Integrand size = 25, antiderivative size = 138 \[ \int (a+a \sec (c+d x))^2 \sqrt {e \sin (c+d x)} \, dx=-\frac {2 a^2 \sqrt {e} \arctan \left (\frac {\sqrt {e \sin (c+d x)}}{\sqrt {e}}\right )}{d}+\frac {2 a^2 \sqrt {e} \text {arctanh}\left (\frac {\sqrt {e \sin (c+d x)}}{\sqrt {e}}\right )}{d}+\frac {a^2 E\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{d \sqrt {\sin (c+d x)}}+\frac {a^2 \sec (c+d x) (e \sin (c+d x))^{3/2}}{d e} \] Output:
-2*a^2*e^(1/2)*arctan((e*sin(d*x+c))^(1/2)/e^(1/2))/d+2*a^2*e^(1/2)*arctan h((e*sin(d*x+c))^(1/2)/e^(1/2))/d-a^2*EllipticE(cos(1/2*c+1/4*Pi+1/2*d*x), 2^(1/2))*(e*sin(d*x+c))^(1/2)/d/sin(d*x+c)^(1/2)+a^2*sec(d*x+c)*(e*sin(d*x +c))^(3/2)/d/e
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 11.53 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.22 \[ \int (a+a \sec (c+d x))^2 \sqrt {e \sin (c+d x)} \, dx=-\frac {2 a^2 \cos ^4\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x) \sec ^4\left (\frac {1}{2} \arcsin (\sin (c+d x))\right ) \sqrt {e \sin (c+d x)} \left (3 \arctan \left (\sqrt {\sin (c+d x)}\right ) \sqrt {\cos ^2(c+d x)}-3 \text {arctanh}\left (\sqrt {\sin (c+d x)}\right ) \sqrt {\cos ^2(c+d x)}-3 \sin ^{\frac {3}{2}}(c+d x)+\sqrt {\cos ^2(c+d x)} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {3}{2},\frac {7}{4},\sin ^2(c+d x)\right ) \sin ^{\frac {3}{2}}(c+d x)\right )}{3 d \sqrt {\sin (c+d x)}} \] Input:
Integrate[(a + a*Sec[c + d*x])^2*Sqrt[e*Sin[c + d*x]],x]
Output:
(-2*a^2*Cos[(c + d*x)/2]^4*Sec[c + d*x]*Sec[ArcSin[Sin[c + d*x]]/2]^4*Sqrt [e*Sin[c + d*x]]*(3*ArcTan[Sqrt[Sin[c + d*x]]]*Sqrt[Cos[c + d*x]^2] - 3*Ar cTanh[Sqrt[Sin[c + d*x]]]*Sqrt[Cos[c + d*x]^2] - 3*Sin[c + d*x]^(3/2) + Sq rt[Cos[c + d*x]^2]*Hypergeometric2F1[3/4, 3/2, 7/4, Sin[c + d*x]^2]*Sin[c + d*x]^(3/2)))/(3*d*Sqrt[Sin[c + d*x]])
Time = 0.58 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {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.
\(\displaystyle \int (a \sec (c+d x)+a)^2 \sqrt {e \sin (c+d x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \left (a-a \csc \left (c+d x-\frac {\pi }{2}\right )\right )^2 \sqrt {e \cos \left (c+d x-\frac {\pi }{2}\right )}dx\) |
\(\Big \downarrow \) 4360 |
\(\displaystyle \int \sec ^2(c+d x) (a (-\cos (c+d x))-a)^2 \sqrt {e \sin (c+d x)}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\left (a \left (-\sin \left (c+d x+\frac {\pi }{2}\right )\right )-a\right )^2 \sqrt {-e \cos \left (c+d x+\frac {\pi }{2}\right )}}{\sin \left (c+d x+\frac {\pi }{2}\right )^2}dx\) |
\(\Big \downarrow \) 3352 |
\(\displaystyle \int \left (a^2 \sqrt {e \sin (c+d x)}+a^2 \sec ^2(c+d x) \sqrt {e \sin (c+d x)}+2 a^2 \sec (c+d x) \sqrt {e \sin (c+d x)}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {2 a^2 \sqrt {e} \arctan \left (\frac {\sqrt {e \sin (c+d x)}}{\sqrt {e}}\right )}{d}+\frac {2 a^2 \sqrt {e} \text {arctanh}\left (\frac {\sqrt {e \sin (c+d x)}}{\sqrt {e}}\right )}{d}+\frac {a^2 \sec (c+d x) (e \sin (c+d x))^{3/2}}{d e}+\frac {a^2 E\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{d \sqrt {\sin (c+d x)}}\) |
Input:
Int[(a + a*Sec[c + d*x])^2*Sqrt[e*Sin[c + d*x]],x]
Output:
(-2*a^2*Sqrt[e]*ArcTan[Sqrt[e*Sin[c + d*x]]/Sqrt[e]])/d + (2*a^2*Sqrt[e]*A rcTanh[Sqrt[e*Sin[c + d*x]]/Sqrt[e]])/d + (a^2*EllipticE[(c - Pi/2 + d*x)/ 2, 2]*Sqrt[e*Sin[c + d*x]])/(d*Sqrt[Sin[c + d*x]]) + (a^2*Sec[c + d*x]*(e* Sin[c + d*x])^(3/2))/(d*e)
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]
Time = 0.32 (sec) , antiderivative size = 220, normalized size of antiderivative = 1.59
method | result | size |
default | \(-\frac {a^{2} \left (2 \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \sqrt {\sin \left (d x +c \right )}\, \operatorname {EllipticE}\left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right ) e -\sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \sqrt {\sin \left (d x +c \right )}\, \operatorname {EllipticF}\left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right ) e +4 \cos \left (d x +c \right ) \sqrt {e}\, \sqrt {e \sin \left (d x +c \right )}\, \arctan \left (\frac {\sqrt {e \sin \left (d x +c \right )}}{\sqrt {e}}\right )-4 \cos \left (d x +c \right ) \sqrt {e}\, \sqrt {e \sin \left (d x +c \right )}\, \operatorname {arctanh}\left (\frac {\sqrt {e \sin \left (d x +c \right )}}{\sqrt {e}}\right )+2 e \cos \left (d x +c \right )^{2}-2 e \right )}{2 \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}\, d}\) | \(220\) |
parts | \(-\frac {a^{2} e \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \sqrt {\sin \left (d x +c \right )}\, \left (2 \operatorname {EllipticE}\left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )-\operatorname {EllipticF}\left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )\right )}{\cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}\, d}+\frac {a^{2} e \sqrt {\cos \left (d x +c \right )^{2} e \sin \left (d x +c \right )}\, \left (2 \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \sqrt {\sin \left (d x +c \right )}\, \operatorname {EllipticE}\left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )-\sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \sqrt {\sin \left (d x +c \right )}\, \operatorname {EllipticF}\left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )-2 \cos \left (d x +c \right )^{2}+2\right )}{2 \sqrt {-e \sin \left (d x +c \right ) \left (\sin \left (d x +c \right )-1\right ) \left (\sin \left (d x +c \right )+1\right )}\, \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}\, d}-\frac {2 a^{2} \sqrt {e}\, \left (\arctan \left (\frac {\sqrt {e \sin \left (d x +c \right )}}{\sqrt {e}}\right )-\operatorname {arctanh}\left (\frac {\sqrt {e \sin \left (d x +c \right )}}{\sqrt {e}}\right )\right )}{d}\) | \(334\) |
Input:
int((a+a*sec(d*x+c))^2*(e*sin(d*x+c))^(1/2),x,method=_RETURNVERBOSE)
Output:
-1/2/cos(d*x+c)/(e*sin(d*x+c))^(1/2)*a^2*(2*(-sin(d*x+c)+1)^(1/2)*(2*sin(d *x+c)+2)^(1/2)*sin(d*x+c)^(1/2)*EllipticE((-sin(d*x+c)+1)^(1/2),1/2*2^(1/2 ))*e-(-sin(d*x+c)+1)^(1/2)*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(1/2)*Ellipti cF((-sin(d*x+c)+1)^(1/2),1/2*2^(1/2))*e+4*cos(d*x+c)*e^(1/2)*(e*sin(d*x+c) )^(1/2)*arctan((e*sin(d*x+c))^(1/2)/e^(1/2))-4*cos(d*x+c)*e^(1/2)*(e*sin(d *x+c))^(1/2)*arctanh((e*sin(d*x+c))^(1/2)/e^(1/2))+2*e*cos(d*x+c)^2-2*e)/d
Result contains complex when optimal does not.
Time = 0.35 (sec) , antiderivative size = 667, normalized size of antiderivative = 4.83 \[ \int (a+a \sec (c+d x))^2 \sqrt {e \sin (c+d x)} \, dx =\text {Too large to display} \] Input:
integrate((a+a*sec(d*x+c))^2*(e*sin(d*x+c))^(1/2),x, algorithm="fricas")
Output:
[-1/4*(2*a^2*sqrt(-e)*arctan(1/4*(cos(d*x + c)^2 - 6*sin(d*x + c) - 2)*sqr t(e*sin(d*x + c))*sqrt(-e)/(e*cos(d*x + c)^2 - e*sin(d*x + c) - e))*cos(d* x + c) - a^2*sqrt(-e)*cos(d*x + c)*log((e*cos(d*x + c)^4 - 72*e*cos(d*x + c)^2 - 8*(7*cos(d*x + c)^2 - (cos(d*x + c)^2 - 8)*sin(d*x + c) - 8)*sqrt(e *sin(d*x + c))*sqrt(-e) + 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)) - 4*I*a^2*sqrt(-1/2*I*e)*cos(d*x + c)*weierstrassZeta(4, 0, weierstr assPInverse(4, 0, cos(d*x + c) + I*sin(d*x + c))) + 4*I*a^2*sqrt(1/2*I*e)* cos(d*x + c)*weierstrassZeta(4, 0, weierstrassPInverse(4, 0, cos(d*x + c) - I*sin(d*x + c))) - 4*sqrt(e*sin(d*x + c))*a^2*sin(d*x + c))/(d*cos(d*x + c)), -1/4*(2*a^2*sqrt(e)*arctan(1/4*(cos(d*x + c)^2 + 6*sin(d*x + c) - 2) *sqrt(e*sin(d*x + c))*sqrt(e)/(e*cos(d*x + c)^2 + e*sin(d*x + c) - e))*cos (d*x + c) - a^2*sqrt(e)*cos(d*x + c)*log((e*cos(d*x + c)^4 - 72*e*cos(d*x + c)^2 - 8*(7*cos(d*x + c)^2 + (cos(d*x + c)^2 - 8)*sin(d*x + c) - 8)*sqrt (e*sin(d*x + c))*sqrt(e) - 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)) - 4*I*a^2*sqrt(-1/2*I*e)*cos(d*x + c)*weierstrassZeta(4, 0, weierst rassPInverse(4, 0, cos(d*x + c) + I*sin(d*x + c))) + 4*I*a^2*sqrt(1/2*I*e) *cos(d*x + c)*weierstrassZeta(4, 0, weierstrassPInverse(4, 0, cos(d*x + c) - I*sin(d*x + c))) - 4*sqrt(e*sin(d*x + c))*a^2*sin(d*x + c))/(d*cos(d...
\[ \int (a+a \sec (c+d x))^2 \sqrt {e \sin (c+d x)} \, dx=a^{2} \left (\int \sqrt {e \sin {\left (c + d x \right )}}\, dx + \int 2 \sqrt {e \sin {\left (c + d x \right )}} \sec {\left (c + d x \right )}\, dx + \int \sqrt {e \sin {\left (c + d x \right )}} \sec ^{2}{\left (c + d x \right )}\, dx\right ) \] Input:
integrate((a+a*sec(d*x+c))**2*(e*sin(d*x+c))**(1/2),x)
Output:
a**2*(Integral(sqrt(e*sin(c + d*x)), x) + Integral(2*sqrt(e*sin(c + d*x))* sec(c + d*x), x) + Integral(sqrt(e*sin(c + d*x))*sec(c + d*x)**2, x))
\[ \int (a+a \sec (c+d x))^2 \sqrt {e \sin (c+d x)} \, dx=\int { {\left (a \sec \left (d x + c\right ) + a\right )}^{2} \sqrt {e \sin \left (d x + c\right )} \,d x } \] Input:
integrate((a+a*sec(d*x+c))^2*(e*sin(d*x+c))^(1/2),x, algorithm="maxima")
Output:
integrate((a*sec(d*x + c) + a)^2*sqrt(e*sin(d*x + c)), x)
\[ \int (a+a \sec (c+d x))^2 \sqrt {e \sin (c+d x)} \, dx=\int { {\left (a \sec \left (d x + c\right ) + a\right )}^{2} \sqrt {e \sin \left (d x + c\right )} \,d x } \] Input:
integrate((a+a*sec(d*x+c))^2*(e*sin(d*x+c))^(1/2),x, algorithm="giac")
Output:
integrate((a*sec(d*x + c) + a)^2*sqrt(e*sin(d*x + c)), x)
Timed out. \[ \int (a+a \sec (c+d x))^2 \sqrt {e \sin (c+d x)} \, dx=\int \sqrt {e\,\sin \left (c+d\,x\right )}\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^2 \,d x \] Input:
int((e*sin(c + d*x))^(1/2)*(a + a/cos(c + d*x))^2,x)
Output:
int((e*sin(c + d*x))^(1/2)*(a + a/cos(c + d*x))^2, x)
\[ \int (a+a \sec (c+d x))^2 \sqrt {e \sin (c+d x)} \, dx=\sqrt {e}\, a^{2} \left (\int \sqrt {\sin \left (d x +c \right )}d x +\int \sqrt {\sin \left (d x +c \right )}\, \sec \left (d x +c \right )^{2}d x +2 \left (\int \sqrt {\sin \left (d x +c \right )}\, \sec \left (d x +c \right )d x \right )\right ) \] Input:
int((a+a*sec(d*x+c))^2*(e*sin(d*x+c))^(1/2),x)
Output:
sqrt(e)*a**2*(int(sqrt(sin(c + d*x)),x) + int(sqrt(sin(c + d*x))*sec(c + d *x)**2,x) + 2*int(sqrt(sin(c + d*x))*sec(c + d*x),x))