3.6.8 \(\int \frac {A+B x}{x^{7/2} (a+b x)^{3/2}} \, dx\)

Optimal. Leaf size=114 \[ -\frac {16 b \sqrt {a+b x} (6 A b-5 a B)}{15 a^4 \sqrt {x}}+\frac {8 \sqrt {a+b x} (6 A b-5 a B)}{15 a^3 x^{3/2}}-\frac {2 (6 A b-5 a B)}{5 a^2 x^{3/2} \sqrt {a+b x}}-\frac {2 A}{5 a x^{5/2} \sqrt {a+b x}} \]

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Rubi [A]  time = 0.04, antiderivative size = 114, normalized size of antiderivative = 1.00, number of steps used = 4, 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 {8 \sqrt {a+b x} (6 A b-5 a B)}{15 a^3 x^{3/2}}-\frac {2 (6 A b-5 a B)}{5 a^2 x^{3/2} \sqrt {a+b x}}-\frac {16 b \sqrt {a+b x} (6 A b-5 a B)}{15 a^4 \sqrt {x}}-\frac {2 A}{5 a x^{5/2} \sqrt {a+b x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(A + B*x)/(x^(7/2)*(a + b*x)^(3/2)),x]

[Out]

(-2*A)/(5*a*x^(5/2)*Sqrt[a + b*x]) - (2*(6*A*b - 5*a*B))/(5*a^2*x^(3/2)*Sqrt[a + b*x]) + (8*(6*A*b - 5*a*B)*Sq
rt[a + b*x])/(15*a^3*x^(3/2)) - (16*b*(6*A*b - 5*a*B)*Sqrt[a + b*x])/(15*a^4*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} (a+b x)^{3/2}} \, dx &=-\frac {2 A}{5 a x^{5/2} \sqrt {a+b x}}+\frac {\left (2 \left (-3 A b+\frac {5 a B}{2}\right )\right ) \int \frac {1}{x^{5/2} (a+b x)^{3/2}} \, dx}{5 a}\\ &=-\frac {2 A}{5 a x^{5/2} \sqrt {a+b x}}-\frac {2 (6 A b-5 a B)}{5 a^2 x^{3/2} \sqrt {a+b x}}-\frac {(4 (6 A b-5 a B)) \int \frac {1}{x^{5/2} \sqrt {a+b x}} \, dx}{5 a^2}\\ &=-\frac {2 A}{5 a x^{5/2} \sqrt {a+b x}}-\frac {2 (6 A b-5 a B)}{5 a^2 x^{3/2} \sqrt {a+b x}}+\frac {8 (6 A b-5 a B) \sqrt {a+b x}}{15 a^3 x^{3/2}}+\frac {(8 b (6 A b-5 a B)) \int \frac {1}{x^{3/2} \sqrt {a+b x}} \, dx}{15 a^3}\\ &=-\frac {2 A}{5 a x^{5/2} \sqrt {a+b x}}-\frac {2 (6 A b-5 a B)}{5 a^2 x^{3/2} \sqrt {a+b x}}+\frac {8 (6 A b-5 a B) \sqrt {a+b x}}{15 a^3 x^{3/2}}-\frac {16 b (6 A b-5 a B) \sqrt {a+b x}}{15 a^4 \sqrt {x}}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 75, normalized size = 0.66 \begin {gather*} -\frac {2 \left (a^3 (3 A+5 B x)-2 a^2 b x (3 A+10 B x)+8 a b^2 x^2 (3 A-5 B x)+48 A b^3 x^3\right )}{15 a^4 x^{5/2} \sqrt {a+b x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x)/(x^(7/2)*(a + b*x)^(3/2)),x]

[Out]

(-2*(48*A*b^3*x^3 + 8*a*b^2*x^2*(3*A - 5*B*x) + a^3*(3*A + 5*B*x) - 2*a^2*b*x*(3*A + 10*B*x)))/(15*a^4*x^(5/2)
*Sqrt[a + b*x])

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IntegrateAlgebraic [A]  time = 0.20, size = 82, normalized size = 0.72 \begin {gather*} \frac {2 \left (-3 a^3 A-5 a^3 B x+6 a^2 A b x+20 a^2 b B x^2-24 a A b^2 x^2+40 a b^2 B x^3-48 A b^3 x^3\right )}{15 a^4 x^{5/2} \sqrt {a+b x}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(A + B*x)/(x^(7/2)*(a + b*x)^(3/2)),x]

[Out]

(2*(-3*a^3*A + 6*a^2*A*b*x - 5*a^3*B*x - 24*a*A*b^2*x^2 + 20*a^2*b*B*x^2 - 48*A*b^3*x^3 + 40*a*b^2*B*x^3))/(15
*a^4*x^(5/2)*Sqrt[a + b*x])

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fricas [A]  time = 1.45, size = 92, normalized size = 0.81 \begin {gather*} -\frac {2 \, {\left (3 \, A a^{3} - 8 \, {\left (5 \, B a b^{2} - 6 \, A b^{3}\right )} x^{3} - 4 \, {\left (5 \, B a^{2} b - 6 \, A a b^{2}\right )} x^{2} + {\left (5 \, B a^{3} - 6 \, A a^{2} b\right )} x\right )} \sqrt {b x + a} \sqrt {x}}{15 \, {\left (a^{4} b x^{4} + a^{5} x^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^(7/2)/(b*x+a)^(3/2),x, algorithm="fricas")

[Out]

-2/15*(3*A*a^3 - 8*(5*B*a*b^2 - 6*A*b^3)*x^3 - 4*(5*B*a^2*b - 6*A*a*b^2)*x^2 + (5*B*a^3 - 6*A*a^2*b)*x)*sqrt(b
*x + a)*sqrt(x)/(a^4*b*x^4 + a^5*x^3)

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giac [B]  time = 1.49, size = 254, normalized size = 2.23 \begin {gather*} \frac {2 \, \sqrt {b x + a} {\left ({\left (b x + a\right )} {\left (\frac {{\left (25 \, B a^{6} b^{7} - 33 \, A a^{5} b^{8}\right )} {\left (b x + a\right )}}{a^{9} b^{2} {\left | b \right |}} - \frac {5 \, {\left (11 \, B a^{7} b^{7} - 15 \, A a^{6} b^{8}\right )}}{a^{9} b^{2} {\left | b \right |}}\right )} + \frac {15 \, {\left (2 \, B a^{8} b^{7} - 3 \, A a^{7} b^{8}\right )}}{a^{9} b^{2} {\left | b \right |}}\right )}}{15 \, {\left ({\left (b x + a\right )} b - a b\right )}^{\frac {5}{2}}} + \frac {4 \, {\left (B^{2} a^{2} b^{7} - 2 \, A B a b^{8} + A^{2} b^{9}\right )}}{{\left (B a {\left (\sqrt {b x + a} \sqrt {b} - \sqrt {{\left (b x + a\right )} b - a b}\right )}^{2} b^{\frac {7}{2}} + B a^{2} b^{\frac {9}{2}} - A {\left (\sqrt {b x + a} \sqrt {b} - \sqrt {{\left (b x + a\right )} b - a b}\right )}^{2} b^{\frac {9}{2}} - A a b^{\frac {11}{2}}\right )} a^{3} {\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)^(3/2),x, algorithm="giac")

[Out]

2/15*sqrt(b*x + a)*((b*x + a)*((25*B*a^6*b^7 - 33*A*a^5*b^8)*(b*x + a)/(a^9*b^2*abs(b)) - 5*(11*B*a^7*b^7 - 15
*A*a^6*b^8)/(a^9*b^2*abs(b))) + 15*(2*B*a^8*b^7 - 3*A*a^7*b^8)/(a^9*b^2*abs(b)))/((b*x + a)*b - a*b)^(5/2) + 4
*(B^2*a^2*b^7 - 2*A*B*a*b^8 + A^2*b^9)/((B*a*(sqrt(b*x + a)*sqrt(b) - sqrt((b*x + a)*b - a*b))^2*b^(7/2) + B*a
^2*b^(9/2) - A*(sqrt(b*x + a)*sqrt(b) - sqrt((b*x + a)*b - a*b))^2*b^(9/2) - A*a*b^(11/2))*a^3*abs(b))

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maple [A]  time = 0.01, size = 77, normalized size = 0.68 \begin {gather*} -\frac {2 \left (48 A \,b^{3} x^{3}-40 B a \,b^{2} x^{3}+24 A a \,b^{2} x^{2}-20 B \,a^{2} b \,x^{2}-6 A \,a^{2} b x +5 B \,a^{3} x +3 A \,a^{3}\right )}{15 \sqrt {b x +a}\, a^{4} 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)^(3/2),x)

[Out]

-2/15*(48*A*b^3*x^3-40*B*a*b^2*x^3+24*A*a*b^2*x^2-20*B*a^2*b*x^2-6*A*a^2*b*x+5*B*a^3*x+3*A*a^3)/(b*x+a)^(1/2)/
x^(5/2)/a^4

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maxima [A]  time = 0.91, size = 142, normalized size = 1.25 \begin {gather*} \frac {16 \, B b^{2} x}{3 \, \sqrt {b x^{2} + a x} a^{3}} - \frac {32 \, A b^{3} x}{5 \, \sqrt {b x^{2} + a x} a^{4}} + \frac {8 \, B b}{3 \, \sqrt {b x^{2} + a x} a^{2}} - \frac {16 \, A b^{2}}{5 \, \sqrt {b x^{2} + a x} a^{3}} - \frac {2 \, B}{3 \, \sqrt {b x^{2} + a x} a x} + \frac {4 \, A b}{5 \, \sqrt {b x^{2} + a x} a^{2} x} - \frac {2 \, A}{5 \, \sqrt {b x^{2} + a x} a x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^(7/2)/(b*x+a)^(3/2),x, algorithm="maxima")

[Out]

16/3*B*b^2*x/(sqrt(b*x^2 + a*x)*a^3) - 32/5*A*b^3*x/(sqrt(b*x^2 + a*x)*a^4) + 8/3*B*b/(sqrt(b*x^2 + a*x)*a^2)
- 16/5*A*b^2/(sqrt(b*x^2 + a*x)*a^3) - 2/3*B/(sqrt(b*x^2 + a*x)*a*x) + 4/5*A*b/(sqrt(b*x^2 + a*x)*a^2*x) - 2/5
*A/(sqrt(b*x^2 + a*x)*a*x^2)

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mupad [B]  time = 0.91, size = 98, normalized size = 0.86 \begin {gather*} -\frac {\sqrt {a+b\,x}\,\left (\frac {2\,A}{5\,a\,b}+\frac {8\,x^2\,\left (6\,A\,b-5\,B\,a\right )}{15\,a^3}+\frac {x^3\,\left (96\,A\,b^3-80\,B\,a\,b^2\right )}{15\,a^4\,b}+\frac {x\,\left (10\,B\,a^3-12\,A\,a^2\,b\right )}{15\,a^4\,b}\right )}{x^{7/2}+\frac {a\,x^{5/2}}{b}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*x)/(x^(7/2)*(a + b*x)^(3/2)),x)

[Out]

-((a + b*x)^(1/2)*((2*A)/(5*a*b) + (8*x^2*(6*A*b - 5*B*a))/(15*a^3) + (x^3*(96*A*b^3 - 80*B*a*b^2))/(15*a^4*b)
+ (x*(10*B*a^3 - 12*A*a^2*b))/(15*a^4*b)))/(x^(7/2) + (a*x^(5/2))/b)

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sympy [B]  time = 158.04, size = 573, normalized size = 5.03 \begin {gather*} A \left (- \frac {2 a^{5} b^{\frac {19}{2}} \sqrt {\frac {a}{b x} + 1}}{5 a^{7} b^{9} x^{2} + 15 a^{6} b^{10} x^{3} + 15 a^{5} b^{11} x^{4} + 5 a^{4} b^{12} x^{5}} - \frac {10 a^{3} b^{\frac {23}{2}} x^{2} \sqrt {\frac {a}{b x} + 1}}{5 a^{7} b^{9} x^{2} + 15 a^{6} b^{10} x^{3} + 15 a^{5} b^{11} x^{4} + 5 a^{4} b^{12} x^{5}} - \frac {60 a^{2} b^{\frac {25}{2}} x^{3} \sqrt {\frac {a}{b x} + 1}}{5 a^{7} b^{9} x^{2} + 15 a^{6} b^{10} x^{3} + 15 a^{5} b^{11} x^{4} + 5 a^{4} b^{12} x^{5}} - \frac {80 a b^{\frac {27}{2}} x^{4} \sqrt {\frac {a}{b x} + 1}}{5 a^{7} b^{9} x^{2} + 15 a^{6} b^{10} x^{3} + 15 a^{5} b^{11} x^{4} + 5 a^{4} b^{12} x^{5}} - \frac {32 b^{\frac {29}{2}} x^{5} \sqrt {\frac {a}{b x} + 1}}{5 a^{7} b^{9} x^{2} + 15 a^{6} b^{10} x^{3} + 15 a^{5} b^{11} x^{4} + 5 a^{4} b^{12} x^{5}}\right ) + B \left (- \frac {2 a^{3} b^{\frac {9}{2}} \sqrt {\frac {a}{b x} + 1}}{3 a^{5} b^{4} x + 6 a^{4} b^{5} x^{2} + 3 a^{3} b^{6} x^{3}} + \frac {6 a^{2} b^{\frac {11}{2}} x \sqrt {\frac {a}{b x} + 1}}{3 a^{5} b^{4} x + 6 a^{4} b^{5} x^{2} + 3 a^{3} b^{6} x^{3}} + \frac {24 a b^{\frac {13}{2}} x^{2} \sqrt {\frac {a}{b x} + 1}}{3 a^{5} b^{4} x + 6 a^{4} b^{5} x^{2} + 3 a^{3} b^{6} x^{3}} + \frac {16 b^{\frac {15}{2}} x^{3} \sqrt {\frac {a}{b x} + 1}}{3 a^{5} b^{4} x + 6 a^{4} b^{5} x^{2} + 3 a^{3} b^{6} x^{3}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x**(7/2)/(b*x+a)**(3/2),x)

[Out]

A*(-2*a**5*b**(19/2)*sqrt(a/(b*x) + 1)/(5*a**7*b**9*x**2 + 15*a**6*b**10*x**3 + 15*a**5*b**11*x**4 + 5*a**4*b*
*12*x**5) - 10*a**3*b**(23/2)*x**2*sqrt(a/(b*x) + 1)/(5*a**7*b**9*x**2 + 15*a**6*b**10*x**3 + 15*a**5*b**11*x*
*4 + 5*a**4*b**12*x**5) - 60*a**2*b**(25/2)*x**3*sqrt(a/(b*x) + 1)/(5*a**7*b**9*x**2 + 15*a**6*b**10*x**3 + 15
*a**5*b**11*x**4 + 5*a**4*b**12*x**5) - 80*a*b**(27/2)*x**4*sqrt(a/(b*x) + 1)/(5*a**7*b**9*x**2 + 15*a**6*b**1
0*x**3 + 15*a**5*b**11*x**4 + 5*a**4*b**12*x**5) - 32*b**(29/2)*x**5*sqrt(a/(b*x) + 1)/(5*a**7*b**9*x**2 + 15*
a**6*b**10*x**3 + 15*a**5*b**11*x**4 + 5*a**4*b**12*x**5)) + B*(-2*a**3*b**(9/2)*sqrt(a/(b*x) + 1)/(3*a**5*b**
4*x + 6*a**4*b**5*x**2 + 3*a**3*b**6*x**3) + 6*a**2*b**(11/2)*x*sqrt(a/(b*x) + 1)/(3*a**5*b**4*x + 6*a**4*b**5
*x**2 + 3*a**3*b**6*x**3) + 24*a*b**(13/2)*x**2*sqrt(a/(b*x) + 1)/(3*a**5*b**4*x + 6*a**4*b**5*x**2 + 3*a**3*b
**6*x**3) + 16*b**(15/2)*x**3*sqrt(a/(b*x) + 1)/(3*a**5*b**4*x + 6*a**4*b**5*x**2 + 3*a**3*b**6*x**3))

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