3.458 \(\int \frac{\sqrt{\sec (c+d x)} (A+B \sec (c+d x))}{\sqrt{a+b \sec (c+d x)}} \, dx\)

Optimal. Leaf size=138 \[ \frac{2 A \sqrt{\sec (c+d x)} \sqrt{\frac{a \cos (c+d x)+b}{a+b}} \text{EllipticF}\left (\frac{1}{2} (c+d x),\frac{2 a}{a+b}\right )}{d \sqrt{a+b \sec (c+d x)}}+\frac{2 B \sqrt{\sec (c+d x)} \sqrt{\frac{a \cos (c+d x)+b}{a+b}} \Pi \left (2;\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{d \sqrt{a+b \sec (c+d x)}} \]

[Out]

(2*A*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*a)/(a + b)]*Sqrt[Sec[c + d*x]])/(d*Sqrt[a +
b*Sec[c + d*x]]) + (2*B*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*EllipticPi[2, (c + d*x)/2, (2*a)/(a + b)]*Sqrt[Sec[
c + d*x]])/(d*Sqrt[a + b*Sec[c + d*x]])

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Rubi [A]  time = 0.391646, antiderivative size = 138, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {4036, 3858, 2663, 2661, 3859, 2807, 2805} \[ \frac{2 A \sqrt{\sec (c+d x)} \sqrt{\frac{a \cos (c+d x)+b}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{d \sqrt{a+b \sec (c+d x)}}+\frac{2 B \sqrt{\sec (c+d x)} \sqrt{\frac{a \cos (c+d x)+b}{a+b}} \Pi \left (2;\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{d \sqrt{a+b \sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*A*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*a)/(a + b)]*Sqrt[Sec[c + d*x]])/(d*Sqrt[a +
b*Sec[c + d*x]]) + (2*B*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*EllipticPi[2, (c + d*x)/2, (2*a)/(a + b)]*Sqrt[Sec[
c + d*x]])/(d*Sqrt[a + b*Sec[c + d*x]])

Rule 4036

Int[(Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/Sqrt[csc[(e_.) + (f_.)*(x_)]*
(b_.) + (a_)], x_Symbol] :> Dist[A, Int[Sqrt[d*Csc[e + f*x]]/Sqrt[a + b*Csc[e + f*x]], x], x] + Dist[B/d, Int[
(d*Csc[e + f*x])^(3/2)/Sqrt[a + b*Csc[e + f*x]], x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a*B, 0
] && NeQ[a^2 - b^2, 0]

Rule 3858

Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[(Sqrt[d*
Csc[e + f*x]]*Sqrt[b + a*Sin[e + f*x]])/Sqrt[a + b*Csc[e + f*x]], Int[1/Sqrt[b + a*Sin[e + f*x]], x], x] /; Fr
eeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 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 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 3859

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(3/2)/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[(d*Sqr
t[d*Csc[e + f*x]]*Sqrt[b + a*Sin[e + f*x]])/Sqrt[a + b*Csc[e + f*x]], Int[1/(Sin[e + f*x]*Sqrt[b + a*Sin[e + f
*x]]), x], x] /; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0]

Rule 2807

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

Rule 2805

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

Rubi steps

\begin{align*} \int \frac{\sqrt{\sec (c+d x)} (A+B \sec (c+d x))}{\sqrt{a+b \sec (c+d x)}} \, dx &=A \int \frac{\sqrt{\sec (c+d x)}}{\sqrt{a+b \sec (c+d x)}} \, dx+B \int \frac{\sec ^{\frac{3}{2}}(c+d x)}{\sqrt{a+b \sec (c+d x)}} \, dx\\ &=\frac{\left (A \sqrt{b+a \cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{b+a \cos (c+d x)}} \, dx}{\sqrt{a+b \sec (c+d x)}}+\frac{\left (B \sqrt{b+a \cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\sec (c+d x)}{\sqrt{b+a \cos (c+d x)}} \, dx}{\sqrt{a+b \sec (c+d x)}}\\ &=\frac{\left (A \sqrt{\frac{b+a \cos (c+d x)}{a+b}} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\frac{b}{a+b}+\frac{a \cos (c+d x)}{a+b}}} \, dx}{\sqrt{a+b \sec (c+d x)}}+\frac{\left (B \sqrt{\frac{b+a \cos (c+d x)}{a+b}} \sqrt{\sec (c+d x)}\right ) \int \frac{\sec (c+d x)}{\sqrt{\frac{b}{a+b}+\frac{a \cos (c+d x)}{a+b}}} \, dx}{\sqrt{a+b \sec (c+d x)}}\\ &=\frac{2 A \sqrt{\frac{b+a \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right ) \sqrt{\sec (c+d x)}}{d \sqrt{a+b \sec (c+d x)}}+\frac{2 B \sqrt{\frac{b+a \cos (c+d x)}{a+b}} \Pi \left (2;\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right ) \sqrt{\sec (c+d x)}}{d \sqrt{a+b \sec (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 0.258445, size = 91, normalized size = 0.66 \[ \frac{2 \sqrt{\sec (c+d x)} \sqrt{\frac{a \cos (c+d x)+b}{a+b}} \left (A \text{EllipticF}\left (\frac{1}{2} (c+d x),\frac{2 a}{a+b}\right )+B \Pi \left (2;\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )\right )}{d \sqrt{a+b \sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*(A*EllipticF[(c + d*x)/2, (2*a)/(a + b)] + B*EllipticPi[2, (c + d*x)/2,
(2*a)/(a + b)])*Sqrt[Sec[c + d*x]])/(d*Sqrt[a + b*Sec[c + d*x]])

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Maple [C]  time = 0.395, size = 283, normalized size = 2.1 \begin{align*} 2\,{\frac{\cos \left ( dx+c \right ) \left ( \sin \left ( dx+c \right ) \right ) ^{2}\sqrt{ \left ( \cos \left ( dx+c \right ) \right ) ^{-1}}\sqrt{ \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}}{d \left ( -1+\cos \left ( dx+c \right ) \right ) \left ( b+a\cos \left ( dx+c \right ) \right ) } \left ( A{\it EllipticF} \left ({\frac{-1+\cos \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }\sqrt{{\frac{a-b}{a+b}}}},\sqrt{-{\frac{a+b}{a-b}}} \right ) -B{\it EllipticF} \left ({\frac{-1+\cos \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }\sqrt{{\frac{a-b}{a+b}}}},\sqrt{-{\frac{a+b}{a-b}}} \right ) +2\,B{\it EllipticPi} \left ({\frac{-1+\cos \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }\sqrt{{\frac{a-b}{a+b}}}},{\frac{a+b}{a-b}},{i{\frac{1}{\sqrt{{\frac{a-b}{a+b}}}}}} \right ) \right ) \sqrt{{\frac{b+a\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) }}}\sqrt{{\frac{b+a\cos \left ( dx+c \right ) }{ \left ( a+b \right ) \left ( \cos \left ( dx+c \right ) +1 \right ) }}}{\frac{1}{\sqrt{{\frac{a-b}{a+b}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/d/((a-b)/(a+b))^(1/2)*(A*EllipticF((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))-B*El
lipticF((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))+2*B*EllipticPi((-1+cos(d*x+c))*((
a-b)/(a+b))^(1/2)/sin(d*x+c),(a+b)/(a-b),I/((a-b)/(a+b))^(1/2)))*cos(d*x+c)*sin(d*x+c)^2*(1/cos(d*x+c))^(1/2)*
((b+a*cos(d*x+c))/cos(d*x+c))^(1/2)*(1/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)/(
-1+cos(d*x+c))/(b+a*cos(d*x+c))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B \sec \left (d x + c\right ) + A\right )} \sqrt{\sec \left (d x + c\right )}}{\sqrt{b \sec \left (d x + c\right ) + a}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((B*sec(d*x + c) + A)*sqrt(sec(d*x + c))/sqrt(b*sec(d*x + c) + a), x)

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Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (A + B \sec{\left (c + d x \right )}\right ) \sqrt{\sec{\left (c + d x \right )}}}{\sqrt{a + b \sec{\left (c + d x \right )}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B \sec \left (d x + c\right ) + A\right )} \sqrt{\sec \left (d x + c\right )}}{\sqrt{b \sec \left (d x + c\right ) + a}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((B*sec(d*x + c) + A)*sqrt(sec(d*x + c))/sqrt(b*sec(d*x + c) + a), x)