3.63 \(\int \frac {\sqrt {x}}{a+b \csc (c+d \sqrt {x})} \, dx\)

Optimal. Leaf size=407 \[ \frac {4 i b \text {Li}_3\left (\frac {i a e^{i \left (c+d \sqrt {x}\right )}}{b-\sqrt {b^2-a^2}}\right )}{a d^3 \sqrt {b^2-a^2}}-\frac {4 i b \text {Li}_3\left (\frac {i a e^{i \left (c+d \sqrt {x}\right )}}{b+\sqrt {b^2-a^2}}\right )}{a d^3 \sqrt {b^2-a^2}}+\frac {4 b \sqrt {x} \text {Li}_2\left (\frac {i a e^{i \left (c+d \sqrt {x}\right )}}{b-\sqrt {b^2-a^2}}\right )}{a d^2 \sqrt {b^2-a^2}}-\frac {4 b \sqrt {x} \text {Li}_2\left (\frac {i a e^{i \left (c+d \sqrt {x}\right )}}{b+\sqrt {b^2-a^2}}\right )}{a d^2 \sqrt {b^2-a^2}}+\frac {2 i b x \log \left (1-\frac {i a e^{i \left (c+d \sqrt {x}\right )}}{b-\sqrt {b^2-a^2}}\right )}{a d \sqrt {b^2-a^2}}-\frac {2 i b x \log \left (1-\frac {i a e^{i \left (c+d \sqrt {x}\right )}}{\sqrt {b^2-a^2}+b}\right )}{a d \sqrt {b^2-a^2}}+\frac {2 x^{3/2}}{3 a} \]

[Out]

2/3*x^(3/2)/a+2*I*b*x*ln(1-I*a*exp(I*(c+d*x^(1/2)))/(b-(-a^2+b^2)^(1/2)))/a/d/(-a^2+b^2)^(1/2)-2*I*b*x*ln(1-I*
a*exp(I*(c+d*x^(1/2)))/(b+(-a^2+b^2)^(1/2)))/a/d/(-a^2+b^2)^(1/2)+4*I*b*polylog(3,I*a*exp(I*(c+d*x^(1/2)))/(b-
(-a^2+b^2)^(1/2)))/a/d^3/(-a^2+b^2)^(1/2)-4*I*b*polylog(3,I*a*exp(I*(c+d*x^(1/2)))/(b+(-a^2+b^2)^(1/2)))/a/d^3
/(-a^2+b^2)^(1/2)+4*b*polylog(2,I*a*exp(I*(c+d*x^(1/2)))/(b-(-a^2+b^2)^(1/2)))*x^(1/2)/a/d^2/(-a^2+b^2)^(1/2)-
4*b*polylog(2,I*a*exp(I*(c+d*x^(1/2)))/(b+(-a^2+b^2)^(1/2)))*x^(1/2)/a/d^2/(-a^2+b^2)^(1/2)

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Rubi [A]  time = 0.87, antiderivative size = 407, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 8, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {4205, 4191, 3323, 2264, 2190, 2531, 2282, 6589} \[ \frac {4 b \sqrt {x} \text {PolyLog}\left (2,\frac {i a e^{i \left (c+d \sqrt {x}\right )}}{b-\sqrt {b^2-a^2}}\right )}{a d^2 \sqrt {b^2-a^2}}-\frac {4 b \sqrt {x} \text {PolyLog}\left (2,\frac {i a e^{i \left (c+d \sqrt {x}\right )}}{\sqrt {b^2-a^2}+b}\right )}{a d^2 \sqrt {b^2-a^2}}+\frac {4 i b \text {PolyLog}\left (3,\frac {i a e^{i \left (c+d \sqrt {x}\right )}}{b-\sqrt {b^2-a^2}}\right )}{a d^3 \sqrt {b^2-a^2}}-\frac {4 i b \text {PolyLog}\left (3,\frac {i a e^{i \left (c+d \sqrt {x}\right )}}{\sqrt {b^2-a^2}+b}\right )}{a d^3 \sqrt {b^2-a^2}}+\frac {2 i b x \log \left (1-\frac {i a e^{i \left (c+d \sqrt {x}\right )}}{b-\sqrt {b^2-a^2}}\right )}{a d \sqrt {b^2-a^2}}-\frac {2 i b x \log \left (1-\frac {i a e^{i \left (c+d \sqrt {x}\right )}}{\sqrt {b^2-a^2}+b}\right )}{a d \sqrt {b^2-a^2}}+\frac {2 x^{3/2}}{3 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*x^(3/2))/(3*a) + ((2*I)*b*x*Log[1 - (I*a*E^(I*(c + d*Sqrt[x])))/(b - Sqrt[-a^2 + b^2])])/(a*Sqrt[-a^2 + b^2
]*d) - ((2*I)*b*x*Log[1 - (I*a*E^(I*(c + d*Sqrt[x])))/(b + Sqrt[-a^2 + b^2])])/(a*Sqrt[-a^2 + b^2]*d) + (4*b*S
qrt[x]*PolyLog[2, (I*a*E^(I*(c + d*Sqrt[x])))/(b - Sqrt[-a^2 + b^2])])/(a*Sqrt[-a^2 + b^2]*d^2) - (4*b*Sqrt[x]
*PolyLog[2, (I*a*E^(I*(c + d*Sqrt[x])))/(b + Sqrt[-a^2 + b^2])])/(a*Sqrt[-a^2 + b^2]*d^2) + ((4*I)*b*PolyLog[3
, (I*a*E^(I*(c + d*Sqrt[x])))/(b - Sqrt[-a^2 + b^2])])/(a*Sqrt[-a^2 + b^2]*d^3) - ((4*I)*b*PolyLog[3, (I*a*E^(
I*(c + d*Sqrt[x])))/(b + Sqrt[-a^2 + b^2])])/(a*Sqrt[-a^2 + b^2]*d^3)

Rule 2190

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/((a_) + (b_.)*((F_)^((g_.)*((e_.) +
(f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[((c + d*x)^m*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), x]
- Dist[(d*m)/(b*f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*Log[1 + (b*(F^(g*(e + f*x)))^n)/a], x], x] /; FreeQ[{F,
a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]

Rule 2264

Int[((F_)^(u_)*((f_.) + (g_.)*(x_))^(m_.))/((a_.) + (b_.)*(F_)^(u_) + (c_.)*(F_)^(v_)), x_Symbol] :> With[{q =
Rt[b^2 - 4*a*c, 2]}, Dist[(2*c)/q, Int[((f + g*x)^m*F^u)/(b - q + 2*c*F^u), x], x] - Dist[(2*c)/q, Int[((f +
g*x)^m*F^u)/(b + q + 2*c*F^u), x], x]] /; FreeQ[{F, a, b, c, f, g}, x] && EqQ[v, 2*u] && LinearQ[u, x] && NeQ[
b^2 - 4*a*c, 0] && IGtQ[m, 0]

Rule 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 2531

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> -Simp[((
f + g*x)^m*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)])/(b*c*n*Log[F]), x] + Dist[(g*m)/(b*c*n*Log[F]), Int[(f + g*x)
^(m - 1)*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)], x], x] /; FreeQ[{F, a, b, c, e, f, g, n}, x] && GtQ[m, 0]

Rule 3323

Int[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[2, Int[((c + d*x)^m*E
^(I*(e + f*x)))/(I*b + 2*a*E^(I*(e + f*x)) - I*b*E^(2*I*(e + f*x))), x], x] /; FreeQ[{a, b, c, d, e, f}, x] &&
NeQ[a^2 - b^2, 0] && IGtQ[m, 0]

Rule 4191

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[
(c + d*x)^m, 1/(Sin[e + f*x]^n/(b + a*Sin[e + f*x])^n), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && ILtQ[n, 0] &
& IGtQ[m, 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]

Rule 6589

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[PolyLog[n + 1, c*(a
+ b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d, e, n, p}, x] && EqQ[b*d, a*e]

Rubi steps

\begin {align*} \int \frac {\sqrt {x}}{a+b \csc \left (c+d \sqrt {x}\right )} \, dx &=2 \operatorname {Subst}\left (\int \frac {x^2}{a+b \csc (c+d x)} \, dx,x,\sqrt {x}\right )\\ &=2 \operatorname {Subst}\left (\int \left (\frac {x^2}{a}-\frac {b x^2}{a (b+a \sin (c+d x))}\right ) \, dx,x,\sqrt {x}\right )\\ &=\frac {2 x^{3/2}}{3 a}-\frac {(2 b) \operatorname {Subst}\left (\int \frac {x^2}{b+a \sin (c+d x)} \, dx,x,\sqrt {x}\right )}{a}\\ &=\frac {2 x^{3/2}}{3 a}-\frac {(4 b) \operatorname {Subst}\left (\int \frac {e^{i (c+d x)} x^2}{i a+2 b e^{i (c+d x)}-i a e^{2 i (c+d x)}} \, dx,x,\sqrt {x}\right )}{a}\\ &=\frac {2 x^{3/2}}{3 a}+\frac {(4 i b) \operatorname {Subst}\left (\int \frac {e^{i (c+d x)} x^2}{2 b-2 \sqrt {-a^2+b^2}-2 i a e^{i (c+d x)}} \, dx,x,\sqrt {x}\right )}{\sqrt {-a^2+b^2}}-\frac {(4 i b) \operatorname {Subst}\left (\int \frac {e^{i (c+d x)} x^2}{2 b+2 \sqrt {-a^2+b^2}-2 i a e^{i (c+d x)}} \, dx,x,\sqrt {x}\right )}{\sqrt {-a^2+b^2}}\\ &=\frac {2 x^{3/2}}{3 a}+\frac {2 i b x \log \left (1-\frac {i a e^{i \left (c+d \sqrt {x}\right )}}{b-\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d}-\frac {2 i b x \log \left (1-\frac {i a e^{i \left (c+d \sqrt {x}\right )}}{b+\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d}-\frac {(4 i b) \operatorname {Subst}\left (\int x \log \left (1-\frac {2 i a e^{i (c+d x)}}{2 b-2 \sqrt {-a^2+b^2}}\right ) \, dx,x,\sqrt {x}\right )}{a \sqrt {-a^2+b^2} d}+\frac {(4 i b) \operatorname {Subst}\left (\int x \log \left (1-\frac {2 i a e^{i (c+d x)}}{2 b+2 \sqrt {-a^2+b^2}}\right ) \, dx,x,\sqrt {x}\right )}{a \sqrt {-a^2+b^2} d}\\ &=\frac {2 x^{3/2}}{3 a}+\frac {2 i b x \log \left (1-\frac {i a e^{i \left (c+d \sqrt {x}\right )}}{b-\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d}-\frac {2 i b x \log \left (1-\frac {i a e^{i \left (c+d \sqrt {x}\right )}}{b+\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d}+\frac {4 b \sqrt {x} \text {Li}_2\left (\frac {i a e^{i \left (c+d \sqrt {x}\right )}}{b-\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d^2}-\frac {4 b \sqrt {x} \text {Li}_2\left (\frac {i a e^{i \left (c+d \sqrt {x}\right )}}{b+\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d^2}-\frac {(4 b) \operatorname {Subst}\left (\int \text {Li}_2\left (\frac {2 i a e^{i (c+d x)}}{2 b-2 \sqrt {-a^2+b^2}}\right ) \, dx,x,\sqrt {x}\right )}{a \sqrt {-a^2+b^2} d^2}+\frac {(4 b) \operatorname {Subst}\left (\int \text {Li}_2\left (\frac {2 i a e^{i (c+d x)}}{2 b+2 \sqrt {-a^2+b^2}}\right ) \, dx,x,\sqrt {x}\right )}{a \sqrt {-a^2+b^2} d^2}\\ &=\frac {2 x^{3/2}}{3 a}+\frac {2 i b x \log \left (1-\frac {i a e^{i \left (c+d \sqrt {x}\right )}}{b-\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d}-\frac {2 i b x \log \left (1-\frac {i a e^{i \left (c+d \sqrt {x}\right )}}{b+\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d}+\frac {4 b \sqrt {x} \text {Li}_2\left (\frac {i a e^{i \left (c+d \sqrt {x}\right )}}{b-\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d^2}-\frac {4 b \sqrt {x} \text {Li}_2\left (\frac {i a e^{i \left (c+d \sqrt {x}\right )}}{b+\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d^2}+\frac {(4 i b) \operatorname {Subst}\left (\int \frac {\text {Li}_2\left (\frac {i a x}{b-\sqrt {-a^2+b^2}}\right )}{x} \, dx,x,e^{i \left (c+d \sqrt {x}\right )}\right )}{a \sqrt {-a^2+b^2} d^3}-\frac {(4 i b) \operatorname {Subst}\left (\int \frac {\text {Li}_2\left (\frac {i a x}{b+\sqrt {-a^2+b^2}}\right )}{x} \, dx,x,e^{i \left (c+d \sqrt {x}\right )}\right )}{a \sqrt {-a^2+b^2} d^3}\\ &=\frac {2 x^{3/2}}{3 a}+\frac {2 i b x \log \left (1-\frac {i a e^{i \left (c+d \sqrt {x}\right )}}{b-\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d}-\frac {2 i b x \log \left (1-\frac {i a e^{i \left (c+d \sqrt {x}\right )}}{b+\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d}+\frac {4 b \sqrt {x} \text {Li}_2\left (\frac {i a e^{i \left (c+d \sqrt {x}\right )}}{b-\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d^2}-\frac {4 b \sqrt {x} \text {Li}_2\left (\frac {i a e^{i \left (c+d \sqrt {x}\right )}}{b+\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d^2}+\frac {4 i b \text {Li}_3\left (\frac {i a e^{i \left (c+d \sqrt {x}\right )}}{b-\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d^3}-\frac {4 i b \text {Li}_3\left (\frac {i a e^{i \left (c+d \sqrt {x}\right )}}{b+\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d^3}\\ \end {align*}

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Mathematica [A]  time = 10.70, size = 512, normalized size = 1.26 \[ \frac {2 \csc \left (c+d \sqrt {x}\right ) \left (a \sin \left (c+d \sqrt {x}\right )+b\right ) \left (x^{3/2}+\frac {3 b e^{3 i c} \left (a^2-b^2\right ) \left (d^2 x \log \left (1+\frac {a e^{i \left (2 c+d \sqrt {x}\right )}}{i b e^{i c}-\sqrt {e^{2 i c} \left (a^2-b^2\right )}}\right )-d^2 x \log \left (1+\frac {a e^{i \left (2 c+d \sqrt {x}\right )}}{\sqrt {e^{2 i c} \left (a^2-b^2\right )}+i b e^{i c}}\right )-2 i d \sqrt {x} \text {Li}_2\left (\frac {i a e^{i \left (2 c+d \sqrt {x}\right )}}{e^{i c} b+i \sqrt {\left (a^2-b^2\right ) e^{2 i c}}}\right )+2 i d \sqrt {x} \text {Li}_2\left (-\frac {a e^{i \left (2 c+d \sqrt {x}\right )}}{i e^{i c} b+\sqrt {\left (a^2-b^2\right ) e^{2 i c}}}\right )+2 \text {Li}_3\left (\frac {i a e^{i \left (2 c+d \sqrt {x}\right )}}{e^{i c} b+i \sqrt {\left (a^2-b^2\right ) e^{2 i c}}}\right )-2 \text {Li}_3\left (-\frac {a e^{i \left (2 c+d \sqrt {x}\right )}}{i e^{i c} b+\sqrt {\left (a^2-b^2\right ) e^{2 i c}}}\right )\right )}{d^3 \sqrt {e^{2 i c} \left (b^2-a^2\right )} \sqrt {e^{4 i c} \left (-\left (a^2-b^2\right )^2\right )}}\right )}{3 a \left (a+b \csc \left (c+d \sqrt {x}\right )\right )} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(2*Csc[c + d*Sqrt[x]]*(x^(3/2) + (3*b*(a^2 - b^2)*E^((3*I)*c)*(d^2*x*Log[1 + (a*E^(I*(2*c + d*Sqrt[x])))/(I*b*
E^(I*c) - Sqrt[(a^2 - b^2)*E^((2*I)*c)])] - d^2*x*Log[1 + (a*E^(I*(2*c + d*Sqrt[x])))/(I*b*E^(I*c) + Sqrt[(a^2
- b^2)*E^((2*I)*c)])] - (2*I)*d*Sqrt[x]*PolyLog[2, (I*a*E^(I*(2*c + d*Sqrt[x])))/(b*E^(I*c) + I*Sqrt[(a^2 - b
^2)*E^((2*I)*c)])] + (2*I)*d*Sqrt[x]*PolyLog[2, -((a*E^(I*(2*c + d*Sqrt[x])))/(I*b*E^(I*c) + Sqrt[(a^2 - b^2)*
E^((2*I)*c)]))] + 2*PolyLog[3, (I*a*E^(I*(2*c + d*Sqrt[x])))/(b*E^(I*c) + I*Sqrt[(a^2 - b^2)*E^((2*I)*c)])] -
2*PolyLog[3, -((a*E^(I*(2*c + d*Sqrt[x])))/(I*b*E^(I*c) + Sqrt[(a^2 - b^2)*E^((2*I)*c)]))]))/(d^3*Sqrt[(-a^2 +
b^2)*E^((2*I)*c)]*Sqrt[-((a^2 - b^2)^2*E^((4*I)*c))]))*(b + a*Sin[c + d*Sqrt[x]]))/(3*a*(a + b*Csc[c + d*Sqrt
[x]]))

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fricas [F]  time = 0.65, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {x}}{b \csc \left (d \sqrt {x} + c\right ) + a}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(x)/(b*csc(d*sqrt(x) + c) + a), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 2.20, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {x}}{a +b \csc \left (c +d \sqrt {x}\right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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