3.4.50 \(\int \frac {(a+b x^2) (A+B x^2)}{x^{7/2}} \, dx\) [350]

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

[Out]

-2/5*a*A/x^(5/2)+2/3*b*B*x^(3/2)-2*(A*b+B*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 = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {459} \begin {gather*} -\frac {2 (a B+A b)}{\sqrt {x}}-\frac {2 a A}{5 x^{5/2}}+\frac {2}{3} b B x^{3/2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((a + b*x^2)*(A + B*x^2))/x^(7/2),x]

[Out]

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

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

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Mathematica [A]
time = 0.03, size = 35, normalized size = 0.95 \begin {gather*} -\frac {2 \left (3 a A+15 A b x^2+15 a B x^2-5 b B x^4\right )}{15 x^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x^2)*(A + B*x^2))/x^(7/2),x]

[Out]

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

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Maple [A]
time = 0.05, size = 28, normalized size = 0.76

method	result	size
derivativedivides	\(-\frac {2 a A}{5 x^{\frac {5}{2}}}+\frac {2 b B \,x^{\frac {3}{2}}}{3}-\frac {2 \left (A b +B a \right )}{\sqrt {x}}\)	\(28\)
default	\(-\frac {2 a A}{5 x^{\frac {5}{2}}}+\frac {2 b B \,x^{\frac {3}{2}}}{3}-\frac {2 \left (A b +B a \right )}{\sqrt {x}}\)	\(28\)
gosper	\(-\frac {2 \left (-5 b B \,x^{4}+15 A b \,x^{2}+15 B a \,x^{2}+3 A a \right )}{15 x^{\frac {5}{2}}}\)	\(32\)
trager	\(-\frac {2 \left (-5 b B \,x^{4}+15 A b \,x^{2}+15 B a \,x^{2}+3 A a \right )}{15 x^{\frac {5}{2}}}\)	\(32\)
risch	\(-\frac {2 \left (-5 b B \,x^{4}+15 A b \,x^{2}+15 B a \,x^{2}+3 A a \right )}{15 x^{\frac {5}{2}}}\)	\(32\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)*(B*x^2+A)/x^(7/2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.32, size = 29, normalized size = 0.78 \begin {gather*} \frac {2}{3} \, B b x^{\frac {3}{2}} - \frac {2 \, {\left (5 \, {\left (B a + A b\right )} x^{2} + A a\right )}}{5 \, x^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)*(B*x^2+A)/x^(7/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 1.26, size = 29, normalized size = 0.78 \begin {gather*} \frac {2 \, {\left (5 \, B b x^{4} - 15 \, {\left (B a + A b\right )} x^{2} - 3 \, A a\right )}}{15 \, x^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)*(B*x^2+A)/x^(7/2),x, algorithm="fricas")

[Out]

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

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Sympy [A]
time = 0.35, size = 42, normalized size = 1.14 \begin {gather*} - \frac {2 A a}{5 x^{\frac {5}{2}}} - \frac {2 A b}{\sqrt {x}} - \frac {2 B a}{\sqrt {x}} + \frac {2 B b x^{\frac {3}{2}}}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)*(B*x**2+A)/x**(7/2),x)

[Out]

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

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Giac [A]
time = 1.74, size = 31, normalized size = 0.84 \begin {gather*} \frac {2}{3} \, B b x^{\frac {3}{2}} - \frac {2 \, {\left (5 \, B a x^{2} + 5 \, A b x^{2} + A a\right )}}{5 \, x^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)*(B*x^2+A)/x^(7/2),x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 0.02, size = 31, normalized size = 0.84 \begin {gather*} -\frac {6\,A\,a+30\,A\,b\,x^2+30\,B\,a\,x^2-10\,B\,b\,x^4}{15\,x^{5/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*x^2)*(a + b*x^2))/x^(7/2),x)

[Out]

-(6*A*a + 30*A*b*x^2 + 30*B*a*x^2 - 10*B*b*x^4)/(15*x^(5/2))

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