3.514 \(\int \frac {\sec ^2(c+d x)}{\sqrt {a+b \sin (c+d x)}} \, dx\)

Optimal. Leaf size=183 \[ -\frac {\sec (c+d x) (b-a \sin (c+d x)) \sqrt {a+b \sin (c+d x)}}{d \left (a^2-b^2\right )}-\frac {a \sqrt {a+b \sin (c+d x)} E\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{d \left (a^2-b^2\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}+\frac {\sqrt {\frac {a+b \sin (c+d x)}{a+b}} F\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{d \sqrt {a+b \sin (c+d x)}} \]

[Out]

-sec(d*x+c)*(b-a*sin(d*x+c))*(a+b*sin(d*x+c))^(1/2)/d/(a^2-b^2)+a*(sin(1/2*c+1/4*Pi+1/2*d*x)^2)^(1/2)/sin(1/2*
c+1/4*Pi+1/2*d*x)*EllipticE(cos(1/2*c+1/4*Pi+1/2*d*x),2^(1/2)*(b/(a+b))^(1/2))*(a+b*sin(d*x+c))^(1/2)/(a^2-b^2
)/d/((a+b*sin(d*x+c))/(a+b))^(1/2)-(sin(1/2*c+1/4*Pi+1/2*d*x)^2)^(1/2)/sin(1/2*c+1/4*Pi+1/2*d*x)*EllipticF(cos
(1/2*c+1/4*Pi+1/2*d*x),2^(1/2)*(b/(a+b))^(1/2))*((a+b*sin(d*x+c))/(a+b))^(1/2)/d/(a+b*sin(d*x+c))^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.20, antiderivative size = 183, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {2696, 2752, 2663, 2661, 2655, 2653} \[ -\frac {\sec (c+d x) (b-a \sin (c+d x)) \sqrt {a+b \sin (c+d x)}}{d \left (a^2-b^2\right )}-\frac {a \sqrt {a+b \sin (c+d x)} E\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{d \left (a^2-b^2\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}+\frac {\sqrt {\frac {a+b \sin (c+d x)}{a+b}} F\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{d \sqrt {a+b \sin (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^2/Sqrt[a + b*Sin[c + d*x]],x]

[Out]

-((Sec[c + d*x]*(b - a*Sin[c + d*x])*Sqrt[a + b*Sin[c + d*x]])/((a^2 - b^2)*d)) - (a*EllipticE[(c - Pi/2 + d*x
)/2, (2*b)/(a + b)]*Sqrt[a + b*Sin[c + d*x]])/((a^2 - b^2)*d*Sqrt[(a + b*Sin[c + d*x])/(a + b)]) + (EllipticF[
(c - Pi/2 + d*x)/2, (2*b)/(a + b)]*Sqrt[(a + b*Sin[c + d*x])/(a + b)])/(d*Sqrt[a + b*Sin[c + d*x]])

Rule 2653

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*Sqrt[a + b]*EllipticE[(1*(c - Pi/2 + d*x)
)/2, (2*b)/(a + b)])/d, x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]

Rule 2655

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[a + b*Sin[c + d*x]]/Sqrt[(a + b*Sin[c +
d*x])/(a + b)], Int[Sqrt[a/(a + b) + (b*Sin[c + d*x])/(a + b)], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 -
b^2, 0] &&  !GtQ[a + b, 0]

Rule 2661

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, (2*b)
/(a + b)])/(d*Sqrt[a + b]), x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]

Rule 2663

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[(a + b*Sin[c + d*x])/(a + b)]/Sqrt[a
+ b*Sin[c + d*x]], Int[1/Sqrt[a/(a + b) + (b*Sin[c + d*x])/(a + b)], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a
^2 - b^2, 0] &&  !GtQ[a + b, 0]

Rule 2696

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[((g*Co
s[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^(m + 1)*(b - a*Sin[e + f*x]))/(f*g*(a^2 - b^2)*(p + 1)), x] + Dist[1/
(g^2*(a^2 - b^2)*(p + 1)), Int[(g*Cos[e + f*x])^(p + 2)*(a + b*Sin[e + f*x])^m*(a^2*(p + 2) - b^2*(m + p + 2)
+ a*b*(m + p + 3)*Sin[e + f*x]), x], x] /; FreeQ[{a, b, e, f, g, m}, x] && NeQ[a^2 - b^2, 0] && LtQ[p, -1] &&
IntegersQ[2*m, 2*p]

Rule 2752

Int[((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[(b*c
- a*d)/b, Int[1/Sqrt[a + b*Sin[e + f*x]], x], x] + Dist[d/b, Int[Sqrt[a + b*Sin[e + f*x]], x], x] /; FreeQ[{a
, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0]

Rubi steps

\begin {align*} \int \frac {\sec ^2(c+d x)}{\sqrt {a+b \sin (c+d x)}} \, dx &=-\frac {\sec (c+d x) (b-a \sin (c+d x)) \sqrt {a+b \sin (c+d x)}}{\left (a^2-b^2\right ) d}+\frac {\int \frac {\frac {b^2}{2}+\frac {1}{2} a b \sin (c+d x)}{\sqrt {a+b \sin (c+d x)}} \, dx}{-a^2+b^2}\\ &=-\frac {\sec (c+d x) (b-a \sin (c+d x)) \sqrt {a+b \sin (c+d x)}}{\left (a^2-b^2\right ) d}+\frac {1}{2} \int \frac {1}{\sqrt {a+b \sin (c+d x)}} \, dx-\frac {a \int \sqrt {a+b \sin (c+d x)} \, dx}{2 \left (a^2-b^2\right )}\\ &=-\frac {\sec (c+d x) (b-a \sin (c+d x)) \sqrt {a+b \sin (c+d x)}}{\left (a^2-b^2\right ) d}-\frac {\left (a \sqrt {a+b \sin (c+d x)}\right ) \int \sqrt {\frac {a}{a+b}+\frac {b \sin (c+d x)}{a+b}} \, dx}{2 \left (a^2-b^2\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}+\frac {\sqrt {\frac {a+b \sin (c+d x)}{a+b}} \int \frac {1}{\sqrt {\frac {a}{a+b}+\frac {b \sin (c+d x)}{a+b}}} \, dx}{2 \sqrt {a+b \sin (c+d x)}}\\ &=-\frac {\sec (c+d x) (b-a \sin (c+d x)) \sqrt {a+b \sin (c+d x)}}{\left (a^2-b^2\right ) d}-\frac {a E\left (\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )|\frac {2 b}{a+b}\right ) \sqrt {a+b \sin (c+d x)}}{\left (a^2-b^2\right ) d \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}+\frac {F\left (\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )|\frac {2 b}{a+b}\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}{d \sqrt {a+b \sin (c+d x)}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.63, size = 177, normalized size = 0.97 \[ \frac {-\left (a^2-b^2\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}} F\left (\frac {1}{4} (-2 c-2 d x+\pi )|\frac {2 b}{a+b}\right )+a^2 \tan (c+d x)-a b \sec (c+d x)+a b \sin (c+d x) \tan (c+d x)+a (a+b) \sqrt {\frac {a+b \sin (c+d x)}{a+b}} E\left (\frac {1}{4} (-2 c-2 d x+\pi )|\frac {2 b}{a+b}\right )-b^2 \tan (c+d x)}{d (a-b) (a+b) \sqrt {a+b \sin (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^2/Sqrt[a + b*Sin[c + d*x]],x]

[Out]

(-(a*b*Sec[c + d*x]) + a*(a + b)*EllipticE[(-2*c + Pi - 2*d*x)/4, (2*b)/(a + b)]*Sqrt[(a + b*Sin[c + d*x])/(a
+ b)] - (a^2 - b^2)*EllipticF[(-2*c + Pi - 2*d*x)/4, (2*b)/(a + b)]*Sqrt[(a + b*Sin[c + d*x])/(a + b)] + a^2*T
an[c + d*x] - b^2*Tan[c + d*x] + a*b*Sin[c + d*x]*Tan[c + d*x])/((a - b)*(a + b)*d*Sqrt[a + b*Sin[c + d*x]])

________________________________________________________________________________________

fricas [F]  time = 0.97, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sec \left (d x + c\right )^{2}}{\sqrt {b \sin \left (d x + c\right ) + a}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2/(a+b*sin(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

integral(sec(d*x + c)^2/sqrt(b*sin(d*x + c) + a), x)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec \left (d x + c\right )^{2}}{\sqrt {b \sin \left (d x + c\right ) + a}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2/(a+b*sin(d*x+c))^(1/2),x, algorithm="giac")

[Out]

integrate(sec(d*x + c)^2/sqrt(b*sin(d*x + c) + a), x)

________________________________________________________________________________________

maple [B]  time = 0.80, size = 640, normalized size = 3.50 \[ \frac {\sqrt {\left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right ) b +\left (\cos ^{2}\left (d x +c \right )\right ) a}\, \left (\sqrt {\frac {b \sin \left (d x +c \right )}{a -b}+\frac {a}{a -b}}\, \sqrt {-\frac {b \sin \left (d x +c \right )}{a +b}+\frac {b}{a +b}}\, \sqrt {-\frac {b \sin \left (d x +c \right )}{a -b}-\frac {b}{a -b}}\, \EllipticE \left (\sqrt {\frac {b \sin \left (d x +c \right )}{a -b}+\frac {a}{a -b}}, \sqrt {\frac {a -b}{a +b}}\right ) a^{3}-\sqrt {\frac {b \sin \left (d x +c \right )}{a -b}+\frac {a}{a -b}}\, \sqrt {-\frac {b \sin \left (d x +c \right )}{a +b}+\frac {b}{a +b}}\, \sqrt {-\frac {b \sin \left (d x +c \right )}{a -b}-\frac {b}{a -b}}\, \EllipticE \left (\sqrt {\frac {b \sin \left (d x +c \right )}{a -b}+\frac {a}{a -b}}, \sqrt {\frac {a -b}{a +b}}\right ) a \,b^{2}-\sqrt {\frac {b \sin \left (d x +c \right )}{a -b}+\frac {a}{a -b}}\, \sqrt {-\frac {b \sin \left (d x +c \right )}{a +b}+\frac {b}{a +b}}\, \sqrt {-\frac {b \sin \left (d x +c \right )}{a -b}-\frac {b}{a -b}}\, \EllipticF \left (\sqrt {\frac {b \sin \left (d x +c \right )}{a -b}+\frac {a}{a -b}}, \sqrt {\frac {a -b}{a +b}}\right ) a^{2} b +\sqrt {\frac {b \sin \left (d x +c \right )}{a -b}+\frac {a}{a -b}}\, \sqrt {-\frac {b \sin \left (d x +c \right )}{a +b}+\frac {b}{a +b}}\, \sqrt {-\frac {b \sin \left (d x +c \right )}{a -b}-\frac {b}{a -b}}\, \EllipticF \left (\sqrt {\frac {b \sin \left (d x +c \right )}{a -b}+\frac {a}{a -b}}, \sqrt {\frac {a -b}{a +b}}\right ) b^{3}-a \,b^{2} \left (\cos ^{2}\left (d x +c \right )\right )+a^{2} b \sin \left (d x +c \right )-b^{3} \sin \left (d x +c \right )\right )}{b \left (a +b \right ) \sqrt {-\left (a +b \sin \left (d x +c \right )\right ) \left (\sin \left (d x +c \right )-1\right ) \left (1+\sin \left (d x +c \right )\right )}\, \left (a -b \right ) \cos \left (d x +c \right ) \sqrt {a +b \sin \left (d x +c \right )}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^2/(a+b*sin(d*x+c))^(1/2),x)

[Out]

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

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec \left (d x + c\right )^{2}}{\sqrt {b \sin \left (d x + c\right ) + a}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2/(a+b*sin(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

integrate(sec(d*x + c)^2/sqrt(b*sin(d*x + c) + a), x)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{{\cos \left (c+d\,x\right )}^2\,\sqrt {a+b\,\sin \left (c+d\,x\right )}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(c + d*x)^2*(a + b*sin(c + d*x))^(1/2)),x)

[Out]

int(1/(cos(c + d*x)^2*(a + b*sin(c + d*x))^(1/2)), x)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec ^{2}{\left (c + d x \right )}}{\sqrt {a + b \sin {\left (c + d x \right )}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**2/(a+b*sin(d*x+c))**(1/2),x)

[Out]

Integral(sec(c + d*x)**2/sqrt(a + b*sin(c + d*x)), x)

________________________________________________________________________________________