3.2.34 \(\int \sqrt {x} (a+b x^3) (A+B x^3) \, dx\) [134]

Optimal. Leaf size=39 \[ \frac {2}{3} a A x^{3/2}+\frac {2}{9} (A b+a B) x^{9/2}+\frac {2}{15} b B x^{15/2} \]

[Out]

2/3*a*A*x^(3/2)+2/9*(A*b+B*a)*x^(9/2)+2/15*b*B*x^(15/2)

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Rubi [A]
time = 0.01, antiderivative size = 39, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {459} \begin {gather*} \frac {2}{9} x^{9/2} (a B+A b)+\frac {2}{3} a A x^{3/2}+\frac {2}{15} b B x^{15/2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[x]*(a + b*x^3)*(A + B*x^3),x]

[Out]

(2*a*A*x^(3/2))/3 + (2*(A*b + a*B)*x^(9/2))/9 + (2*b*B*x^(15/2))/15

Rule 459

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Int[ExpandI
ntegrand[(e*x)^m*(a + b*x^n)^p*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0] &
& IGtQ[p, 0] && IGtQ[q, 0]

Rubi steps

\begin {align*} \int \sqrt {x} \left (a+b x^3\right ) \left (A+B x^3\right ) \, dx &=\int \left (a A \sqrt {x}+(A b+a B) x^{7/2}+b B x^{13/2}\right ) \, dx\\ &=\frac {2}{3} a A x^{3/2}+\frac {2}{9} (A b+a B) x^{9/2}+\frac {2}{15} b B x^{15/2}\\ \end {align*}

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Mathematica [A]
time = 0.02, size = 35, normalized size = 0.90 \begin {gather*} \frac {2}{45} x^{3/2} \left (15 a A+5 A b x^3+5 a B x^3+3 b B x^6\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[x]*(a + b*x^3)*(A + B*x^3),x]

[Out]

(2*x^(3/2)*(15*a*A + 5*A*b*x^3 + 5*a*B*x^3 + 3*b*B*x^6))/45

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Maple [A]
time = 0.11, size = 28, normalized size = 0.72

method	result	size
derivativedivides	\(\frac {2 a A \,x^{\frac {3}{2}}}{3}+\frac {2 \left (A b +B a \right ) x^{\frac {9}{2}}}{9}+\frac {2 b B \,x^{\frac {15}{2}}}{15}\)	\(28\)
default	\(\frac {2 a A \,x^{\frac {3}{2}}}{3}+\frac {2 \left (A b +B a \right ) x^{\frac {9}{2}}}{9}+\frac {2 b B \,x^{\frac {15}{2}}}{15}\)	\(28\)
gosper	\(\frac {2 x^{\frac {3}{2}} \left (3 b B \,x^{6}+5 A b \,x^{3}+5 B a \,x^{3}+15 A a \right )}{45}\)	\(32\)
trager	\(\frac {2 x^{\frac {3}{2}} \left (3 b B \,x^{6}+5 A b \,x^{3}+5 B a \,x^{3}+15 A a \right )}{45}\)	\(32\)
risch	\(\frac {2 x^{\frac {3}{2}} \left (3 b B \,x^{6}+5 A b \,x^{3}+5 B a \,x^{3}+15 A a \right )}{45}\)	\(32\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^3+a)*(B*x^3+A)*x^(1/2),x,method=_RETURNVERBOSE)

[Out]

2/3*a*A*x^(3/2)+2/9*(A*b+B*a)*x^(9/2)+2/15*b*B*x^(15/2)

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Maxima [A]
time = 0.30, size = 27, normalized size = 0.69 \begin {gather*} \frac {2}{15} \, B b x^{\frac {15}{2}} + \frac {2}{9} \, {\left (B a + A b\right )} x^{\frac {9}{2}} + \frac {2}{3} \, A a x^{\frac {3}{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)*(B*x^3+A)*x^(1/2),x, algorithm="maxima")

[Out]

2/15*B*b*x^(15/2) + 2/9*(B*a + A*b)*x^(9/2) + 2/3*A*a*x^(3/2)

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Fricas [A]
time = 2.18, size = 30, normalized size = 0.77 \begin {gather*} \frac {2}{45} \, {\left (3 \, B b x^{7} + 5 \, {\left (B a + A b\right )} x^{4} + 15 \, A a x\right )} \sqrt {x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)*(B*x^3+A)*x^(1/2),x, algorithm="fricas")

[Out]

2/45*(3*B*b*x^7 + 5*(B*a + A*b)*x^4 + 15*A*a*x)*sqrt(x)

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Sympy [A]
time = 1.22, size = 46, normalized size = 1.18 \begin {gather*} \frac {2 A a x^{\frac {3}{2}}}{3} + \frac {2 A b x^{\frac {9}{2}}}{9} + \frac {2 B a x^{\frac {9}{2}}}{9} + \frac {2 B b x^{\frac {15}{2}}}{15} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**3+a)*(B*x**3+A)*x**(1/2),x)

[Out]

2*A*a*x**(3/2)/3 + 2*A*b*x**(9/2)/9 + 2*B*a*x**(9/2)/9 + 2*B*b*x**(15/2)/15

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Giac [A]
time = 0.58, size = 29, normalized size = 0.74 \begin {gather*} \frac {2}{15} \, B b x^{\frac {15}{2}} + \frac {2}{9} \, B a x^{\frac {9}{2}} + \frac {2}{9} \, A b x^{\frac {9}{2}} + \frac {2}{3} \, A a x^{\frac {3}{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)*(B*x^3+A)*x^(1/2),x, algorithm="giac")

[Out]

2/15*B*b*x^(15/2) + 2/9*B*a*x^(9/2) + 2/9*A*b*x^(9/2) + 2/3*A*a*x^(3/2)

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Mupad [B]
time = 0.04, size = 31, normalized size = 0.79 \begin {gather*} \frac {2\,x^{3/2}\,\left (15\,A\,a+5\,A\,b\,x^3+5\,B\,a\,x^3+3\,B\,b\,x^6\right )}{45} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*x^(3/2)*(15*A*a + 5*A*b*x^3 + 5*B*a*x^3 + 3*B*b*x^6))/45

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