Integrand size = 25, antiderivative size = 236 \[ \int \frac {(a+a \sec (c+d x))^2}{(e \csc (c+d x))^{5/2}} \, dx=-\frac {2 a^2 \arctan \left (\sqrt {\sin (c+d x)}\right )}{d e^2 \sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}}+\frac {2 a^2 \text {arctanh}\left (\sqrt {\sin (c+d x)}\right )}{d e^2 \sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}}-\frac {9 a^2 E\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right )}{5 d e^2 \sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}}-\frac {4 a^2 \sin (c+d x)}{3 d e^2 \sqrt {e \csc (c+d x)}}-\frac {2 a^2 \cos (c+d x) \sin (c+d x)}{5 d e^2 \sqrt {e \csc (c+d x)}}+\frac {a^2 \tan (c+d x)}{d e^2 \sqrt {e \csc (c+d x)}} \]
-4/3*a^2*sin(d*x+c)/d/e^2/(e*csc(d*x+c))^(1/2)-2/5*a^2*cos(d*x+c)*sin(d*x+ c)/d/e^2/(e*csc(d*x+c))^(1/2)-2*a^2*arctan(sin(d*x+c)^(1/2))/d/e^2/(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^2/(e*cs c(d*x+c))^(1/2)/sin(d*x+c)^(1/2)+9/5*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)*EllipticE(cos(1/2*c+1/4*Pi+1/2*d*x),2^(1/2))/ d/e^2/(e*csc(d*x+c))^(1/2)/sin(d*x+c)^(1/2)+a^2*tan(d*x+c)/d/e^2/(e*csc(d* x+c))^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 15.03 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.64 \[ \int \frac {(a+a \sec (c+d x))^2}{(e \csc (c+d x))^{5/2}} \, dx=\frac {2 a^2 \cos ^4\left (\frac {1}{2} (c+d x)\right ) \sec ^4\left (\frac {1}{2} \csc ^{-1}(\csc (c+d x))\right ) \left (-10 \sqrt {\cos ^2(c+d x)} \operatorname {Hypergeometric2F1}\left (-\frac {3}{4},1,\frac {1}{4},\csc ^2(c+d x)\right )+3 \sqrt {-\cot ^2(c+d x)} \left (-10 \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {3}{2},\frac {3}{4},\csc ^2(c+d x)\right )+\operatorname {Hypergeometric2F1}\left (-\frac {5}{4},\frac {3}{2},-\frac {1}{4},\csc ^2(c+d x)\right ) \sin ^2(c+d x)\right )\right ) \tan (c+d x)}{15 d e^2 \sqrt {e \csc (c+d x)}} \]
(2*a^2*Cos[(c + d*x)/2]^4*Sec[ArcCsc[Csc[c + d*x]]/2]^4*(-10*Sqrt[Cos[c + d*x]^2]*Hypergeometric2F1[-3/4, 1, 1/4, Csc[c + d*x]^2] + 3*Sqrt[-Cot[c + d*x]^2]*(-10*Hypergeometric2F1[-1/4, 3/2, 3/4, Csc[c + d*x]^2] + Hypergeom etric2F1[-5/4, 3/2, -1/4, Csc[c + d*x]^2]*Sin[c + d*x]^2))*Tan[c + d*x])/( 15*d*e^2*Sqrt[e*Csc[c + d*x]])
Time = 0.62 (sec) , antiderivative size = 160, normalized size of antiderivative = 0.68, 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.
\(\displaystyle \int \frac {(a \sec (c+d x)+a)^2}{(e \csc (c+d x))^{5/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\left (a-a \csc \left (c+d x-\frac {\pi }{2}\right )\right )^2}{\left (e \sec \left (c+d x-\frac {\pi }{2}\right )\right )^{5/2}}dx\) |
\(\Big \downarrow \) 4366 |
\(\displaystyle \frac {\int (\sec (c+d x) a+a)^2 \sin ^{\frac {5}{2}}(c+d x)dx}{e^2 \sqrt {\sin (c+d x)} \sqrt {e \csc (c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \cos \left (c+d x-\frac {\pi }{2}\right )^{5/2} \left (a-a \csc \left (c+d x-\frac {\pi }{2}\right )\right )^2dx}{e^2 \sqrt {\sin (c+d x)} \sqrt {e \csc (c+d x)}}\) |
\(\Big \downarrow \) 4360 |
\(\displaystyle \frac {\int (-\cos (c+d x) a-a)^2 \sec ^2(c+d x) \sin ^{\frac {5}{2}}(c+d x)dx}{e^2 \sqrt {\sin (c+d x)} \sqrt {e \csc (c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {\left (-\cos \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2} \left (-\sin \left (c+d x+\frac {\pi }{2}\right ) a-a\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right )^2}dx}{e^2 \sqrt {\sin (c+d x)} \sqrt {e \csc (c+d x)}}\) |
\(\Big \downarrow \) 3352 |
\(\displaystyle \frac {\int \left (a^2 \sin ^{\frac {5}{2}}(c+d x)+a^2 \sec ^2(c+d x) \sin ^{\frac {5}{2}}(c+d x)+2 a^2 \sec (c+d x) \sin ^{\frac {5}{2}}(c+d x)\right )dx}{e^2 \sqrt {\sin (c+d x)} \sqrt {e \csc (c+d x)}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {-\frac {2 a^2 \arctan \left (\sqrt {\sin (c+d x)}\right )}{d}+\frac {2 a^2 \text {arctanh}\left (\sqrt {\sin (c+d x)}\right )}{d}-\frac {4 a^2 \sin ^{\frac {3}{2}}(c+d x)}{3 d}-\frac {9 a^2 E\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right )}{5 d}-\frac {2 a^2 \sin ^{\frac {3}{2}}(c+d x) \cos (c+d x)}{5 d}+\frac {a^2 \sin ^{\frac {3}{2}}(c+d x) \sec (c+d x)}{d}}{e^2 \sqrt {\sin (c+d x)} \sqrt {e \csc (c+d x)}}\) |
((-2*a^2*ArcTan[Sqrt[Sin[c + d*x]]])/d + (2*a^2*ArcTanh[Sqrt[Sin[c + d*x]] ])/d - (9*a^2*EllipticE[(c - Pi/2 + d*x)/2, 2])/(5*d) - (4*a^2*Sin[c + d*x ]^(3/2))/(3*d) - (2*a^2*Cos[c + d*x]*Sin[c + d*x]^(3/2))/(5*d) + (a^2*Sec[ c + d*x]*Sin[c + d*x]^(3/2))/d)/(e^2*Sqrt[e*Csc[c + d*x]]*Sqrt[Sin[c + d*x ]])
3.3.92.3.1 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 = 13.64 (sec) , antiderivative size = 1069, normalized size of antiderivative = 4.53
method | result | size |
parts | \(\text {Expression too large to display}\) | \(1069\) |
default | \(\text {Expression too large to display}\) | \(1333\) |
1/5*a^2/d*2^(1/2)*(-6*(-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)*EllipticE((-I*(I- cot(d*x+c)+csc(d*x+c)))^(1/2),1/2*2^(1/2))*cos(d*x+c)+3*(-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)+cos(d*x+c)^3*2^(1/2)-6*(-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)*El lipticE((-I*(I-cot(d*x+c)+csc(d*x+c)))^(1/2),1/2*2^(1/2))+3*(-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))-4*2^(1/2)*cos(d*x+c)+3*2^(1/2))/(e*csc(d*x+c))^(1/2)/e^2*csc(d*x+c) +1/2*a^2/d*2^(1/2)/(e*csc(d*x+c))^(1/2)/e^2*(6*(-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)*EllipticE((-I*(I-cot(d*x+c)+csc(d*x+c)))^(1/2),1/2*2^(1/2))*cot(d* x+c)-3*(-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))*cot(d*x+c)+6*(-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 )*EllipticE((-I*(I-cot(d*x+c)+csc(d*x+c)))^(1/2),1/2*2^(1/2))*csc(d*x+c)-3 *(-I*(I-cot(d*x+c)+csc(d*x+c)))^(1/2)*(-I*(I+cot(d*x+c)-csc(d*x+c)))^(1...
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.46 (sec) , antiderivative size = 776, normalized size of antiderivative = 3.29 \[ \int \frac {(a+a \sec (c+d x))^2}{(e \csc (c+d x))^{5/2}} \, dx=\text {Too large to display} \]
[-1/60*(30*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) + 15*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)*sin(d*x + c) + 8)) + 54*a^2*sqrt(2*I*e)*cos(d*x + c)*weierstrassZeta(4, 0, weiers trassPInverse(4, 0, cos(d*x + c) + I*sin(d*x + c))) + 54*a^2*sqrt(-2*I*e)* cos(d*x + c)*weierstrassZeta(4, 0, weierstrassPInverse(4, 0, cos(d*x + c) - I*sin(d*x + c))) - 4*(6*a^2*cos(d*x + c)^4 + 20*a^2*cos(d*x + c)^3 - 21* a^2*cos(d*x + c)^2 - 20*a^2*cos(d*x + c) + 15*a^2)*sqrt(e/sin(d*x + c)))/( d*e^3*cos(d*x + c)), 1/60*(30*a^2*sqrt(e)*arctan(1/4*(cos(d*x + c)^2 + 6*s in(d*x + c) - 2)*sqrt(e)*sqrt(e/sin(d*x + c))/(e*sin(d*x + c) - e))*cos(d* x + c) + 15*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)*sin(d*x + c) + 8)) - 54*a^2*sqrt(2*I*e)*cos(d*x + c)*weierstrassZet a(4, 0, weierstrassPInverse(4, 0, cos(d*x + c) + I*sin(d*x + c))) - 54*a^2 *sqrt(-2*I*e)*cos(d*x + c)*weierstrassZeta(4, 0, weierstrassPInverse(4,...
Timed out. \[ \int \frac {(a+a \sec (c+d x))^2}{(e \csc (c+d x))^{5/2}} \, dx=\text {Timed out} \]
\[ \int \frac {(a+a \sec (c+d x))^2}{(e \csc (c+d x))^{5/2}} \, dx=\int { \frac {{\left (a \sec \left (d x + c\right ) + a\right )}^{2}}{\left (e \csc \left (d x + c\right )\right )^{\frac {5}{2}}} \,d x } \]
\[ \int \frac {(a+a \sec (c+d x))^2}{(e \csc (c+d x))^{5/2}} \, dx=\int { \frac {{\left (a \sec \left (d x + c\right ) + a\right )}^{2}}{\left (e \csc \left (d x + c\right )\right )^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {(a+a \sec (c+d x))^2}{(e \csc (c+d x))^{5/2}} \, dx=\int \frac {{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^2}{{\left (\frac {e}{\sin \left (c+d\,x\right )}\right )}^{5/2}} \,d x \]