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

Optimal. Leaf size=127 \[ -\frac{4 b \left (2 a^2-b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} \left (c+d \sqrt{x}\right )\right )}{\sqrt{a+b}}\right )}{a^2 d (a-b)^{3/2} (a+b)^{3/2}}+\frac{2 b^2 \tan \left (c+d \sqrt{x}\right )}{a d \left (a^2-b^2\right ) \left (a+b \sec \left (c+d \sqrt{x}\right )\right )}+\frac{2 \sqrt{x}}{a^2} \]

[Out]

(2*Sqrt[x])/a^2 - (4*b*(2*a^2 - b^2)*ArcTanh[(Sqrt[a - b]*Tan[(c + d*Sqrt[x])/2])/Sqrt[a + b]])/(a^2*(a - b)^(
3/2)*(a + b)^(3/2)*d) + (2*b^2*Tan[c + d*Sqrt[x]])/(a*(a^2 - b^2)*d*(a + b*Sec[c + d*Sqrt[x]]))

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Rubi [A]  time = 0.198528, antiderivative size = 127, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.273, Rules used = {4204, 3785, 3919, 3831, 2659, 208} \[ -\frac{4 b \left (2 a^2-b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} \left (c+d \sqrt{x}\right )\right )}{\sqrt{a+b}}\right )}{a^2 d (a-b)^{3/2} (a+b)^{3/2}}+\frac{2 b^2 \tan \left (c+d \sqrt{x}\right )}{a d \left (a^2-b^2\right ) \left (a+b \sec \left (c+d \sqrt{x}\right )\right )}+\frac{2 \sqrt{x}}{a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[x])/a^2 - (4*b*(2*a^2 - b^2)*ArcTanh[(Sqrt[a - b]*Tan[(c + d*Sqrt[x])/2])/Sqrt[a + b]])/(a^2*(a - b)^(
3/2)*(a + b)^(3/2)*d) + (2*b^2*Tan[c + d*Sqrt[x]])/(a*(a^2 - b^2)*d*(a + b*Sec[c + d*Sqrt[x]]))

Rule 4204

Int[(x_)^(m_.)*((a_.) + (b_.)*Sec[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplif
y[(m + 1)/n] - 1)*(a + b*Sec[c + d*x])^p, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p}, x] && IGtQ[Simplify[
(m + 1)/n], 0] && IntegerQ[p]

Rule 3785

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Simp[(b^2*Cot[c + d*x]*(a + b*Csc[c + d*x])^(n +
1))/(a*d*(n + 1)*(a^2 - b^2)), x] + Dist[1/(a*(n + 1)*(a^2 - b^2)), Int[(a + b*Csc[c + d*x])^(n + 1)*Simp[(a^
2 - b^2)*(n + 1) - a*b*(n + 1)*Csc[c + d*x] + b^2*(n + 2)*Csc[c + d*x]^2, x], x], x] /; FreeQ[{a, b, c, d}, x]
&& NeQ[a^2 - b^2, 0] && LtQ[n, -1] && IntegerQ[2*n]

Rule 3919

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

Rule 3831

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

Rule 2659

Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x
]}, Dist[(2*e)/d, Subst[Int[1/(a + b + (a - b)*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}
, x] && NeQ[a^2 - b^2, 0]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rubi steps

\begin{align*} \int \frac{1}{\sqrt{x} \left (a+b \sec \left (c+d \sqrt{x}\right )\right )^2} \, dx &=2 \operatorname{Subst}\left (\int \frac{1}{(a+b \sec (c+d x))^2} \, dx,x,\sqrt{x}\right )\\ &=\frac{2 b^2 \tan \left (c+d \sqrt{x}\right )}{a \left (a^2-b^2\right ) d \left (a+b \sec \left (c+d \sqrt{x}\right )\right )}-\frac{2 \operatorname{Subst}\left (\int \frac{-a^2+b^2+a b \sec (c+d x)}{a+b \sec (c+d x)} \, dx,x,\sqrt{x}\right )}{a \left (a^2-b^2\right )}\\ &=\frac{2 \sqrt{x}}{a^2}+\frac{2 b^2 \tan \left (c+d \sqrt{x}\right )}{a \left (a^2-b^2\right ) d \left (a+b \sec \left (c+d \sqrt{x}\right )\right )}-\frac{\left (2 b \left (2 a^2-b^2\right )\right ) \operatorname{Subst}\left (\int \frac{\sec (c+d x)}{a+b \sec (c+d x)} \, dx,x,\sqrt{x}\right )}{a^2 \left (a^2-b^2\right )}\\ &=\frac{2 \sqrt{x}}{a^2}+\frac{2 b^2 \tan \left (c+d \sqrt{x}\right )}{a \left (a^2-b^2\right ) d \left (a+b \sec \left (c+d \sqrt{x}\right )\right )}-\frac{\left (2 \left (2 a^2-b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{1+\frac{a \cos (c+d x)}{b}} \, dx,x,\sqrt{x}\right )}{a^2 \left (a^2-b^2\right )}\\ &=\frac{2 \sqrt{x}}{a^2}+\frac{2 b^2 \tan \left (c+d \sqrt{x}\right )}{a \left (a^2-b^2\right ) d \left (a+b \sec \left (c+d \sqrt{x}\right )\right )}-\frac{\left (4 \left (2 a^2-b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{1+\frac{a}{b}+\left (1-\frac{a}{b}\right ) x^2} \, dx,x,\tan \left (\frac{1}{2} \left (c+d \sqrt{x}\right )\right )\right )}{a^2 \left (a^2-b^2\right ) d}\\ &=\frac{2 \sqrt{x}}{a^2}-\frac{4 b \left (2 a^2-b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} \left (c+d \sqrt{x}\right )\right )}{\sqrt{a+b}}\right )}{a^2 (a-b)^{3/2} (a+b)^{3/2} d}+\frac{2 b^2 \tan \left (c+d \sqrt{x}\right )}{a \left (a^2-b^2\right ) d \left (a+b \sec \left (c+d \sqrt{x}\right )\right )}\\ \end{align*}

Mathematica [A]  time = 0.754276, size = 163, normalized size = 1.28 \[ \frac{2 \left (\frac{b \left (\left (a^2-b^2\right ) \left (c+d \sqrt{x}\right )+a b \sin \left (c+d \sqrt{x}\right )\right )+a \left (a^2-b^2\right ) \left (c+d \sqrt{x}\right ) \cos \left (c+d \sqrt{x}\right )}{a \cos \left (c+d \sqrt{x}\right )+b}-\frac{2 b \left (b^2-2 a^2\right ) \tanh ^{-1}\left (\frac{(b-a) \tan \left (\frac{1}{2} \left (c+d \sqrt{x}\right )\right )}{\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2}}\right )}{a^2 d (a-b) (a+b)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*((-2*b*(-2*a^2 + b^2)*ArcTanh[((-a + b)*Tan[(c + d*Sqrt[x])/2])/Sqrt[a^2 - b^2]])/Sqrt[a^2 - b^2] + (a*(a^2
- b^2)*(c + d*Sqrt[x])*Cos[c + d*Sqrt[x]] + b*((a^2 - b^2)*(c + d*Sqrt[x]) + a*b*Sin[c + d*Sqrt[x]]))/(b + a*
Cos[c + d*Sqrt[x]])))/(a^2*(a - b)*(a + b)*d)

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Maple [A]  time = 0.075, size = 216, normalized size = 1.7 \begin{align*} 4\,{\frac{\arctan \left ( \tan \left ( c/2+1/2\,d\sqrt{x} \right ) \right ) }{d{a}^{2}}}-4\,{\frac{{b}^{2}\tan \left ( c/2+1/2\,d\sqrt{x} \right ) }{da \left ({a}^{2}-{b}^{2} \right ) \left ( \left ( \tan \left ( c/2+1/2\,d\sqrt{x} \right ) \right ) ^{2}a- \left ( \tan \left ( c/2+1/2\,d\sqrt{x} \right ) \right ) ^{2}b-a-b \right ) }}-8\,{\frac{b}{d \left ( a+b \right ) \left ( a-b \right ) \sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}{\it Artanh} \left ({\frac{ \left ( a-b \right ) \tan \left ( c/2+1/2\,d\sqrt{x} \right ) }{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}} \right ) }+4\,{\frac{{b}^{3}}{d{a}^{2} \left ( a+b \right ) \left ( a-b \right ) \sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}{\it Artanh} \left ({\frac{ \left ( a-b \right ) \tan \left ( c/2+1/2\,d\sqrt{x} \right ) }{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4/d/a^2*arctan(tan(1/2*c+1/2*d*x^(1/2)))-4/d*b^2/a/(a^2-b^2)*tan(1/2*c+1/2*d*x^(1/2))/(tan(1/2*c+1/2*d*x^(1/2)
)^2*a-tan(1/2*c+1/2*d*x^(1/2))^2*b-a-b)-8/d*b/(a+b)/(a-b)/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tan(1/2*c+1/2*d*x^
(1/2))/((a+b)*(a-b))^(1/2))+4/d*b^3/a^2/(a+b)/(a-b)/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tan(1/2*c+1/2*d*x^(1/2))
/((a+b)*(a-b))^(1/2))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.9442, size = 1307, normalized size = 10.29 \begin{align*} \left [\frac{2 \,{\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} d \sqrt{x} \cos \left (d \sqrt{x} + c\right ) + 2 \,{\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} d \sqrt{x} +{\left ({\left (2 \, a^{3} b - a b^{3}\right )} \sqrt{a^{2} - b^{2}} \cos \left (d \sqrt{x} + c\right ) +{\left (2 \, a^{2} b^{2} - b^{4}\right )} \sqrt{a^{2} - b^{2}}\right )} \log \left (\frac{2 \, a b \cos \left (d \sqrt{x} + c\right ) -{\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d \sqrt{x} + c\right )^{2} + 2 \, a^{2} - b^{2} - 2 \,{\left (\sqrt{a^{2} - b^{2}} b \cos \left (d \sqrt{x} + c\right ) + \sqrt{a^{2} - b^{2}} a\right )} \sin \left (d \sqrt{x} + c\right )}{a^{2} \cos \left (d \sqrt{x} + c\right )^{2} + 2 \, a b \cos \left (d \sqrt{x} + c\right ) + b^{2}}\right ) + 2 \,{\left (a^{3} b^{2} - a b^{4}\right )} \sin \left (d \sqrt{x} + c\right )}{{\left (a^{7} - 2 \, a^{5} b^{2} + a^{3} b^{4}\right )} d \cos \left (d \sqrt{x} + c\right ) +{\left (a^{6} b - 2 \, a^{4} b^{3} + a^{2} b^{5}\right )} d}, \frac{2 \,{\left ({\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} d \sqrt{x} \cos \left (d \sqrt{x} + c\right ) +{\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} d \sqrt{x} -{\left ({\left (2 \, a^{3} b - a b^{3}\right )} \sqrt{-a^{2} + b^{2}} \cos \left (d \sqrt{x} + c\right ) +{\left (2 \, a^{2} b^{2} - b^{4}\right )} \sqrt{-a^{2} + b^{2}}\right )} \arctan \left (-\frac{\sqrt{-a^{2} + b^{2}} b \cos \left (d \sqrt{x} + c\right ) + \sqrt{-a^{2} + b^{2}} a}{{\left (a^{2} - b^{2}\right )} \sin \left (d \sqrt{x} + c\right )}\right ) +{\left (a^{3} b^{2} - a b^{4}\right )} \sin \left (d \sqrt{x} + c\right )\right )}}{{\left (a^{7} - 2 \, a^{5} b^{2} + a^{3} b^{4}\right )} d \cos \left (d \sqrt{x} + c\right ) +{\left (a^{6} b - 2 \, a^{4} b^{3} + a^{2} b^{5}\right )} d}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[(2*(a^5 - 2*a^3*b^2 + a*b^4)*d*sqrt(x)*cos(d*sqrt(x) + c) + 2*(a^4*b - 2*a^2*b^3 + b^5)*d*sqrt(x) + ((2*a^3*b
- a*b^3)*sqrt(a^2 - b^2)*cos(d*sqrt(x) + c) + (2*a^2*b^2 - b^4)*sqrt(a^2 - b^2))*log((2*a*b*cos(d*sqrt(x) + c
) - (a^2 - 2*b^2)*cos(d*sqrt(x) + c)^2 + 2*a^2 - b^2 - 2*(sqrt(a^2 - b^2)*b*cos(d*sqrt(x) + c) + sqrt(a^2 - b^
2)*a)*sin(d*sqrt(x) + c))/(a^2*cos(d*sqrt(x) + c)^2 + 2*a*b*cos(d*sqrt(x) + c) + b^2)) + 2*(a^3*b^2 - a*b^4)*s
in(d*sqrt(x) + c))/((a^7 - 2*a^5*b^2 + a^3*b^4)*d*cos(d*sqrt(x) + c) + (a^6*b - 2*a^4*b^3 + a^2*b^5)*d), 2*((a
^5 - 2*a^3*b^2 + a*b^4)*d*sqrt(x)*cos(d*sqrt(x) + c) + (a^4*b - 2*a^2*b^3 + b^5)*d*sqrt(x) - ((2*a^3*b - a*b^3
)*sqrt(-a^2 + b^2)*cos(d*sqrt(x) + c) + (2*a^2*b^2 - b^4)*sqrt(-a^2 + b^2))*arctan(-(sqrt(-a^2 + b^2)*b*cos(d*
sqrt(x) + c) + sqrt(-a^2 + b^2)*a)/((a^2 - b^2)*sin(d*sqrt(x) + c))) + (a^3*b^2 - a*b^4)*sin(d*sqrt(x) + c))/(
(a^7 - 2*a^5*b^2 + a^3*b^4)*d*cos(d*sqrt(x) + c) + (a^6*b - 2*a^4*b^3 + a^2*b^5)*d)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.2749, size = 265, normalized size = 2.09 \begin{align*} -\frac{4 \, b^{2} \tan \left (\frac{1}{2} \, d \sqrt{x} + \frac{1}{2} \, c\right )}{{\left (a^{3} d - a b^{2} d\right )}{\left (a \tan \left (\frac{1}{2} \, d \sqrt{x} + \frac{1}{2} \, c\right )^{2} - b \tan \left (\frac{1}{2} \, d \sqrt{x} + \frac{1}{2} \, c\right )^{2} - a - b\right )}} + \frac{4 \,{\left (2 \, a^{2} b - b^{3}\right )}{\left (\pi \left \lfloor \frac{d \sqrt{x} + c}{2 \, \pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (2 \, a - 2 \, b\right ) + \arctan \left (\frac{a \tan \left (\frac{1}{2} \, d \sqrt{x} + \frac{1}{2} \, c\right ) - b \tan \left (\frac{1}{2} \, d \sqrt{x} + \frac{1}{2} \, c\right )}{\sqrt{-a^{2} + b^{2}}}\right )\right )}}{{\left (a^{4} d - a^{2} b^{2} d\right )} \sqrt{-a^{2} + b^{2}}} + \frac{2 \,{\left (d \sqrt{x} + c\right )}}{a^{2} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4*b^2*tan(1/2*d*sqrt(x) + 1/2*c)/((a^3*d - a*b^2*d)*(a*tan(1/2*d*sqrt(x) + 1/2*c)^2 - b*tan(1/2*d*sqrt(x) + 1
/2*c)^2 - a - b)) + 4*(2*a^2*b - b^3)*(pi*floor(1/2*(d*sqrt(x) + c)/pi + 1/2)*sgn(2*a - 2*b) + arctan((a*tan(1
/2*d*sqrt(x) + 1/2*c) - b*tan(1/2*d*sqrt(x) + 1/2*c))/sqrt(-a^2 + b^2)))/((a^4*d - a^2*b^2*d)*sqrt(-a^2 + b^2)
) + 2*(d*sqrt(x) + c)/(a^2*d)