\(\int \sqrt {x} (a+b \csc (c+d \sqrt {x})) \, dx\) [52]

Optimal result

Integrand size = 20, antiderivative size = 144 \[ \int \sqrt {x} \left (a+b \csc \left (c+d \sqrt {x}\right )\right ) \, dx=\frac {2}{3} a x^{3/2}-\frac {4 b x \text {arctanh}\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {4 i b \sqrt {x} \operatorname {PolyLog}\left (2,-e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {4 i b \sqrt {x} \operatorname {PolyLog}\left (2,e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {4 b \operatorname {PolyLog}\left (3,-e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {4 b \operatorname {PolyLog}\left (3,e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {14, 4290, 4268, 2611, 2320, 6724} \[ \int \sqrt {x} \left (a+b \csc \left (c+d \sqrt {x}\right )\right ) \, dx=\frac {2}{3} a x^{3/2}-\frac {4 b x \text {arctanh}\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d}-\frac {4 b \operatorname {PolyLog}\left (3,-e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {4 b \operatorname {PolyLog}\left (3,e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {4 i b \sqrt {x} \operatorname {PolyLog}\left (2,-e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {4 i b \sqrt {x} \operatorname {PolyLog}\left (2,e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2} \]

[In]

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

[Out]

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

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 2320

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 2611

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 4268

Int[csc[(e_.) + (f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[-2*(c + d*x)^m*(ArcTanh[E^(I*(e + f*
x))]/f), x] + (-Dist[d*(m/f), Int[(c + d*x)^(m - 1)*Log[1 - E^(I*(e + f*x))], x], x] + Dist[d*(m/f), Int[(c +
d*x)^(m - 1)*Log[1 + E^(I*(e + f*x))], x], x]) /; FreeQ[{c, d, e, f}, x] && IGtQ[m, 0]

Rule 4290

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 6724

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

Mathematica [A] (verified)

Time = 4.27 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.33 \[ \int \sqrt {x} \left (a+b \csc \left (c+d \sqrt {x}\right )\right ) \, dx=\frac {2 \left (a d^3 x^{3/2}-6 b d^2 x \text {arctanh}\left (\cos \left (c+d \sqrt {x}\right )+i \sin \left (c+d \sqrt {x}\right )\right )+6 i b d \sqrt {x} \operatorname {PolyLog}\left (2,-\cos \left (c+d \sqrt {x}\right )-i \sin \left (c+d \sqrt {x}\right )\right )-6 i b d \sqrt {x} \operatorname {PolyLog}\left (2,\cos \left (c+d \sqrt {x}\right )+i \sin \left (c+d \sqrt {x}\right )\right )-6 b \operatorname {PolyLog}\left (3,-\cos \left (c+d \sqrt {x}\right )-i \sin \left (c+d \sqrt {x}\right )\right )+6 b \operatorname {PolyLog}\left (3,\cos \left (c+d \sqrt {x}\right )+i \sin \left (c+d \sqrt {x}\right )\right )\right )}{3 d^3} \]

[In]

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

[Out]

(2*(a*d^3*x^(3/2) - 6*b*d^2*x*ArcTanh[Cos[c + d*Sqrt[x]] + I*Sin[c + d*Sqrt[x]]] + (6*I)*b*d*Sqrt[x]*PolyLog[2
, -Cos[c + d*Sqrt[x]] - I*Sin[c + d*Sqrt[x]]] - (6*I)*b*d*Sqrt[x]*PolyLog[2, Cos[c + d*Sqrt[x]] + I*Sin[c + d*
Sqrt[x]]] - 6*b*PolyLog[3, -Cos[c + d*Sqrt[x]] - I*Sin[c + d*Sqrt[x]]] + 6*b*PolyLog[3, Cos[c + d*Sqrt[x]] + I
*Sin[c + d*Sqrt[x]]]))/(3*d^3)

Maple [F]

\[\int \left (a +b \csc \left (c +d \sqrt {x}\right )\right ) \sqrt {x}d x\]

[In]

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

[Out]

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

Fricas [F]

\[ \int \sqrt {x} \left (a+b \csc \left (c+d \sqrt {x}\right )\right ) \, dx=\int { {\left (b \csc \left (d \sqrt {x} + c\right ) + a\right )} \sqrt {x} \,d x } \]

[In]

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

[Out]

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

Sympy [F]

\[ \int \sqrt {x} \left (a+b \csc \left (c+d \sqrt {x}\right )\right ) \, dx=\int \sqrt {x} \left (a + b \csc {\left (c + d \sqrt {x} \right )}\right )\, dx \]

[In]

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

[Out]

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

Maxima [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 370 vs. \(2 (110) = 220\).

Time = 0.27 (sec) , antiderivative size = 370, normalized size of antiderivative = 2.57 \[ \int \sqrt {x} \left (a+b \csc \left (c+d \sqrt {x}\right )\right ) \, dx=\frac {2 \, {\left (d \sqrt {x} + c\right )}^{3} a - 6 \, {\left (d \sqrt {x} + c\right )}^{2} a c + 6 \, {\left (d \sqrt {x} + c\right )} a c^{2} - 6 \, b c^{2} \log \left (\cot \left (d \sqrt {x} + c\right ) + \csc \left (d \sqrt {x} + c\right )\right ) + 6 \, {\left (-i \, {\left (d \sqrt {x} + c\right )}^{2} b + 2 i \, {\left (d \sqrt {x} + c\right )} b c\right )} \arctan \left (\sin \left (d \sqrt {x} + c\right ), \cos \left (d \sqrt {x} + c\right ) + 1\right ) + 6 \, {\left (-i \, {\left (d \sqrt {x} + c\right )}^{2} b + 2 i \, {\left (d \sqrt {x} + c\right )} b c\right )} \arctan \left (\sin \left (d \sqrt {x} + c\right ), -\cos \left (d \sqrt {x} + c\right ) + 1\right ) + 12 \, {\left (i \, {\left (d \sqrt {x} + c\right )} b - i \, b c\right )} {\rm Li}_2\left (-e^{\left (i \, d \sqrt {x} + i \, c\right )}\right ) + 12 \, {\left (-i \, {\left (d \sqrt {x} + c\right )} b + i \, b c\right )} {\rm Li}_2\left (e^{\left (i \, d \sqrt {x} + i \, c\right )}\right ) - 3 \, {\left ({\left (d \sqrt {x} + c\right )}^{2} b - 2 \, {\left (d \sqrt {x} + c\right )} b c\right )} \log \left (\cos \left (d \sqrt {x} + c\right )^{2} + \sin \left (d \sqrt {x} + c\right )^{2} + 2 \, \cos \left (d \sqrt {x} + c\right ) + 1\right ) + 3 \, {\left ({\left (d \sqrt {x} + c\right )}^{2} b - 2 \, {\left (d \sqrt {x} + c\right )} b c\right )} \log \left (\cos \left (d \sqrt {x} + c\right )^{2} + \sin \left (d \sqrt {x} + c\right )^{2} - 2 \, \cos \left (d \sqrt {x} + c\right ) + 1\right ) - 12 \, b {\rm Li}_{3}(-e^{\left (i \, d \sqrt {x} + i \, c\right )}) + 12 \, b {\rm Li}_{3}(e^{\left (i \, d \sqrt {x} + i \, c\right )})}{3 \, d^{3}} \]

[In]

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

[Out]

1/3*(2*(d*sqrt(x) + c)^3*a - 6*(d*sqrt(x) + c)^2*a*c + 6*(d*sqrt(x) + c)*a*c^2 - 6*b*c^2*log(cot(d*sqrt(x) + c
) + csc(d*sqrt(x) + c)) + 6*(-I*(d*sqrt(x) + c)^2*b + 2*I*(d*sqrt(x) + c)*b*c)*arctan2(sin(d*sqrt(x) + c), cos
(d*sqrt(x) + c) + 1) + 6*(-I*(d*sqrt(x) + c)^2*b + 2*I*(d*sqrt(x) + c)*b*c)*arctan2(sin(d*sqrt(x) + c), -cos(d
*sqrt(x) + c) + 1) + 12*(I*(d*sqrt(x) + c)*b - I*b*c)*dilog(-e^(I*d*sqrt(x) + I*c)) + 12*(-I*(d*sqrt(x) + c)*b
+ I*b*c)*dilog(e^(I*d*sqrt(x) + I*c)) - 3*((d*sqrt(x) + c)^2*b - 2*(d*sqrt(x) + c)*b*c)*log(cos(d*sqrt(x) + c
)^2 + sin(d*sqrt(x) + c)^2 + 2*cos(d*sqrt(x) + c) + 1) + 3*((d*sqrt(x) + c)^2*b - 2*(d*sqrt(x) + c)*b*c)*log(c
os(d*sqrt(x) + c)^2 + sin(d*sqrt(x) + c)^2 - 2*cos(d*sqrt(x) + c) + 1) - 12*b*polylog(3, -e^(I*d*sqrt(x) + I*c
)) + 12*b*polylog(3, e^(I*d*sqrt(x) + I*c)))/d^3

Giac [F]

\[ \int \sqrt {x} \left (a+b \csc \left (c+d \sqrt {x}\right )\right ) \, dx=\int { {\left (b \csc \left (d \sqrt {x} + c\right ) + a\right )} \sqrt {x} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \sqrt {x} \left (a+b \csc \left (c+d \sqrt {x}\right )\right ) \, dx=\int \sqrt {x}\,\left (a+\frac {b}{\sin \left (c+d\,\sqrt {x}\right )}\right ) \,d x \]

[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)