3.658 \(\int \frac{1}{\sqrt{\sec (c+d x)} (a+b \sec (c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=214 \[ -\frac{4 b \sqrt{\sec (c+d x)} \sqrt{\frac{a \cos (c+d x)+b}{a+b}} \text{EllipticF}\left (\frac{1}{2} (c+d x),\frac{2 a}{a+b}\right )}{a^2 d \sqrt{a+b \sec (c+d x)}}+\frac{2 b^2 \sin (c+d x) \sqrt{\sec (c+d x)}}{a d \left (a^2-b^2\right ) \sqrt{a+b \sec (c+d x)}}+\frac{2 \left (a^2-2 b^2\right ) \sqrt{a+b \sec (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{a^2 d \left (a^2-b^2\right ) \sqrt{\sec (c+d x)} \sqrt{\frac{a \cos (c+d x)+b}{a+b}}} \]

[Out]

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

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Rubi [A]  time = 0.434004, antiderivative size = 214, 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 = {3847, 4035, 3856, 2655, 2653, 3858, 2663, 2661} \[ \frac{2 b^2 \sin (c+d x) \sqrt{\sec (c+d x)}}{a d \left (a^2-b^2\right ) \sqrt{a+b \sec (c+d x)}}+\frac{2 \left (a^2-2 b^2\right ) \sqrt{a+b \sec (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{a^2 d \left (a^2-b^2\right ) \sqrt{\sec (c+d x)} \sqrt{\frac{a \cos (c+d x)+b}{a+b}}}-\frac{4 b \sqrt{\sec (c+d x)} \sqrt{\frac{a \cos (c+d x)+b}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{a^2 d \sqrt{a+b \sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[Sec[c + d*x]]*(a + b*Sec[c + d*x])^(3/2)),x]

[Out]

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

Rule 3847

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

Mathematica [A]  time = 0.654781, size = 165, normalized size = 0.77 \[ \frac{2 \sqrt{\sec (c+d x)} \left (b \left (a b \sin (c+d x)-2 \left (a^2-b^2\right ) \sqrt{\frac{a \cos (c+d x)+b}{a+b}} \text{EllipticF}\left (\frac{1}{2} (c+d x),\frac{2 a}{a+b}\right )\right )+\left (a^2 b+a^3-2 a b^2-2 b^3\right ) \sqrt{\frac{a \cos (c+d x)+b}{a+b}} E\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )\right )}{a^2 d (a-b) (a+b) \sqrt{a+b \sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(Sqrt[Sec[c + d*x]]*(a + b*Sec[c + d*x])^(3/2)),x]

[Out]

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

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Maple [B]  time = 0.296, size = 999, normalized size = 4.7 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/d/((a-b)/(a+b))^(1/2)/(a+b)/a^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+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*b-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^2+2*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)+2*EllipticF((-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)-(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^2*sin(d*x+c)+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*(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)-((a-b)/
(a+b))^(1/2)*cos(d*x+c)^2*a^2-((a-b)/(a+b))^(1/2)*cos(d*x+c)^2*a*b+a^2*((a-b)/(a+b))^(1/2)*cos(d*x+c)-2*((a-b)
/(a+b))^(1/2)*cos(d*x+c)*b^2+a*b*((a-b)/(a+b))^(1/2)+2*b^2*((a-b)/(a+b))^(1/2))*((b+a*cos(d*x+c))/cos(d*x+c))^
(1/2)/(1/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{1}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac{3}{2}} \sqrt{\sec \left (d x + c\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/((b*sec(d*x + c) + a)^(3/2)*sqrt(sec(d*x + c))), 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 )^{3} + 2 \, a b \sec \left (d x + c\right )^{2} + a^{2} \sec \left (d x + c\right )}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (a + b \sec{\left (c + d x \right )}\right )^{\frac{3}{2}} \sqrt{\sec{\left (c + d x \right )}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac{3}{2}} \sqrt{\sec \left (d x + c\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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