3.51 \(\int x^{3/2} (a+b \sec (c+d \sqrt {x})) \, dx\)

Optimal. Leaf size=284 \[ \frac {2}{5} a x^{5/2}+\frac {48 b \text {Li}_5\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^5}-\frac {48 b \text {Li}_5\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^5}-\frac {48 i b \sqrt {x} \text {Li}_4\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {48 i b \sqrt {x} \text {Li}_4\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}-\frac {24 b x \text {Li}_3\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {24 b x \text {Li}_3\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {8 i b x^{3/2} \text {Li}_2\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {8 i b x^{3/2} \text {Li}_2\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {4 i b x^2 \tan ^{-1}\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d} \]

[Out]

2/5*a*x^(5/2)-4*I*b*x^2*arctan(exp(I*(c+d*x^(1/2))))/d+8*I*b*x^(3/2)*polylog(2,-I*exp(I*(c+d*x^(1/2))))/d^2-8*
I*b*x^(3/2)*polylog(2,I*exp(I*(c+d*x^(1/2))))/d^2-24*b*x*polylog(3,-I*exp(I*(c+d*x^(1/2))))/d^3+24*b*x*polylog
(3,I*exp(I*(c+d*x^(1/2))))/d^3+48*b*polylog(5,-I*exp(I*(c+d*x^(1/2))))/d^5-48*b*polylog(5,I*exp(I*(c+d*x^(1/2)
)))/d^5-48*I*b*polylog(4,-I*exp(I*(c+d*x^(1/2))))*x^(1/2)/d^4+48*I*b*polylog(4,I*exp(I*(c+d*x^(1/2))))*x^(1/2)
/d^4

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Rubi [A]  time = 0.25, antiderivative size = 284, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 7, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {14, 4204, 4181, 2531, 6609, 2282, 6589} \[ \frac {8 i b x^{3/2} \text {PolyLog}\left (2,-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {8 i b x^{3/2} \text {PolyLog}\left (2,i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {24 b x \text {PolyLog}\left (3,-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {24 b x \text {PolyLog}\left (3,i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}-\frac {48 i b \sqrt {x} \text {PolyLog}\left (4,-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {48 i b \sqrt {x} \text {PolyLog}\left (4,i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {48 b \text {PolyLog}\left (5,-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^5}-\frac {48 b \text {PolyLog}\left (5,i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^5}+\frac {2}{5} a x^{5/2}-\frac {4 i b x^2 \tan ^{-1}\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d} \]

Antiderivative was successfully verified.

[In]

Int[x^(3/2)*(a + b*Sec[c + d*Sqrt[x]]),x]

[Out]

(2*a*x^(5/2))/5 - ((4*I)*b*x^2*ArcTan[E^(I*(c + d*Sqrt[x]))])/d + ((8*I)*b*x^(3/2)*PolyLog[2, (-I)*E^(I*(c + d
*Sqrt[x]))])/d^2 - ((8*I)*b*x^(3/2)*PolyLog[2, I*E^(I*(c + d*Sqrt[x]))])/d^2 - (24*b*x*PolyLog[3, (-I)*E^(I*(c
+ d*Sqrt[x]))])/d^3 + (24*b*x*PolyLog[3, I*E^(I*(c + d*Sqrt[x]))])/d^3 - ((48*I)*b*Sqrt[x]*PolyLog[4, (-I)*E^
(I*(c + d*Sqrt[x]))])/d^4 + ((48*I)*b*Sqrt[x]*PolyLog[4, I*E^(I*(c + d*Sqrt[x]))])/d^4 + (48*b*PolyLog[5, (-I)
*E^(I*(c + d*Sqrt[x]))])/d^5 - (48*b*PolyLog[5, I*E^(I*(c + d*Sqrt[x]))])/d^5

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 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 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 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 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]

Rule 6609

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

Rubi steps

\begin {align*} \int x^{3/2} \left (a+b \sec \left (c+d \sqrt {x}\right )\right ) \, dx &=\int \left (a x^{3/2}+b x^{3/2} \sec \left (c+d \sqrt {x}\right )\right ) \, dx\\ &=\frac {2}{5} a x^{5/2}+b \int x^{3/2} \sec \left (c+d \sqrt {x}\right ) \, dx\\ &=\frac {2}{5} a x^{5/2}+(2 b) \operatorname {Subst}\left (\int x^4 \sec (c+d x) \, dx,x,\sqrt {x}\right )\\ &=\frac {2}{5} a x^{5/2}-\frac {4 i b x^2 \tan ^{-1}\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d}-\frac {(8 b) \operatorname {Subst}\left (\int x^3 \log \left (1-i e^{i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d}+\frac {(8 b) \operatorname {Subst}\left (\int x^3 \log \left (1+i e^{i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d}\\ &=\frac {2}{5} a x^{5/2}-\frac {4 i b x^2 \tan ^{-1}\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {8 i b x^{3/2} \text {Li}_2\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {8 i b x^{3/2} \text {Li}_2\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {(24 i b) \operatorname {Subst}\left (\int x^2 \text {Li}_2\left (-i e^{i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^2}+\frac {(24 i b) \operatorname {Subst}\left (\int x^2 \text {Li}_2\left (i e^{i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^2}\\ &=\frac {2}{5} a x^{5/2}-\frac {4 i b x^2 \tan ^{-1}\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {8 i b x^{3/2} \text {Li}_2\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {8 i b x^{3/2} \text {Li}_2\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {24 b x \text {Li}_3\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {24 b x \text {Li}_3\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {(48 b) \operatorname {Subst}\left (\int x \text {Li}_3\left (-i e^{i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^3}-\frac {(48 b) \operatorname {Subst}\left (\int x \text {Li}_3\left (i e^{i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^3}\\ &=\frac {2}{5} a x^{5/2}-\frac {4 i b x^2 \tan ^{-1}\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {8 i b x^{3/2} \text {Li}_2\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {8 i b x^{3/2} \text {Li}_2\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {24 b x \text {Li}_3\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {24 b x \text {Li}_3\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}-\frac {48 i b \sqrt {x} \text {Li}_4\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {48 i b \sqrt {x} \text {Li}_4\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {(48 i b) \operatorname {Subst}\left (\int \text {Li}_4\left (-i e^{i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^4}-\frac {(48 i b) \operatorname {Subst}\left (\int \text {Li}_4\left (i e^{i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^4}\\ &=\frac {2}{5} a x^{5/2}-\frac {4 i b x^2 \tan ^{-1}\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {8 i b x^{3/2} \text {Li}_2\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {8 i b x^{3/2} \text {Li}_2\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {24 b x \text {Li}_3\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {24 b x \text {Li}_3\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}-\frac {48 i b \sqrt {x} \text {Li}_4\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {48 i b \sqrt {x} \text {Li}_4\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {(48 b) \operatorname {Subst}\left (\int \frac {\text {Li}_4(-i x)}{x} \, dx,x,e^{i \left (c+d \sqrt {x}\right )}\right )}{d^5}-\frac {(48 b) \operatorname {Subst}\left (\int \frac {\text {Li}_4(i x)}{x} \, dx,x,e^{i \left (c+d \sqrt {x}\right )}\right )}{d^5}\\ &=\frac {2}{5} a x^{5/2}-\frac {4 i b x^2 \tan ^{-1}\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {8 i b x^{3/2} \text {Li}_2\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {8 i b x^{3/2} \text {Li}_2\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {24 b x \text {Li}_3\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {24 b x \text {Li}_3\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}-\frac {48 i b \sqrt {x} \text {Li}_4\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {48 i b \sqrt {x} \text {Li}_4\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {48 b \text {Li}_5\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^5}-\frac {48 b \text {Li}_5\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^5}\\ \end {align*}

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Mathematica [A]  time = 0.25, size = 281, normalized size = 0.99 \[ \frac {2 \left (a d^5 x^{5/2}-10 i b d^4 x^2 \tan ^{-1}\left (e^{i \left (c+d \sqrt {x}\right )}\right )+20 i b d^3 x^{3/2} \text {Li}_2\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )-20 i b d^3 x^{3/2} \text {Li}_2\left (i e^{i \left (c+d \sqrt {x}\right )}\right )-60 b d^2 x \text {Li}_3\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )+60 b d^2 x \text {Li}_3\left (i e^{i \left (c+d \sqrt {x}\right )}\right )-120 i b d \sqrt {x} \text {Li}_4\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )+120 i b d \sqrt {x} \text {Li}_4\left (i e^{i \left (c+d \sqrt {x}\right )}\right )+120 b \text {Li}_5\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )-120 b \text {Li}_5\left (i e^{i \left (c+d \sqrt {x}\right )}\right )\right )}{5 d^5} \]

Antiderivative was successfully verified.

[In]

Integrate[x^(3/2)*(a + b*Sec[c + d*Sqrt[x]]),x]

[Out]

(2*(a*d^5*x^(5/2) - (10*I)*b*d^4*x^2*ArcTan[E^(I*(c + d*Sqrt[x]))] + (20*I)*b*d^3*x^(3/2)*PolyLog[2, (-I)*E^(I
*(c + d*Sqrt[x]))] - (20*I)*b*d^3*x^(3/2)*PolyLog[2, I*E^(I*(c + d*Sqrt[x]))] - 60*b*d^2*x*PolyLog[3, (-I)*E^(
I*(c + d*Sqrt[x]))] + 60*b*d^2*x*PolyLog[3, I*E^(I*(c + d*Sqrt[x]))] - (120*I)*b*d*Sqrt[x]*PolyLog[4, (-I)*E^(
I*(c + d*Sqrt[x]))] + (120*I)*b*d*Sqrt[x]*PolyLog[4, I*E^(I*(c + d*Sqrt[x]))] + 120*b*PolyLog[5, (-I)*E^(I*(c
+ d*Sqrt[x]))] - 120*b*PolyLog[5, I*E^(I*(c + d*Sqrt[x]))]))/(5*d^5)

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fricas [F]  time = 0.60, size = 0, normalized size = 0.00 \[ {\rm integral}\left (b x^{\frac {3}{2}} \sec \left (d \sqrt {x} + c\right ) + a x^{\frac {3}{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(3/2)*(a+b*sec(c+d*x^(1/2))),x, algorithm="fricas")

[Out]

integral(b*x^(3/2)*sec(d*sqrt(x) + c) + a*x^(3/2), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \sec \left (d \sqrt {x} + c\right ) + a\right )} x^{\frac {3}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(3/2)*(a+b*sec(c+d*x^(1/2))),x, algorithm="giac")

[Out]

integrate((b*sec(d*sqrt(x) + c) + a)*x^(3/2), x)

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maple [F]  time = 1.20, size = 0, normalized size = 0.00 \[ \int x^{\frac {3}{2}} \left (a +b \sec \left (c +d \sqrt {x}\right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(3/2)*(a+b*sec(c+d*x^(1/2))),x)

[Out]

int(x^(3/2)*(a+b*sec(c+d*x^(1/2))),x)

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maxima [B]  time = 0.97, size = 738, normalized size = 2.60 \[ \frac {2 \, {\left (d \sqrt {x} + c\right )}^{5} a - 10 \, {\left (d \sqrt {x} + c\right )}^{4} a c + 20 \, {\left (d \sqrt {x} + c\right )}^{3} a c^{2} - 20 \, {\left (d \sqrt {x} + c\right )}^{2} a c^{3} + 10 \, {\left (d \sqrt {x} + c\right )} a c^{4} + 10 \, b c^{4} \log \left (\sec \left (d \sqrt {x} + c\right ) + \tan \left (d \sqrt {x} + c\right )\right ) + 5 \, {\left (-2 i \, {\left (d \sqrt {x} + c\right )}^{4} b + 8 i \, {\left (d \sqrt {x} + c\right )}^{3} b c - 12 i \, {\left (d \sqrt {x} + c\right )}^{2} b c^{2} + 8 i \, {\left (d \sqrt {x} + c\right )} b c^{3}\right )} \arctan \left (\cos \left (d \sqrt {x} + c\right ), \sin \left (d \sqrt {x} + c\right ) + 1\right ) + 5 \, {\left (-2 i \, {\left (d \sqrt {x} + c\right )}^{4} b + 8 i \, {\left (d \sqrt {x} + c\right )}^{3} b c - 12 i \, {\left (d \sqrt {x} + c\right )}^{2} b c^{2} + 8 i \, {\left (d \sqrt {x} + c\right )} b c^{3}\right )} \arctan \left (\cos \left (d \sqrt {x} + c\right ), -\sin \left (d \sqrt {x} + c\right ) + 1\right ) + 5 \, {\left (-8 i \, {\left (d \sqrt {x} + c\right )}^{3} b + 24 i \, {\left (d \sqrt {x} + c\right )}^{2} b c - 24 i \, {\left (d \sqrt {x} + c\right )} b c^{2} + 8 i \, b c^{3}\right )} {\rm Li}_2\left (i \, e^{\left (i \, d \sqrt {x} + i \, c\right )}\right ) + 5 \, {\left (8 i \, {\left (d \sqrt {x} + c\right )}^{3} b - 24 i \, {\left (d \sqrt {x} + c\right )}^{2} b c + 24 i \, {\left (d \sqrt {x} + c\right )} b c^{2} - 8 i \, b c^{3}\right )} {\rm Li}_2\left (-i \, e^{\left (i \, d \sqrt {x} + i \, c\right )}\right ) + 5 \, {\left ({\left (d \sqrt {x} + c\right )}^{4} b - 4 \, {\left (d \sqrt {x} + c\right )}^{3} b c + 6 \, {\left (d \sqrt {x} + c\right )}^{2} b c^{2} - 4 \, {\left (d \sqrt {x} + c\right )} b c^{3}\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 ) - 5 \, {\left ({\left (d \sqrt {x} + c\right )}^{4} b - 4 \, {\left (d \sqrt {x} + c\right )}^{3} b c + 6 \, {\left (d \sqrt {x} + c\right )}^{2} b c^{2} - 4 \, {\left (d \sqrt {x} + c\right )} b c^{3}\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 ) - 240 \, b {\rm Li}_{5}(i \, e^{\left (i \, d \sqrt {x} + i \, c\right )}) + 240 \, b {\rm Li}_{5}(-i \, e^{\left (i \, d \sqrt {x} + i \, c\right )}) + 5 \, {\left (48 i \, {\left (d \sqrt {x} + c\right )} b - 48 i \, b c\right )} {\rm Li}_{4}(i \, e^{\left (i \, d \sqrt {x} + i \, c\right )}) + 5 \, {\left (-48 i \, {\left (d \sqrt {x} + c\right )} b + 48 i \, b c\right )} {\rm Li}_{4}(-i \, e^{\left (i \, d \sqrt {x} + i \, c\right )}) + 120 \, {\left ({\left (d \sqrt {x} + c\right )}^{2} b - 2 \, {\left (d \sqrt {x} + c\right )} b c + b c^{2}\right )} {\rm Li}_{3}(i \, e^{\left (i \, d \sqrt {x} + i \, c\right )}) - 120 \, {\left ({\left (d \sqrt {x} + c\right )}^{2} b - 2 \, {\left (d \sqrt {x} + c\right )} b c + b c^{2}\right )} {\rm Li}_{3}(-i \, e^{\left (i \, d \sqrt {x} + i \, c\right )})}{5 \, d^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(3/2)*(a+b*sec(c+d*x^(1/2))),x, algorithm="maxima")

[Out]

1/5*(2*(d*sqrt(x) + c)^5*a - 10*(d*sqrt(x) + c)^4*a*c + 20*(d*sqrt(x) + c)^3*a*c^2 - 20*(d*sqrt(x) + c)^2*a*c^
3 + 10*(d*sqrt(x) + c)*a*c^4 + 10*b*c^4*log(sec(d*sqrt(x) + c) + tan(d*sqrt(x) + c)) + 5*(-2*I*(d*sqrt(x) + c)
^4*b + 8*I*(d*sqrt(x) + c)^3*b*c - 12*I*(d*sqrt(x) + c)^2*b*c^2 + 8*I*(d*sqrt(x) + c)*b*c^3)*arctan2(cos(d*sqr
t(x) + c), sin(d*sqrt(x) + c) + 1) + 5*(-2*I*(d*sqrt(x) + c)^4*b + 8*I*(d*sqrt(x) + c)^3*b*c - 12*I*(d*sqrt(x)
+ c)^2*b*c^2 + 8*I*(d*sqrt(x) + c)*b*c^3)*arctan2(cos(d*sqrt(x) + c), -sin(d*sqrt(x) + c) + 1) + 5*(-8*I*(d*s
qrt(x) + c)^3*b + 24*I*(d*sqrt(x) + c)^2*b*c - 24*I*(d*sqrt(x) + c)*b*c^2 + 8*I*b*c^3)*dilog(I*e^(I*d*sqrt(x)
+ I*c)) + 5*(8*I*(d*sqrt(x) + c)^3*b - 24*I*(d*sqrt(x) + c)^2*b*c + 24*I*(d*sqrt(x) + c)*b*c^2 - 8*I*b*c^3)*di
log(-I*e^(I*d*sqrt(x) + I*c)) + 5*((d*sqrt(x) + c)^4*b - 4*(d*sqrt(x) + c)^3*b*c + 6*(d*sqrt(x) + c)^2*b*c^2 -
4*(d*sqrt(x) + c)*b*c^3)*log(cos(d*sqrt(x) + c)^2 + sin(d*sqrt(x) + c)^2 + 2*sin(d*sqrt(x) + c) + 1) - 5*((d*
sqrt(x) + c)^4*b - 4*(d*sqrt(x) + c)^3*b*c + 6*(d*sqrt(x) + c)^2*b*c^2 - 4*(d*sqrt(x) + c)*b*c^3)*log(cos(d*sq
rt(x) + c)^2 + sin(d*sqrt(x) + c)^2 - 2*sin(d*sqrt(x) + c) + 1) - 240*b*polylog(5, I*e^(I*d*sqrt(x) + I*c)) +
240*b*polylog(5, -I*e^(I*d*sqrt(x) + I*c)) + 5*(48*I*(d*sqrt(x) + c)*b - 48*I*b*c)*polylog(4, I*e^(I*d*sqrt(x)
+ I*c)) + 5*(-48*I*(d*sqrt(x) + c)*b + 48*I*b*c)*polylog(4, -I*e^(I*d*sqrt(x) + I*c)) + 120*((d*sqrt(x) + c)^
2*b - 2*(d*sqrt(x) + c)*b*c + b*c^2)*polylog(3, I*e^(I*d*sqrt(x) + I*c)) - 120*((d*sqrt(x) + c)^2*b - 2*(d*sqr
t(x) + c)*b*c + b*c^2)*polylog(3, -I*e^(I*d*sqrt(x) + I*c)))/d^5

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int x^{3/2}\,\left (a+\frac {b}{\cos \left (c+d\,\sqrt {x}\right )}\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(3/2)*(a + b/cos(c + d*x^(1/2))),x)

[Out]

int(x^(3/2)*(a + b/cos(c + d*x^(1/2))), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{\frac {3}{2}} \left (a + b \sec {\left (c + d \sqrt {x} \right )}\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(3/2)*(a+b*sec(c+d*x**(1/2))),x)

[Out]

Integral(x**(3/2)*(a + b*sec(c + d*sqrt(x))), x)

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