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]

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

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Rubi [A]  time = 0.20, antiderivative size = 127, normalized size of antiderivative = 1.00, 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 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]

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 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 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 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 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]

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*}

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Mathematica [A]  time = 0.79, 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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fricas [B]  time = 0.58, size = 574, normalized size = 4.52 \[ \left [\frac {2 \, {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} d \sqrt {x} \cos \left (d \sqrt {x} + c\right ) + 2 \, {\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} d \sqrt {x} + {\left ({\left (2 \, a^{3} b - a b^{3}\right )} \sqrt {a^{2} - b^{2}} \cos \left (d \sqrt {x} + c\right ) + {\left (2 \, a^{2} b^{2} - b^{4}\right )} \sqrt {a^{2} - b^{2}}\right )} \log \left (\frac {2 \, a b \cos \left (d \sqrt {x} + c\right ) - {\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d \sqrt {x} + c\right )^{2} + 2 \, a^{2} - b^{2} - 2 \, {\left (\sqrt {a^{2} - b^{2}} b \cos \left (d \sqrt {x} + c\right ) + \sqrt {a^{2} - b^{2}} a\right )} \sin \left (d \sqrt {x} + c\right )}{a^{2} \cos \left (d \sqrt {x} + c\right )^{2} + 2 \, a b \cos \left (d \sqrt {x} + c\right ) + b^{2}}\right ) + 2 \, {\left (a^{3} b^{2} - a b^{4}\right )} \sin \left (d \sqrt {x} + c\right )}{{\left (a^{7} - 2 \, a^{5} b^{2} + a^{3} b^{4}\right )} d \cos \left (d \sqrt {x} + c\right ) + {\left (a^{6} b - 2 \, a^{4} b^{3} + a^{2} b^{5}\right )} d}, \frac {2 \, {\left ({\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} d \sqrt {x} \cos \left (d \sqrt {x} + c\right ) + {\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} d \sqrt {x} - {\left ({\left (2 \, a^{3} b - a b^{3}\right )} \sqrt {-a^{2} + b^{2}} \cos \left (d \sqrt {x} + c\right ) + {\left (2 \, a^{2} b^{2} - b^{4}\right )} \sqrt {-a^{2} + b^{2}}\right )} \arctan \left (-\frac {\sqrt {-a^{2} + b^{2}} b \cos \left (d \sqrt {x} + c\right ) + \sqrt {-a^{2} + b^{2}} a}{{\left (a^{2} - b^{2}\right )} \sin \left (d \sqrt {x} + c\right )}\right ) + {\left (a^{3} b^{2} - a b^{4}\right )} \sin \left (d \sqrt {x} + c\right )\right )}}{{\left (a^{7} - 2 \, a^{5} b^{2} + a^{3} b^{4}\right )} d \cos \left (d \sqrt {x} + c\right ) + {\left (a^{6} b - 2 \, a^{4} b^{3} + a^{2} b^{5}\right )} d}\right ] \]

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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giac [A]  time = 0.74, size = 196, normalized size = 1.54 \[ -\frac {4 \, b^{2} \tan \left (\frac {1}{2} \, d \sqrt {x} + \frac {1}{2} \, c\right )}{{\left (a^{3} d - a b^{2} d\right )} {\left (a \tan \left (\frac {1}{2} \, d \sqrt {x} + \frac {1}{2} \, c\right )^{2} - b \tan \left (\frac {1}{2} \, d \sqrt {x} + \frac {1}{2} \, c\right )^{2} - a - b\right )}} + \frac {4 \, {\left (2 \, a^{2} b - b^{3}\right )} {\left (\pi \left \lfloor \frac {d \sqrt {x} + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (2 \, a - 2 \, b\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d \sqrt {x} + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d \sqrt {x} + \frac {1}{2} \, c\right )}{\sqrt {-a^{2} + b^{2}}}\right )\right )}}{{\left (a^{4} d - a^{2} b^{2} d\right )} \sqrt {-a^{2} + b^{2}}} + \frac {2 \, {\left (d \sqrt {x} + c\right )}}{a^{2} d} \]

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)

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maple [A]  time = 0.85, size = 216, normalized size = 1.70 \[ -\frac {4 b^{2} \tan \left (\frac {c}{2}+\frac {d \sqrt {x}}{2}\right )}{d a \left (a^{2}-b^{2}\right ) \left (a \left (\tan ^{2}\left (\frac {c}{2}+\frac {d \sqrt {x}}{2}\right )\right )-\left (\tan ^{2}\left (\frac {c}{2}+\frac {d \sqrt {x}}{2}\right )\right ) b -a -b \right )}-\frac {8 b \arctanh \left (\frac {\left (a -b \right ) \tan \left (\frac {c}{2}+\frac {d \sqrt {x}}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{d \left (a -b \right ) \left (a +b \right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}+\frac {4 b^{3} \arctanh \left (\frac {\left (a -b \right ) \tan \left (\frac {c}{2}+\frac {d \sqrt {x}}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{d \,a^{2} \left (a -b \right ) \left (a +b \right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}+\frac {4 \arctan \left (\tan \left (\frac {c}{2}+\frac {d \sqrt {x}}{2}\right )\right )}{d \,a^{2}} \]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

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 >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a^2-4*b^2>0)', see `assume?`
for more details)Is 4*a^2-4*b^2 positive or negative?

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mupad [B]  time = 5.25, size = 330, normalized size = 2.60 \[ \frac {\frac {b^2\,4{}\mathrm {i}}{a\,d\,\left (a^2-b^2\right )}+\frac {b^3\,{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,\sqrt {x}\,1{}\mathrm {i}}\,4{}\mathrm {i}}{a^2\,d\,\left (a^2-b^2\right )}}{a+a\,{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,\sqrt {x}\,2{}\mathrm {i}}+2\,b\,{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,\sqrt {x}\,1{}\mathrm {i}}}+\frac {2\,\sqrt {x}}{a^2}+\frac {\ln \left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,\sqrt {x}\,1{}\mathrm {i}}\,\left (4\,a^2\,b-2\,b^3\right )-\frac {\left (4\,a^2\,b-2\,b^3\right )\,\left (a^2-b^2\right )\,\left (a+b\,{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,\sqrt {x}\,1{}\mathrm {i}}\right )\,1{}\mathrm {i}}{{\left (a+b\right )}^{3/2}\,{\left (a-b\right )}^{3/2}}\right )\,\left (4\,a^2\,b-2\,b^3\right )}{a^2\,d\,{\left (a+b\right )}^{3/2}\,{\left (a-b\right )}^{3/2}}-\frac {2\,b\,\ln \left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,\sqrt {x}\,1{}\mathrm {i}}\,\left (4\,a^2\,b-2\,b^3\right )+\frac {b\,\left (a^2-b^2\right )\,\left (2\,a^2-b^2\right )\,\left (a+b\,{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,\sqrt {x}\,1{}\mathrm {i}}\right )\,2{}\mathrm {i}}{{\left (a+b\right )}^{3/2}\,{\left (a-b\right )}^{3/2}}\right )\,\left (2\,a^2-b^2\right )}{a^2\,d\,{\left (a+b\right )}^{3/2}\,{\left (a-b\right )}^{3/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((b^2*4i)/(a*d*(a^2 - b^2)) + (b^3*exp(c*1i + d*x^(1/2)*1i)*4i)/(a^2*d*(a^2 - b^2)))/(a + a*exp(c*2i + d*x^(1/
2)*2i) + 2*b*exp(c*1i + d*x^(1/2)*1i)) + (2*x^(1/2))/a^2 + (log(exp(c*1i + d*x^(1/2)*1i)*(4*a^2*b - 2*b^3) - (
(4*a^2*b - 2*b^3)*(a^2 - b^2)*(a + b*exp(c*1i + d*x^(1/2)*1i))*1i)/((a + b)^(3/2)*(a - b)^(3/2)))*(4*a^2*b - 2
*b^3))/(a^2*d*(a + b)^(3/2)*(a - b)^(3/2)) - (2*b*log(exp(c*1i + d*x^(1/2)*1i)*(4*a^2*b - 2*b^3) + (b*(a^2 - b
^2)*(2*a^2 - b^2)*(a + b*exp(c*1i + d*x^(1/2)*1i))*2i)/((a + b)^(3/2)*(a - b)^(3/2)))*(2*a^2 - b^2))/(a^2*d*(a
+ b)^(3/2)*(a - b)^(3/2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {x} \left (a + b \sec {\left (c + d \sqrt {x} \right )}\right )^{2}}\, dx \]

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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