], x] /; FreeQ [{a, b, e, f, g}, x] && EqQ[a^2 - b^2, 0] && GtQ[m, 1] && LtQ[p, -1] && Int egersQ[2*m, 2*p]
Int[Sqrt[cos[(e_.) + (f_.)*(x_)]*(g_.)]/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.) *(x_)]], x_Symbol] :> Simp[g*Sqrt[1 + Cos[e + f*x]]*(Sqrt[a + b*Sin[e + f*x ]]/(a + a*Cos[e + f*x] + b*Sin[e + f*x])) Int[Sqrt[1 + Cos[e + f*x]]/Sqrt [g*Cos[e + f*x]], x], x] - Simp[g*Sqrt[1 + Cos[e + f*x]]*(Sqrt[a + b*Sin[e + f*x]]/(b + b*Cos[e + f*x] + a*Sin[e + f*x])) Int[Sin[e + f*x]/(Sqrt[g*C os[e + f*x]]*Sqrt[1 + Cos[e + f*x]]), x], x] /; FreeQ[{a, b, e, f, g}, x] & & EqQ[a^2 - b^2, 0]
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Simp[-2*(b/f) Subst[Int[1/(b + d*x^2), x], x, b*(Cos[e + f*x]/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[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[cos[(e_.) + (f_.)*(x_)]*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*(( c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[1/(b*f) Su bst[Int[(a + x)^m*(c + (d/b)*x)^n, x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x]
Time = 2.58 (sec) , antiderivative size = 321, normalized size of antiderivative = 1.53
| method | result | size |
| default | \(\frac {\left (\cos \left (d x +c \right ) \operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\left (1+\cos \left (d x +c \right )\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\right )-\sin \left (d x +c \right ) \operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\left (1+\cos \left (d x +c \right )\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\right )-\cos \left (d x +c \right ) \arctan \left (\sqrt {-\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\right )+\sin \left (d x +c \right ) \arctan \left (\sqrt {-\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\right )+4 \cos \left (d x +c \right ) \sqrt {-\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+\operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\left (1+\cos \left (d x +c \right )\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\right )-\arctan \left (\sqrt {-\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\right )+4 \sqrt {-\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\right ) \sqrt {a \left (1+\sin \left (d x +c \right )\right )}\, a}{d \left (1+\cos \left (d x +c \right )\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, e \sqrt {e \cos \left (d x +c \right )}}\) | \(321\) |
1/d*(cos(d*x+c)*arctanh(sin(d*x+c)/(1+cos(d*x+c))/(-cos(d*x+c)/(1+cos(d*x+ c)))^(1/2))-sin(d*x+c)*arctanh(sin(d*x+c)/(1+cos(d*x+c))/(-cos(d*x+c)/(1+c os(d*x+c)))^(1/2))-cos(d*x+c)*arctan((-cos(d*x+c)/(1+cos(d*x+c)))^(1/2))+s in(d*x+c)*arctan((-cos(d*x+c)/(1+cos(d*x+c)))^(1/2))+4*cos(d*x+c)*(-cos(d* x+c)/(1+cos(d*x+c)))^(1/2)+arctanh(sin(d*x+c)/(1+cos(d*x+c))/(-cos(d*x+c)/ (1+cos(d*x+c)))^(1/2))-arctan((-cos(d*x+c)/(1+cos(d*x+c)))^(1/2))+4*(-cos( d*x+c)/(1+cos(d*x+c)))^(1/2))*(a*(1+sin(d*x+c)))^(1/2)*a/(1+cos(d*x+c))/(- cos(d*x+c)/(1+cos(d*x+c)))^(1/2)/e/(e*cos(d*x+c))^(1/2)
Result contains complex when optimal does not.
Time = 0.46 (sec) , antiderivative size = 1053, normalized size of antiderivative = 5.01 \[ \int \frac {(a+a \sin (c+d x))^{3/2}}{(e \cos (c+d x))^{3/2}} \, dx=\text {Too large to display} \]
-1/2*(d*e^2*(-a^6/(d^4*e^6))^(1/4)*cos(d*x + c)*log((2*(a^4*sin(d*x + c) + (a*d^2*e^3*cos(d*x + c) + a*d^2*e^3)*sqrt(-a^6/(d^4*e^6)))*sqrt(e*cos(d*x + c))*sqrt(a*sin(d*x + c) + a) + (2*d^3*e^5*cos(d*x + c)^2 + d^3*e^5*cos( d*x + c) - d^3*e^5*sin(d*x + c) - d^3*e^5)*(-a^6/(d^4*e^6))^(3/4) + (a^3*d *e^2*cos(d*x + c) + a^3*d*e^2 + (2*a^3*d*e^2*cos(d*x + c) + a^3*d*e^2)*sin (d*x + c))*(-a^6/(d^4*e^6))^(1/4))/(cos(d*x + c) + sin(d*x + c) + 1)) - d* e^2*(-a^6/(d^4*e^6))^(1/4)*cos(d*x + c)*log((2*(a^4*sin(d*x + c) + (a*d^2* e^3*cos(d*x + c) + a*d^2*e^3)*sqrt(-a^6/(d^4*e^6)))*sqrt(e*cos(d*x + c))*s qrt(a*sin(d*x + c) + a) - (2*d^3*e^5*cos(d*x + c)^2 + d^3*e^5*cos(d*x + c) - d^3*e^5*sin(d*x + c) - d^3*e^5)*(-a^6/(d^4*e^6))^(3/4) - (a^3*d*e^2*cos (d*x + c) + a^3*d*e^2 + (2*a^3*d*e^2*cos(d*x + c) + a^3*d*e^2)*sin(d*x + c ))*(-a^6/(d^4*e^6))^(1/4))/(cos(d*x + c) + sin(d*x + c) + 1)) - I*d*e^2*(- a^6/(d^4*e^6))^(1/4)*cos(d*x + c)*log((2*(a^4*sin(d*x + c) - (a*d^2*e^3*co s(d*x + c) + a*d^2*e^3)*sqrt(-a^6/(d^4*e^6)))*sqrt(e*cos(d*x + c))*sqrt(a* sin(d*x + c) + a) + (2*I*d^3*e^5*cos(d*x + c)^2 + I*d^3*e^5*cos(d*x + c) - I*d^3*e^5*sin(d*x + c) - I*d^3*e^5)*(-a^6/(d^4*e^6))^(3/4) + (-I*a^3*d*e^ 2*cos(d*x + c) - I*a^3*d*e^2 + (-2*I*a^3*d*e^2*cos(d*x + c) - I*a^3*d*e^2) *sin(d*x + c))*(-a^6/(d^4*e^6))^(1/4))/(cos(d*x + c) + sin(d*x + c) + 1)) + I*d*e^2*(-a^6/(d^4*e^6))^(1/4)*cos(d*x + c)*log((2*(a^4*sin(d*x + c) - ( a*d^2*e^3*cos(d*x + c) + a*d^2*e^3)*sqrt(-a^6/(d^4*e^6)))*sqrt(e*cos(d*...
\[ \int \frac {(a+a \sin (c+d x))^{3/2}}{(e \cos (c+d x))^{3/2}} \, dx=\int \frac {\left (a \left (\sin {\left (c + d x \right )} + 1\right )\right )^{\frac {3}{2}}}{\left (e \cos {\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx \]
\[ \int \frac {(a+a \sin (c+d x))^{3/2}}{(e \cos (c+d x))^{3/2}} \, dx=\int { \frac {{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}}}{\left (e \cos \left (d x + c\right )\right )^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {(a+a \sin (c+d x))^{3/2}}{(e \cos (c+d x))^{3/2}} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {(a+a \sin (c+d x))^{3/2}}{(e \cos (c+d x))^{3/2}} \, dx=\int \frac {{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{3/2}}{{\left (e\,\cos \left (c+d\,x\right )\right )}^{3/2}} \,d x \]