Integrand size = 25, antiderivative size = 222 \[ \int \frac {(a+a \sec (c+d x))^2}{(e \csc (c+d x))^{3/2}} \, dx=-\frac {4 a^2}{d e \sqrt {e \csc (c+d x)}}-\frac {2 a^2 \cos (c+d x)}{3 d e \sqrt {e \csc (c+d x)}}+\frac {a^2 \sec (c+d x)}{d e \sqrt {e \csc (c+d x)}}+\frac {2 a^2 \arctan \left (\sqrt {\sin (c+d x)}\right )}{d e \sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}}+\frac {2 a^2 \text {arctanh}\left (\sqrt {\sin (c+d x)}\right )}{d e \sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}}-\frac {a^2 \operatorname {EllipticF}\left (\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),2\right )}{3 d e \sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}} \]
-4*a^2/d/e/(e*csc(d*x+c))^(1/2)-2/3*a^2*cos(d*x+c)/d/e/(e*csc(d*x+c))^(1/2 )+a^2*sec(d*x+c)/d/e/(e*csc(d*x+c))^(1/2)+2*a^2*arctan(sin(d*x+c)^(1/2))/d /e/(e*csc(d*x+c))^(1/2)/sin(d*x+c)^(1/2)+2*a^2*arctanh(sin(d*x+c)^(1/2))/d /e/(e*csc(d*x+c))^(1/2)/sin(d*x+c)^(1/2)+1/3*a^2*(sin(1/2*c+1/4*Pi+1/2*d*x )^2)^(1/2)/sin(1/2*c+1/4*Pi+1/2*d*x)*EllipticF(cos(1/2*c+1/4*Pi+1/2*d*x),2 ^(1/2))/d/e/(e*csc(d*x+c))^(1/2)/sin(d*x+c)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 15.71 (sec) , antiderivative size = 164, normalized size of antiderivative = 0.74 \[ \int \frac {(a+a \sec (c+d x))^2}{(e \csc (c+d x))^{3/2}} \, dx=\frac {2 a^2 \cos ^4\left (\frac {1}{2} (c+d x)\right ) \sqrt {e \csc (c+d x)} \sec ^4\left (\frac {1}{2} \csc ^{-1}(\csc (c+d x))\right ) \left (3-6 \sqrt {\cos ^2(c+d x)} \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},1,\frac {3}{4},\csc ^2(c+d x)\right )+3 \sqrt {-\cot ^2(c+d x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},\csc ^2(c+d x)\right )+\sqrt {-\cot ^2(c+d x)} \operatorname {Hypergeometric2F1}\left (-\frac {3}{4},\frac {3}{2},\frac {1}{4},\csc ^2(c+d x)\right ) \sin ^2(c+d x)\right ) \tan (c+d x)}{3 d e^2} \]
(2*a^2*Cos[(c + d*x)/2]^4*Sqrt[e*Csc[c + d*x]]*Sec[ArcCsc[Csc[c + d*x]]/2] ^4*(3 - 6*Sqrt[Cos[c + d*x]^2]*Hypergeometric2F1[-1/4, 1, 3/4, Csc[c + d*x ]^2] + 3*Sqrt[-Cot[c + d*x]^2]*Hypergeometric2F1[1/4, 1/2, 5/4, Csc[c + d* x]^2] + Sqrt[-Cot[c + d*x]^2]*Hypergeometric2F1[-3/4, 3/2, 1/4, Csc[c + d* x]^2]*Sin[c + d*x]^2)*Tan[c + d*x])/(3*d*e^2)
Time = 0.62 (sec) , antiderivative size = 158, normalized size of antiderivative = 0.71, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {3042, 4366, 3042, 4360, 3042, 3352, 2009}
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.
| \(\displaystyle \int \frac {(a \sec (c+d x)+a)^2}{(e \csc (c+d x))^{3/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a-a \csc \left (c+d x-\frac {\pi }{2}\right )\right )^2}{\left (e \sec \left (c+d x-\frac {\pi }{2}\right )\right )^{3/2}}dx\) |
| \(\Big \downarrow \) 4366 |
| \(\displaystyle \frac {\int (\sec (c+d x) a+a)^2 \sin ^{\frac {3}{2}}(c+d x)dx}{e \sqrt {\sin (c+d x)} \sqrt {e \csc (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \cos \left (c+d x-\frac {\pi }{2}\right )^{3/2} \left (a-a \csc \left (c+d x-\frac {\pi }{2}\right )\right )^2dx}{e \sqrt {\sin (c+d x)} \sqrt {e \csc (c+d x)}}\) |
| \(\Big \downarrow \) 4360 |
| \(\displaystyle \frac {\int (-\cos (c+d x) a-a)^2 \sec ^2(c+d x) \sin ^{\frac {3}{2}}(c+d x)dx}{e \sqrt {\sin (c+d x)} \sqrt {e \csc (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\left (-\cos \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2} \left (-\sin \left (c+d x+\frac {\pi }{2}\right ) a-a\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right )^2}dx}{e \sqrt {\sin (c+d x)} \sqrt {e \csc (c+d x)}}\) |
| \(\Big \downarrow \) 3352 |
| \(\displaystyle \frac {\int \left (\sec ^2(c+d x) \sin ^{\frac {3}{2}}(c+d x) a^2+2 \sec (c+d x) \sin ^{\frac {3}{2}}(c+d x) a^2+\sin ^{\frac {3}{2}}(c+d x) a^2\right )dx}{e \sqrt {\sin (c+d x)} \sqrt {e \csc (c+d x)}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {2 a^2 \arctan \left (\sqrt {\sin (c+d x)}\right )}{d}+\frac {2 a^2 \text {arctanh}\left (\sqrt {\sin (c+d x)}\right )}{d}-\frac {4 a^2 \sqrt {\sin (c+d x)}}{d}-\frac {a^2 \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{3 d}-\frac {2 a^2 \sqrt {\sin (c+d x)} \cos (c+d x)}{3 d}+\frac {a^2 \sqrt {\sin (c+d x)} \sec (c+d x)}{d}}{e \sqrt {\sin (c+d x)} \sqrt {e \csc (c+d x)}}\) |
((2*a^2*ArcTan[Sqrt[Sin[c + d*x]]])/d + (2*a^2*ArcTanh[Sqrt[Sin[c + d*x]]] )/d - (a^2*EllipticF[(c - Pi/2 + d*x)/2, 2])/(3*d) - (4*a^2*Sqrt[Sin[c + d *x]])/d - (2*a^2*Cos[c + d*x]*Sqrt[Sin[c + d*x]])/(3*d) + (a^2*Sec[c + d*x ]*Sqrt[Sin[c + d*x]])/d)/(e*Sqrt[e*Csc[c + d*x]]*Sqrt[Sin[c + d*x]])
3.3.91.3.1 Defintions of rubi rules used
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n _)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Int[ExpandTrig [(g*cos[e + f*x])^p, (d*sin[e + f*x])^n*(a + b*sin[e + f*x])^m, x], x] /; F reeQ[{a, b, d, e, f, g, n, p}, x] && EqQ[a^2 - b^2, 0] && IGtQ[m, 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 = 12.70 (sec) , antiderivative size = 810, normalized size of antiderivative = 3.65
| method | result | size |
| parts | \(\text {Expression too large to display}\) | \(810\) |
| default | \(\text {Expression too large to display}\) | \(1083\) |
-1/3*a^2/d*2^(1/2)*(I*(-I*(I-cot(d*x+c)+csc(d*x+c)))^(1/2)*(-I*(I+cot(d*x+ c)-csc(d*x+c)))^(1/2)*(-I*(cot(d*x+c)-csc(d*x+c)))^(1/2)*EllipticF((-I*(I- cot(d*x+c)+csc(d*x+c)))^(1/2),1/2*2^(1/2))*cos(d*x+c)+I*(-I*(I-cot(d*x+c)+ csc(d*x+c)))^(1/2)*(-I*(I+cot(d*x+c)-csc(d*x+c)))^(1/2)*(-I*(cot(d*x+c)-cs c(d*x+c)))^(1/2)*EllipticF((-I*(I-cot(d*x+c)+csc(d*x+c)))^(1/2),1/2*2^(1/2 ))-cos(d*x+c)*sin(d*x+c)*2^(1/2))/(e*csc(d*x+c))^(1/2)/(cos(d*x+c)-1)/e/(c os(d*x+c)+1)*sin(d*x+c)+1/2*a^2/d*2^(1/2)*(I*(-I*(I-cot(d*x+c)+csc(d*x+c)) )^(1/2)*(-I*(I+cot(d*x+c)-csc(d*x+c)))^(1/2)*(-I*(cot(d*x+c)-csc(d*x+c)))^ (1/2)*EllipticF((-I*(I-cot(d*x+c)+csc(d*x+c)))^(1/2),1/2*2^(1/2))*cos(d*x+ c)^2+I*(-I*(I-cot(d*x+c)+csc(d*x+c)))^(1/2)*(-I*(I+cot(d*x+c)-csc(d*x+c))) ^(1/2)*(-I*(cot(d*x+c)-csc(d*x+c)))^(1/2)*EllipticF((-I*(I-cot(d*x+c)+csc( d*x+c)))^(1/2),1/2*2^(1/2))*cos(d*x+c)-2^(1/2)*sin(d*x+c))/(e*csc(d*x+c))^ (1/2)/e/(cos(d*x+c)^2-1)*tan(d*x+c)+2*a^2/d*(arctan((sin(d*x+c)/(cos(d*x+c )+1)^2)^(1/2)*(cot(d*x+c)+csc(d*x+c)))*(sin(d*x+c)/(cos(d*x+c)+1)^2)^(1/2) *cos(d*x+c)-(sin(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*arctanh((sin(d*x+c)/(cos(d *x+c)+1)^2)^(1/2)*(cot(d*x+c)+csc(d*x+c)))*cos(d*x+c)+arctan((sin(d*x+c)/( cos(d*x+c)+1)^2)^(1/2)*(cot(d*x+c)+csc(d*x+c)))*(sin(d*x+c)/(cos(d*x+c)+1) ^2)^(1/2)-(sin(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*arctanh((sin(d*x+c)/(cos(d*x +c)+1)^2)^(1/2)*(cot(d*x+c)+csc(d*x+c)))+2*sin(d*x+c))/(e*csc(d*x+c))^(1/2 )/(cos(d*x+c)-1)/e/(cos(d*x+c)+1)*sin(d*x+c)
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.43 (sec) , antiderivative size = 724, normalized size of antiderivative = 3.26 \[ \int \frac {(a+a \sec (c+d x))^2}{(e \csc (c+d x))^{3/2}} \, dx=\left [-\frac {6 \, a^{2} \sqrt {-e} \arctan \left (-\frac {{\left (\cos \left (d x + c\right )^{2} - 6 \, \sin \left (d x + c\right ) - 2\right )} \sqrt {-e} \sqrt {\frac {e}{\sin \left (d x + c\right )}}}{4 \, {\left (e \sin \left (d x + c\right ) + e\right )}}\right ) \cos \left (d x + c\right ) + 3 \, a^{2} \sqrt {-e} \cos \left (d x + c\right ) \log \left (\frac {e \cos \left (d x + c\right )^{4} - 72 \, e \cos \left (d x + c\right )^{2} - 8 \, {\left (\cos \left (d x + c\right )^{4} - 9 \, \cos \left (d x + c\right )^{2} + {\left (7 \, \cos \left (d x + c\right )^{2} - 8\right )} \sin \left (d x + c\right ) + 8\right )} \sqrt {-e} \sqrt {\frac {e}{\sin \left (d x + c\right )}} + 28 \, {\left (e \cos \left (d x + c\right )^{2} - 2 \, e\right )} \sin \left (d x + c\right ) + 72 \, e}{\cos \left (d x + c\right )^{4} - 8 \, \cos \left (d x + c\right )^{2} - 4 \, {\left (\cos \left (d x + c\right )^{2} - 2\right )} \sin \left (d x + c\right ) + 8}\right ) - 2 i \, a^{2} \sqrt {2 i \, e} \cos \left (d x + c\right ) {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 2 i \, a^{2} \sqrt {-2 i \, e} \cos \left (d x + c\right ) {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 4 \, {\left (2 \, a^{2} \cos \left (d x + c\right )^{2} + 12 \, a^{2} \cos \left (d x + c\right ) - 3 \, a^{2}\right )} \sqrt {\frac {e}{\sin \left (d x + c\right )}} \sin \left (d x + c\right )}{12 \, d e^{2} \cos \left (d x + c\right )}, -\frac {6 \, a^{2} \sqrt {e} \arctan \left (\frac {{\left (\cos \left (d x + c\right )^{2} + 6 \, \sin \left (d x + c\right ) - 2\right )} \sqrt {e} \sqrt {\frac {e}{\sin \left (d x + c\right )}}}{4 \, {\left (e \sin \left (d x + c\right ) - e\right )}}\right ) \cos \left (d x + c\right ) - 3 \, a^{2} \sqrt {e} \cos \left (d x + c\right ) \log \left (\frac {e \cos \left (d x + c\right )^{4} - 72 \, e \cos \left (d x + c\right )^{2} + 8 \, {\left (\cos \left (d x + c\right )^{4} - 9 \, \cos \left (d x + c\right )^{2} - {\left (7 \, \cos \left (d x + c\right )^{2} - 8\right )} \sin \left (d x + c\right ) + 8\right )} \sqrt {e} \sqrt {\frac {e}{\sin \left (d x + c\right )}} - 28 \, {\left (e \cos \left (d x + c\right )^{2} - 2 \, e\right )} \sin \left (d x + c\right ) + 72 \, e}{\cos \left (d x + c\right )^{4} - 8 \, \cos \left (d x + c\right )^{2} + 4 \, {\left (\cos \left (d x + c\right )^{2} - 2\right )} \sin \left (d x + c\right ) + 8}\right ) - 2 i \, a^{2} \sqrt {2 i \, e} \cos \left (d x + c\right ) {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 2 i \, a^{2} \sqrt {-2 i \, e} \cos \left (d x + c\right ) {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 4 \, {\left (2 \, a^{2} \cos \left (d x + c\right )^{2} + 12 \, a^{2} \cos \left (d x + c\right ) - 3 \, a^{2}\right )} \sqrt {\frac {e}{\sin \left (d x + c\right )}} \sin \left (d x + c\right )}{12 \, d e^{2} \cos \left (d x + c\right )}\right ] \]
[-1/12*(6*a^2*sqrt(-e)*arctan(-1/4*(cos(d*x + c)^2 - 6*sin(d*x + c) - 2)*s qrt(-e)*sqrt(e/sin(d*x + c))/(e*sin(d*x + c) + e))*cos(d*x + c) + 3*a^2*sq rt(-e)*cos(d*x + c)*log((e*cos(d*x + c)^4 - 72*e*cos(d*x + c)^2 - 8*(cos(d *x + c)^4 - 9*cos(d*x + c)^2 + (7*cos(d*x + c)^2 - 8)*sin(d*x + c) + 8)*sq rt(-e)*sqrt(e/sin(d*x + c)) + 28*(e*cos(d*x + c)^2 - 2*e)*sin(d*x + c) + 7 2*e)/(cos(d*x + c)^4 - 8*cos(d*x + c)^2 - 4*(cos(d*x + c)^2 - 2)*sin(d*x + c) + 8)) - 2*I*a^2*sqrt(2*I*e)*cos(d*x + c)*weierstrassPInverse(4, 0, cos (d*x + c) + I*sin(d*x + c)) + 2*I*a^2*sqrt(-2*I*e)*cos(d*x + c)*weierstras sPInverse(4, 0, cos(d*x + c) - I*sin(d*x + c)) + 4*(2*a^2*cos(d*x + c)^2 + 12*a^2*cos(d*x + c) - 3*a^2)*sqrt(e/sin(d*x + c))*sin(d*x + c))/(d*e^2*co s(d*x + c)), -1/12*(6*a^2*sqrt(e)*arctan(1/4*(cos(d*x + c)^2 + 6*sin(d*x + c) - 2)*sqrt(e)*sqrt(e/sin(d*x + c))/(e*sin(d*x + c) - e))*cos(d*x + c) - 3*a^2*sqrt(e)*cos(d*x + c)*log((e*cos(d*x + c)^4 - 72*e*cos(d*x + c)^2 + 8*(cos(d*x + c)^4 - 9*cos(d*x + c)^2 - (7*cos(d*x + c)^2 - 8)*sin(d*x + c) + 8)*sqrt(e)*sqrt(e/sin(d*x + c)) - 28*(e*cos(d*x + c)^2 - 2*e)*sin(d*x + c) + 72*e)/(cos(d*x + c)^4 - 8*cos(d*x + c)^2 + 4*(cos(d*x + c)^2 - 2)*si n(d*x + c) + 8)) - 2*I*a^2*sqrt(2*I*e)*cos(d*x + c)*weierstrassPInverse(4, 0, cos(d*x + c) + I*sin(d*x + c)) + 2*I*a^2*sqrt(-2*I*e)*cos(d*x + c)*wei erstrassPInverse(4, 0, cos(d*x + c) - I*sin(d*x + c)) + 4*(2*a^2*cos(d*x + c)^2 + 12*a^2*cos(d*x + c) - 3*a^2)*sqrt(e/sin(d*x + c))*sin(d*x + c))...
\[ \int \frac {(a+a \sec (c+d x))^2}{(e \csc (c+d x))^{3/2}} \, dx=a^{2} \left (\int \frac {1}{\left (e \csc {\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx + \int \frac {2 \sec {\left (c + d x \right )}}{\left (e \csc {\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx + \int \frac {\sec ^{2}{\left (c + d x \right )}}{\left (e \csc {\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx\right ) \]
a**2*(Integral((e*csc(c + d*x))**(-3/2), x) + Integral(2*sec(c + d*x)/(e*c sc(c + d*x))**(3/2), x) + Integral(sec(c + d*x)**2/(e*csc(c + d*x))**(3/2) , x))
\[ \int \frac {(a+a \sec (c+d x))^2}{(e \csc (c+d x))^{3/2}} \, dx=\int { \frac {{\left (a \sec \left (d x + c\right ) + a\right )}^{2}}{\left (e \csc \left (d x + c\right )\right )^{\frac {3}{2}}} \,d x } \]
\[ \int \frac {(a+a \sec (c+d x))^2}{(e \csc (c+d x))^{3/2}} \, dx=\int { \frac {{\left (a \sec \left (d x + c\right ) + a\right )}^{2}}{\left (e \csc \left (d x + c\right )\right )^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {(a+a \sec (c+d x))^2}{(e \csc (c+d x))^{3/2}} \, dx=\int \frac {{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^2}{{\left (\frac {e}{\sin \left (c+d\,x\right )}\right )}^{3/2}} \,d x \]