3.323 \(\int \frac{(a+b x) (A+B x)}{\sqrt{x}} \, dx\)

Optimal. Leaf size=37 \[ \frac{2}{3} x^{3/2} (a B+A b)+2 a A \sqrt{x}+\frac{2}{5} b B x^{5/2} \]

[Out]

2*a*A*Sqrt[x] + (2*(A*b + a*B)*x^(3/2))/3 + (2*b*B*x^(5/2))/5

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Rubi [A]  time = 0.0135054, antiderivative size = 37, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 1, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.062, Rules used = {76} \[ \frac{2}{3} x^{3/2} (a B+A b)+2 a A \sqrt{x}+\frac{2}{5} b B x^{5/2} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*x)*(A + B*x))/Sqrt[x],x]

[Out]

2*a*A*Sqrt[x] + (2*(A*b + a*B)*x^(3/2))/3 + (2*b*B*x^(5/2))/5

Rule 76

Int[((d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_))*((e_) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*
x)*(d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && IGtQ[p, 0] && (NeQ[n, -1] || EqQ[p, 1]) && N
eQ[b*e + a*f, 0] && ( !IntegerQ[n] || LtQ[9*p + 5*n, 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && Rational
Q[a, b, d, e, f])) && (NeQ[n + p + 3, 0] || EqQ[p, 1])

Rubi steps

\begin{align*} \int \frac{(a+b x) (A+B x)}{\sqrt{x}} \, dx &=\int \left (\frac{a A}{\sqrt{x}}+(A b+a B) \sqrt{x}+b B x^{3/2}\right ) \, dx\\ &=2 a A \sqrt{x}+\frac{2}{3} (A b+a B) x^{3/2}+\frac{2}{5} b B x^{5/2}\\ \end{align*}

Mathematica [A]  time = 0.0110247, size = 31, normalized size = 0.84 \[ \frac{2}{15} \sqrt{x} (5 a (3 A+B x)+b x (5 A+3 B x)) \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x)*(A + B*x))/Sqrt[x],x]

[Out]

(2*Sqrt[x]*(5*a*(3*A + B*x) + b*x*(5*A + 3*B*x)))/15

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Maple [A]  time = 0.003, size = 28, normalized size = 0.8 \begin{align*}{\frac{6\,bB{x}^{2}+10\,Abx+10\,Bax+30\,Aa}{15}\sqrt{x}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)*(B*x+A)/x^(1/2),x)

[Out]

2/15*x^(1/2)*(3*B*b*x^2+5*A*b*x+5*B*a*x+15*A*a)

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Maxima [A]  time = 1.05287, size = 36, normalized size = 0.97 \begin{align*} \frac{2}{5} \, B b x^{\frac{5}{2}} + 2 \, A a \sqrt{x} + \frac{2}{3} \,{\left (B a + A b\right )} x^{\frac{3}{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)*(B*x+A)/x^(1/2),x, algorithm="maxima")

[Out]

2/5*B*b*x^(5/2) + 2*A*a*sqrt(x) + 2/3*(B*a + A*b)*x^(3/2)

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Fricas [A]  time = 2.48563, size = 72, normalized size = 1.95 \begin{align*} \frac{2}{15} \,{\left (3 \, B b x^{2} + 15 \, A a + 5 \,{\left (B a + A b\right )} x\right )} \sqrt{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)*(B*x+A)/x^(1/2),x, algorithm="fricas")

[Out]

2/15*(3*B*b*x^2 + 15*A*a + 5*(B*a + A*b)*x)*sqrt(x)

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Sympy [A]  time = 0.411172, size = 44, normalized size = 1.19 \begin{align*} 2 A a \sqrt{x} + \frac{2 A b x^{\frac{3}{2}}}{3} + \frac{2 B a x^{\frac{3}{2}}}{3} + \frac{2 B b x^{\frac{5}{2}}}{5} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)*(B*x+A)/x**(1/2),x)

[Out]

2*A*a*sqrt(x) + 2*A*b*x**(3/2)/3 + 2*B*a*x**(3/2)/3 + 2*B*b*x**(5/2)/5

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Giac [A]  time = 1.19712, size = 39, normalized size = 1.05 \begin{align*} \frac{2}{5} \, B b x^{\frac{5}{2}} + \frac{2}{3} \, B a x^{\frac{3}{2}} + \frac{2}{3} \, A b x^{\frac{3}{2}} + 2 \, A a \sqrt{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)*(B*x+A)/x^(1/2),x, algorithm="giac")

[Out]

2/5*B*b*x^(5/2) + 2/3*B*a*x^(3/2) + 2/3*A*b*x^(3/2) + 2*A*a*sqrt(x)