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

Optimal. Leaf size=125 \[ \frac {4 b \left (2 a^2-b^2\right ) \tanh ^{-1}\left (\frac {a+b \tan \left (\frac {1}{2} \left (c+d \sqrt {x}\right )\right )}{\sqrt {a^2-b^2}}\right )}{a^2 d \left (a^2-b^2\right )^{3/2}}-\frac {2 b^2 \cot \left (c+d \sqrt {x}\right )}{a d \left (a^2-b^2\right ) \left (a+b \csc \left (c+d \sqrt {x}\right )\right )}+\frac {2 \sqrt {x}}{a^2} \]

[Out]

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

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Rubi [A]  time = 0.21, antiderivative size = 125, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.318, Rules used = {4205, 3785, 3919, 3831, 2660, 618, 206} \[ \frac {4 b \left (2 a^2-b^2\right ) \tanh ^{-1}\left (\frac {a+b \tan \left (\frac {1}{2} \left (c+d \sqrt {x}\right )\right )}{\sqrt {a^2-b^2}}\right )}{a^2 d \left (a^2-b^2\right )^{3/2}}-\frac {2 b^2 \cot \left (c+d \sqrt {x}\right )}{a d \left (a^2-b^2\right ) \left (a+b \csc \left (c+d \sqrt {x}\right )\right )}+\frac {2 \sqrt {x}}{a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 2660

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x]}, Dis
t[(2*e)/d, Subst[Int[1/(a + 2*b*e*x + a*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 4205

Int[((a_.) + Csc[(c_.) + (d_.)*(x_)^(n_)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplif
y[(m + 1)/n] - 1)*(a + b*Csc[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 \csc \left (c+d \sqrt {x}\right )\right )^2} \, dx &=2 \operatorname {Subst}\left (\int \frac {1}{(a+b \csc (c+d x))^2} \, dx,x,\sqrt {x}\right )\\ &=-\frac {2 b^2 \cot \left (c+d \sqrt {x}\right )}{a \left (a^2-b^2\right ) d \left (a+b \csc \left (c+d \sqrt {x}\right )\right )}-\frac {2 \operatorname {Subst}\left (\int \frac {-a^2+b^2+a b \csc (c+d x)}{a+b \csc (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 \cot \left (c+d \sqrt {x}\right )}{a \left (a^2-b^2\right ) d \left (a+b \csc \left (c+d \sqrt {x}\right )\right )}-\frac {\left (2 b \left (2 a^2-b^2\right )\right ) \operatorname {Subst}\left (\int \frac {\csc (c+d x)}{a+b \csc (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 \cot \left (c+d \sqrt {x}\right )}{a \left (a^2-b^2\right ) d \left (a+b \csc \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 \sin (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 \cot \left (c+d \sqrt {x}\right )}{a \left (a^2-b^2\right ) d \left (a+b \csc \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 {2 a x}{b}+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 {2 b^2 \cot \left (c+d \sqrt {x}\right )}{a \left (a^2-b^2\right ) d \left (a+b \csc \left (c+d \sqrt {x}\right )\right )}+\frac {\left (8 \left (2 a^2-b^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{-4 \left (1-\frac {a^2}{b^2}\right )-x^2} \, dx,x,\frac {2 a}{b}+2 \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 {b \left (\frac {a}{b}+\tan \left (\frac {1}{2} \left (c+d \sqrt {x}\right )\right )\right )}{\sqrt {a^2-b^2}}\right )}{a^2 \left (a^2-b^2\right )^{3/2} d}-\frac {2 b^2 \cot \left (c+d \sqrt {x}\right )}{a \left (a^2-b^2\right ) d \left (a+b \csc \left (c+d \sqrt {x}\right )\right )}\\ \end {align*}

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Mathematica [A]  time = 0.69, size = 172, normalized size = 1.38 \[ \frac {2 \csc \left (c+d \sqrt {x}\right ) \left (a \sin \left (c+d \sqrt {x}\right )+b\right ) \left (-\frac {2 b \left (b^2-2 a^2\right ) \tan ^{-1}\left (\frac {a+b \tan \left (\frac {1}{2} \left (c+d \sqrt {x}\right )\right )}{\sqrt {b^2-a^2}}\right ) \left (a+b \csc \left (c+d \sqrt {x}\right )\right )}{\left (b^2-a^2\right )^{3/2}}+\frac {a b^2 \cot \left (c+d \sqrt {x}\right )}{(b-a) (a+b)}+\left (c+d \sqrt {x}\right ) \left (a+b \csc \left (c+d \sqrt {x}\right )\right )\right )}{a^2 d \left (a+b \csc \left (c+d \sqrt {x}\right )\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Csc[c + d*Sqrt[x]]*((a*b^2*Cot[c + d*Sqrt[x]])/((-a + b)*(a + b)) + (c + d*Sqrt[x])*(a + b*Csc[c + d*Sqrt[x
]]) - (2*b*(-2*a^2 + b^2)*ArcTan[(a + b*Tan[(c + d*Sqrt[x])/2])/Sqrt[-a^2 + b^2]]*(a + b*Csc[c + d*Sqrt[x]]))/
(-a^2 + b^2)^(3/2))*(b + a*Sin[c + d*Sqrt[x]]))/(a^2*d*(a + b*Csc[c + d*Sqrt[x]])^2)

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fricas [B]  time = 0.56, size = 576, normalized size = 4.61 \[ \left [\frac {2 \, {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} d \sqrt {x} \sin \left (d \sqrt {x} + c\right ) + 2 \, {\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} d \sqrt {x} - 2 \, {\left (a^{3} b^{2} - a b^{4}\right )} \cos \left (d \sqrt {x} + c\right ) + {\left ({\left (2 \, a^{3} b - a b^{3}\right )} \sqrt {a^{2} - b^{2}} \sin \left (d \sqrt {x} + c\right ) + {\left (2 \, a^{2} b^{2} - b^{4}\right )} \sqrt {a^{2} - b^{2}}\right )} \log \left (\frac {{\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d \sqrt {x} + c\right )^{2} + 2 \, \sqrt {a^{2} - b^{2}} a \cos \left (d \sqrt {x} + c\right ) + a^{2} + b^{2} + 2 \, {\left (\sqrt {a^{2} - b^{2}} b \cos \left (d \sqrt {x} + c\right ) + a b\right )} \sin \left (d \sqrt {x} + c\right )}{a^{2} \cos \left (d \sqrt {x} + c\right )^{2} - 2 \, a b \sin \left (d \sqrt {x} + c\right ) - a^{2} - b^{2}}\right )}{{\left (a^{7} - 2 \, a^{5} b^{2} + a^{3} b^{4}\right )} d \sin \left (d \sqrt {x} + c\right ) + {\left (a^{6} b - 2 \, a^{4} b^{3} + a^{2} b^{5}\right )} d}, \frac {2 \, {\left ({\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} d \sqrt {x} \sin \left (d \sqrt {x} + c\right ) + {\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} d \sqrt {x} + {\left ({\left (2 \, a^{3} b - a b^{3}\right )} \sqrt {-a^{2} + b^{2}} \sin \left (d \sqrt {x} + c\right ) + {\left (2 \, a^{2} b^{2} - b^{4}\right )} \sqrt {-a^{2} + b^{2}}\right )} \arctan \left (-\frac {\sqrt {-a^{2} + b^{2}} b \sin \left (d \sqrt {x} + c\right ) + \sqrt {-a^{2} + b^{2}} a}{{\left (a^{2} - b^{2}\right )} \cos \left (d \sqrt {x} + c\right )}\right ) - {\left (a^{3} b^{2} - a b^{4}\right )} \cos \left (d \sqrt {x} + c\right )\right )}}{{\left (a^{7} - 2 \, a^{5} b^{2} + a^{3} b^{4}\right )} d \sin \left (d \sqrt {x} + c\right ) + {\left (a^{6} b - 2 \, a^{4} b^{3} + a^{2} b^{5}\right )} d}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*csc(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)*sin(d*sqrt(x) + c) + 2*(a^4*b - 2*a^2*b^3 + b^5)*d*sqrt(x) - 2*(a^3*b^
2 - a*b^4)*cos(d*sqrt(x) + c) + ((2*a^3*b - a*b^3)*sqrt(a^2 - b^2)*sin(d*sqrt(x) + c) + (2*a^2*b^2 - b^4)*sqrt
(a^2 - b^2))*log(((a^2 - 2*b^2)*cos(d*sqrt(x) + c)^2 + 2*sqrt(a^2 - b^2)*a*cos(d*sqrt(x) + c) + a^2 + b^2 + 2*
(sqrt(a^2 - b^2)*b*cos(d*sqrt(x) + c) + a*b)*sin(d*sqrt(x) + c))/(a^2*cos(d*sqrt(x) + c)^2 - 2*a*b*sin(d*sqrt(
x) + c) - a^2 - b^2)))/((a^7 - 2*a^5*b^2 + a^3*b^4)*d*sin(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)*sin(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)*sin(d*sqrt(x) + c) + (2*a^2*b^2 - b^4)*sqrt(-a^2 + b^2))*arctan(-(sqrt(-a^2 + b^2)*b*si
n(d*sqrt(x) + c) + sqrt(-a^2 + b^2)*a)/((a^2 - b^2)*cos(d*sqrt(x) + c))) - (a^3*b^2 - a*b^4)*cos(d*sqrt(x) + c
))/((a^7 - 2*a^5*b^2 + a^3*b^4)*d*sin(d*sqrt(x) + c) + (a^6*b - 2*a^4*b^3 + a^2*b^5)*d)]

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giac [A]  time = 0.30, size = 174, normalized size = 1.39 \[ -\frac {4 \, {\left (2 \, a^{2} b - b^{3}\right )} {\left (\pi \left \lfloor \frac {d \sqrt {x} + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (b) + \arctan \left (\frac {b \tan \left (\frac {1}{2} \, d \sqrt {x} + \frac {1}{2} \, c\right ) + a}{\sqrt {-a^{2} + b^{2}}}\right )\right )}}{{\left (a^{4} d - a^{2} b^{2} d\right )} \sqrt {-a^{2} + b^{2}}} - \frac {4 \, {\left (a b \tan \left (\frac {1}{2} \, d \sqrt {x} + \frac {1}{2} \, c\right ) + b^{2}\right )}}{{\left (a^{3} d - a b^{2} d\right )} {\left (b \tan \left (\frac {1}{2} \, d \sqrt {x} + \frac {1}{2} \, c\right )^{2} + 2 \, a \tan \left (\frac {1}{2} \, d \sqrt {x} + \frac {1}{2} \, c\right ) + b\right )}} + \frac {2 \, {\left (d \sqrt {x} + c\right )}}{a^{2} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 1.53, size = 263, normalized size = 2.10 \[ -\frac {4 b \tan \left (\frac {c}{2}+\frac {d \sqrt {x}}{2}\right )}{d \left (\left (\tan ^{2}\left (\frac {c}{2}+\frac {d \sqrt {x}}{2}\right )\right ) b +2 a \tan \left (\frac {c}{2}+\frac {d \sqrt {x}}{2}\right )+b \right ) \left (a^{2}-b^{2}\right )}-\frac {4 b^{2}}{d a \left (\left (\tan ^{2}\left (\frac {c}{2}+\frac {d \sqrt {x}}{2}\right )\right ) b +2 a \tan \left (\frac {c}{2}+\frac {d \sqrt {x}}{2}\right )+b \right ) \left (a^{2}-b^{2}\right )}-\frac {8 b \arctan \left (\frac {2 b \tan \left (\frac {c}{2}+\frac {d \sqrt {x}}{2}\right )+2 a}{2 \sqrt {-a^{2}+b^{2}}}\right )}{d \left (a^{2}-b^{2}\right ) \sqrt {-a^{2}+b^{2}}}+\frac {4 b^{3} \arctan \left (\frac {2 b \tan \left (\frac {c}{2}+\frac {d \sqrt {x}}{2}\right )+2 a}{2 \sqrt {-a^{2}+b^{2}}}\right )}{d \,a^{2} \left (a^{2}-b^{2}\right ) \sqrt {-a^{2}+b^{2}}}+\frac {4 \arctan \left (\tan \left (\frac {c}{2}+\frac {d \sqrt {x}}{2}\right )\right )}{d \,a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4/d*b/(tan(1/2*c+1/2*d*x^(1/2))^2*b+2*a*tan(1/2*c+1/2*d*x^(1/2))+b)/(a^2-b^2)*tan(1/2*c+1/2*d*x^(1/2))-4/d*b^
2/a/(tan(1/2*c+1/2*d*x^(1/2))^2*b+2*a*tan(1/2*c+1/2*d*x^(1/2))+b)/(a^2-b^2)-8/d*b/(a^2-b^2)/(-a^2+b^2)^(1/2)*a
rctan(1/2*(2*b*tan(1/2*c+1/2*d*x^(1/2))+2*a)/(-a^2+b^2)^(1/2))+4/d*b^3/a^2/(a^2-b^2)/(-a^2+b^2)^(1/2)*arctan(1
/2*(2*b*tan(1/2*c+1/2*d*x^(1/2))+2*a)/(-a^2+b^2)^(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*csc(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 = 5.35, size = 2737, normalized size = 21.90 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- (4*atan((512*a^3*b^3*tan(c/2 + (d*x^(1/2))/2))/((512*a^3*b^9)/(a^6 + a^2*b^4 - 2*a^4*b^2) - (1536*a^5*b^7)/(
a^6 + a^2*b^4 - 2*a^4*b^2) + (1024*a^7*b^5)/(a^6 + a^2*b^4 - 2*a^4*b^2) + (512*a^9*b^3)/(a^6 + a^2*b^4 - 2*a^4
*b^2) - (512*a^11*b)/(a^6 + a^2*b^4 - 2*a^4*b^2)) - (512*a*b^5*tan(c/2 + (d*x^(1/2))/2))/((512*a^3*b^9)/(a^6 +
a^2*b^4 - 2*a^4*b^2) - (1536*a^5*b^7)/(a^6 + a^2*b^4 - 2*a^4*b^2) + (1024*a^7*b^5)/(a^6 + a^2*b^4 - 2*a^4*b^2
) + (512*a^9*b^3)/(a^6 + a^2*b^4 - 2*a^4*b^2) - (512*a^11*b)/(a^6 + a^2*b^4 - 2*a^4*b^2)) + (512*a^5*b*tan(c/2
+ (d*x^(1/2))/2))/((512*a^3*b^9)/(a^6 + a^2*b^4 - 2*a^4*b^2) - (1536*a^5*b^7)/(a^6 + a^2*b^4 - 2*a^4*b^2) + (
1024*a^7*b^5)/(a^6 + a^2*b^4 - 2*a^4*b^2) + (512*a^9*b^3)/(a^6 + a^2*b^4 - 2*a^4*b^2) - (512*a^11*b)/(a^6 + a^
2*b^4 - 2*a^4*b^2))))/(a^2*d) - ((4*b^2)/(a*(a^2 - b^2)) + (4*b*tan(c/2 + (d*x^(1/2))/2))/(a^2 - b^2))/(d*(b +
b*tan(c/2 + (d*x^(1/2))/2)^2 + 2*a*tan(c/2 + (d*x^(1/2))/2))) - (b*atan(((b*(2*a^2 - b^2)*((a + b)^3*(a - b)^
3)^(1/2)*((32*tan(c/2 + (d*x^(1/2))/2)*(8*a*b^7 - 8*a^7*b - 32*a^3*b^5 + 36*a^5*b^3))/(a^7 + a^3*b^4 - 2*a^5*b
^2) - (32*(4*a*b^6 - 8*a^3*b^4 + 4*a^5*b^2))/(a^6 + a^2*b^4 - 2*a^4*b^2) + (2*b*(2*a^2 - b^2)*((a + b)^3*(a -
b)^3)^(1/2)*((32*(2*a^8*b - 2*a^6*b^3))/(a^6 + a^2*b^4 - 2*a^4*b^2) + (32*tan(c/2 + (d*x^(1/2))/2)*(4*a^4*b^6
- 12*a^6*b^4 + 8*a^8*b^2))/(a^7 + a^3*b^4 - 2*a^5*b^2) - (2*b*((32*(a^5*b^6 - 2*a^7*b^4 + a^9*b^2))/(a^6 + a^2
*b^4 - 2*a^4*b^2) + (32*tan(c/2 + (d*x^(1/2))/2)*(3*a^11*b - 2*a^5*b^7 + 7*a^7*b^5 - 8*a^9*b^3))/(a^7 + a^3*b^
4 - 2*a^5*b^2))*(2*a^2 - b^2)*((a + b)^3*(a - b)^3)^(1/2))/(a^8 - a^2*b^6 + 3*a^4*b^4 - 3*a^6*b^2)))/(a^8 - a^
2*b^6 + 3*a^4*b^4 - 3*a^6*b^2))*2i)/(a^8 - a^2*b^6 + 3*a^4*b^4 - 3*a^6*b^2) - (b*(2*a^2 - b^2)*((a + b)^3*(a -
b)^3)^(1/2)*((32*(4*a*b^6 - 8*a^3*b^4 + 4*a^5*b^2))/(a^6 + a^2*b^4 - 2*a^4*b^2) - (32*tan(c/2 + (d*x^(1/2))/2
)*(8*a*b^7 - 8*a^7*b - 32*a^3*b^5 + 36*a^5*b^3))/(a^7 + a^3*b^4 - 2*a^5*b^2) + (2*b*(2*a^2 - b^2)*((a + b)^3*(
a - b)^3)^(1/2)*((32*(2*a^8*b - 2*a^6*b^3))/(a^6 + a^2*b^4 - 2*a^4*b^2) + (32*tan(c/2 + (d*x^(1/2))/2)*(4*a^4*
b^6 - 12*a^6*b^4 + 8*a^8*b^2))/(a^7 + a^3*b^4 - 2*a^5*b^2) + (2*b*((32*(a^5*b^6 - 2*a^7*b^4 + a^9*b^2))/(a^6 +
a^2*b^4 - 2*a^4*b^2) + (32*tan(c/2 + (d*x^(1/2))/2)*(3*a^11*b - 2*a^5*b^7 + 7*a^7*b^5 - 8*a^9*b^3))/(a^7 + a^
3*b^4 - 2*a^5*b^2))*(2*a^2 - b^2)*((a + b)^3*(a - b)^3)^(1/2))/(a^8 - a^2*b^6 + 3*a^4*b^4 - 3*a^6*b^2)))/(a^8
- a^2*b^6 + 3*a^4*b^4 - 3*a^6*b^2))*2i)/(a^8 - a^2*b^6 + 3*a^4*b^4 - 3*a^6*b^2))/((64*(8*b^5 - 16*a^2*b^3))/(a
^6 + a^2*b^4 - 2*a^4*b^2) + (64*tan(c/2 + (d*x^(1/2))/2)*(16*b^6 - 48*a^2*b^4 + 32*a^4*b^2))/(a^7 + a^3*b^4 -
2*a^5*b^2) + (2*b*(2*a^2 - b^2)*((a + b)^3*(a - b)^3)^(1/2)*((32*tan(c/2 + (d*x^(1/2))/2)*(8*a*b^7 - 8*a^7*b -
32*a^3*b^5 + 36*a^5*b^3))/(a^7 + a^3*b^4 - 2*a^5*b^2) - (32*(4*a*b^6 - 8*a^3*b^4 + 4*a^5*b^2))/(a^6 + a^2*b^4
- 2*a^4*b^2) + (2*b*(2*a^2 - b^2)*((a + b)^3*(a - b)^3)^(1/2)*((32*(2*a^8*b - 2*a^6*b^3))/(a^6 + a^2*b^4 - 2*
a^4*b^2) + (32*tan(c/2 + (d*x^(1/2))/2)*(4*a^4*b^6 - 12*a^6*b^4 + 8*a^8*b^2))/(a^7 + a^3*b^4 - 2*a^5*b^2) - (2
*b*((32*(a^5*b^6 - 2*a^7*b^4 + a^9*b^2))/(a^6 + a^2*b^4 - 2*a^4*b^2) + (32*tan(c/2 + (d*x^(1/2))/2)*(3*a^11*b
- 2*a^5*b^7 + 7*a^7*b^5 - 8*a^9*b^3))/(a^7 + a^3*b^4 - 2*a^5*b^2))*(2*a^2 - b^2)*((a + b)^3*(a - b)^3)^(1/2))/
(a^8 - a^2*b^6 + 3*a^4*b^4 - 3*a^6*b^2)))/(a^8 - a^2*b^6 + 3*a^4*b^4 - 3*a^6*b^2)))/(a^8 - a^2*b^6 + 3*a^4*b^4
- 3*a^6*b^2) + (2*b*(2*a^2 - b^2)*((a + b)^3*(a - b)^3)^(1/2)*((32*(4*a*b^6 - 8*a^3*b^4 + 4*a^5*b^2))/(a^6 +
a^2*b^4 - 2*a^4*b^2) - (32*tan(c/2 + (d*x^(1/2))/2)*(8*a*b^7 - 8*a^7*b - 32*a^3*b^5 + 36*a^5*b^3))/(a^7 + a^3*
b^4 - 2*a^5*b^2) + (2*b*(2*a^2 - b^2)*((a + b)^3*(a - b)^3)^(1/2)*((32*(2*a^8*b - 2*a^6*b^3))/(a^6 + a^2*b^4 -
2*a^4*b^2) + (32*tan(c/2 + (d*x^(1/2))/2)*(4*a^4*b^6 - 12*a^6*b^4 + 8*a^8*b^2))/(a^7 + a^3*b^4 - 2*a^5*b^2) +
(2*b*((32*(a^5*b^6 - 2*a^7*b^4 + a^9*b^2))/(a^6 + a^2*b^4 - 2*a^4*b^2) + (32*tan(c/2 + (d*x^(1/2))/2)*(3*a^11
*b - 2*a^5*b^7 + 7*a^7*b^5 - 8*a^9*b^3))/(a^7 + a^3*b^4 - 2*a^5*b^2))*(2*a^2 - b^2)*((a + b)^3*(a - b)^3)^(1/2
))/(a^8 - a^2*b^6 + 3*a^4*b^4 - 3*a^6*b^2)))/(a^8 - a^2*b^6 + 3*a^4*b^4 - 3*a^6*b^2)))/(a^8 - a^2*b^6 + 3*a^4*
b^4 - 3*a^6*b^2)))*(2*a^2 - b^2)*((a + b)^3*(a - b)^3)^(1/2)*4i)/(d*(a^8 - a^2*b^6 + 3*a^4*b^4 - 3*a^6*b^2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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