Integrand size = 21, antiderivative size = 57 \[ \int \frac {A+B \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=-\frac {2 A E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}+\frac {2 B \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}+\frac {2 A \sin (c+d x)}{d \sqrt {\cos (c+d x)}} \] Output:
-2*A*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/d+2*B*InverseJacobiAM(1/2*d*x+1 /2*c,2^(1/2))/d+2*A*sin(d*x+c)/d/cos(d*x+c)^(1/2)
Time = 0.22 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.89 \[ \int \frac {A+B \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\frac {2 \left (-A E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+B \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+\frac {A \sin (c+d x)}{\sqrt {\cos (c+d x)}}\right )}{d} \] Input:
Integrate[(A + B*Cos[c + d*x])/Cos[c + d*x]^(3/2),x]
Output:
(2*(-(A*EllipticE[(c + d*x)/2, 2]) + B*EllipticF[(c + d*x)/2, 2] + (A*Sin[ c + d*x])/Sqrt[Cos[c + d*x]]))/d
Time = 0.36 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.02, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 3227, 3042, 3116, 3042, 3119, 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.
\(\displaystyle \int \frac {A+B \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {A+B \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx\) |
\(\Big \downarrow \) 3227 |
\(\displaystyle A \int \frac {1}{\cos ^{\frac {3}{2}}(c+d x)}dx+B \int \frac {1}{\sqrt {\cos (c+d x)}}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle A \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx+B \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx\) |
\(\Big \downarrow \) 3116 |
\(\displaystyle A \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\int \sqrt {\cos (c+d x)}dx\right )+B \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle A \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx\right )+B \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx\) |
\(\Big \downarrow \) 3119 |
\(\displaystyle B \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+A \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )\) |
\(\Big \downarrow \) 3120 |
\(\displaystyle A \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )+\frac {2 B \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}\) |
Input:
Int[(A + B*Cos[c + d*x])/Cos[c + d*x]^(3/2),x]
Output:
(2*B*EllipticF[(c + d*x)/2, 2])/d + A*((-2*EllipticE[(c + d*x)/2, 2])/d + (2*Sin[c + d*x])/(d*Sqrt[Cos[c + d*x]]))
Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]*(( b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1))), x] + Simp[(n + 2)/(b^2*(n + 1)) I nt[(b*Sin[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1] && IntegerQ[2*n]
Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)* (c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
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_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x _)]), x_Symbol] :> Simp[c Int[(b*Sin[e + f*x])^m, x], x] + Simp[d/b Int [(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]
Leaf count of result is larger than twice the leaf count of optimal. \(149\) vs. \(2(58)=116\).
Time = 2.98 (sec) , antiderivative size = 150, normalized size of antiderivative = 2.63
method | result | size |
default | \(\frac {4 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-2 A \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-2 B \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )}{\sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}\) | \(150\) |
parts | \(-\frac {2 A \left (-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{\sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}+\frac {2 B \,\operatorname {InverseJacobiAM}\left (\frac {d x}{2}+\frac {c}{2}, \sqrt {2}\right )}{d}\) | \(202\) |
Input:
int((A+B*cos(d*x+c))/cos(d*x+c)^(3/2),x,method=_RETURNVERBOSE)
Output:
2*(2*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2-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)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)) -B*(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)))/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 = 156, normalized size of antiderivative = 2.74 \[ \int \frac {A+B \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\frac {-i \, \sqrt {2} B \cos \left (d x + c\right ) {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + i \, \sqrt {2} B \cos \left (d x + c\right ) {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - i \, \sqrt {2} A \cos \left (d x + c\right ) {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + i \, \sqrt {2} A \cos \left (d x + c\right ) {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) + 2 \, A \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{d \cos \left (d x + c\right )} \] Input:
integrate((A+B*cos(d*x+c))/cos(d*x+c)^(3/2),x, algorithm="fricas")
Output:
(-I*sqrt(2)*B*cos(d*x + c)*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin (d*x + c)) + I*sqrt(2)*B*cos(d*x + c)*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) - I*sqrt(2)*A*cos(d*x + c)*weierstrassZeta(-4, 0, we ierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + I*sqrt(2)*A*cos (d*x + c)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) + 2*A*sqrt(cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c))
\[ \int \frac {A+B \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\int \frac {A + B \cos {\left (c + d x \right )}}{\cos ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx \] Input:
integrate((A+B*cos(d*x+c))/cos(d*x+c)**(3/2),x)
Output:
Integral((A + B*cos(c + d*x))/cos(c + d*x)**(3/2), x)
\[ \int \frac {A+B \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\int { \frac {B \cos \left (d x + c\right ) + A}{\cos \left (d x + c\right )^{\frac {3}{2}}} \,d x } \] Input:
integrate((A+B*cos(d*x+c))/cos(d*x+c)^(3/2),x, algorithm="maxima")
Output:
integrate((B*cos(d*x + c) + A)/cos(d*x + c)^(3/2), x)
\[ \int \frac {A+B \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\int { \frac {B \cos \left (d x + c\right ) + A}{\cos \left (d x + c\right )^{\frac {3}{2}}} \,d x } \] Input:
integrate((A+B*cos(d*x+c))/cos(d*x+c)^(3/2),x, algorithm="giac")
Output:
integrate((B*cos(d*x + c) + A)/cos(d*x + c)^(3/2), x)
Time = 43.05 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.05 \[ \int \frac {A+B \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\frac {2\,B\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {2\,A\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{d\,\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {{\sin \left (c+d\,x\right )}^2}} \] Input:
int((A + B*cos(c + d*x))/cos(c + d*x)^(3/2),x)
Output:
(2*B*ellipticF(c/2 + (d*x)/2, 2))/d + (2*A*sin(c + d*x)*hypergeom([-1/4, 1 /2], 3/4, cos(c + d*x)^2))/(d*cos(c + d*x)^(1/2)*(sin(c + d*x)^2)^(1/2))
\[ \int \frac {A+B \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\left (\int \frac {\sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )}d x \right ) b +\left (\int \frac {\sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{2}}d x \right ) a \] Input:
int((A+B*cos(d*x+c))/cos(d*x+c)^(3/2),x)
Output:
int(sqrt(cos(c + d*x))/cos(c + d*x),x)*b + int(sqrt(cos(c + d*x))/cos(c + d*x)**2,x)*a