3.52 \(\int \sqrt{x} (a+b \sec (c+d \sqrt{x})) \, dx\)

Optimal. Leaf size=158 \[ \frac{4 i b \sqrt{x} \text{PolyLog}\left (2,-i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{4 i b \sqrt{x} \text{PolyLog}\left (2,i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{4 b \text{PolyLog}\left (3,-i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{4 b \text{PolyLog}\left (3,i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{2}{3} a x^{3/2}-\frac{4 i b x \tan ^{-1}\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d} \]

[Out]

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

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Rubi [A]  time = 0.131913, antiderivative size = 158, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 6, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.3, Rules used = {14, 4204, 4181, 2531, 2282, 6589} \[ \frac{4 i b \sqrt{x} \text{PolyLog}\left (2,-i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{4 i b \sqrt{x} \text{PolyLog}\left (2,i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{4 b \text{PolyLog}\left (3,-i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{4 b \text{PolyLog}\left (3,i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{2}{3} a x^{3/2}-\frac{4 i b x \tan ^{-1}\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a*x^(3/2))/3 - ((4*I)*b*x*ArcTan[E^(I*(c + d*Sqrt[x]))])/d + ((4*I)*b*Sqrt[x]*PolyLog[2, (-I)*E^(I*(c + d*S
qrt[x]))])/d^2 - ((4*I)*b*Sqrt[x]*PolyLog[2, I*E^(I*(c + d*Sqrt[x]))])/d^2 - (4*b*PolyLog[3, (-I)*E^(I*(c + d*
Sqrt[x]))])/d^3 + (4*b*PolyLog[3, I*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 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 4181

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

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

Mathematica [A]  time = 0.0993399, size = 155, normalized size = 0.98 \[ \frac{2 \left (6 i b d \sqrt{x} \text{PolyLog}\left (2,-i e^{i \left (c+d \sqrt{x}\right )}\right )-6 i b d \sqrt{x} \text{PolyLog}\left (2,i e^{i \left (c+d \sqrt{x}\right )}\right )-6 b \text{PolyLog}\left (3,-i e^{i \left (c+d \sqrt{x}\right )}\right )+6 b \text{PolyLog}\left (3,i e^{i \left (c+d \sqrt{x}\right )}\right )+a d^3 x^{3/2}-6 i b d^2 x \tan ^{-1}\left (e^{i \left (c+d \sqrt{x}\right )}\right )\right )}{3 d^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.106, size = 0, normalized size = 0. \begin{align*} \int \left ( a+b\sec \left ( c+d\sqrt{x} \right ) \right ) \sqrt{x}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 2.08502, size = 505, normalized size = 3.2 \begin{align*} \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 (\sec \left (d \sqrt{x} + c\right ) + \tan \left (d \sqrt{x} + c\right )\right ) + 3 \,{\left (-2 i \,{\left (d \sqrt{x} + c\right )}^{2} b + 4 i \,{\left (d \sqrt{x} + c\right )} b c\right )} \arctan \left (\cos \left (d \sqrt{x} + c\right ), \sin \left (d \sqrt{x} + c\right ) + 1\right ) + 3 \,{\left (-2 i \,{\left (d \sqrt{x} + c\right )}^{2} b + 4 i \,{\left (d \sqrt{x} + c\right )} b c\right )} \arctan \left (\cos \left (d \sqrt{x} + c\right ), -\sin \left (d \sqrt{x} + c\right ) + 1\right ) + 3 \,{\left (-4 i \,{\left (d \sqrt{x} + c\right )} b + 4 i \, b c\right )}{\rm Li}_2\left (i \, e^{\left (i \, d \sqrt{x} + i \, c\right )}\right ) + 3 \,{\left (4 i \,{\left (d \sqrt{x} + c\right )} b - 4 i \, b c\right )}{\rm Li}_2\left (-i \, 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 \, \sin \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 \, \sin \left (d \sqrt{x} + c\right ) + 1\right ) + 12 \, b{\rm Li}_{3}(i \, e^{\left (i \, d \sqrt{x} + i \, c\right )}) - 12 \, b{\rm Li}_{3}(-i \, e^{\left (i \, d \sqrt{x} + i \, c\right )})}{3 \, d^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(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(sec(d*sqrt(x) + c
) + tan(d*sqrt(x) + c)) + 3*(-2*I*(d*sqrt(x) + c)^2*b + 4*I*(d*sqrt(x) + c)*b*c)*arctan2(cos(d*sqrt(x) + c), s
in(d*sqrt(x) + c) + 1) + 3*(-2*I*(d*sqrt(x) + c)^2*b + 4*I*(d*sqrt(x) + c)*b*c)*arctan2(cos(d*sqrt(x) + c), -s
in(d*sqrt(x) + c) + 1) + 3*(-4*I*(d*sqrt(x) + c)*b + 4*I*b*c)*dilog(I*e^(I*d*sqrt(x) + I*c)) + 3*(4*I*(d*sqrt(
x) + c)*b - 4*I*b*c)*dilog(-I*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*sin(d*sqrt(x) + c) + 1) - 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*sin(d*sqrt(x) + c) + 1) + 12*b*polylog(3, I*e^(I*
d*sqrt(x) + I*c)) - 12*b*polylog(3, -I*e^(I*d*sqrt(x) + I*c)))/d^3

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (b \sqrt{x} \sec \left (d \sqrt{x} + c\right ) + a \sqrt{x}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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