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

Optimal. Leaf size=66 \[ \frac{i b \text{PolyLog}\left (2,-e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^2}+a x-\frac{2 b \sqrt{x} \log \left (1+e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d}+i b x \]

[Out]

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

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Rubi [A]  time = 0.102825, antiderivative size = 66, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.357, Rules used = {3739, 3719, 2190, 2279, 2391} \[ a x+\frac{i b \text{Li}_2\left (-e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{2 b \sqrt{x} \log \left (1+e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d}+i b x \]

Antiderivative was successfully verified.

[In]

Int[a + b*Tan[c + d*Sqrt[x]],x]

[Out]

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

Rule 3739

Int[((a_.) + (b_.)*Tan[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(1/n - 1)*(a + b*Ta
n[c + d*x])^p, x], x, x^n], x] /; FreeQ[{a, b, c, d, p}, x] && IGtQ[1/n, 0] && IntegerQ[p]

Rule 3719

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

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 2279

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

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rubi steps

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

Mathematica [A]  time = 0.023226, size = 66, normalized size = 1. \[ \frac{i b \text{PolyLog}\left (2,-e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^2}+a x-\frac{2 b \sqrt{x} \log \left (1+e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d}+i b x \]

Antiderivative was successfully verified.

[In]

Integrate[a + b*Tan[c + d*Sqrt[x]],x]

[Out]

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

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Maple [F]  time = 0.166, size = 0, normalized size = 0. \begin{align*} \int a+b\tan \left ( c+d\sqrt{x} \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} a x + 2 \, b \int \frac{\sin \left (2 \, d \sqrt{x} + 2 \, c\right )}{\cos \left (2 \, d \sqrt{x} + 2 \, c\right )^{2} + \sin \left (2 \, d \sqrt{x} + 2 \, c\right )^{2} + 2 \, \cos \left (2 \, d \sqrt{x} + 2 \, c\right ) + 1}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*x + 2*b*integrate(sin(2*d*sqrt(x) + 2*c)/(cos(2*d*sqrt(x) + 2*c)^2 + sin(2*d*sqrt(x) + 2*c)^2 + 2*cos(2*d*sq
rt(x) + 2*c) + 1), x)

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Fricas [B]  time = 1.70473, size = 439, normalized size = 6.65 \begin{align*} \frac{2 \, a d^{2} x - 2 \, b d \sqrt{x} \log \left (-\frac{2 \,{\left (i \, \tan \left (d \sqrt{x} + c\right ) - 1\right )}}{\tan \left (d \sqrt{x} + c\right )^{2} + 1}\right ) - 2 \, b d \sqrt{x} \log \left (-\frac{2 \,{\left (-i \, \tan \left (d \sqrt{x} + c\right ) - 1\right )}}{\tan \left (d \sqrt{x} + c\right )^{2} + 1}\right ) - i \, b{\rm Li}_2\left (\frac{2 \,{\left (i \, \tan \left (d \sqrt{x} + c\right ) - 1\right )}}{\tan \left (d \sqrt{x} + c\right )^{2} + 1} + 1\right ) + i \, b{\rm Li}_2\left (\frac{2 \,{\left (-i \, \tan \left (d \sqrt{x} + c\right ) - 1\right )}}{\tan \left (d \sqrt{x} + c\right )^{2} + 1} + 1\right )}{2 \, d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*a*d^2*x - 2*b*d*sqrt(x)*log(-2*(I*tan(d*sqrt(x) + c) - 1)/(tan(d*sqrt(x) + c)^2 + 1)) - 2*b*d*sqrt(x)*l
og(-2*(-I*tan(d*sqrt(x) + c) - 1)/(tan(d*sqrt(x) + c)^2 + 1)) - I*b*dilog(2*(I*tan(d*sqrt(x) + c) - 1)/(tan(d*
sqrt(x) + c)^2 + 1) + 1) + I*b*dilog(2*(-I*tan(d*sqrt(x) + c) - 1)/(tan(d*sqrt(x) + c)^2 + 1) + 1))/d^2

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(a + b*tan(c + d*sqrt(x)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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