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

Optimal. Leaf size=187 \[ \frac{2 \left (a^2-b^2\right ) \sqrt{\sec (c+d x)} \sqrt{\frac{a \cos (c+d x)+b}{a+b}} \text{EllipticF}\left (\frac{1}{2} (c+d x),\frac{2 a}{a+b}\right )}{3 d \sqrt{a+b \sec (c+d x)}}+\frac{2 a \sin (c+d x) \sqrt{a+b \sec (c+d x)}}{3 d \sqrt{\sec (c+d x)}}+\frac{8 b \sqrt{a+b \sec (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{3 d \sqrt{\sec (c+d x)} \sqrt{\frac{a \cos (c+d x)+b}{a+b}}} \]

[Out]

(2*(a^2 - b^2)*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*a)/(a + b)]*Sqrt[Sec[c + d*x]])/(3
*d*Sqrt[a + b*Sec[c + d*x]]) + (8*b*EllipticE[(c + d*x)/2, (2*a)/(a + b)]*Sqrt[a + b*Sec[c + d*x]])/(3*d*Sqrt[
(b + a*Cos[c + d*x])/(a + b)]*Sqrt[Sec[c + d*x]]) + (2*a*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x])/(3*d*Sqrt[Sec[
c + d*x]])

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Rubi [A]  time = 0.405912, antiderivative size = 187, 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 = {3864, 4035, 3856, 2655, 2653, 3858, 2663, 2661} \[ \frac{2 \left (a^2-b^2\right ) \sqrt{\sec (c+d x)} \sqrt{\frac{a \cos (c+d x)+b}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{3 d \sqrt{a+b \sec (c+d x)}}+\frac{2 a \sin (c+d x) \sqrt{a+b \sec (c+d x)}}{3 d \sqrt{\sec (c+d x)}}+\frac{8 b \sqrt{a+b \sec (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{3 d \sqrt{\sec (c+d x)} \sqrt{\frac{a \cos (c+d x)+b}{a+b}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(a^2 - b^2)*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*a)/(a + b)]*Sqrt[Sec[c + d*x]])/(3
*d*Sqrt[a + b*Sec[c + d*x]]) + (8*b*EllipticE[(c + d*x)/2, (2*a)/(a + b)]*Sqrt[a + b*Sec[c + d*x]])/(3*d*Sqrt[
(b + a*Cos[c + d*x])/(a + b)]*Sqrt[Sec[c + d*x]]) + (2*a*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x])/(3*d*Sqrt[Sec[
c + d*x]])

Rule 3864

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(3/2), x_Symbol] :> Simp[(a*Co
t[e + f*x]*Sqrt[a + b*Csc[e + f*x]]*(d*Csc[e + f*x])^n)/(f*n), x] + Dist[1/(2*d*n), Int[((d*Csc[e + f*x])^(n +
1)*Simp[a*b*(2*n - 1) + 2*(b^2*n + a^2*(n + 1))*Csc[e + f*x] + a*b*(2*n + 3)*Csc[e + f*x]^2, x])/Sqrt[a + b*C
sc[e + f*x]], x], x] /; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0] && LeQ[n, -1] && IntegersQ[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{(a+b \sec (c+d x))^{3/2}}{\sec ^{\frac{3}{2}}(c+d x)} \, dx &=\frac{2 a \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{3 d \sqrt{\sec (c+d x)}}-\frac{1}{3} \int \frac{-4 a b-\left (a^2+3 b^2\right ) \sec (c+d x)}{\sqrt{\sec (c+d x)} \sqrt{a+b \sec (c+d x)}} \, dx\\ &=\frac{2 a \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{3 d \sqrt{\sec (c+d x)}}+\frac{1}{3} (4 b) \int \frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{\sec (c+d x)}} \, dx-\frac{1}{3} \left (-a^2+b^2\right ) \int \frac{\sqrt{\sec (c+d x)}}{\sqrt{a+b \sec (c+d x)}} \, dx\\ &=\frac{2 a \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{3 d \sqrt{\sec (c+d x)}}-\frac{\left (\left (-a^2+b^2\right ) \sqrt{b+a \cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{b+a \cos (c+d x)}} \, dx}{3 \sqrt{a+b \sec (c+d x)}}+\frac{\left (4 b \sqrt{a+b \sec (c+d x)}\right ) \int \sqrt{b+a \cos (c+d x)} \, dx}{3 \sqrt{b+a \cos (c+d x)} \sqrt{\sec (c+d x)}}\\ &=\frac{2 a \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{3 d \sqrt{\sec (c+d x)}}-\frac{\left (\left (-a^2+b^2\right ) \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}{3 \sqrt{a+b \sec (c+d x)}}+\frac{\left (4 b \sqrt{a+b \sec (c+d x)}\right ) \int \sqrt{\frac{b}{a+b}+\frac{a \cos (c+d x)}{a+b}} \, dx}{3 \sqrt{\frac{b+a \cos (c+d x)}{a+b}} \sqrt{\sec (c+d x)}}\\ &=\frac{2 \left (a^2-b^2\right ) \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)}}{3 d \sqrt{a+b \sec (c+d x)}}+\frac{8 b E\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right ) \sqrt{a+b \sec (c+d x)}}{3 d \sqrt{\frac{b+a \cos (c+d x)}{a+b}} \sqrt{\sec (c+d x)}}+\frac{2 a \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{3 d \sqrt{\sec (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 0.653467, size = 156, normalized size = 0.83 \[ \frac{(a+b \sec (c+d x))^{3/2} \left (2 \left (a^2-b^2\right ) \sqrt{\frac{a \cos (c+d x)+b}{a+b}} \text{EllipticF}\left (\frac{1}{2} (c+d x),\frac{2 a}{a+b}\right )+2 a \sin (c+d x) (a \cos (c+d x)+b)+8 b (a+b) \sqrt{\frac{a \cos (c+d x)+b}{a+b}} E\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )\right )}{3 d \sec ^{\frac{3}{2}}(c+d x) (a \cos (c+d x)+b)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a + b*Sec[c + d*x])^(3/2)*(8*b*(a + b)*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*EllipticE[(c + d*x)/2, (2*a)/(a +
b)] + 2*(a^2 - b^2)*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*a)/(a + b)] + 2*a*(b + a*Cos[
c + d*x])*Sin[c + d*x]))/(3*d*(b + a*Cos[c + d*x])^2*Sec[c + d*x]^(3/2))

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Maple [B]  time = 0.311, size = 1219, normalized size = 6.5 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3/d/((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^2-4*cos(d*x+c)*s
in(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*b+3*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)*EllipticF((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a
-b))^(1/2))*b^2+4*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))*a*b-4*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^2+EllipticF((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)
/(a-b))^(1/2))*a^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)*sin(d*x+c)-4*Ellip
ticF((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*a*b*(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)+3*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(1/(cos(
d*x+c)+1))^(1/2)*EllipticF((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*b^2*sin(d*x+c)
+4*EllipticE((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*a*b*(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)-4*EllipticE((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/si
n(d*x+c),(-(a+b)/(a-b))^(1/2))*b^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)*si
n(d*x+c)+((a-b)/(a+b))^(1/2)*cos(d*x+c)^3*a^2+5*((a-b)/(a+b))^(1/2)*cos(d*x+c)^2*a*b-a^2*((a-b)/(a+b))^(1/2)*c
os(d*x+c)-4*((a-b)/(a+b))^(1/2)*cos(d*x+c)*a*b+4*((a-b)/(a+b))^(1/2)*cos(d*x+c)*b^2-a*b*((a-b)/(a+b))^(1/2)-4*
b^2*((a-b)/(a+b))^(1/2))*((b+a*cos(d*x+c))/cos(d*x+c))^(1/2)*cos(d*x+c)^2*(1/cos(d*x+c))^(3/2)/sin(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 )}^{\frac{3}{2}}}{\sec \left (d x + c\right )^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [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))**(3/2)/sec(d*x+c)**(3/2),x)

[Out]

Timed out

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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 )}^{\frac{3}{2}}}{\sec \left (d x + c\right )^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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