Integrand size = 23, antiderivative size = 48 \[ \int \cos ^{\frac {3}{2}}(c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {2 (A+3 C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {2 A \sqrt {\cos (c+d x)} \sin (c+d x)}{3 d} \] Output:
2/3*(A+3*C)*InverseJacobiAM(1/2*d*x+1/2*c,2^(1/2))/d+2/3*A*cos(d*x+c)^(1/2 )*sin(d*x+c)/d
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 0.83 (sec) , antiderivative size = 124, normalized size of antiderivative = 2.58 \[ \int \cos ^{\frac {3}{2}}(c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx=-\frac {4 \sqrt {\cos (c+d x)} \left (C+A \cos ^2(c+d x)\right ) \sin (c) \left ((A+3 C) \sqrt {\cos ^2(d x-\arctan (\cot (c)))} \sqrt {\csc ^2(c)} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};\sin ^2(d x-\arctan (\cot (c)))\right ) \sec (d x-\arctan (\cot (c)))-A \csc (c) \sin (c+d x)\right )}{3 d (A+2 C+A \cos (2 (c+d x)))} \] Input:
Integrate[Cos[c + d*x]^(3/2)*(A + C*Sec[c + d*x]^2),x]
Output:
(-4*Sqrt[Cos[c + d*x]]*(C + A*Cos[c + d*x]^2)*Sin[c]*((A + 3*C)*Sqrt[Cos[d *x - ArcTan[Cot[c]]]^2]*Sqrt[Csc[c]^2]*HypergeometricPFQ[{1/4, 1/2}, {5/4} , Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[d*x - ArcTan[Cot[c]]] - A*Csc[c]*Sin[c + d*x]))/(3*d*(A + 2*C + A*Cos[2*(c + d*x)]))
Time = 0.33 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3042, 4554, 3042, 3493, 3042, 3120}
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.
\(\displaystyle \int \cos ^{\frac {3}{2}}(c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \cos (c+d x)^{3/2} \left (A+C \sec (c+d x)^2\right )dx\) |
\(\Big \downarrow \) 4554 |
\(\displaystyle \int \frac {A \cos ^2(c+d x)+C}{\sqrt {\cos (c+d x)}}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {A \sin \left (c+d x+\frac {\pi }{2}\right )^2+C}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx\) |
\(\Big \downarrow \) 3493 |
\(\displaystyle \frac {1}{3} (A+3 C) \int \frac {1}{\sqrt {\cos (c+d x)}}dx+\frac {2 A \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{3} (A+3 C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 A \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\) |
\(\Big \downarrow \) 3120 |
\(\displaystyle \frac {2 (A+3 C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {2 A \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\) |
Input:
Int[Cos[c + d*x]^(3/2)*(A + C*Sec[c + d*x]^2),x]
Output:
(2*(A + 3*C)*EllipticF[(c + d*x)/2, 2])/(3*d) + (2*A*Sqrt[Cos[c + d*x]]*Si n[c + d*x])/(3*d)
Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2 )*(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_) + (C_.)*sin[(e_.) + (f_.)*( x_)]^2), x_Symbol] :> Simp[(-C)*Cos[e + f*x]*((b*Sin[e + f*x])^(m + 1)/(b*f *(m + 2))), x] + Simp[(A*(m + 2) + C*(m + 1))/(m + 2) Int[(b*Sin[e + f*x] )^m, x], x] /; FreeQ[{b, e, f, A, C, m}, x] && !LtQ[m, -1]
Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(m_)*((A_.) + (C_.)*sec[(e_.) + (f_.)*( x_)]^2), x_Symbol] :> Simp[b^2 Int[(b*Cos[e + f*x])^(m - 2)*(C + A*Cos[e + f*x]^2), x], x] /; FreeQ[{b, e, f, A, C, m}, x] && !IntegerQ[m]
Leaf count of result is larger than twice the leaf count of optimal. \(227\) vs. \(2(43)=86\).
Time = 2.47 (sec) , antiderivative size = 228, normalized size of antiderivative = 4.75
method | result | size |
default | \(-\frac {2 \sqrt {\left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (4 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) A +A \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+3 C \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{3 \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}\) | \(228\) |
Input:
int(cos(d*x+c)^(3/2)*(A+C*sec(d*x+c)^2),x,method=_RETURNVERBOSE)
Output:
-2/3*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(4*A*cos(1/2* d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4-2*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)* A+A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*Elliptic F(cos(1/2*d*x+1/2*c),2^(1/2))+3*C*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2* d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2)))/(-2*sin(1/2*d *x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/sin(1/2*d*x+1/2*c)/(2*cos(1/2*d*x+ 1/2*c)^2-1)^(1/2)/d
Result contains complex when optimal does not.
Time = 0.09 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.69 \[ \int \cos ^{\frac {3}{2}}(c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {2 \, A \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) + \sqrt {2} {\left (-i \, A - 3 i \, C\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + \sqrt {2} {\left (i \, A + 3 i \, C\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )}{3 \, d} \] Input:
integrate(cos(d*x+c)^(3/2)*(A+C*sec(d*x+c)^2),x, algorithm="fricas")
Output:
1/3*(2*A*sqrt(cos(d*x + c))*sin(d*x + c) + sqrt(2)*(-I*A - 3*I*C)*weierstr assPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + sqrt(2)*(I*A + 3*I*C)* weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)))/d
Timed out. \[ \int \cos ^{\frac {3}{2}}(c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \] Input:
integrate(cos(d*x+c)**(3/2)*(A+C*sec(d*x+c)**2),x)
Output:
Timed out
\[ \int \cos ^{\frac {3}{2}}(c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + A\right )} \cos \left (d x + c\right )^{\frac {3}{2}} \,d x } \] Input:
integrate(cos(d*x+c)^(3/2)*(A+C*sec(d*x+c)^2),x, algorithm="maxima")
Output:
integrate((C*sec(d*x + c)^2 + A)*cos(d*x + c)^(3/2), x)
\[ \int \cos ^{\frac {3}{2}}(c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + A\right )} \cos \left (d x + c\right )^{\frac {3}{2}} \,d x } \] Input:
integrate(cos(d*x+c)^(3/2)*(A+C*sec(d*x+c)^2),x, algorithm="giac")
Output:
integrate((C*sec(d*x + c)^2 + A)*cos(d*x + c)^(3/2), x)
Time = 0.20 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.10 \[ \int \cos ^{\frac {3}{2}}(c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {2\,A\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{3\,d}+\frac {2\,C\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {2\,A\,\sqrt {\cos \left (c+d\,x\right )}\,\sin \left (c+d\,x\right )}{3\,d} \] Input:
int(cos(c + d*x)^(3/2)*(A + C/cos(c + d*x)^2),x)
Output:
(2*A*ellipticF(c/2 + (d*x)/2, 2))/(3*d) + (2*C*ellipticF(c/2 + (d*x)/2, 2) )/d + (2*A*cos(c + d*x)^(1/2)*sin(c + d*x))/(3*d)
\[ \int \cos ^{\frac {3}{2}}(c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx=\left (\int \sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right ) \sec \left (d x +c \right )^{2}d x \right ) c +\left (\int \sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )d x \right ) a \] Input:
int(cos(d*x+c)^(3/2)*(A+C*sec(d*x+c)^2),x)
Output:
int(sqrt(cos(c + d*x))*cos(c + d*x)*sec(c + d*x)**2,x)*c + int(sqrt(cos(c + d*x))*cos(c + d*x),x)*a