\(\int \frac {(A+B \cos (c+d x)) \sec ^2(c+d x)}{\sqrt {b \cos (c+d x)}} \, dx\) [825]

Optimal result

Integrand size = 31, antiderivative size = 135 \[ \int \frac {(A+B \cos (c+d x)) \sec ^2(c+d x)}{\sqrt {b \cos (c+d x)}} \, dx=-\frac {2 B \sqrt {b \cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b d \sqrt {\cos (c+d x)}}+\frac {2 A \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d \sqrt {b \cos (c+d x)}}+\frac {2 A b \sin (c+d x)}{3 d (b \cos (c+d x))^{3/2}}+\frac {2 B \sin (c+d x)}{d \sqrt {b \cos (c+d x)}} \]

[Out]

2/3*A*b*sin(d*x+c)/d/(b*cos(d*x+c))^(3/2)+2*B*sin(d*x+c)/d/(b*cos(d*x+c))^(1/2)+2/3*A*(cos(1/2*d*x+1/2*c)^2)^(
1/2)/cos(1/2*d*x+1/2*c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)/d/(b*cos(d*x+c))^(1/2)-2*B*(cos
(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))*(b*cos(d*x+c))^(1/2)/b/d/cos
(d*x+c)^(1/2)

Rubi [A] (verified)

Time = 0.18 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {16, 2827, 2716, 2721, 2720, 2719} \[ \int \frac {(A+B \cos (c+d x)) \sec ^2(c+d x)}{\sqrt {b \cos (c+d x)}} \, dx=\frac {2 A b \sin (c+d x)}{3 d (b \cos (c+d x))^{3/2}}+\frac {2 A \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d \sqrt {b \cos (c+d x)}}+\frac {2 B \sin (c+d x)}{d \sqrt {b \cos (c+d x)}}-\frac {2 B E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {b \cos (c+d x)}}{b d \sqrt {\cos (c+d x)}} \]

[In]

Int[((A + B*Cos[c + d*x])*Sec[c + d*x]^2)/Sqrt[b*Cos[c + d*x]],x]

[Out]

(-2*B*Sqrt[b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2])/(b*d*Sqrt[Cos[c + d*x]]) + (2*A*Sqrt[Cos[c + d*x]]*Ellip
ticF[(c + d*x)/2, 2])/(3*d*Sqrt[b*Cos[c + d*x]]) + (2*A*b*Sin[c + d*x])/(3*d*(b*Cos[c + d*x])^(3/2)) + (2*B*Si
n[c + d*x])/(d*Sqrt[b*Cos[c + d*x]])

Rule 16

Int[(u_.)*(v_)^(m_.)*((b_)*(v_))^(n_), x_Symbol] :> Dist[1/b^m, Int[u*(b*v)^(m + n), x], x] /; FreeQ[{b, n}, x
] && IntegerQ[m]

Rule 2716

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] + Dist[(n + 2)/(b^2*(n + 1)), Int[(b*Sin[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1
] && IntegerQ[2*n]

Rule 2719

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]

Rule 2720

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]

Rule 2721

Int[((b_)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[(b*Sin[c + d*x])^n/Sin[c + d*x]^n, Int[Sin[c + d*x]
^n, x], x] /; FreeQ[{b, c, d}, x] && LtQ[-1, n, 1] && IntegerQ[2*n]

Rule 2827

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

Rubi steps \begin{align*} \text {integral}& = b^2 \int \frac {A+B \cos (c+d x)}{(b \cos (c+d x))^{5/2}} \, dx \\ & = \left (A b^2\right ) \int \frac {1}{(b \cos (c+d x))^{5/2}} \, dx+(b B) \int \frac {1}{(b \cos (c+d x))^{3/2}} \, dx \\ & = \frac {2 A b \sin (c+d x)}{3 d (b \cos (c+d x))^{3/2}}+\frac {2 B \sin (c+d x)}{d \sqrt {b \cos (c+d x)}}+\frac {1}{3} A \int \frac {1}{\sqrt {b \cos (c+d x)}} \, dx-\frac {B \int \sqrt {b \cos (c+d x)} \, dx}{b} \\ & = \frac {2 A b \sin (c+d x)}{3 d (b \cos (c+d x))^{3/2}}+\frac {2 B \sin (c+d x)}{d \sqrt {b \cos (c+d x)}}+\frac {\left (A \sqrt {\cos (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{3 \sqrt {b \cos (c+d x)}}-\frac {\left (B \sqrt {b \cos (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx}{b \sqrt {\cos (c+d x)}} \\ & = -\frac {2 B \sqrt {b \cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b d \sqrt {\cos (c+d x)}}+\frac {2 A \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d \sqrt {b \cos (c+d x)}}+\frac {2 A b \sin (c+d x)}{3 d (b \cos (c+d x))^{3/2}}+\frac {2 B \sin (c+d x)}{d \sqrt {b \cos (c+d x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.24 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.62 \[ \int \frac {(A+B \cos (c+d x)) \sec ^2(c+d x)}{\sqrt {b \cos (c+d x)}} \, dx=\frac {2 \left (-3 B \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+A \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+3 B \sin (c+d x)+A \tan (c+d x)\right )}{3 d \sqrt {b \cos (c+d x)}} \]

[In]

Integrate[((A + B*Cos[c + d*x])*Sec[c + d*x]^2)/Sqrt[b*Cos[c + d*x]],x]

[Out]

(2*(-3*B*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2] + A*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2] + 3*B*S
in[c + d*x] + A*Tan[c + d*x]))/(3*d*Sqrt[b*Cos[c + d*x]])

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(405\) vs. \(2(171)=342\).

Time = 6.36 (sec) , antiderivative size = 406, normalized size of antiderivative = 3.01

method	result	size
default	\(\frac {2 \sqrt {-\left (-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1\right ) b \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (2 A \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-12 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+6 B \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-A \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+6 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-3 B \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right ) \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +b \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}}{3 b \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} \left (4 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-4 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1\right ) \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) b}\, d}\)	\(406\)
parts	\(-\frac {2 A \left (-2 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right ) \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) b \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}}{3 \sqrt {-b \left (2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}\, \left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) b}\, d}-\frac {2 B \left (-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +b \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +b \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{\sqrt {-b \left (2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) b}\, d}\)	\(435\)

[In]

int((A+B*cos(d*x+c))*sec(d*x+c)^2/(cos(d*x+c)*b)^(1/2),x,method=_RETURNVERBOSE)

[Out]

2/3*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*b*sin(1/2*d*x+1/2*c)^2)^(1/2)/b/sin(1/2*d*x+1/2*c)^3/(4*sin(1/2*d*x+1/2*c)^4
-4*sin(1/2*d*x+1/2*c)^2+1)*(2*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*
d*x+1/2*c)^2-1)^(1/2)*sin(1/2*d*x+1/2*c)^2-12*B*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4+6*B*(sin(1/2*d*x+1/2*c
)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*sin(1/2*d*x+1/2*c)^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)*EllipticF
(cos(1/2*d*x+1/2*c),2^(1/2))+6*B*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2-3*B*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*s
in(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))*(-2*sin(1/2*d*x+1/2*c)^4*b+b*sin(1/2*d*x+1
/2*c)^2)^(1/2)/((2*cos(1/2*d*x+1/2*c)^2-1)*b)^(1/2)/d

Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.10 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.42 \[ \int \frac {(A+B \cos (c+d x)) \sec ^2(c+d x)}{\sqrt {b \cos (c+d x)}} \, dx=\frac {-i \, \sqrt {2} A \sqrt {b} \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + i \, \sqrt {2} A \sqrt {b} \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 3 i \, \sqrt {2} B \sqrt {b} \cos \left (d x + c\right )^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 3 i \, \sqrt {2} B \sqrt {b} \cos \left (d x + c\right )^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) + 2 \, {\left (3 \, B \cos \left (d x + c\right ) + A\right )} \sqrt {b \cos \left (d x + c\right )} \sin \left (d x + c\right )}{3 \, b d \cos \left (d x + c\right )^{2}} \]

[In]

integrate((A+B*cos(d*x+c))*sec(d*x+c)^2/(b*cos(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

1/3*(-I*sqrt(2)*A*sqrt(b)*cos(d*x + c)^2*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + I*sqrt(2)
*A*sqrt(b)*cos(d*x + c)^2*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) - 3*I*sqrt(2)*B*sqrt(b)*co
s(d*x + c)^2*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 3*I*sqrt(2)*B
*sqrt(b)*cos(d*x + c)^2*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) + 2*
(3*B*cos(d*x + c) + A)*sqrt(b*cos(d*x + c))*sin(d*x + c))/(b*d*cos(d*x + c)^2)

Sympy [F]

\[ \int \frac {(A+B \cos (c+d x)) \sec ^2(c+d x)}{\sqrt {b \cos (c+d x)}} \, dx=\int \frac {\left (A + B \cos {\left (c + d x \right )}\right ) \sec ^{2}{\left (c + d x \right )}}{\sqrt {b \cos {\left (c + d x \right )}}}\, dx \]

[In]

integrate((A+B*cos(d*x+c))*sec(d*x+c)**2/(b*cos(d*x+c))**(1/2),x)

[Out]

Integral((A + B*cos(c + d*x))*sec(c + d*x)**2/sqrt(b*cos(c + d*x)), x)

Maxima [F]

\[ \int \frac {(A+B \cos (c+d x)) \sec ^2(c+d x)}{\sqrt {b \cos (c+d x)}} \, dx=\int { \frac {{\left (B \cos \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{2}}{\sqrt {b \cos \left (d x + c\right )}} \,d x } \]

[In]

integrate((A+B*cos(d*x+c))*sec(d*x+c)^2/(b*cos(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

integrate((B*cos(d*x + c) + A)*sec(d*x + c)^2/sqrt(b*cos(d*x + c)), x)

Giac [F]

\[ \int \frac {(A+B \cos (c+d x)) \sec ^2(c+d x)}{\sqrt {b \cos (c+d x)}} \, dx=\int { \frac {{\left (B \cos \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{2}}{\sqrt {b \cos \left (d x + c\right )}} \,d x } \]

[In]

integrate((A+B*cos(d*x+c))*sec(d*x+c)^2/(b*cos(d*x+c))^(1/2),x, algorithm="giac")

[Out]

integrate((B*cos(d*x + c) + A)*sec(d*x + c)^2/sqrt(b*cos(d*x + c)), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {(A+B \cos (c+d x)) \sec ^2(c+d x)}{\sqrt {b \cos (c+d x)}} \, dx=\int \frac {A+B\,\cos \left (c+d\,x\right )}{{\cos \left (c+d\,x\right )}^2\,\sqrt {b\,\cos \left (c+d\,x\right )}} \,d x \]

[In]

int((A + B*cos(c + d*x))/(cos(c + d*x)^2*(b*cos(c + d*x))^(1/2)),x)

[Out]

int((A + B*cos(c + d*x))/(cos(c + d*x)^2*(b*cos(c + d*x))^(1/2)), x)