Integrand size = 25, antiderivative size = 192 \[ \int (a+a \sec (c+d x))^2 (e \sin (c+d x))^{3/2} \, dx=\frac {2 a^2 e^{3/2} \arctan \left (\frac {\sqrt {e \sin (c+d x)}}{\sqrt {e}}\right )}{d}+\frac {2 a^2 e^{3/2} \text {arctanh}\left (\frac {\sqrt {e \sin (c+d x)}}{\sqrt {e}}\right )}{d}-\frac {a^2 e^2 \operatorname {EllipticF}\left (\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),2\right ) \sqrt {\sin (c+d x)}}{3 d \sqrt {e \sin (c+d x)}}-\frac {4 a^2 e \sqrt {e \sin (c+d x)}}{d}-\frac {2 a^2 e \cos (c+d x) \sqrt {e \sin (c+d x)}}{3 d}+\frac {a^2 e \sec (c+d x) \sqrt {e \sin (c+d x)}}{d} \] Output:
2*a^2*e^(3/2)*arctan((e*sin(d*x+c))^(1/2)/e^(1/2))/d+2*a^2*e^(3/2)*arctanh ((e*sin(d*x+c))^(1/2)/e^(1/2))/d-1/3*a^2*e^2*InverseJacobiAM(1/2*c-1/4*Pi+ 1/2*d*x,2^(1/2))*sin(d*x+c)^(1/2)/d/(e*sin(d*x+c))^(1/2)-4*a^2*e*(e*sin(d* x+c))^(1/2)/d-2/3*a^2*e*cos(d*x+c)*(e*sin(d*x+c))^(1/2)/d+a^2*e*sec(d*x+c) *(e*sin(d*x+c))^(1/2)/d
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 26.37 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.06 \[ \int (a+a \sec (c+d x))^2 (e \sin (c+d x))^{3/2} \, dx=\frac {16 a^2 e \cos ^4\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x) \sqrt {e \sin (c+d x)} \left (6 \arctan \left (\sqrt {\sin (c+d x)}\right ) \sqrt {\cos ^2(c+d x)}+6 \text {arctanh}\left (\sqrt {\sin (c+d x)}\right ) \sqrt {\cos ^2(c+d x)}+\sqrt {\sin (c+d x)}-12 \sqrt {\cos ^2(c+d x)} \sqrt {\sin (c+d x)}-\sqrt {\cos ^2(c+d x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},\sin ^2(c+d x)\right ) \sqrt {\sin (c+d x)}+2 \sin ^{\frac {5}{2}}(c+d x)\right ) \sin ^4\left (\frac {1}{2} \arcsin (\sin (c+d x))\right )}{3 d \sin ^{\frac {9}{2}}(c+d x)} \] Input:
Integrate[(a + a*Sec[c + d*x])^2*(e*Sin[c + d*x])^(3/2),x]
Output:
(16*a^2*e*Cos[(c + d*x)/2]^4*Sec[c + d*x]*Sqrt[e*Sin[c + d*x]]*(6*ArcTan[S qrt[Sin[c + d*x]]]*Sqrt[Cos[c + d*x]^2] + 6*ArcTanh[Sqrt[Sin[c + d*x]]]*Sq rt[Cos[c + d*x]^2] + Sqrt[Sin[c + d*x]] - 12*Sqrt[Cos[c + d*x]^2]*Sqrt[Sin [c + d*x]] - Sqrt[Cos[c + d*x]^2]*Hypergeometric2F1[1/4, 1/2, 5/4, Sin[c + d*x]^2]*Sqrt[Sin[c + d*x]] + 2*Sin[c + d*x]^(5/2))*Sin[ArcSin[Sin[c + d*x ]]/2]^4)/(3*d*Sin[c + d*x]^(9/2))
Time = 0.68 (sec) , antiderivative size = 192, 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.
\(\displaystyle \int (a \sec (c+d x)+a)^2 (e \sin (c+d x))^{3/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \left (a-a \csc \left (c+d x-\frac {\pi }{2}\right )\right )^2 \left (e \cos \left (c+d x-\frac {\pi }{2}\right )\right )^{3/2}dx\) |
\(\Big \downarrow \) 4360 |
\(\displaystyle \int \sec ^2(c+d x) (a (-\cos (c+d x))-a)^2 (e \sin (c+d x))^{3/2}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\left (a \left (-\sin \left (c+d x+\frac {\pi }{2}\right )\right )-a\right )^2 \left (-e \cos \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}}{\sin \left (c+d x+\frac {\pi }{2}\right )^2}dx\) |
\(\Big \downarrow \) 3352 |
\(\displaystyle \int \left (a^2 (e \sin (c+d x))^{3/2}+a^2 \sec ^2(c+d x) (e \sin (c+d x))^{3/2}+2 a^2 \sec (c+d x) (e \sin (c+d x))^{3/2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {2 a^2 e^{3/2} \arctan \left (\frac {\sqrt {e \sin (c+d x)}}{\sqrt {e}}\right )}{d}+\frac {2 a^2 e^{3/2} \text {arctanh}\left (\frac {\sqrt {e \sin (c+d x)}}{\sqrt {e}}\right )}{d}-\frac {a^2 e^2 \sqrt {\sin (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{3 d \sqrt {e \sin (c+d x)}}-\frac {4 a^2 e \sqrt {e \sin (c+d x)}}{d}-\frac {2 a^2 e \cos (c+d x) \sqrt {e \sin (c+d x)}}{3 d}+\frac {a^2 e \sec (c+d x) \sqrt {e \sin (c+d x)}}{d}\) |
Input:
Int[(a + a*Sec[c + d*x])^2*(e*Sin[c + d*x])^(3/2),x]
Output:
(2*a^2*e^(3/2)*ArcTan[Sqrt[e*Sin[c + d*x]]/Sqrt[e]])/d + (2*a^2*e^(3/2)*Ar cTanh[Sqrt[e*Sin[c + d*x]]/Sqrt[e]])/d - (a^2*e^2*EllipticF[(c - Pi/2 + d* x)/2, 2]*Sqrt[Sin[c + d*x]])/(3*d*Sqrt[e*Sin[c + d*x]]) - (4*a^2*e*Sqrt[e* Sin[c + d*x]])/d - (2*a^2*e*Cos[c + d*x]*Sqrt[e*Sin[c + d*x]])/(3*d) + (a^ 2*e*Sec[c + d*x]*Sqrt[e*Sin[c + d*x]])/d
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 = 0.33 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.05
method | result | size |
default | \(\frac {a^{2} \left (12 \cos \left (d x +c \right ) e^{\frac {3}{2}} \sqrt {e \sin \left (d x +c \right )}\, \arctan \left (\frac {\sqrt {e \sin \left (d x +c \right )}}{\sqrt {e}}\right )+12 \cos \left (d x +c \right ) e^{\frac {3}{2}} \sqrt {e \sin \left (d x +c \right )}\, \operatorname {arctanh}\left (\frac {\sqrt {e \sin \left (d x +c \right )}}{\sqrt {e}}\right )+\sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \sqrt {\sin \left (d x +c \right )}\, \operatorname {EllipticF}\left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right ) e^{2}-4 e^{2} \cos \left (d x +c \right )^{2} \sin \left (d x +c \right )-24 e^{2} \sin \left (d x +c \right ) \cos \left (d x +c \right )+6 e^{2} \sin \left (d x +c \right )\right )}{6 \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}\, d}\) | \(201\) |
parts | \(-\frac {a^{2} e^{2} \left (\sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \sqrt {\sin \left (d x +c \right )}\, \operatorname {EllipticF}\left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )-2 \sin \left (d x +c \right )^{3}+2 \sin \left (d x +c \right )\right )}{3 \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}\, d}+\frac {a^{2} e^{2} \sqrt {\cos \left (d x +c \right )^{2} e \sin \left (d x +c \right )}\, \left (\sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \sqrt {\sin \left (d x +c \right )}\, \operatorname {EllipticF}\left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )+2 \sin \left (d x +c \right )\right )}{2 \sqrt {-e \sin \left (d x +c \right ) \left (\sin \left (d x +c \right )-1\right ) \left (\sin \left (d x +c \right )+1\right )}\, \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}\, d}+\frac {2 a^{2} \left (e^{\frac {3}{2}} \arctan \left (\frac {\sqrt {e \sin \left (d x +c \right )}}{\sqrt {e}}\right )+e^{\frac {3}{2}} \operatorname {arctanh}\left (\frac {\sqrt {e \sin \left (d x +c \right )}}{\sqrt {e}}\right )-2 e \sqrt {e \sin \left (d x +c \right )}\right )}{d}\) | \(295\) |
Input:
int((a+a*sec(d*x+c))^2*(e*sin(d*x+c))^(3/2),x,method=_RETURNVERBOSE)
Output:
1/6/cos(d*x+c)/(e*sin(d*x+c))^(1/2)*a^2*(12*cos(d*x+c)*e^(3/2)*(e*sin(d*x+ c))^(1/2)*arctan((e*sin(d*x+c))^(1/2)/e^(1/2))+12*cos(d*x+c)*e^(3/2)*(e*si n(d*x+c))^(1/2)*arctanh((e*sin(d*x+c))^(1/2)/e^(1/2))+(-sin(d*x+c)+1)^(1/2 )*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(1/2)*EllipticF((-sin(d*x+c)+1)^(1/2), 1/2*2^(1/2))*e^2-4*e^2*cos(d*x+c)^2*sin(d*x+c)-24*e^2*sin(d*x+c)*cos(d*x+c )+6*e^2*sin(d*x+c))/d
Result contains complex when optimal does not.
Time = 0.35 (sec) , antiderivative size = 709, normalized size of antiderivative = 3.69 \[ \int (a+a \sec (c+d x))^2 (e \sin (c+d x))^{3/2} \, dx =\text {Too large to display} \] Input:
integrate((a+a*sec(d*x+c))^2*(e*sin(d*x+c))^(3/2),x, algorithm="fricas")
Output:
[-1/12*(6*a^2*sqrt(-e)*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) - 3*a^2*sqrt(-e)*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*a^2*sqrt(-1/2*I*e)*e*cos(d*x + c)*weierstrassPInverse(4, 0, cos(d*x + c) + I*sin(d*x + c)) + 4*a^2*sqrt(1/2*I*e)*e*cos(d*x + c)*wei erstrassPInverse(4, 0, cos(d*x + c) - I*sin(d*x + c)) + 4*(2*a^2*e*cos(d*x + c)^2 + 12*a^2*e*cos(d*x + c) - 3*a^2*e)*sqrt(e*sin(d*x + c)))/(d*cos(d* x + c)), 1/12*(6*a^2*e^(3/2)*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) + 3*a^2*e^(3/2)*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*a^2*sqrt(-1/2*I*e)*e*cos(d*x + c)*weierstrassPInverse(4, 0 , cos(d*x + c) + I*sin(d*x + c)) - 4*a^2*sqrt(1/2*I*e)*e*cos(d*x + c)*weie rstrassPInverse(4, 0, cos(d*x + c) - I*sin(d*x + c)) - 4*(2*a^2*e*cos(d*x + c)^2 + 12*a^2*e*cos(d*x + c) - 3*a^2*e)*sqrt(e*sin(d*x + c)))/(d*cos(...
Timed out. \[ \int (a+a \sec (c+d x))^2 (e \sin (c+d x))^{3/2} \, dx=\text {Timed out} \] Input:
integrate((a+a*sec(d*x+c))**2*(e*sin(d*x+c))**(3/2),x)
Output:
Timed out
\[ \int (a+a \sec (c+d x))^2 (e \sin (c+d x))^{3/2} \, dx=\int { {\left (a \sec \left (d x + c\right ) + a\right )}^{2} \left (e \sin \left (d x + c\right )\right )^{\frac {3}{2}} \,d x } \] Input:
integrate((a+a*sec(d*x+c))^2*(e*sin(d*x+c))^(3/2),x, algorithm="maxima")
Output:
integrate((a*sec(d*x + c) + a)^2*(e*sin(d*x + c))^(3/2), x)
\[ \int (a+a \sec (c+d x))^2 (e \sin (c+d x))^{3/2} \, dx=\int { {\left (a \sec \left (d x + c\right ) + a\right )}^{2} \left (e \sin \left (d x + c\right )\right )^{\frac {3}{2}} \,d x } \] Input:
integrate((a+a*sec(d*x+c))^2*(e*sin(d*x+c))^(3/2),x, algorithm="giac")
Output:
integrate((a*sec(d*x + c) + a)^2*(e*sin(d*x + c))^(3/2), x)
Timed out. \[ \int (a+a \sec (c+d x))^2 (e \sin (c+d x))^{3/2} \, dx=\int {\left (e\,\sin \left (c+d\,x\right )\right )}^{3/2}\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^2 \,d x \] Input:
int((e*sin(c + d*x))^(3/2)*(a + a/cos(c + d*x))^2,x)
Output:
int((e*sin(c + d*x))^(3/2)*(a + a/cos(c + d*x))^2, x)
\[ \int (a+a \sec (c+d x))^2 (e \sin (c+d x))^{3/2} \, dx=\sqrt {e}\, a^{2} e \left (\int \sqrt {\sin \left (d x +c \right )}\, \sec \left (d x +c \right )^{2} \sin \left (d x +c \right )d x +2 \left (\int \sqrt {\sin \left (d x +c \right )}\, \sec \left (d x +c \right ) \sin \left (d x +c \right )d x \right )+\int \sqrt {\sin \left (d x +c \right )}\, \sin \left (d x +c \right )d x \right ) \] Input:
int((a+a*sec(d*x+c))^2*(e*sin(d*x+c))^(3/2),x)
Output:
sqrt(e)*a**2*e*(int(sqrt(sin(c + d*x))*sec(c + d*x)**2*sin(c + d*x),x) + 2 *int(sqrt(sin(c + d*x))*sec(c + d*x)*sin(c + d*x),x) + int(sqrt(sin(c + d* x))*sin(c + d*x),x))