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

Optimal. Leaf size=234 \[ \frac {3 i b \text {Li}_4\left (-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{2 d^4 \left (a^2+b^2\right )}+\frac {3 b \sqrt {x} \text {Li}_3\left (-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{d^3 \left (a^2+b^2\right )}-\frac {3 i b x \text {Li}_2\left (-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{d^2 \left (a^2+b^2\right )}+\frac {2 b x^{3/2} \log \left (1+\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{d \left (a^2+b^2\right )}+\frac {x^2}{2 (a+i b)} \]

[Out]

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

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Rubi [A]  time = 0.34, antiderivative size = 234, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.389, Rules used = {3747, 3732, 2190, 2531, 6609, 2282, 6589} \[ -\frac {3 i b x \text {Li}_2\left (-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{d^2 \left (a^2+b^2\right )}+\frac {3 b \sqrt {x} \text {Li}_3\left (-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{d^3 \left (a^2+b^2\right )}+\frac {3 i b \text {Li}_4\left (-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{2 d^4 \left (a^2+b^2\right )}+\frac {2 b x^{3/2} \log \left (1+\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{d \left (a^2+b^2\right )}+\frac {x^2}{2 (a+i b)} \]

Antiderivative was successfully verified.

[In]

Int[x/(a + b*Tan[c + d*Sqrt[x]]),x]

[Out]

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

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 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 3732

Int[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(c + d*x)^(m + 1)/(d*
(m + 1)*(a + I*b)), x] + Dist[2*I*b, Int[((c + d*x)^m*E^Simp[2*I*(e + f*x), x])/((a + I*b)^2 + (a^2 + b^2)*E^S
imp[2*I*(e + f*x), x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[a^2 + b^2, 0] && IGtQ[m, 0]

Rule 3747

Int[(x_)^(m_.)*((a_.) + (b_.)*Tan[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplif
y[(m + 1)/n] - 1)*(a + b*Tan[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 \frac {x}{a+b \tan \left (c+d \sqrt {x}\right )} \, dx &=2 \operatorname {Subst}\left (\int \frac {x^3}{a+b \tan (c+d x)} \, dx,x,\sqrt {x}\right )\\ &=\frac {x^2}{2 (a+i b)}+(4 i b) \operatorname {Subst}\left (\int \frac {e^{2 i (c+d x)} x^3}{(a+i b)^2+\left (a^2+b^2\right ) e^{2 i (c+d x)}} \, dx,x,\sqrt {x}\right )\\ &=\frac {x^2}{2 (a+i b)}+\frac {2 b x^{3/2} \log \left (1+\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{\left (a^2+b^2\right ) d}-\frac {(6 b) \operatorname {Subst}\left (\int x^2 \log \left (1+\frac {\left (a^2+b^2\right ) e^{2 i (c+d x)}}{(a+i b)^2}\right ) \, dx,x,\sqrt {x}\right )}{\left (a^2+b^2\right ) d}\\ &=\frac {x^2}{2 (a+i b)}+\frac {2 b x^{3/2} \log \left (1+\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{\left (a^2+b^2\right ) d}-\frac {3 i b x \text {Li}_2\left (-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{\left (a^2+b^2\right ) d^2}+\frac {(6 i b) \operatorname {Subst}\left (\int x \text {Li}_2\left (-\frac {\left (a^2+b^2\right ) e^{2 i (c+d x)}}{(a+i b)^2}\right ) \, dx,x,\sqrt {x}\right )}{\left (a^2+b^2\right ) d^2}\\ &=\frac {x^2}{2 (a+i b)}+\frac {2 b x^{3/2} \log \left (1+\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{\left (a^2+b^2\right ) d}-\frac {3 i b x \text {Li}_2\left (-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{\left (a^2+b^2\right ) d^2}+\frac {3 b \sqrt {x} \text {Li}_3\left (-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{\left (a^2+b^2\right ) d^3}-\frac {(3 b) \operatorname {Subst}\left (\int \text {Li}_3\left (-\frac {\left (a^2+b^2\right ) e^{2 i (c+d x)}}{(a+i b)^2}\right ) \, dx,x,\sqrt {x}\right )}{\left (a^2+b^2\right ) d^3}\\ &=\frac {x^2}{2 (a+i b)}+\frac {2 b x^{3/2} \log \left (1+\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{\left (a^2+b^2\right ) d}-\frac {3 i b x \text {Li}_2\left (-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{\left (a^2+b^2\right ) d^2}+\frac {3 b \sqrt {x} \text {Li}_3\left (-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{\left (a^2+b^2\right ) d^3}+\frac {(3 i b) \operatorname {Subst}\left (\int \frac {\text {Li}_3\left (-\frac {\left (a^2+b^2\right ) x}{(a+i b)^2}\right )}{x} \, dx,x,e^{2 i \left (c+d \sqrt {x}\right )}\right )}{2 \left (a^2+b^2\right ) d^4}\\ &=\frac {x^2}{2 (a+i b)}+\frac {2 b x^{3/2} \log \left (1+\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{\left (a^2+b^2\right ) d}-\frac {3 i b x \text {Li}_2\left (-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{\left (a^2+b^2\right ) d^2}+\frac {3 b \sqrt {x} \text {Li}_3\left (-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{\left (a^2+b^2\right ) d^3}+\frac {3 i b \text {Li}_4\left (-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{2 \left (a^2+b^2\right ) d^4}\\ \end {align*}

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Mathematica [A]  time = 1.28, size = 213, normalized size = 0.91 \[ \frac {4 b d^3 x^{3/2} \log \left (1+\frac {(a+i b) e^{-2 i \left (c+d \sqrt {x}\right )}}{a-i b}\right )+6 i b d^2 x \text {Li}_2\left (\frac {(-a-i b) e^{-2 i \left (c+d \sqrt {x}\right )}}{a-i b}\right )+6 b d \sqrt {x} \text {Li}_3\left (\frac {(-a-i b) e^{-2 i \left (c+d \sqrt {x}\right )}}{a-i b}\right )-3 i b \text {Li}_4\left (\frac {(-a-i b) e^{-2 i \left (c+d \sqrt {x}\right )}}{a-i b}\right )+a d^4 x^2+i b d^4 x^2}{2 d^4 \left (a^2+b^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*d^4*x^2 + I*b*d^4*x^2 + 4*b*d^3*x^(3/2)*Log[1 + (a + I*b)/((a - I*b)*E^((2*I)*(c + d*Sqrt[x])))] + (6*I)*b*
d^2*x*PolyLog[2, (-a - I*b)/((a - I*b)*E^((2*I)*(c + d*Sqrt[x])))] + 6*b*d*Sqrt[x]*PolyLog[3, (-a - I*b)/((a -
I*b)*E^((2*I)*(c + d*Sqrt[x])))] - (3*I)*b*PolyLog[4, (-a - I*b)/((a - I*b)*E^((2*I)*(c + d*Sqrt[x])))])/(2*(
a^2 + b^2)*d^4)

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fricas [F]  time = 0.46, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {x}{b \tan \left (d \sqrt {x} + c\right ) + a}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 1.03, size = 0, normalized size = 0.00 \[ \int \frac {x}{a +b \tan \left (c +d \sqrt {x}\right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 1.02, size = 553, normalized size = 2.36 \[ -\frac {6 \, {\left (\frac {2 \, {\left (d \sqrt {x} + c\right )} a}{a^{2} + b^{2}} + \frac {2 \, b \log \left (b \tan \left (d \sqrt {x} + c\right ) + a\right )}{a^{2} + b^{2}} - \frac {b \log \left (\tan \left (d \sqrt {x} + c\right )^{2} + 1\right )}{a^{2} + b^{2}}\right )} c^{3} - \frac {3 \, {\left (d \sqrt {x} + c\right )}^{4} {\left (a - i \, b\right )} - 12 \, {\left (d \sqrt {x} + c\right )}^{3} {\left (a - i \, b\right )} c + 18 \, {\left (d \sqrt {x} + c\right )}^{2} {\left (a - i \, b\right )} c^{2} + {\left (-16 i \, {\left (d \sqrt {x} + c\right )}^{3} b + 36 i \, {\left (d \sqrt {x} + c\right )}^{2} b c - 36 i \, {\left (d \sqrt {x} + c\right )} b c^{2}\right )} \arctan \left (\frac {2 \, a b \cos \left (2 \, d \sqrt {x} + 2 \, c\right ) - {\left (a^{2} - b^{2}\right )} \sin \left (2 \, d \sqrt {x} + 2 \, c\right )}{a^{2} + b^{2}}, \frac {2 \, a b \sin \left (2 \, d \sqrt {x} + 2 \, c\right ) + a^{2} + b^{2} + {\left (a^{2} - b^{2}\right )} \cos \left (2 \, d \sqrt {x} + 2 \, c\right )}{a^{2} + b^{2}}\right ) + {\left (-24 i \, {\left (d \sqrt {x} + c\right )}^{2} b + 36 i \, {\left (d \sqrt {x} + c\right )} b c - 18 i \, b c^{2}\right )} {\rm Li}_2\left (\frac {{\left (i \, a + b\right )} e^{\left (2 i \, d \sqrt {x} + 2 i \, c\right )}}{-i \, a + b}\right ) + 2 \, {\left (4 \, {\left (d \sqrt {x} + c\right )}^{3} b - 9 \, {\left (d \sqrt {x} + c\right )}^{2} b c + 9 \, {\left (d \sqrt {x} + c\right )} b c^{2}\right )} \log \left (\frac {{\left (a^{2} + b^{2}\right )} \cos \left (2 \, d \sqrt {x} + 2 \, c\right )^{2} + 4 \, a b \sin \left (2 \, d \sqrt {x} + 2 \, c\right ) + {\left (a^{2} + b^{2}\right )} \sin \left (2 \, d \sqrt {x} + 2 \, c\right )^{2} + a^{2} + b^{2} + 2 \, {\left (a^{2} - b^{2}\right )} \cos \left (2 \, d \sqrt {x} + 2 \, c\right )}{a^{2} + b^{2}}\right ) + 12 i \, b {\rm Li}_{4}(\frac {{\left (i \, a + b\right )} e^{\left (2 i \, d \sqrt {x} + 2 i \, c\right )}}{-i \, a + b}) + 6 \, {\left (4 \, {\left (d \sqrt {x} + c\right )} b - 3 \, b c\right )} {\rm Li}_{3}(\frac {{\left (i \, a + b\right )} e^{\left (2 i \, d \sqrt {x} + 2 i \, c\right )}}{-i \, a + b})}{a^{2} + b^{2}}}{6 \, d^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(6*(2*(d*sqrt(x) + c)*a/(a^2 + b^2) + 2*b*log(b*tan(d*sqrt(x) + c) + a)/(a^2 + b^2) - b*log(tan(d*sqrt(x)
+ c)^2 + 1)/(a^2 + b^2))*c^3 - (3*(d*sqrt(x) + c)^4*(a - I*b) - 12*(d*sqrt(x) + c)^3*(a - I*b)*c + 18*(d*sqrt
(x) + c)^2*(a - I*b)*c^2 + (-16*I*(d*sqrt(x) + c)^3*b + 36*I*(d*sqrt(x) + c)^2*b*c - 36*I*(d*sqrt(x) + c)*b*c^
2)*arctan2((2*a*b*cos(2*d*sqrt(x) + 2*c) - (a^2 - b^2)*sin(2*d*sqrt(x) + 2*c))/(a^2 + b^2), (2*a*b*sin(2*d*sqr
t(x) + 2*c) + a^2 + b^2 + (a^2 - b^2)*cos(2*d*sqrt(x) + 2*c))/(a^2 + b^2)) + (-24*I*(d*sqrt(x) + c)^2*b + 36*I
*(d*sqrt(x) + c)*b*c - 18*I*b*c^2)*dilog((I*a + b)*e^(2*I*d*sqrt(x) + 2*I*c)/(-I*a + b)) + 2*(4*(d*sqrt(x) + c
)^3*b - 9*(d*sqrt(x) + c)^2*b*c + 9*(d*sqrt(x) + c)*b*c^2)*log(((a^2 + b^2)*cos(2*d*sqrt(x) + 2*c)^2 + 4*a*b*s
in(2*d*sqrt(x) + 2*c) + (a^2 + b^2)*sin(2*d*sqrt(x) + 2*c)^2 + a^2 + b^2 + 2*(a^2 - b^2)*cos(2*d*sqrt(x) + 2*c
))/(a^2 + b^2)) + 12*I*b*polylog(4, (I*a + b)*e^(2*I*d*sqrt(x) + 2*I*c)/(-I*a + b)) + 6*(4*(d*sqrt(x) + c)*b -
3*b*c)*polylog(3, (I*a + b)*e^(2*I*d*sqrt(x) + 2*I*c)/(-I*a + b)))/(a^2 + b^2))/d^4

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(a + b*tan(c + d*x^(1/2))),x)

[Out]

int(x/(a + b*tan(c + d*x^(1/2))), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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