3.4.30 \(\int \sqrt {x} (a+b x)^2 (A+B x) \, dx\) [330]

Optimal. Leaf size=63 \[ \frac {2}{3} a^2 A x^{3/2}+\frac {2}{5} a (2 A b+a B) x^{5/2}+\frac {2}{7} b (A b+2 a B) x^{7/2}+\frac {2}{9} b^2 B x^{9/2} \]

[Out]

2/3*a^2*A*x^(3/2)+2/5*a*(2*A*b+B*a)*x^(5/2)+2/7*b*(A*b+2*B*a)*x^(7/2)+2/9*b^2*B*x^(9/2)

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Rubi [A]
time = 0.02, antiderivative size = 63, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {77} \begin {gather*} \frac {2}{3} a^2 A x^{3/2}+\frac {2}{7} b x^{7/2} (2 a B+A b)+\frac {2}{5} a x^{5/2} (a B+2 A b)+\frac {2}{9} b^2 B x^{9/2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[x]*(a + b*x)^2*(A + B*x),x]

[Out]

(2*a^2*A*x^(3/2))/3 + (2*a*(2*A*b + a*B)*x^(5/2))/5 + (2*b*(A*b + 2*a*B)*x^(7/2))/7 + (2*b^2*B*x^(9/2))/9

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 \sqrt {x} (a+b x)^2 (A+B x) \, dx &=\int \left (a^2 A \sqrt {x}+a (2 A b+a B) x^{3/2}+b (A b+2 a B) x^{5/2}+b^2 B x^{7/2}\right ) \, dx\\ &=\frac {2}{3} a^2 A x^{3/2}+\frac {2}{5} a (2 A b+a B) x^{5/2}+\frac {2}{7} b (A b+2 a B) x^{7/2}+\frac {2}{9} b^2 B x^{9/2}\\ \end {align*}

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Mathematica [A]
time = 0.04, size = 52, normalized size = 0.83 \begin {gather*} \frac {2}{315} x^{3/2} \left (21 a^2 (5 A+3 B x)+18 a b x (7 A+5 B x)+5 b^2 x^2 (9 A+7 B x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[x]*(a + b*x)^2*(A + B*x),x]

[Out]

(2*x^(3/2)*(21*a^2*(5*A + 3*B*x) + 18*a*b*x*(7*A + 5*B*x) + 5*b^2*x^2*(9*A + 7*B*x)))/315

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Maple [A]
time = 0.07, size = 52, normalized size = 0.83

method	result	size
gosper	\(\frac {2 x^{\frac {3}{2}} \left (35 b^{2} B \,x^{3}+45 A \,b^{2} x^{2}+90 B a b \,x^{2}+126 a A b x +63 a^{2} B x +105 a^{2} A \right )}{315}\)	\(52\)
derivativedivides	\(\frac {2 b^{2} B \,x^{\frac {9}{2}}}{9}+\frac {2 \left (b^{2} A +2 a b B \right ) x^{\frac {7}{2}}}{7}+\frac {2 \left (2 a b A +a^{2} B \right ) x^{\frac {5}{2}}}{5}+\frac {2 a^{2} A \,x^{\frac {3}{2}}}{3}\)	\(52\)
default	\(\frac {2 b^{2} B \,x^{\frac {9}{2}}}{9}+\frac {2 \left (b^{2} A +2 a b B \right ) x^{\frac {7}{2}}}{7}+\frac {2 \left (2 a b A +a^{2} B \right ) x^{\frac {5}{2}}}{5}+\frac {2 a^{2} A \,x^{\frac {3}{2}}}{3}\)	\(52\)
trager	\(\frac {2 x^{\frac {3}{2}} \left (35 b^{2} B \,x^{3}+45 A \,b^{2} x^{2}+90 B a b \,x^{2}+126 a A b x +63 a^{2} B x +105 a^{2} A \right )}{315}\)	\(52\)
risch	\(\frac {2 x^{\frac {3}{2}} \left (35 b^{2} B \,x^{3}+45 A \,b^{2} x^{2}+90 B a b \,x^{2}+126 a A b x +63 a^{2} B x +105 a^{2} A \right )}{315}\)	\(52\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/9*b^2*B*x^(9/2)+2/7*(A*b^2+2*B*a*b)*x^(7/2)+2/5*(2*A*a*b+B*a^2)*x^(5/2)+2/3*a^2*A*x^(3/2)

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Maxima [A]
time = 0.27, size = 51, normalized size = 0.81 \begin {gather*} \frac {2}{9} \, B b^{2} x^{\frac {9}{2}} + \frac {2}{3} \, A a^{2} x^{\frac {3}{2}} + \frac {2}{7} \, {\left (2 \, B a b + A b^{2}\right )} x^{\frac {7}{2}} + \frac {2}{5} \, {\left (B a^{2} + 2 \, A a b\right )} x^{\frac {5}{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/9*B*b^2*x^(9/2) + 2/3*A*a^2*x^(3/2) + 2/7*(2*B*a*b + A*b^2)*x^(7/2) + 2/5*(B*a^2 + 2*A*a*b)*x^(5/2)

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Fricas [A]
time = 1.40, size = 54, normalized size = 0.86 \begin {gather*} \frac {2}{315} \, {\left (35 \, B b^{2} x^{4} + 105 \, A a^{2} x + 45 \, {\left (2 \, B a b + A b^{2}\right )} x^{3} + 63 \, {\left (B a^{2} + 2 \, A a b\right )} x^{2}\right )} \sqrt {x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/315*(35*B*b^2*x^4 + 105*A*a^2*x + 45*(2*B*a*b + A*b^2)*x^3 + 63*(B*a^2 + 2*A*a*b)*x^2)*sqrt(x)

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Sympy [A]
time = 1.31, size = 66, normalized size = 1.05 \begin {gather*} \frac {2 A a^{2} x^{\frac {3}{2}}}{3} + \frac {2 B b^{2} x^{\frac {9}{2}}}{9} + \frac {2 x^{\frac {7}{2}} \left (A b^{2} + 2 B a b\right )}{7} + \frac {2 x^{\frac {5}{2}} \cdot \left (2 A a b + B a^{2}\right )}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*A*a**2*x**(3/2)/3 + 2*B*b**2*x**(9/2)/9 + 2*x**(7/2)*(A*b**2 + 2*B*a*b)/7 + 2*x**(5/2)*(2*A*a*b + B*a**2)/5

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Giac [A]
time = 2.31, size = 53, normalized size = 0.84 \begin {gather*} \frac {2}{9} \, B b^{2} x^{\frac {9}{2}} + \frac {4}{7} \, B a b x^{\frac {7}{2}} + \frac {2}{7} \, A b^{2} x^{\frac {7}{2}} + \frac {2}{5} \, B a^{2} x^{\frac {5}{2}} + \frac {4}{5} \, A a b x^{\frac {5}{2}} + \frac {2}{3} \, A a^{2} x^{\frac {3}{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/9*B*b^2*x^(9/2) + 4/7*B*a*b*x^(7/2) + 2/7*A*b^2*x^(7/2) + 2/5*B*a^2*x^(5/2) + 4/5*A*a*b*x^(5/2) + 2/3*A*a^2*
x^(3/2)

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Mupad [B]
time = 0.06, size = 51, normalized size = 0.81 \begin {gather*} x^{5/2}\,\left (\frac {2\,B\,a^2}{5}+\frac {4\,A\,b\,a}{5}\right )+x^{7/2}\,\left (\frac {2\,A\,b^2}{7}+\frac {4\,B\,a\,b}{7}\right )+\frac {2\,A\,a^2\,x^{3/2}}{3}+\frac {2\,B\,b^2\,x^{9/2}}{9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(1/2)*(A + B*x)*(a + b*x)^2,x)

[Out]

x^(5/2)*((2*B*a^2)/5 + (4*A*a*b)/5) + x^(7/2)*((2*A*b^2)/7 + (4*B*a*b)/7) + (2*A*a^2*x^(3/2))/3 + (2*B*b^2*x^(
9/2))/9

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