3.5.96 \(\int \frac {A+B x}{x^{7/2} \sqrt {a+b x}} \, dx\)

Optimal. Leaf size=84 \[ -\frac {4 b \sqrt {a+b x} (4 A b-5 a B)}{15 a^3 \sqrt {x}}+\frac {2 \sqrt {a+b x} (4 A b-5 a B)}{15 a^2 x^{3/2}}-\frac {2 A \sqrt {a+b x}}{5 a x^{5/2}} \]

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Rubi [A]  time = 0.03, antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {78, 45, 37} \begin {gather*} \frac {2 \sqrt {a+b x} (4 A b-5 a B)}{15 a^2 x^{3/2}}-\frac {4 b \sqrt {a+b x} (4 A b-5 a B)}{15 a^3 \sqrt {x}}-\frac {2 A \sqrt {a+b x}}{5 a x^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n +
1))/((b*c - a*d)*(m + 1)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*Simplify[m + n + 2])/((b*c - a*d)*(m + 1)), Int[(a + b*x)^Simplify[m +
1]*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && ILtQ[Simplify[m + n + 2], 0] &&
NeQ[m, -1] &&  !(LtQ[m, -1] && LtQ[n, -1] && (EqQ[a, 0] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && (
SumSimplerQ[m, 1] ||  !SumSimplerQ[n, 1])

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ
[p, n]))))

Rubi steps

\begin {align*} \int \frac {A+B x}{x^{7/2} \sqrt {a+b x}} \, dx &=-\frac {2 A \sqrt {a+b x}}{5 a x^{5/2}}+\frac {\left (2 \left (-2 A b+\frac {5 a B}{2}\right )\right ) \int \frac {1}{x^{5/2} \sqrt {a+b x}} \, dx}{5 a}\\ &=-\frac {2 A \sqrt {a+b x}}{5 a x^{5/2}}+\frac {2 (4 A b-5 a B) \sqrt {a+b x}}{15 a^2 x^{3/2}}+\frac {(2 b (4 A b-5 a B)) \int \frac {1}{x^{3/2} \sqrt {a+b x}} \, dx}{15 a^2}\\ &=-\frac {2 A \sqrt {a+b x}}{5 a x^{5/2}}+\frac {2 (4 A b-5 a B) \sqrt {a+b x}}{15 a^2 x^{3/2}}-\frac {4 b (4 A b-5 a B) \sqrt {a+b x}}{15 a^3 \sqrt {x}}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 56, normalized size = 0.67 \begin {gather*} -\frac {2 \sqrt {a+b x} \left (a^2 (3 A+5 B x)-2 a b x (2 A+5 B x)+8 A b^2 x^2\right )}{15 a^3 x^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sqrt[a + b*x]*(8*A*b^2*x^2 - 2*a*b*x*(2*A + 5*B*x) + a^2*(3*A + 5*B*x)))/(15*a^3*x^(5/2))

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IntegrateAlgebraic [A]  time = 0.15, size = 58, normalized size = 0.69 \begin {gather*} \frac {2 \sqrt {a+b x} \left (-3 a^2 A-5 a^2 B x+4 a A b x+10 a b B x^2-8 A b^2 x^2\right )}{15 a^3 x^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[a + b*x]*(-3*a^2*A + 4*a*A*b*x - 5*a^2*B*x - 8*A*b^2*x^2 + 10*a*b*B*x^2))/(15*a^3*x^(5/2))

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fricas [A]  time = 1.11, size = 53, normalized size = 0.63 \begin {gather*} -\frac {2 \, {\left (3 \, A a^{2} - 2 \, {\left (5 \, B a b - 4 \, A b^{2}\right )} x^{2} + {\left (5 \, B a^{2} - 4 \, A a b\right )} x\right )} \sqrt {b x + a}}{15 \, a^{3} x^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^(7/2)/(b*x+a)^(1/2),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 1.91, size = 105, normalized size = 1.25 \begin {gather*} \frac {2 \, \sqrt {b x + a} {\left ({\left (b x + a\right )} {\left (\frac {2 \, {\left (5 \, B a b^{4} - 4 \, A b^{5}\right )} {\left (b x + a\right )}}{a^{3}} - \frac {5 \, {\left (5 \, B a^{2} b^{4} - 4 \, A a b^{5}\right )}}{a^{3}}\right )} + \frac {15 \, {\left (B a^{3} b^{4} - A a^{2} b^{5}\right )}}{a^{3}}\right )} b}{15 \, {\left ({\left (b x + a\right )} b - a b\right )}^{\frac {5}{2}} {\left | b \right |}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^(7/2)/(b*x+a)^(1/2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.00, size = 53, normalized size = 0.63 \begin {gather*} -\frac {2 \sqrt {b x +a}\, \left (8 A \,b^{2} x^{2}-10 B a b \,x^{2}-4 A a b x +5 B \,a^{2} x +3 A \,a^{2}\right )}{15 a^{3} x^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)/x^(7/2)/(b*x+a)^(1/2),x)

[Out]

-2/15*(b*x+a)^(1/2)*(8*A*b^2*x^2-10*B*a*b*x^2-4*A*a*b*x+5*B*a^2*x+3*A*a^2)/x^(5/2)/a^3

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maxima [A]  time = 0.92, size = 106, normalized size = 1.26 \begin {gather*} \frac {4 \, \sqrt {b x^{2} + a x} B b}{3 \, a^{2} x} - \frac {16 \, \sqrt {b x^{2} + a x} A b^{2}}{15 \, a^{3} x} - \frac {2 \, \sqrt {b x^{2} + a x} B}{3 \, a x^{2}} + \frac {8 \, \sqrt {b x^{2} + a x} A b}{15 \, a^{2} x^{2}} - \frac {2 \, \sqrt {b x^{2} + a x} A}{5 \, a x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^(7/2)/(b*x+a)^(1/2),x, algorithm="maxima")

[Out]

4/3*sqrt(b*x^2 + a*x)*B*b/(a^2*x) - 16/15*sqrt(b*x^2 + a*x)*A*b^2/(a^3*x) - 2/3*sqrt(b*x^2 + a*x)*B/(a*x^2) +
8/15*sqrt(b*x^2 + a*x)*A*b/(a^2*x^2) - 2/5*sqrt(b*x^2 + a*x)*A/(a*x^3)

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mupad [B]  time = 0.87, size = 57, normalized size = 0.68 \begin {gather*} -\frac {\sqrt {a+b\,x}\,\left (\frac {2\,A}{5\,a}+\frac {x^2\,\left (16\,A\,b^2-20\,B\,a\,b\right )}{15\,a^3}+\frac {x\,\left (10\,B\,a^2-8\,A\,a\,b\right )}{15\,a^3}\right )}{x^{5/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*x)/(x^(7/2)*(a + b*x)^(1/2)),x)

[Out]

-((a + b*x)^(1/2)*((2*A)/(5*a) + (x^2*(16*A*b^2 - 20*B*a*b))/(15*a^3) + (x*(10*B*a^2 - 8*A*a*b))/(15*a^3)))/x^
(5/2)

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sympy [B]  time = 35.72, size = 342, normalized size = 4.07 \begin {gather*} - \frac {6 A a^{4} b^{\frac {9}{2}} \sqrt {\frac {a}{b x} + 1}}{15 a^{5} b^{4} x^{2} + 30 a^{4} b^{5} x^{3} + 15 a^{3} b^{6} x^{4}} - \frac {4 A a^{3} b^{\frac {11}{2}} x \sqrt {\frac {a}{b x} + 1}}{15 a^{5} b^{4} x^{2} + 30 a^{4} b^{5} x^{3} + 15 a^{3} b^{6} x^{4}} - \frac {6 A a^{2} b^{\frac {13}{2}} x^{2} \sqrt {\frac {a}{b x} + 1}}{15 a^{5} b^{4} x^{2} + 30 a^{4} b^{5} x^{3} + 15 a^{3} b^{6} x^{4}} - \frac {24 A a b^{\frac {15}{2}} x^{3} \sqrt {\frac {a}{b x} + 1}}{15 a^{5} b^{4} x^{2} + 30 a^{4} b^{5} x^{3} + 15 a^{3} b^{6} x^{4}} - \frac {16 A b^{\frac {17}{2}} x^{4} \sqrt {\frac {a}{b x} + 1}}{15 a^{5} b^{4} x^{2} + 30 a^{4} b^{5} x^{3} + 15 a^{3} b^{6} x^{4}} - \frac {2 B \sqrt {b} \sqrt {\frac {a}{b x} + 1}}{3 a x} + \frac {4 B b^{\frac {3}{2}} \sqrt {\frac {a}{b x} + 1}}{3 a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x**(7/2)/(b*x+a)**(1/2),x)

[Out]

-6*A*a**4*b**(9/2)*sqrt(a/(b*x) + 1)/(15*a**5*b**4*x**2 + 30*a**4*b**5*x**3 + 15*a**3*b**6*x**4) - 4*A*a**3*b*
*(11/2)*x*sqrt(a/(b*x) + 1)/(15*a**5*b**4*x**2 + 30*a**4*b**5*x**3 + 15*a**3*b**6*x**4) - 6*A*a**2*b**(13/2)*x
**2*sqrt(a/(b*x) + 1)/(15*a**5*b**4*x**2 + 30*a**4*b**5*x**3 + 15*a**3*b**6*x**4) - 24*A*a*b**(15/2)*x**3*sqrt
(a/(b*x) + 1)/(15*a**5*b**4*x**2 + 30*a**4*b**5*x**3 + 15*a**3*b**6*x**4) - 16*A*b**(17/2)*x**4*sqrt(a/(b*x) +
1)/(15*a**5*b**4*x**2 + 30*a**4*b**5*x**3 + 15*a**3*b**6*x**4) - 2*B*sqrt(b)*sqrt(a/(b*x) + 1)/(3*a*x) + 4*B*
b**(3/2)*sqrt(a/(b*x) + 1)/(3*a**2)

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