3.39 \(\int \frac {1}{a+b \tan (c+d \sqrt {x})} \, dx\)

Optimal. Leaf size=119 \[ -\frac {i b \text {Li}_2\left (-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{d^2 \left (a^2+b^2\right )}+\frac {2 b \sqrt {x} \log \left (1+\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{d \left (a^2+b^2\right )}+\frac {x}{a+i b} \]

[Out]

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

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Rubi [A]  time = 0.17, antiderivative size = 119, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {3739, 3732, 2190, 2279, 2391} \[ -\frac {i b \text {Li}_2\left (-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{d^2 \left (a^2+b^2\right )}+\frac {2 b \sqrt {x} \log \left (1+\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{d \left (a^2+b^2\right )}+\frac {x}{a+i b} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Tan[c + d*Sqrt[x]])^(-1),x]

[Out]

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

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]

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 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]

Rubi steps

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

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Mathematica [A]  time = 0.26, size = 111, normalized size = 0.93 \[ \frac {i b \text {Li}_2\left (\frac {(-a-i b) e^{-2 i \left (c+d \sqrt {x}\right )}}{a-i b}\right )+2 b d \sqrt {x} \log \left (1+\frac {(a+i b) e^{-2 i \left (c+d \sqrt {x}\right )}}{a-i b}\right )+d^2 x (a+i b)}{d^2 \left (a^2+b^2\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Tan[c + d*Sqrt[x]])^(-1),x]

[Out]

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

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fricas [B]  time = 0.60, size = 540, normalized size = 4.54 \[ \frac {2 \, a d^{2} x - 2 \, b c \log \left (\frac {{\left (i \, a b + b^{2}\right )} \tan \left (d \sqrt {x} + c\right )^{2} - a^{2} + i \, a b + {\left (i \, a^{2} + i \, b^{2}\right )} \tan \left (d \sqrt {x} + c\right )}{\tan \left (d \sqrt {x} + c\right )^{2} + 1}\right ) - 2 \, b c \log \left (\frac {{\left (i \, a b - b^{2}\right )} \tan \left (d \sqrt {x} + c\right )^{2} + a^{2} + i \, a b + {\left (i \, a^{2} + i \, b^{2}\right )} \tan \left (d \sqrt {x} + c\right )}{\tan \left (d \sqrt {x} + c\right )^{2} + 1}\right ) + i \, b {\rm Li}_2\left (\frac {2 \, {\left (i \, a b - b^{2}\right )} \tan \left (d \sqrt {x} + c\right )^{2} - 2 \, a^{2} - 2 i \, a b - {\left (-2 i \, a^{2} + 4 \, a b + 2 i \, b^{2}\right )} \tan \left (d \sqrt {x} + c\right )}{{\left (a^{2} + b^{2}\right )} \tan \left (d \sqrt {x} + c\right )^{2} + a^{2} + b^{2}} + 1\right ) - i \, b {\rm Li}_2\left (\frac {2 \, {\left (-i \, a b - b^{2}\right )} \tan \left (d \sqrt {x} + c\right )^{2} - 2 \, a^{2} + 2 i \, a b - {\left (2 i \, a^{2} + 4 \, a b - 2 i \, b^{2}\right )} \tan \left (d \sqrt {x} + c\right )}{{\left (a^{2} + b^{2}\right )} \tan \left (d \sqrt {x} + c\right )^{2} + a^{2} + b^{2}} + 1\right ) + 2 \, {\left (b d \sqrt {x} + b c\right )} \log \left (-\frac {2 \, {\left (i \, a b - b^{2}\right )} \tan \left (d \sqrt {x} + c\right )^{2} - 2 \, a^{2} - 2 i \, a b - {\left (-2 i \, a^{2} + 4 \, a b + 2 i \, b^{2}\right )} \tan \left (d \sqrt {x} + c\right )}{{\left (a^{2} + b^{2}\right )} \tan \left (d \sqrt {x} + c\right )^{2} + a^{2} + b^{2}}\right ) + 2 \, {\left (b d \sqrt {x} + b c\right )} \log \left (-\frac {2 \, {\left (-i \, a b - b^{2}\right )} \tan \left (d \sqrt {x} + c\right )^{2} - 2 \, a^{2} + 2 i \, a b - {\left (2 i \, a^{2} + 4 \, a b - 2 i \, b^{2}\right )} \tan \left (d \sqrt {x} + c\right )}{{\left (a^{2} + b^{2}\right )} \tan \left (d \sqrt {x} + c\right )^{2} + a^{2} + b^{2}}\right )}{2 \, {\left (a^{2} + b^{2}\right )} d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*a*d^2*x - 2*b*c*log(((I*a*b + b^2)*tan(d*sqrt(x) + c)^2 - a^2 + I*a*b + (I*a^2 + I*b^2)*tan(d*sqrt(x) +
c))/(tan(d*sqrt(x) + c)^2 + 1)) - 2*b*c*log(((I*a*b - b^2)*tan(d*sqrt(x) + c)^2 + a^2 + I*a*b + (I*a^2 + I*b^
2)*tan(d*sqrt(x) + c))/(tan(d*sqrt(x) + c)^2 + 1)) + I*b*dilog((2*(I*a*b - b^2)*tan(d*sqrt(x) + c)^2 - 2*a^2 -
2*I*a*b - (-2*I*a^2 + 4*a*b + 2*I*b^2)*tan(d*sqrt(x) + c))/((a^2 + b^2)*tan(d*sqrt(x) + c)^2 + a^2 + b^2) + 1
) - I*b*dilog((2*(-I*a*b - b^2)*tan(d*sqrt(x) + c)^2 - 2*a^2 + 2*I*a*b - (2*I*a^2 + 4*a*b - 2*I*b^2)*tan(d*sqr
t(x) + c))/((a^2 + b^2)*tan(d*sqrt(x) + c)^2 + a^2 + b^2) + 1) + 2*(b*d*sqrt(x) + b*c)*log(-(2*(I*a*b - b^2)*t
an(d*sqrt(x) + c)^2 - 2*a^2 - 2*I*a*b - (-2*I*a^2 + 4*a*b + 2*I*b^2)*tan(d*sqrt(x) + c))/((a^2 + b^2)*tan(d*sq
rt(x) + c)^2 + a^2 + b^2)) + 2*(b*d*sqrt(x) + b*c)*log(-(2*(-I*a*b - b^2)*tan(d*sqrt(x) + c)^2 - 2*a^2 + 2*I*a
*b - (2*I*a^2 + 4*a*b - 2*I*b^2)*tan(d*sqrt(x) + c))/((a^2 + b^2)*tan(d*sqrt(x) + c)^2 + a^2 + b^2)))/((a^2 +
b^2)*d^2)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{b \tan \left (d \sqrt {x} + c\right ) + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.96, size = 0, normalized size = 0.00 \[ \int \frac {1}{a +b \tan \left (c +d \sqrt {x}\right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.74, size = 264, normalized size = 2.22 \[ \frac {{\left (a - i \, b\right )} d^{2} x - 2 i \, b d \sqrt {x} \arctan \left (\frac {2 \, a b \cos \left (2 \, d \sqrt {x} + 2 \, c\right ) - {\left (a^{2} - b^{2}\right )} \sin \left (2 \, d \sqrt {x} + 2 \, c\right )}{a^{2} + b^{2}}, \frac {2 \, a b \sin \left (2 \, d \sqrt {x} + 2 \, c\right ) + a^{2} + b^{2} + {\left (a^{2} - b^{2}\right )} \cos \left (2 \, d \sqrt {x} + 2 \, c\right )}{a^{2} + b^{2}}\right ) + b d \sqrt {x} \log \left (\frac {{\left (a^{2} + b^{2}\right )} \cos \left (2 \, d \sqrt {x} + 2 \, c\right )^{2} + 4 \, a b \sin \left (2 \, d \sqrt {x} + 2 \, c\right ) + {\left (a^{2} + b^{2}\right )} \sin \left (2 \, d \sqrt {x} + 2 \, c\right )^{2} + a^{2} + b^{2} + 2 \, {\left (a^{2} - b^{2}\right )} \cos \left (2 \, d \sqrt {x} + 2 \, c\right )}{a^{2} + b^{2}}\right ) - i \, b {\rm Li}_2\left (\frac {{\left (i \, a + b\right )} e^{\left (2 i \, d \sqrt {x} + 2 i \, c\right )}}{-i \, a + b}\right )}{{\left (a^{2} + b^{2}\right )} d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((a - I*b)*d^2*x - 2*I*b*d*sqrt(x)*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*sqrt(x) + 2*c) + a^2 + b^2 + (a^2 - b^2)*cos(2*d*sqrt(x) + 2*c))/(a^2 + b^2)) + b
*d*sqrt(x)*log(((a^2 + b^2)*cos(2*d*sqrt(x) + 2*c)^2 + 4*a*b*sin(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)) - I*b*dilog((I*a + b)*e^(2*I*d*s
qrt(x) + 2*I*c)/(-I*a + b)))/((a^2 + b^2)*d^2)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{a+b\,\mathrm {tan}\left (c+d\,\sqrt {x}\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + b*tan(c + d*x^(1/2))),x)

[Out]

int(1/(a + b*tan(c + d*x^(1/2))), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{a + b \tan {\left (c + d \sqrt {x} \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/(a + b*tan(c + d*sqrt(x))), x)

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