3.657 \(\int \frac{\sqrt{\sec (c+d x)}}{(a+b \sec (c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=200 \[ \frac{2 \sqrt{\sec (c+d x)} \sqrt{\frac{a \cos (c+d x)+b}{a+b}} \text{EllipticF}\left (\frac{1}{2} (c+d x),\frac{2 a}{a+b}\right )}{a d \sqrt{a+b \sec (c+d x)}}-\frac{2 b \sin (c+d x) \sqrt{\sec (c+d x)}}{d \left (a^2-b^2\right ) \sqrt{a+b \sec (c+d x)}}+\frac{2 b \sqrt{a+b \sec (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{a d \left (a^2-b^2\right ) \sqrt{\sec (c+d x)} \sqrt{\frac{a \cos (c+d x)+b}{a+b}}} \]

[Out]

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

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Rubi [A]  time = 0.379514, antiderivative size = 200, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.32, Rules used = {3843, 4035, 3856, 2655, 2653, 3858, 2663, 2661} \[ -\frac{2 b \sin (c+d x) \sqrt{\sec (c+d x)}}{d \left (a^2-b^2\right ) \sqrt{a+b \sec (c+d x)}}+\frac{2 b \sqrt{a+b \sec (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{a d \left (a^2-b^2\right ) \sqrt{\sec (c+d x)} \sqrt{\frac{a \cos (c+d x)+b}{a+b}}}+\frac{2 \sqrt{\sec (c+d x)} \sqrt{\frac{a \cos (c+d x)+b}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{a d \sqrt{a+b \sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[Sec[c + d*x]]/(a + b*Sec[c + d*x])^(3/2),x]

[Out]

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

Rule 3843

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> -Simp[(b*d*
Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^(n - 1))/(f*(m + 1)*(a^2 - b^2)), x] + Dist[1/((m +
1)*(a^2 - b^2)), Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^(n - 1)*Simp[b*d*(n - 1) + a*d*(m + 1)*Csc
[e + f*x] - b*d*(m + n + 1)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0] && Lt
Q[m, -1] && LtQ[0, n, 1] && IntegersQ[2*m, 2*n]

Rule 4035

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

Rule 3856

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

Rubi steps

\begin{align*} \int \frac{\sqrt{\sec (c+d x)}}{(a+b \sec (c+d x))^{3/2}} \, dx &=-\frac{2 b \sqrt{\sec (c+d x)} \sin (c+d x)}{\left (a^2-b^2\right ) d \sqrt{a+b \sec (c+d x)}}-\frac{2 \int \frac{-\frac{b}{2}-\frac{1}{2} a \sec (c+d x)}{\sqrt{\sec (c+d x)} \sqrt{a+b \sec (c+d x)}} \, dx}{a^2-b^2}\\ &=-\frac{2 b \sqrt{\sec (c+d x)} \sin (c+d x)}{\left (a^2-b^2\right ) d \sqrt{a+b \sec (c+d x)}}+\frac{\int \frac{\sqrt{\sec (c+d x)}}{\sqrt{a+b \sec (c+d x)}} \, dx}{a}+\frac{b \int \frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{\sec (c+d x)}} \, dx}{a \left (a^2-b^2\right )}\\ &=-\frac{2 b \sqrt{\sec (c+d x)} \sin (c+d x)}{\left (a^2-b^2\right ) d \sqrt{a+b \sec (c+d x)}}+\frac{\left (\sqrt{b+a \cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{b+a \cos (c+d x)}} \, dx}{a \sqrt{a+b \sec (c+d x)}}+\frac{\left (b \sqrt{a+b \sec (c+d x)}\right ) \int \sqrt{b+a \cos (c+d x)} \, dx}{a \left (a^2-b^2\right ) \sqrt{b+a \cos (c+d x)} \sqrt{\sec (c+d x)}}\\ &=-\frac{2 b \sqrt{\sec (c+d x)} \sin (c+d x)}{\left (a^2-b^2\right ) d \sqrt{a+b \sec (c+d x)}}+\frac{\left (\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}{a \sqrt{a+b \sec (c+d x)}}+\frac{\left (b \sqrt{a+b \sec (c+d x)}\right ) \int \sqrt{\frac{b}{a+b}+\frac{a \cos (c+d x)}{a+b}} \, dx}{a \left (a^2-b^2\right ) \sqrt{\frac{b+a \cos (c+d x)}{a+b}} \sqrt{\sec (c+d x)}}\\ &=\frac{2 \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)}}{a d \sqrt{a+b \sec (c+d x)}}+\frac{2 b E\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right ) \sqrt{a+b \sec (c+d x)}}{a \left (a^2-b^2\right ) d \sqrt{\frac{b+a \cos (c+d x)}{a+b}} \sqrt{\sec (c+d x)}}-\frac{2 b \sqrt{\sec (c+d x)} \sin (c+d x)}{\left (a^2-b^2\right ) d \sqrt{a+b \sec (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 0.581527, size = 156, normalized size = 0.78 \[ \frac{2 \sec ^{\frac{3}{2}}(c+d x) (a \cos (c+d x)+b) \left (\left (a^2-b^2\right ) \sqrt{\frac{a \cos (c+d x)+b}{a+b}} \text{EllipticF}\left (\frac{1}{2} (c+d x),\frac{2 a}{a+b}\right )-a b \sin (c+d x)+b (a+b) \sqrt{\frac{a \cos (c+d x)+b}{a+b}} E\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )\right )}{a d (a-b) (a+b) (a+b \sec (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[Sec[c + d*x]]/(a + b*Sec[c + d*x])^(3/2),x]

[Out]

(2*(b + a*Cos[c + d*x])*Sec[c + d*x]^(3/2)*(b*(a + b)*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*EllipticE[(c + d*x)/2
, (2*a)/(a + b)] + (a^2 - b^2)*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*a)/(a + b)] - a*b*
Sin[c + d*x]))/(a*(a - b)*(a + b)*d*(a + b*Sec[c + d*x])^(3/2))

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Maple [B]  time = 0.265, size = 510, normalized size = 2.6 \begin{align*} 2\,{\frac{\cos \left ( dx+c \right ) \sqrt{ \left ( \cos \left ( dx+c \right ) \right ) ^{-1}}}{da \left ( a+b \right ) \left ( b+a\cos \left ( dx+c \right ) \right ) \sin \left ( dx+c \right ) } \left ( -\cos \left ( dx+c \right ) \sin \left ( dx+c \right ){\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 ) \sqrt{{\frac{b+a\cos \left ( dx+c \right ) }{ \left ( a+b \right ) \left ( \cos \left ( dx+c \right ) +1 \right ) }}}\sqrt{ \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}a-\cos \left ( dx+c \right ) \sin \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 ) }}}\sqrt{ \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}{\it EllipticE} \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 ) a\sqrt{{\frac{b+a\cos \left ( dx+c \right ) }{ \left ( a+b \right ) \left ( \cos \left ( dx+c \right ) +1 \right ) }}}\sqrt{ \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}\sin \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 ) }}}\sqrt{ \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}{\it EllipticE} \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\sin \left ( dx+c \right ) +\cos \left ( dx+c \right ) \sqrt{{\frac{a-b}{a+b}}}b-\sqrt{{\frac{a-b}{a+b}}}b \right ) \sqrt{{\frac{b+a\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) }}}{\frac{1}{\sqrt{{\frac{a-b}{a+b}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^(1/2)/(a+b*sec(d*x+c))^(3/2),x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{\sec \left (d x + c\right )}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^(1/2)/(a+b*sec(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

integrate(sqrt(sec(d*x + c))/(b*sec(d*x + c) + a)^(3/2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{b \sec \left (d x + c\right ) + a} \sqrt{\sec \left (d x + c\right )}}{b^{2} \sec \left (d x + c\right )^{2} + 2 \, a b \sec \left (d x + c\right ) + a^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^(1/2)/(a+b*sec(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{\sec{\left (c + d x \right )}}}{\left (a + b \sec{\left (c + d x \right )}\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**(1/2)/(a+b*sec(d*x+c))**(3/2),x)

[Out]

Integral(sqrt(sec(c + d*x))/(a + b*sec(c + d*x))**(3/2), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{\sec \left (d x + c\right )}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^(1/2)/(a+b*sec(d*x+c))^(3/2),x, algorithm="giac")

[Out]

integrate(sqrt(sec(d*x + c))/(b*sec(d*x + c) + a)^(3/2), x)