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

Optimal. Leaf size=26 \[ 2 a \sqrt {x}+\frac {2 b \tanh ^{-1}\left (\sin \left (c+d \sqrt {x}\right )\right )}{d} \]

[Out]

2*b*arctanh(sin(c+d*x^(1/2)))/d+2*a*x^(1/2)

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Rubi [A]  time = 0.02, antiderivative size = 26, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {14, 4204, 3770} \[ 2 a \sqrt {x}+\frac {2 b \tanh ^{-1}\left (\sin \left (c+d \sqrt {x}\right )\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

2*a*Sqrt[x] + (2*b*ArcTanh[Sin[c + d*Sqrt[x]]])/d

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 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

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]

Rubi steps

\begin {align*} \int \frac {a+b \sec \left (c+d \sqrt {x}\right )}{\sqrt {x}} \, dx &=\int \left (\frac {a}{\sqrt {x}}+\frac {b \sec \left (c+d \sqrt {x}\right )}{\sqrt {x}}\right ) \, dx\\ &=2 a \sqrt {x}+b \int \frac {\sec \left (c+d \sqrt {x}\right )}{\sqrt {x}} \, dx\\ &=2 a \sqrt {x}+(2 b) \operatorname {Subst}\left (\int \sec (c+d x) \, dx,x,\sqrt {x}\right )\\ &=2 a \sqrt {x}+\frac {2 b \tanh ^{-1}\left (\sin \left (c+d \sqrt {x}\right )\right )}{d}\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 26, normalized size = 1.00 \[ 2 a \sqrt {x}+\frac {2 b \tanh ^{-1}\left (\sin \left (c+d \sqrt {x}\right )\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

2*a*Sqrt[x] + (2*b*ArcTanh[Sin[c + d*Sqrt[x]]])/d

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fricas [A]  time = 0.66, size = 41, normalized size = 1.58 \[ \frac {2 \, a d \sqrt {x} + b \log \left (\sin \left (d \sqrt {x} + c\right ) + 1\right ) - b \log \left (-\sin \left (d \sqrt {x} + c\right ) + 1\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*a*d*sqrt(x) + b*log(sin(d*sqrt(x) + c) + 1) - b*log(-sin(d*sqrt(x) + c) + 1))/d

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giac [B]  time = 0.85, size = 50, normalized size = 1.92 \[ \frac {2 \, {\left ({\left (d \sqrt {x} + c\right )} a + b \log \left ({\left | \tan \left (\frac {1}{2} \, d \sqrt {x} + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - b \log \left ({\left | \tan \left (\frac {1}{2} \, d \sqrt {x} + \frac {1}{2} \, c\right ) - 1 \right |}\right )\right )}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*((d*sqrt(x) + c)*a + b*log(abs(tan(1/2*d*sqrt(x) + 1/2*c) + 1)) - b*log(abs(tan(1/2*d*sqrt(x) + 1/2*c) - 1))
)/d

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maple [A]  time = 0.26, size = 32, normalized size = 1.23 \[ 2 a \sqrt {x}+\frac {2 b \ln \left (\sec \left (c +d \sqrt {x}\right )+\tan \left (c +d \sqrt {x}\right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*a*x^(1/2)+2*b/d*ln(sec(c+d*x^(1/2))+tan(c+d*x^(1/2)))

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maxima [A]  time = 0.37, size = 31, normalized size = 1.19 \[ 2 \, a \sqrt {x} + \frac {2 \, b \log \left (\sec \left (d \sqrt {x} + c\right ) + \tan \left (d \sqrt {x} + c\right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*a*sqrt(x) + 2*b*log(sec(d*sqrt(x) + c) + tan(d*sqrt(x) + c))/d

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mupad [B]  time = 2.94, size = 71, normalized size = 2.73 \[ 2\,a\,\sqrt {x}-\frac {2\,b\,\ln \left (\frac {b\,2{}\mathrm {i}-2\,b\,{\mathrm {e}}^{d\,\sqrt {x}\,1{}\mathrm {i}}\,{\mathrm {e}}^{c\,1{}\mathrm {i}}}{\sqrt {x}}\right )}{d}+\frac {2\,b\,\ln \left (\frac {b\,2{}\mathrm {i}+2\,b\,{\mathrm {e}}^{d\,\sqrt {x}\,1{}\mathrm {i}}\,{\mathrm {e}}^{c\,1{}\mathrm {i}}}{\sqrt {x}}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*a*x^(1/2) - (2*b*log((b*2i - 2*b*exp(d*x^(1/2)*1i)*exp(c*1i))/x^(1/2)))/d + (2*b*log((b*2i + 2*b*exp(d*x^(1/
2)*1i)*exp(c*1i))/x^(1/2)))/d

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sympy [A]  time = 4.74, size = 56, normalized size = 2.15 \[ \begin {cases} \frac {2 a \left (c + d \sqrt {x}\right ) + 2 b \log {\left (\tan {\left (c + d \sqrt {x} \right )} + \sec {\left (c + d \sqrt {x} \right )} \right )}}{d} & \text {for}\: d \neq 0 \\- \sqrt {x} \left (- 2 a - 2 b \sec {\relax (c )}\right ) & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise(((2*a*(c + d*sqrt(x)) + 2*b*log(tan(c + d*sqrt(x)) + sec(c + d*sqrt(x))))/d, Ne(d, 0)), (-sqrt(x)*(-
2*a - 2*b*sec(c)), True))

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