3.1.39 \(\int \frac {1}{a+b \tan (c+d \sqrt {x})} \, dx\) [39]

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

[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 = 0.12, 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 = {3824, 3813, 2221, 2317, 2438} \begin {gather*} -\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} \end {gather*}

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 2221

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/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], 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 2317

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 2438

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 3813

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^Si
mp[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 3824

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 \text {Subst}\left (\int \frac {x}{a+b \tan (c+d x)} \, dx,x,\sqrt {x}\right )\\ &=\frac {x}{a+i b}+(4 i b) \text {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) \text {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) \text {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 = 0.28, size = 110, normalized size = 0.92 \begin {gather*} \frac {(a-i b) d^2 x+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 )-i b \text {PolyLog}\left (2,-\frac {(a-i b) e^{2 i \left (c+d \sqrt {x}\right )}}{a+i b}\right )}{\left (a^2+b^2\right ) d^2} \end {gather*}

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

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Maple [F]
time = 0.83, 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] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 264 vs. \(2 (100) = 200\).
time = 0.59, size = 264, normalized size = 2.22 \begin {gather*} \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}} \end {gather*}

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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Fricas [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 534 vs. \(2 (100) = 200\).
time = 0.43, size = 534, normalized size = 4.49 \begin {gather*} \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 ({\left (i \, a b - b^{2}\right )} \tan \left (d \sqrt {x} + c\right )^{2} - a^{2} - i \, a b + {\left (i \, a^{2} - 2 \, a b - i \, b^{2}\right )} \tan \left (d \sqrt {x} + c\right )\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 ({\left (-i \, a b - b^{2}\right )} \tan \left (d \sqrt {x} + c\right )^{2} - a^{2} + i \, a b + {\left (-i \, a^{2} - 2 \, a b + i \, b^{2}\right )} \tan \left (d \sqrt {x} + c\right )\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 ({\left (i \, a b - b^{2}\right )} \tan \left (d \sqrt {x} + c\right )^{2} - a^{2} - i \, a b + {\left (i \, a^{2} - 2 \, a b - i \, b^{2}\right )} \tan \left (d \sqrt {x} + c\right )\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 ({\left (-i \, a b - b^{2}\right )} \tan \left (d \sqrt {x} + c\right )^{2} - a^{2} + i \, a b + {\left (-i \, a^{2} - 2 \, a b + i \, b^{2}\right )} \tan \left (d \sqrt {x} + c\right )\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}} \end {gather*}

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 - a^2 - I
*a*b + (I*a^2 - 2*a*b - 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*d
ilog(2*((-I*a*b - b^2)*tan(d*sqrt(x) + c)^2 - a^2 + I*a*b + (-I*a^2 - 2*a*b + I*b^2)*tan(d*sqrt(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)*tan(d*sqrt(x) + c
)^2 - a^2 - I*a*b + (I*a^2 - 2*a*b - I*b^2)*tan(d*sqrt(x) + c))/((a^2 + b^2)*tan(d*sqrt(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 - a^2 + I*a*b + (-I*a^2 - 2*a*b + 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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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{a + b \tan {\left (c + d \sqrt {x} \right )}}\, dx \end {gather*}

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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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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

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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