3.4.23 \(\int \frac {(a+b x) (A+B x)}{\sqrt {x}} \, dx\) [323]

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

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 37, normalized size of antiderivative = 1.00, 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 = {77} \begin {gather*} \frac {2}{3} x^{3/2} (a B+A b)+2 a A \sqrt {x}+\frac {2}{5} b B x^{5/2} \end {gather*}

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 77

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*}

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Mathematica [A]
time = 0.03, size = 31, normalized size = 0.84 \begin {gather*} \frac {2}{15} \sqrt {x} (5 a (3 A+B x)+b x (5 A+3 B x)) \end {gather*}

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 = 0.06, size = 28, normalized size = 0.76

method	result	size
trager	\(\left (\frac {2}{5} b B \,x^{2}+\frac {2}{3} A b x +\frac {2}{3} B a x +2 A a \right ) \sqrt {x}\)	\(27\)
gosper	\(\frac {2 \sqrt {x}\, \left (3 b B \,x^{2}+5 A b x +5 B a x +15 A a \right )}{15}\)	\(28\)
derivativedivides	\(\frac {2 \left (A b +B a \right ) x^{\frac {3}{2}}}{3}+\frac {2 b B \,x^{\frac {5}{2}}}{5}+2 a A \sqrt {x}\)	\(28\)
default	\(\frac {2 \left (A b +B a \right ) x^{\frac {3}{2}}}{3}+\frac {2 b B \,x^{\frac {5}{2}}}{5}+2 a A \sqrt {x}\)	\(28\)
risch	\(\frac {2 \sqrt {x}\, \left (3 b B \,x^{2}+5 A b x +5 B a x +15 A a \right )}{15}\)	\(28\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.27, size = 27, normalized size = 0.73 \begin {gather*} \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 {gather*}

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 = 1.08, size = 27, normalized size = 0.73 \begin {gather*} \frac {2}{15} \, {\left (3 \, B b x^{2} + 15 \, A a + 5 \, {\left (B a + A b\right )} x\right )} \sqrt {x} \end {gather*}

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 = 0.08, size = 44, normalized size = 1.19 \begin {gather*} 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 {gather*}

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 = 1.76, size = 29, normalized size = 0.78 \begin {gather*} \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 {gather*}

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)

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Mupad [B]
time = 0.05, size = 27, normalized size = 0.73 \begin {gather*} \frac {2\,\sqrt {x}\,\left (15\,A\,a+5\,A\,b\,x+5\,B\,a\,x+3\,B\,b\,x^2\right )}{15} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*x)*(a + b*x))/x^(1/2),x)

[Out]

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

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