3.535 \(\int \frac {A+B x}{x^{11/2} (a+b x)^{3/2}} \, dx\)

Optimal. Leaf size=180 \[ -\frac {256 b^3 \sqrt {a+b x} (10 A b-9 a B)}{315 a^6 \sqrt {x}}+\frac {128 b^2 \sqrt {a+b x} (10 A b-9 a B)}{315 a^5 x^{3/2}}-\frac {32 b \sqrt {a+b x} (10 A b-9 a B)}{105 a^4 x^{5/2}}+\frac {16 \sqrt {a+b x} (10 A b-9 a B)}{63 a^3 x^{7/2}}-\frac {2 (10 A b-9 a B)}{9 a^2 x^{7/2} \sqrt {a+b x}}-\frac {2 A}{9 a x^{9/2} \sqrt {a+b x}} \]

[Out]

-2/9*A/a/x^(9/2)/(b*x+a)^(1/2)-2/9*(10*A*b-9*B*a)/a^2/x^(7/2)/(b*x+a)^(1/2)+16/63*(10*A*b-9*B*a)*(b*x+a)^(1/2)
/a^3/x^(7/2)-32/105*b*(10*A*b-9*B*a)*(b*x+a)^(1/2)/a^4/x^(5/2)+128/315*b^2*(10*A*b-9*B*a)*(b*x+a)^(1/2)/a^5/x^
(3/2)-256/315*b^3*(10*A*b-9*B*a)*(b*x+a)^(1/2)/a^6/x^(1/2)

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Rubi [A]  time = 0.07, antiderivative size = 180, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {78, 45, 37} \[ \frac {128 b^2 \sqrt {a+b x} (10 A b-9 a B)}{315 a^5 x^{3/2}}-\frac {256 b^3 \sqrt {a+b x} (10 A b-9 a B)}{315 a^6 \sqrt {x}}-\frac {32 b \sqrt {a+b x} (10 A b-9 a B)}{105 a^4 x^{5/2}}+\frac {16 \sqrt {a+b x} (10 A b-9 a B)}{63 a^3 x^{7/2}}-\frac {2 (10 A b-9 a B)}{9 a^2 x^{7/2} \sqrt {a+b x}}-\frac {2 A}{9 a x^{9/2} \sqrt {a+b x}} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*x)/(x^(11/2)*(a + b*x)^(3/2)),x]

[Out]

(-2*A)/(9*a*x^(9/2)*Sqrt[a + b*x]) - (2*(10*A*b - 9*a*B))/(9*a^2*x^(7/2)*Sqrt[a + b*x]) + (16*(10*A*b - 9*a*B)
*Sqrt[a + b*x])/(63*a^3*x^(7/2)) - (32*b*(10*A*b - 9*a*B)*Sqrt[a + b*x])/(105*a^4*x^(5/2)) + (128*b^2*(10*A*b
- 9*a*B)*Sqrt[a + b*x])/(315*a^5*x^(3/2)) - (256*b^3*(10*A*b - 9*a*B)*Sqrt[a + b*x])/(315*a^6*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^{11/2} (a+b x)^{3/2}} \, dx &=-\frac {2 A}{9 a x^{9/2} \sqrt {a+b x}}+\frac {\left (2 \left (-5 A b+\frac {9 a B}{2}\right )\right ) \int \frac {1}{x^{9/2} (a+b x)^{3/2}} \, dx}{9 a}\\ &=-\frac {2 A}{9 a x^{9/2} \sqrt {a+b x}}-\frac {2 (10 A b-9 a B)}{9 a^2 x^{7/2} \sqrt {a+b x}}-\frac {(8 (10 A b-9 a B)) \int \frac {1}{x^{9/2} \sqrt {a+b x}} \, dx}{9 a^2}\\ &=-\frac {2 A}{9 a x^{9/2} \sqrt {a+b x}}-\frac {2 (10 A b-9 a B)}{9 a^2 x^{7/2} \sqrt {a+b x}}+\frac {16 (10 A b-9 a B) \sqrt {a+b x}}{63 a^3 x^{7/2}}+\frac {(16 b (10 A b-9 a B)) \int \frac {1}{x^{7/2} \sqrt {a+b x}} \, dx}{21 a^3}\\ &=-\frac {2 A}{9 a x^{9/2} \sqrt {a+b x}}-\frac {2 (10 A b-9 a B)}{9 a^2 x^{7/2} \sqrt {a+b x}}+\frac {16 (10 A b-9 a B) \sqrt {a+b x}}{63 a^3 x^{7/2}}-\frac {32 b (10 A b-9 a B) \sqrt {a+b x}}{105 a^4 x^{5/2}}-\frac {\left (64 b^2 (10 A b-9 a B)\right ) \int \frac {1}{x^{5/2} \sqrt {a+b x}} \, dx}{105 a^4}\\ &=-\frac {2 A}{9 a x^{9/2} \sqrt {a+b x}}-\frac {2 (10 A b-9 a B)}{9 a^2 x^{7/2} \sqrt {a+b x}}+\frac {16 (10 A b-9 a B) \sqrt {a+b x}}{63 a^3 x^{7/2}}-\frac {32 b (10 A b-9 a B) \sqrt {a+b x}}{105 a^4 x^{5/2}}+\frac {128 b^2 (10 A b-9 a B) \sqrt {a+b x}}{315 a^5 x^{3/2}}+\frac {\left (128 b^3 (10 A b-9 a B)\right ) \int \frac {1}{x^{3/2} \sqrt {a+b x}} \, dx}{315 a^5}\\ &=-\frac {2 A}{9 a x^{9/2} \sqrt {a+b x}}-\frac {2 (10 A b-9 a B)}{9 a^2 x^{7/2} \sqrt {a+b x}}+\frac {16 (10 A b-9 a B) \sqrt {a+b x}}{63 a^3 x^{7/2}}-\frac {32 b (10 A b-9 a B) \sqrt {a+b x}}{105 a^4 x^{5/2}}+\frac {128 b^2 (10 A b-9 a B) \sqrt {a+b x}}{315 a^5 x^{3/2}}-\frac {256 b^3 (10 A b-9 a B) \sqrt {a+b x}}{315 a^6 \sqrt {x}}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 114, normalized size = 0.63 \[ -\frac {2 \left (5 a^5 (7 A+9 B x)-2 a^4 b x (25 A+36 B x)+16 a^3 b^2 x^2 (5 A+9 B x)-32 a^2 b^3 x^3 (5 A+18 B x)+128 a b^4 x^4 (5 A-9 B x)+1280 A b^5 x^5\right )}{315 a^6 x^{9/2} \sqrt {a+b x}} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x)/(x^(11/2)*(a + b*x)^(3/2)),x]

[Out]

(-2*(1280*A*b^5*x^5 + 128*a*b^4*x^4*(5*A - 9*B*x) + 16*a^3*b^2*x^2*(5*A + 9*B*x) + 5*a^5*(7*A + 9*B*x) - 32*a^
2*b^3*x^3*(5*A + 18*B*x) - 2*a^4*b*x*(25*A + 36*B*x)))/(315*a^6*x^(9/2)*Sqrt[a + b*x])

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fricas [A]  time = 0.62, size = 141, normalized size = 0.78 \[ -\frac {2 \, {\left (35 \, A a^{5} - 128 \, {\left (9 \, B a b^{4} - 10 \, A b^{5}\right )} x^{5} - 64 \, {\left (9 \, B a^{2} b^{3} - 10 \, A a b^{4}\right )} x^{4} + 16 \, {\left (9 \, B a^{3} b^{2} - 10 \, A a^{2} b^{3}\right )} x^{3} - 8 \, {\left (9 \, B a^{4} b - 10 \, A a^{3} b^{2}\right )} x^{2} + 5 \, {\left (9 \, B a^{5} - 10 \, A a^{4} b\right )} x\right )} \sqrt {b x + a} \sqrt {x}}{315 \, {\left (a^{6} b x^{6} + a^{7} x^{5}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^(11/2)/(b*x+a)^(3/2),x, algorithm="fricas")

[Out]

-2/315*(35*A*a^5 - 128*(9*B*a*b^4 - 10*A*b^5)*x^5 - 64*(9*B*a^2*b^3 - 10*A*a*b^4)*x^4 + 16*(9*B*a^3*b^2 - 10*A
*a^2*b^3)*x^3 - 8*(9*B*a^4*b - 10*A*a^3*b^2)*x^2 + 5*(9*B*a^5 - 10*A*a^4*b)*x)*sqrt(b*x + a)*sqrt(x)/(a^6*b*x^
6 + a^7*x^5)

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giac [B]  time = 1.51, size = 330, normalized size = 1.83 \[ \frac {2 \, {\left ({\left ({\left (b x + a\right )} {\left ({\left (b x + a\right )} {\left (\frac {{\left (837 \, B a^{15} b^{13} - 965 \, A a^{14} b^{14}\right )} {\left (b x + a\right )}}{a^{20} b^{4} {\left | b \right |}} - \frac {9 \, {\left (401 \, B a^{16} b^{13} - 465 \, A a^{15} b^{14}\right )}}{a^{20} b^{4} {\left | b \right |}}\right )} + \frac {126 \, {\left (47 \, B a^{17} b^{13} - 55 \, A a^{16} b^{14}\right )}}{a^{20} b^{4} {\left | b \right |}}\right )} - \frac {210 \, {\left (21 \, B a^{18} b^{13} - 25 \, A a^{17} b^{14}\right )}}{a^{20} b^{4} {\left | b \right |}}\right )} {\left (b x + a\right )} + \frac {315 \, {\left (4 \, B a^{19} b^{13} - 5 \, A a^{18} b^{14}\right )}}{a^{20} b^{4} {\left | b \right |}}\right )} \sqrt {b x + a}}{315 \, {\left ({\left (b x + a\right )} b - a b\right )}^{\frac {9}{2}}} + \frac {4 \, {\left (B^{2} a^{2} b^{11} - 2 \, A B a b^{12} + A^{2} b^{13}\right )}}{{\left (B a {\left (\sqrt {b x + a} \sqrt {b} - \sqrt {{\left (b x + a\right )} b - a b}\right )}^{2} b^{\frac {11}{2}} + B a^{2} b^{\frac {13}{2}} - A {\left (\sqrt {b x + a} \sqrt {b} - \sqrt {{\left (b x + a\right )} b - a b}\right )}^{2} b^{\frac {13}{2}} - A a b^{\frac {15}{2}}\right )} a^{5} {\left | b \right |}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^(11/2)/(b*x+a)^(3/2),x, algorithm="giac")

[Out]

2/315*(((b*x + a)*((b*x + a)*((837*B*a^15*b^13 - 965*A*a^14*b^14)*(b*x + a)/(a^20*b^4*abs(b)) - 9*(401*B*a^16*
b^13 - 465*A*a^15*b^14)/(a^20*b^4*abs(b))) + 126*(47*B*a^17*b^13 - 55*A*a^16*b^14)/(a^20*b^4*abs(b))) - 210*(2
1*B*a^18*b^13 - 25*A*a^17*b^14)/(a^20*b^4*abs(b)))*(b*x + a) + 315*(4*B*a^19*b^13 - 5*A*a^18*b^14)/(a^20*b^4*a
bs(b)))*sqrt(b*x + a)/((b*x + a)*b - a*b)^(9/2) + 4*(B^2*a^2*b^11 - 2*A*B*a*b^12 + A^2*b^13)/((B*a*(sqrt(b*x +
a)*sqrt(b) - sqrt((b*x + a)*b - a*b))^2*b^(11/2) + B*a^2*b^(13/2) - A*(sqrt(b*x + a)*sqrt(b) - sqrt((b*x + a)
*b - a*b))^2*b^(13/2) - A*a*b^(15/2))*a^5*abs(b))

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maple [A]  time = 0.01, size = 125, normalized size = 0.69 \[ -\frac {2 \left (1280 A \,b^{5} x^{5}-1152 B a \,b^{4} x^{5}+640 A a \,b^{4} x^{4}-576 B \,a^{2} b^{3} x^{4}-160 A \,a^{2} b^{3} x^{3}+144 B \,a^{3} b^{2} x^{3}+80 A \,a^{3} b^{2} x^{2}-72 B \,a^{4} b \,x^{2}-50 A \,a^{4} b x +45 B \,a^{5} x +35 A \,a^{5}\right )}{315 \sqrt {b x +a}\, a^{6} x^{\frac {9}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)/x^(11/2)/(b*x+a)^(3/2),x)

[Out]

-2/315*(1280*A*b^5*x^5-1152*B*a*b^4*x^5+640*A*a*b^4*x^4-576*B*a^2*b^3*x^4-160*A*a^2*b^3*x^3+144*B*a^3*b^2*x^3+
80*A*a^3*b^2*x^2-72*B*a^4*b*x^2-50*A*a^4*b*x+45*B*a^5*x+35*A*a^5)/(b*x+a)^(1/2)/x^(9/2)/a^6

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maxima [A]  time = 0.93, size = 234, normalized size = 1.30 \[ \frac {256 \, B b^{4} x}{35 \, \sqrt {b x^{2} + a x} a^{5}} - \frac {512 \, A b^{5} x}{63 \, \sqrt {b x^{2} + a x} a^{6}} + \frac {128 \, B b^{3}}{35 \, \sqrt {b x^{2} + a x} a^{4}} - \frac {256 \, A b^{4}}{63 \, \sqrt {b x^{2} + a x} a^{5}} - \frac {32 \, B b^{2}}{35 \, \sqrt {b x^{2} + a x} a^{3} x} + \frac {64 \, A b^{3}}{63 \, \sqrt {b x^{2} + a x} a^{4} x} + \frac {16 \, B b}{35 \, \sqrt {b x^{2} + a x} a^{2} x^{2}} - \frac {32 \, A b^{2}}{63 \, \sqrt {b x^{2} + a x} a^{3} x^{2}} - \frac {2 \, B}{7 \, \sqrt {b x^{2} + a x} a x^{3}} + \frac {20 \, A b}{63 \, \sqrt {b x^{2} + a x} a^{2} x^{3}} - \frac {2 \, A}{9 \, \sqrt {b x^{2} + a x} a x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^(11/2)/(b*x+a)^(3/2),x, algorithm="maxima")

[Out]

256/35*B*b^4*x/(sqrt(b*x^2 + a*x)*a^5) - 512/63*A*b^5*x/(sqrt(b*x^2 + a*x)*a^6) + 128/35*B*b^3/(sqrt(b*x^2 + a
*x)*a^4) - 256/63*A*b^4/(sqrt(b*x^2 + a*x)*a^5) - 32/35*B*b^2/(sqrt(b*x^2 + a*x)*a^3*x) + 64/63*A*b^3/(sqrt(b*
x^2 + a*x)*a^4*x) + 16/35*B*b/(sqrt(b*x^2 + a*x)*a^2*x^2) - 32/63*A*b^2/(sqrt(b*x^2 + a*x)*a^3*x^2) - 2/7*B/(s
qrt(b*x^2 + a*x)*a*x^3) + 20/63*A*b/(sqrt(b*x^2 + a*x)*a^2*x^3) - 2/9*A/(sqrt(b*x^2 + a*x)*a*x^4)

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mupad [B]  time = 1.00, size = 131, normalized size = 0.73 \[ -\frac {\sqrt {a+b\,x}\,\left (\frac {2\,A}{9\,a\,b}+\frac {16\,x^2\,\left (10\,A\,b-9\,B\,a\right )}{315\,a^3}+\frac {128\,b^2\,x^4\,\left (10\,A\,b-9\,B\,a\right )}{315\,a^5}+\frac {256\,b^3\,x^5\,\left (10\,A\,b-9\,B\,a\right )}{315\,a^6}-\frac {32\,b\,x^3\,\left (10\,A\,b-9\,B\,a\right )}{315\,a^4}+\frac {x\,\left (90\,B\,a^5-100\,A\,a^4\,b\right )}{315\,a^6\,b}\right )}{x^{11/2}+\frac {a\,x^{9/2}}{b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*x)/(x^(11/2)*(a + b*x)^(3/2)),x)

[Out]

-((a + b*x)^(1/2)*((2*A)/(9*a*b) + (16*x^2*(10*A*b - 9*B*a))/(315*a^3) + (128*b^2*x^4*(10*A*b - 9*B*a))/(315*a
^5) + (256*b^3*x^5*(10*A*b - 9*B*a))/(315*a^6) - (32*b*x^3*(10*A*b - 9*B*a))/(315*a^4) + (x*(90*B*a^5 - 100*A*
a^4*b))/(315*a^6*b)))/(x^(11/2) + (a*x^(9/2))/b)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x**(11/2)/(b*x+a)**(3/2),x)

[Out]

Timed out

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