Integrand size = 25, antiderivative size = 145 \[ \int \frac {(e \csc (c+d x))^{3/2}}{a+a \sec (c+d x)} \, dx=-\frac {4 e \cos (c+d x) \sqrt {e \csc (c+d x)}}{5 a d}+\frac {2 e \cot (c+d x) \csc (c+d x) \sqrt {e \csc (c+d x)}}{5 a d}-\frac {2 e \csc ^2(c+d x) \sqrt {e \csc (c+d x)}}{5 a d}-\frac {4 e \sqrt {e \csc (c+d x)} E\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {\sin (c+d x)}}{5 a d} \] Output:
-4/5*e*cos(d*x+c)*(e*csc(d*x+c))^(1/2)/a/d+2/5*e*cot(d*x+c)*csc(d*x+c)*(e* csc(d*x+c))^(1/2)/a/d-2/5*e*csc(d*x+c)^2*(e*csc(d*x+c))^(1/2)/a/d+4/5*e*(e *csc(d*x+c))^(1/2)*EllipticE(cos(1/2*c+1/4*Pi+1/2*d*x),2^(1/2))*sin(d*x+c) ^(1/2)/a/d
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 1.56 (sec) , antiderivative size = 230, normalized size of antiderivative = 1.59 \[ \int \frac {(e \csc (c+d x))^{3/2}}{a+a \sec (c+d x)} \, dx=\frac {\cos ^2\left (\frac {1}{2} (c+d x)\right ) (e \csc (c+d x))^{3/2} \left (\frac {8 \sqrt {2} e^{i (c-d x)} \sqrt {\frac {i e^{i (c+d x)}}{-1+e^{2 i (c+d x)}}} \left (3-3 e^{2 i (c+d x)}+e^{2 i d x} \left (1+e^{2 i c}\right ) \sqrt {1-e^{2 i (c+d x)}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},e^{2 i (c+d x)}\right )\right ) \sec (c+d x)}{d \left (1+e^{2 i c}\right ) \csc ^{\frac {3}{2}}(c+d x)}-\frac {6 \left (4 \cos (d x) \sec (c)+\sec ^2\left (\frac {1}{2} (c+d x)\right )\right ) \tan (c+d x)}{d}\right )}{15 a (1+\sec (c+d x))} \] Input:
Integrate[(e*Csc[c + d*x])^(3/2)/(a + a*Sec[c + d*x]),x]
Output:
(Cos[(c + d*x)/2]^2*(e*Csc[c + d*x])^(3/2)*((8*Sqrt[2]*E^(I*(c - d*x))*Sqr t[(I*E^(I*(c + d*x)))/(-1 + E^((2*I)*(c + d*x)))]*(3 - 3*E^((2*I)*(c + d*x )) + E^((2*I)*d*x)*(1 + E^((2*I)*c))*Sqrt[1 - E^((2*I)*(c + d*x))]*Hyperge ometric2F1[1/2, 3/4, 7/4, E^((2*I)*(c + d*x))])*Sec[c + d*x])/(d*(1 + E^(( 2*I)*c))*Csc[c + d*x]^(3/2)) - (6*(4*Cos[d*x]*Sec[c] + Sec[(c + d*x)/2]^2) *Tan[c + d*x])/d))/(15*a*(1 + Sec[c + d*x]))
Time = 0.81 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.83, number of steps used = 19, number of rules used = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.720, Rules used = {3042, 4366, 3042, 4360, 25, 25, 3042, 25, 3318, 25, 3042, 3044, 15, 3047, 3042, 3116, 3042, 3119}
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.
\(\displaystyle \int \frac {(e \csc (c+d x))^{3/2}}{a \sec (c+d x)+a} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\left (e \sec \left (c+d x-\frac {\pi }{2}\right )\right )^{3/2}}{a-a \csc \left (c+d x-\frac {\pi }{2}\right )}dx\) |
\(\Big \downarrow \) 4366 |
\(\displaystyle e \sqrt {\sin (c+d x)} \sqrt {e \csc (c+d x)} \int \frac {1}{(\sec (c+d x) a+a) \sin ^{\frac {3}{2}}(c+d x)}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle e \sqrt {\sin (c+d x)} \sqrt {e \csc (c+d x)} \int \frac {1}{\cos \left (c+d x-\frac {\pi }{2}\right )^{3/2} \left (a-a \csc \left (c+d x-\frac {\pi }{2}\right )\right )}dx\) |
\(\Big \downarrow \) 4360 |
\(\displaystyle e \sqrt {\sin (c+d x)} \sqrt {e \csc (c+d x)} \int -\frac {\cos (c+d x)}{(-\cos (c+d x) a-a) \sin ^{\frac {3}{2}}(c+d x)}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -e \sqrt {\sin (c+d x)} \sqrt {e \csc (c+d x)} \int -\frac {\cos (c+d x)}{(\cos (c+d x) a+a) \sin ^{\frac {3}{2}}(c+d x)}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle e \sqrt {\sin (c+d x)} \sqrt {e \csc (c+d x)} \int \frac {\cos (c+d x)}{(\cos (c+d x) a+a) \sin ^{\frac {3}{2}}(c+d x)}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle e \sqrt {\sin (c+d x)} \sqrt {e \csc (c+d x)} \int -\frac {\sin \left (c+d x-\frac {\pi }{2}\right )}{\cos \left (c+d x-\frac {\pi }{2}\right )^{3/2} \left (a-a \sin \left (c+d x-\frac {\pi }{2}\right )\right )}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -e \sqrt {\sin (c+d x)} \sqrt {e \csc (c+d x)} \int \frac {\sin \left (\frac {1}{2} (2 c-\pi )+d x\right )}{\cos \left (\frac {1}{2} (2 c-\pi )+d x\right )^{3/2} \left (a-a \sin \left (\frac {1}{2} (2 c-\pi )+d x\right )\right )}dx\) |
\(\Big \downarrow \) 3318 |
\(\displaystyle -e \sqrt {\sin (c+d x)} \sqrt {e \csc (c+d x)} \left (\frac {\int \frac {\cos ^2(c+d x)}{\sin ^{\frac {7}{2}}(c+d x)}dx}{a}+\frac {\int -\frac {\cos (c+d x)}{\sin ^{\frac {7}{2}}(c+d x)}dx}{a}\right )\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -e \sqrt {\sin (c+d x)} \sqrt {e \csc (c+d x)} \left (\frac {\int \frac {\cos ^2(c+d x)}{\sin ^{\frac {7}{2}}(c+d x)}dx}{a}-\frac {\int \frac {\cos (c+d x)}{\sin ^{\frac {7}{2}}(c+d x)}dx}{a}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -e \sqrt {\sin (c+d x)} \sqrt {e \csc (c+d x)} \left (\frac {\int \frac {\cos (c+d x)^2}{\sin (c+d x)^{7/2}}dx}{a}-\frac {\int \frac {\cos (c+d x)}{\sin (c+d x)^{7/2}}dx}{a}\right )\) |
\(\Big \downarrow \) 3044 |
\(\displaystyle -e \sqrt {\sin (c+d x)} \sqrt {e \csc (c+d x)} \left (\frac {\int \frac {\cos (c+d x)^2}{\sin (c+d x)^{7/2}}dx}{a}-\frac {\int \frac {1}{\sin ^{\frac {7}{2}}(c+d x)}d\sin (c+d x)}{a d}\right )\) |
\(\Big \downarrow \) 15 |
\(\displaystyle -e \sqrt {\sin (c+d x)} \sqrt {e \csc (c+d x)} \left (\frac {\int \frac {\cos (c+d x)^2}{\sin (c+d x)^{7/2}}dx}{a}+\frac {2}{5 a d \sin ^{\frac {5}{2}}(c+d x)}\right )\) |
\(\Big \downarrow \) 3047 |
\(\displaystyle -e \sqrt {\sin (c+d x)} \sqrt {e \csc (c+d x)} \left (\frac {-\frac {2}{5} \int \frac {1}{\sin ^{\frac {3}{2}}(c+d x)}dx-\frac {2 \cos (c+d x)}{5 d \sin ^{\frac {5}{2}}(c+d x)}}{a}+\frac {2}{5 a d \sin ^{\frac {5}{2}}(c+d x)}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -e \sqrt {\sin (c+d x)} \sqrt {e \csc (c+d x)} \left (\frac {-\frac {2}{5} \int \frac {1}{\sin (c+d x)^{3/2}}dx-\frac {2 \cos (c+d x)}{5 d \sin ^{\frac {5}{2}}(c+d x)}}{a}+\frac {2}{5 a d \sin ^{\frac {5}{2}}(c+d x)}\right )\) |
\(\Big \downarrow \) 3116 |
\(\displaystyle -e \sqrt {\sin (c+d x)} \sqrt {e \csc (c+d x)} \left (\frac {-\frac {2}{5} \left (-\int \sqrt {\sin (c+d x)}dx-\frac {2 \cos (c+d x)}{d \sqrt {\sin (c+d x)}}\right )-\frac {2 \cos (c+d x)}{5 d \sin ^{\frac {5}{2}}(c+d x)}}{a}+\frac {2}{5 a d \sin ^{\frac {5}{2}}(c+d x)}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -e \sqrt {\sin (c+d x)} \sqrt {e \csc (c+d x)} \left (\frac {-\frac {2}{5} \left (-\int \sqrt {\sin (c+d x)}dx-\frac {2 \cos (c+d x)}{d \sqrt {\sin (c+d x)}}\right )-\frac {2 \cos (c+d x)}{5 d \sin ^{\frac {5}{2}}(c+d x)}}{a}+\frac {2}{5 a d \sin ^{\frac {5}{2}}(c+d x)}\right )\) |
\(\Big \downarrow \) 3119 |
\(\displaystyle -e \sqrt {\sin (c+d x)} \sqrt {e \csc (c+d x)} \left (\frac {2}{5 a d \sin ^{\frac {5}{2}}(c+d x)}+\frac {-\frac {2 \cos (c+d x)}{5 d \sin ^{\frac {5}{2}}(c+d x)}-\frac {2}{5} \left (-\frac {2 E\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right )}{d}-\frac {2 \cos (c+d x)}{d \sqrt {\sin (c+d x)}}\right )}{a}\right )\) |
Input:
Int[(e*Csc[c + d*x])^(3/2)/(a + a*Sec[c + d*x]),x]
Output:
-(e*Sqrt[e*Csc[c + d*x]]*(((-2*((-2*EllipticE[(c - Pi/2 + d*x)/2, 2])/d - (2*Cos[c + d*x])/(d*Sqrt[Sin[c + d*x]])))/5 - (2*Cos[c + d*x])/(5*d*Sin[c + d*x]^(5/2)))/a + 2/(5*a*d*Sin[c + d*x]^(5/2)))*Sqrt[Sin[c + d*x]])
Int[(a_.)*(x_)^(m_.), x_Symbol] :> Simp[a*(x^(m + 1)/(m + 1)), x] /; FreeQ[ {a, m}, x] && NeQ[m, -1]
Int[cos[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_ Symbol] :> Simp[1/(a*f) Subst[Int[x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a *Sin[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2] && !(I ntegerQ[(m - 1)/2] && LtQ[0, m, n])
Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n _), x_Symbol] :> Simp[a*(a*Cos[e + f*x])^(m - 1)*((b*Sin[e + f*x])^(n + 1)/ (b*f*(n + 1))), x] + Simp[a^2*((m - 1)/(b^2*(n + 1))) Int[(a*Cos[e + f*x] )^(m - 2)*(b*Sin[e + f*x])^(n + 2), x], x] /; FreeQ[{a, b, e, f}, x] && GtQ [m, 1] && LtQ[n, -1] && (IntegersQ[2*m, 2*n] || EqQ[m + n, 0])
Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]*(( b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1))), x] + Simp[(n + 2)/(b^2*(n + 1)) I nt[(b*Sin[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1] && IntegerQ[2*n]
Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)* (c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
Int[((cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^( n_.))/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[g^2/a Int [(g*Cos[e + f*x])^(p - 2)*(d*Sin[e + f*x])^n, x], x] - Simp[g^2/(b*d) Int [(g*Cos[e + f*x])^(p - 2)*(d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, g, n, p}, x] && EqQ[a^2 - b^2, 0]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[(g*Cos[e + f*x])^p*((b + a*Sin[e + f*x])^m/Si n[e + f*x]^m), x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.)*((g_.)*sec[(e_.) + (f_.)*( x_)])^(p_), x_Symbol] :> Simp[g^IntPart[p]*(g*Sec[e + f*x])^FracPart[p]*Cos [e + f*x]^FracPart[p] Int[(a + b*Csc[e + f*x])^m/Cos[e + f*x]^p, x], x] / ; FreeQ[{a, b, e, f, g, m, p}, x] && !IntegerQ[p]
Result contains complex when optimal does not.
Time = 0.95 (sec) , antiderivative size = 274, normalized size of antiderivative = 1.89
method | result | size |
default | \(-\frac {\sqrt {2}\, \left (\left (-4 \cos \left (d x +c \right )^{2}-8 \cos \left (d x +c \right )-4\right ) \sqrt {1-i \cot \left (d x +c \right )+i \csc \left (d x +c \right )}\, \sqrt {i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}\, \operatorname {EllipticE}\left (\sqrt {1+i \cot \left (d x +c \right )-i \csc \left (d x +c \right )}, \frac {\sqrt {2}}{2}\right ) \sqrt {1+i \cot \left (d x +c \right )-i \csc \left (d x +c \right )}+\left (2 \cos \left (d x +c \right )^{2}+4 \cos \left (d x +c \right )+2\right ) \sqrt {1-i \cot \left (d x +c \right )+i \csc \left (d x +c \right )}\, \sqrt {i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}\, \operatorname {EllipticF}\left (\sqrt {1+i \cot \left (d x +c \right )-i \csc \left (d x +c \right )}, \frac {\sqrt {2}}{2}\right ) \sqrt {1+i \cot \left (d x +c \right )-i \csc \left (d x +c \right )}+\left (2 \cos \left (d x +c \right )+3\right ) \sqrt {2}\right ) \sqrt {e \csc \left (d x +c \right )}\, e}{5 a d \left (1+\cos \left (d x +c \right )\right )}\) | \(274\) |
Input:
int((e*csc(d*x+c))^(3/2)/(a+a*sec(d*x+c)),x,method=_RETURNVERBOSE)
Output:
-1/5/a/d*2^(1/2)*((-4*cos(d*x+c)^2-8*cos(d*x+c)-4)*(1-I*cot(d*x+c)+I*csc(d *x+c))^(1/2)*(I*(csc(d*x+c)-cot(d*x+c)))^(1/2)*EllipticE((1+I*cot(d*x+c)-I *csc(d*x+c))^(1/2),1/2*2^(1/2))*(1+I*cot(d*x+c)-I*csc(d*x+c))^(1/2)+(2*cos (d*x+c)^2+4*cos(d*x+c)+2)*(1-I*cot(d*x+c)+I*csc(d*x+c))^(1/2)*(I*(csc(d*x+ c)-cot(d*x+c)))^(1/2)*EllipticF((1+I*cot(d*x+c)-I*csc(d*x+c))^(1/2),1/2*2^ (1/2))*(1+I*cot(d*x+c)-I*csc(d*x+c))^(1/2)+(2*cos(d*x+c)+3)*2^(1/2))*(e*cs c(d*x+c))^(1/2)*e/(1+cos(d*x+c))
Result contains complex when optimal does not.
Time = 0.08 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.88 \[ \int \frac {(e \csc (c+d x))^{3/2}}{a+a \sec (c+d x)} \, dx=-\frac {2 \, {\left ({\left (e \cos \left (d x + c\right ) + e\right )} \sqrt {2 i \, e} {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + {\left (e \cos \left (d x + c\right ) + e\right )} \sqrt {-2 i \, e} {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) + {\left (2 \, e \cos \left (d x + c\right )^{2} + 2 \, e \cos \left (d x + c\right ) + e\right )} \sqrt {\frac {e}{\sin \left (d x + c\right )}}\right )}}{5 \, {\left (a d \cos \left (d x + c\right ) + a d\right )}} \] Input:
integrate((e*csc(d*x+c))^(3/2)/(a+a*sec(d*x+c)),x, algorithm="fricas")
Output:
-2/5*((e*cos(d*x + c) + e)*sqrt(2*I*e)*weierstrassZeta(4, 0, weierstrassPI nverse(4, 0, cos(d*x + c) + I*sin(d*x + c))) + (e*cos(d*x + c) + e)*sqrt(- 2*I*e)*weierstrassZeta(4, 0, weierstrassPInverse(4, 0, cos(d*x + c) - I*si n(d*x + c))) + (2*e*cos(d*x + c)^2 + 2*e*cos(d*x + c) + e)*sqrt(e/sin(d*x + c)))/(a*d*cos(d*x + c) + a*d)
\[ \int \frac {(e \csc (c+d x))^{3/2}}{a+a \sec (c+d x)} \, dx=\frac {\int \frac {\left (e \csc {\left (c + d x \right )}\right )^{\frac {3}{2}}}{\sec {\left (c + d x \right )} + 1}\, dx}{a} \] Input:
integrate((e*csc(d*x+c))**(3/2)/(a+a*sec(d*x+c)),x)
Output:
Integral((e*csc(c + d*x))**(3/2)/(sec(c + d*x) + 1), x)/a
Timed out. \[ \int \frac {(e \csc (c+d x))^{3/2}}{a+a \sec (c+d x)} \, dx=\text {Timed out} \] Input:
integrate((e*csc(d*x+c))^(3/2)/(a+a*sec(d*x+c)),x, algorithm="maxima")
Output:
Timed out
\[ \int \frac {(e \csc (c+d x))^{3/2}}{a+a \sec (c+d x)} \, dx=\int { \frac {\left (e \csc \left (d x + c\right )\right )^{\frac {3}{2}}}{a \sec \left (d x + c\right ) + a} \,d x } \] Input:
integrate((e*csc(d*x+c))^(3/2)/(a+a*sec(d*x+c)),x, algorithm="giac")
Output:
integrate((e*csc(d*x + c))^(3/2)/(a*sec(d*x + c) + a), x)
Timed out. \[ \int \frac {(e \csc (c+d x))^{3/2}}{a+a \sec (c+d x)} \, dx=\int \frac {\cos \left (c+d\,x\right )\,{\left (\frac {e}{\sin \left (c+d\,x\right )}\right )}^{3/2}}{a\,\left (\cos \left (c+d\,x\right )+1\right )} \,d x \] Input:
int((e/sin(c + d*x))^(3/2)/(a + a/cos(c + d*x)),x)
Output:
int((cos(c + d*x)*(e/sin(c + d*x))^(3/2))/(a*(cos(c + d*x) + 1)), x)
\[ \int \frac {(e \csc (c+d x))^{3/2}}{a+a \sec (c+d x)} \, dx=\frac {\sqrt {e}\, \left (\int \frac {\sqrt {\csc \left (d x +c \right )}\, \csc \left (d x +c \right )}{\sec \left (d x +c \right )+1}d x \right ) e}{a} \] Input:
int((e*csc(d*x+c))^(3/2)/(a+a*sec(d*x+c)),x)
Output:
(sqrt(e)*int((sqrt(csc(c + d*x))*csc(c + d*x))/(sec(c + d*x) + 1),x)*e)/a