3.722 \(\int \frac {q+p x}{\sqrt {b+a x} (f+\sqrt {b+a x})} \, dx\)

Optimal. Leaf size=54 \[ -\frac {2 \left (-a q+b p+f^2 (-p)\right ) \log \left (\sqrt {a x+b}+f\right )}{a^2}-\frac {2 f p \sqrt {a x+b}}{a^2}+\frac {p x}{a} \]

[Out]

p*x/a-2*(-f^2*p-a*q+b*p)*ln(f+(a*x+b)^(1/2))/a^2-2*f*p*(a*x+b)^(1/2)/a^2

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Rubi [A]  time = 0.38, antiderivative size = 54, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 1, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.036, Rules used = {697} \[ -\frac {2 \left (-a q+b p+f^2 (-p)\right ) \log \left (\sqrt {a x+b}+f\right )}{a^2}-\frac {2 f p \sqrt {a x+b}}{a^2}+\frac {p x}{a} \]

Antiderivative was successfully verified.

[In]

Int[(q + p*x)/(Sqrt[b + a*x]*(f + Sqrt[b + a*x])),x]

[Out]

(p*x)/a - (2*f*p*Sqrt[b + a*x])/a^2 - (2*(b*p - f^2*p - a*q)*Log[f + Sqrt[b + a*x]])/a^2

Rule 697

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*(a + c*
x^2)^p, x], x] /; FreeQ[{a, c, d, e, m}, x] && NeQ[c*d^2 + a*e^2, 0] && IGtQ[p, 0]

Rubi steps

\begin {align*} \int \frac {q+p x}{\sqrt {b+a x} \left (f+\sqrt {b+a x}\right )} \, dx &=\frac {2 \operatorname {Subst}\left (\int \frac {-b p+a q+p x^2}{f+x} \, dx,x,\sqrt {b+a x}\right )}{a^2}\\ &=\frac {2 \operatorname {Subst}\left (\int \left (-f p+p x+\frac {-b p+f^2 p+a q}{f+x}\right ) \, dx,x,\sqrt {b+a x}\right )}{a^2}\\ &=\frac {p x}{a}-\frac {2 f p \sqrt {b+a x}}{a^2}-\frac {2 \left (b p-f^2 p-a q\right ) \log \left (f+\sqrt {b+a x}\right )}{a^2}\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 50, normalized size = 0.93 \[ \frac {2 \left (a q-b p+f^2 p\right ) \log \left (\sqrt {a x+b}+f\right )+p \left (a x-2 f \sqrt {a x+b}\right )}{a^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(q + p*x)/(Sqrt[b + a*x]*(f + Sqrt[b + a*x])),x]

[Out]

(p*(a*x - 2*f*Sqrt[b + a*x]) + 2*(-(b*p) + f^2*p + a*q)*Log[f + Sqrt[b + a*x]])/a^2

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fricas [A]  time = 0.47, size = 45, normalized size = 0.83 \[ \frac {a p x - 2 \, \sqrt {a x + b} f p + 2 \, {\left ({\left (f^{2} - b\right )} p + a q\right )} \log \left (f + \sqrt {a x + b}\right )}{a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((p*x+q)/(a*x+b)^(1/2)/(f+(a*x+b)^(1/2)),x, algorithm="fricas")

[Out]

(a*p*x - 2*sqrt(a*x + b)*f*p + 2*((f^2 - b)*p + a*q)*log(f + sqrt(a*x + b)))/a^2

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giac [A]  time = 0.32, size = 61, normalized size = 1.13 \[ \frac {2 \, {\left (f^{2} p - b p + a q\right )} \log \left ({\left | f + \sqrt {a x + b} \right |}\right )}{a^{2}} - \frac {2 \, \sqrt {a x + b} a^{2} f p - {\left (a x + b\right )} a^{2} p}{a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((p*x+q)/(a*x+b)^(1/2)/(f+(a*x+b)^(1/2)),x, algorithm="giac")

[Out]

2*(f^2*p - b*p + a*q)*log(abs(f + sqrt(a*x + b)))/a^2 - (2*sqrt(a*x + b)*a^2*f*p - (a*x + b)*a^2*p)/a^4

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maple [A]  time = 0.00, size = 80, normalized size = 1.48 \[ \frac {2 f^{2} p \ln \left (f +\sqrt {a x +b}\right )}{a^{2}}+\frac {p x}{a}+\frac {2 q \ln \left (f +\sqrt {a x +b}\right )}{a}-\frac {2 b p \ln \left (f +\sqrt {a x +b}\right )}{a^{2}}+\frac {b p}{a^{2}}-\frac {2 \sqrt {a x +b}\, f p}{a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((p*x+q)/(a*x+b)^(1/2)/(f+(a*x+b)^(1/2)),x)

[Out]

p*x/a+1/a^2*b*p-2*f*p*(a*x+b)^(1/2)/a^2+2/a^2*ln(f+(a*x+b)^(1/2))*f^2*p+2/a*ln(f+(a*x+b)^(1/2))*q-2/a^2*ln(f+(
a*x+b)^(1/2))*b*p

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maxima [A]  time = 0.88, size = 58, normalized size = 1.07 \[ \frac {\frac {2 \, {\left ({\left (f^{2} - b\right )} p + a q\right )} \log \left (f + \sqrt {a x + b}\right )}{a} - \frac {2 \, \sqrt {a x + b} f p - {\left (a x + b\right )} p}{a}}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((p*x+q)/(a*x+b)^(1/2)/(f+(a*x+b)^(1/2)),x, algorithm="maxima")

[Out]

(2*((f^2 - b)*p + a*q)*log(f + sqrt(a*x + b))/a - (2*sqrt(a*x + b)*f*p - (a*x + b)*p)/a)/a

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mupad [B]  time = 3.10, size = 50, normalized size = 0.93 \[ \frac {\ln \left (f+\sqrt {b+a\,x}\right )\,\left (2\,p\,f^2+2\,a\,q-2\,b\,p\right )}{a^2}+\frac {p\,x}{a}-\frac {2\,f\,p\,\sqrt {b+a\,x}}{a^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((q + p*x)/((f + (b + a*x)^(1/2))*(b + a*x)^(1/2)),x)

[Out]

(log(f + (b + a*x)^(1/2))*(2*a*q - 2*b*p + 2*f^2*p))/a^2 + (p*x)/a - (2*f*p*(b + a*x)^(1/2))/a^2

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sympy [A]  time = 32.90, size = 99, normalized size = 1.83 \[ - \frac {2 f p \sqrt {a x + b}}{a^{2}} - \frac {2 f \left (- a q + b p - f^{2} p\right ) \left (\begin {cases} \frac {1}{\sqrt {a x + b}} & \text {for}\: f = 0 \\\frac {\log {\left (\frac {f}{\sqrt {a x + b}} + 1 \right )}}{f} & \text {otherwise} \end {cases}\right )}{a^{2}} + \frac {p \left (a x + b\right )}{a^{2}} + \frac {2 \left (- a q + b p - f^{2} p\right ) \log {\left (\frac {1}{\sqrt {a x + b}} \right )}}{a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((p*x+q)/(a*x+b)**(1/2)/(f+(a*x+b)**(1/2)),x)

[Out]

-2*f*p*sqrt(a*x + b)/a**2 - 2*f*(-a*q + b*p - f**2*p)*Piecewise((1/sqrt(a*x + b), Eq(f, 0)), (log(f/sqrt(a*x +
b) + 1)/f, True))/a**2 + p*(a*x + b)/a**2 + 2*(-a*q + b*p - f**2*p)*log(1/sqrt(a*x + b))/a**2

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