3.6.20 \(\int \frac {A+B x}{x^{9/2} (a+b x)^{5/2}} \, dx\)

Optimal. Leaf size=177 \[ \frac {256 b^2 \sqrt {a+b x} (10 A b-7 a B)}{105 a^6 \sqrt {x}}-\frac {128 b \sqrt {a+b x} (10 A b-7 a B)}{105 a^5 x^{3/2}}+\frac {32 \sqrt {a+b x} (10 A b-7 a B)}{35 a^4 x^{5/2}}-\frac {16 (10 A b-7 a B)}{21 a^3 x^{5/2} \sqrt {a+b x}}-\frac {2 (10 A b-7 a B)}{21 a^2 x^{5/2} (a+b x)^{3/2}}-\frac {2 A}{7 a x^{7/2} (a+b x)^{3/2}} \]

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Rubi [A]  time = 0.07, antiderivative size = 177, 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} \begin {gather*} \frac {256 b^2 \sqrt {a+b x} (10 A b-7 a B)}{105 a^6 \sqrt {x}}-\frac {128 b \sqrt {a+b x} (10 A b-7 a B)}{105 a^5 x^{3/2}}+\frac {32 \sqrt {a+b x} (10 A b-7 a B)}{35 a^4 x^{5/2}}-\frac {16 (10 A b-7 a B)}{21 a^3 x^{5/2} \sqrt {a+b x}}-\frac {2 (10 A b-7 a B)}{21 a^2 x^{5/2} (a+b x)^{3/2}}-\frac {2 A}{7 a x^{7/2} (a+b x)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(A + B*x)/(x^(9/2)*(a + b*x)^(5/2)),x]

[Out]

(-2*A)/(7*a*x^(7/2)*(a + b*x)^(3/2)) - (2*(10*A*b - 7*a*B))/(21*a^2*x^(5/2)*(a + b*x)^(3/2)) - (16*(10*A*b - 7
*a*B))/(21*a^3*x^(5/2)*Sqrt[a + b*x]) + (32*(10*A*b - 7*a*B)*Sqrt[a + b*x])/(35*a^4*x^(5/2)) - (128*b*(10*A*b
- 7*a*B)*Sqrt[a + b*x])/(105*a^5*x^(3/2)) + (256*b^2*(10*A*b - 7*a*B)*Sqrt[a + b*x])/(105*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^{9/2} (a+b x)^{5/2}} \, dx &=-\frac {2 A}{7 a x^{7/2} (a+b x)^{3/2}}+\frac {\left (2 \left (-5 A b+\frac {7 a B}{2}\right )\right ) \int \frac {1}{x^{7/2} (a+b x)^{5/2}} \, dx}{7 a}\\ &=-\frac {2 A}{7 a x^{7/2} (a+b x)^{3/2}}-\frac {2 (10 A b-7 a B)}{21 a^2 x^{5/2} (a+b x)^{3/2}}-\frac {(8 (10 A b-7 a B)) \int \frac {1}{x^{7/2} (a+b x)^{3/2}} \, dx}{21 a^2}\\ &=-\frac {2 A}{7 a x^{7/2} (a+b x)^{3/2}}-\frac {2 (10 A b-7 a B)}{21 a^2 x^{5/2} (a+b x)^{3/2}}-\frac {16 (10 A b-7 a B)}{21 a^3 x^{5/2} \sqrt {a+b x}}-\frac {(16 (10 A b-7 a B)) \int \frac {1}{x^{7/2} \sqrt {a+b x}} \, dx}{7 a^3}\\ &=-\frac {2 A}{7 a x^{7/2} (a+b x)^{3/2}}-\frac {2 (10 A b-7 a B)}{21 a^2 x^{5/2} (a+b x)^{3/2}}-\frac {16 (10 A b-7 a B)}{21 a^3 x^{5/2} \sqrt {a+b x}}+\frac {32 (10 A b-7 a B) \sqrt {a+b x}}{35 a^4 x^{5/2}}+\frac {(64 b (10 A b-7 a B)) \int \frac {1}{x^{5/2} \sqrt {a+b x}} \, dx}{35 a^4}\\ &=-\frac {2 A}{7 a x^{7/2} (a+b x)^{3/2}}-\frac {2 (10 A b-7 a B)}{21 a^2 x^{5/2} (a+b x)^{3/2}}-\frac {16 (10 A b-7 a B)}{21 a^3 x^{5/2} \sqrt {a+b x}}+\frac {32 (10 A b-7 a B) \sqrt {a+b x}}{35 a^4 x^{5/2}}-\frac {128 b (10 A b-7 a B) \sqrt {a+b x}}{105 a^5 x^{3/2}}-\frac {\left (128 b^2 (10 A b-7 a B)\right ) \int \frac {1}{x^{3/2} \sqrt {a+b x}} \, dx}{105 a^5}\\ &=-\frac {2 A}{7 a x^{7/2} (a+b x)^{3/2}}-\frac {2 (10 A b-7 a B)}{21 a^2 x^{5/2} (a+b x)^{3/2}}-\frac {16 (10 A b-7 a B)}{21 a^3 x^{5/2} \sqrt {a+b x}}+\frac {32 (10 A b-7 a B) \sqrt {a+b x}}{35 a^4 x^{5/2}}-\frac {128 b (10 A b-7 a B) \sqrt {a+b x}}{105 a^5 x^{3/2}}+\frac {256 b^2 (10 A b-7 a B) \sqrt {a+b x}}{105 a^6 \sqrt {x}}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 114, normalized size = 0.64 \begin {gather*} -\frac {2 \left (3 a^5 (5 A+7 B x)-2 a^4 b x (15 A+28 B x)+16 a^3 b^2 x^2 (5 A+21 B x)+96 a^2 b^3 x^3 (14 B x-5 A)+128 a b^4 x^4 (7 B x-15 A)-1280 A b^5 x^5\right )}{105 a^6 x^{7/2} (a+b x)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x)/(x^(9/2)*(a + b*x)^(5/2)),x]

[Out]

(-2*(-1280*A*b^5*x^5 + 128*a*b^4*x^4*(-15*A + 7*B*x) + 3*a^5*(5*A + 7*B*x) + 96*a^2*b^3*x^3*(-5*A + 14*B*x) +
16*a^3*b^2*x^2*(5*A + 21*B*x) - 2*a^4*b*x*(15*A + 28*B*x)))/(105*a^6*x^(7/2)*(a + b*x)^(3/2))

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IntegrateAlgebraic [A]  time = 0.31, size = 130, normalized size = 0.73 \begin {gather*} -\frac {2 \left (15 a^5 A+21 a^5 B x-30 a^4 A b x-56 a^4 b B x^2+80 a^3 A b^2 x^2+336 a^3 b^2 B x^3-480 a^2 A b^3 x^3+1344 a^2 b^3 B x^4-1920 a A b^4 x^4+896 a b^4 B x^5-1280 A b^5 x^5\right )}{105 a^6 x^{7/2} (a+b x)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(A + B*x)/(x^(9/2)*(a + b*x)^(5/2)),x]

[Out]

(-2*(15*a^5*A - 30*a^4*A*b*x + 21*a^5*B*x + 80*a^3*A*b^2*x^2 - 56*a^4*b*B*x^2 - 480*a^2*A*b^3*x^3 + 336*a^3*b^
2*B*x^3 - 1920*a*A*b^4*x^4 + 1344*a^2*b^3*B*x^4 - 1280*A*b^5*x^5 + 896*a*b^4*B*x^5))/(105*a^6*x^(7/2)*(a + b*x
)^(3/2))

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fricas [A]  time = 1.21, size = 152, normalized size = 0.86 \begin {gather*} -\frac {2 \, {\left (15 \, A a^{5} + 128 \, {\left (7 \, B a b^{4} - 10 \, A b^{5}\right )} x^{5} + 192 \, {\left (7 \, B a^{2} b^{3} - 10 \, A a b^{4}\right )} x^{4} + 48 \, {\left (7 \, B a^{3} b^{2} - 10 \, A a^{2} b^{3}\right )} x^{3} - 8 \, {\left (7 \, B a^{4} b - 10 \, A a^{3} b^{2}\right )} x^{2} + 3 \, {\left (7 \, B a^{5} - 10 \, A a^{4} b\right )} x\right )} \sqrt {b x + a} \sqrt {x}}{105 \, {\left (a^{6} b^{2} x^{6} + 2 \, a^{7} b x^{5} + a^{8} x^{4}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^(9/2)/(b*x+a)^(5/2),x, algorithm="fricas")

[Out]

-2/105*(15*A*a^5 + 128*(7*B*a*b^4 - 10*A*b^5)*x^5 + 192*(7*B*a^2*b^3 - 10*A*a*b^4)*x^4 + 48*(7*B*a^3*b^2 - 10*
A*a^2*b^3)*x^3 - 8*(7*B*a^4*b - 10*A*a^3*b^2)*x^2 + 3*(7*B*a^5 - 10*A*a^4*b)*x)*sqrt(b*x + a)*sqrt(x)/(a^6*b^2
*x^6 + 2*a^7*b*x^5 + a^8*x^4)

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giac [B]  time = 2.63, size = 380, normalized size = 2.15 \begin {gather*} -\frac {2 \, {\left ({\left (b x + a\right )} {\left ({\left (b x + a\right )} {\left (\frac {{\left (511 \, B a^{13} b^{9} {\left | b \right |} - 790 \, A a^{12} b^{10} {\left | b \right |}\right )} {\left (b x + a\right )}}{a^{18} b^{4}} - \frac {7 \, {\left (233 \, B a^{14} b^{9} {\left | b \right |} - 365 \, A a^{13} b^{10} {\left | b \right |}\right )}}{a^{18} b^{4}}\right )} + \frac {350 \, {\left (5 \, B a^{15} b^{9} {\left | b \right |} - 8 \, A a^{14} b^{10} {\left | b \right |}\right )}}{a^{18} b^{4}}\right )} - \frac {210 \, {\left (3 \, B a^{16} b^{9} {\left | b \right |} - 5 \, A a^{15} b^{10} {\left | b \right |}\right )}}{a^{18} b^{4}}\right )} \sqrt {b x + a}}{105 \, {\left ({\left (b x + a\right )} b - a b\right )}^{\frac {7}{2}}} - \frac {4 \, {\left (9 \, B a {\left (\sqrt {b x + a} \sqrt {b} - \sqrt {{\left (b x + a\right )} b - a b}\right )}^{4} b^{\frac {9}{2}} + 24 \, B a^{2} {\left (\sqrt {b x + a} \sqrt {b} - \sqrt {{\left (b x + a\right )} b - a b}\right )}^{2} b^{\frac {11}{2}} - 12 \, A {\left (\sqrt {b x + a} \sqrt {b} - \sqrt {{\left (b x + a\right )} b - a b}\right )}^{4} b^{\frac {11}{2}} + 11 \, B a^{3} b^{\frac {13}{2}} - 30 \, A a {\left (\sqrt {b x + a} \sqrt {b} - \sqrt {{\left (b x + a\right )} b - a b}\right )}^{2} b^{\frac {13}{2}} - 14 \, A a^{2} b^{\frac {15}{2}}\right )}}{3 \, {\left ({\left (\sqrt {b x + a} \sqrt {b} - \sqrt {{\left (b x + a\right )} b - a b}\right )}^{2} + a b\right )}^{3} a^{5} {\left | b \right |}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^(9/2)/(b*x+a)^(5/2),x, algorithm="giac")

[Out]

-2/105*((b*x + a)*((b*x + a)*((511*B*a^13*b^9*abs(b) - 790*A*a^12*b^10*abs(b))*(b*x + a)/(a^18*b^4) - 7*(233*B
*a^14*b^9*abs(b) - 365*A*a^13*b^10*abs(b))/(a^18*b^4)) + 350*(5*B*a^15*b^9*abs(b) - 8*A*a^14*b^10*abs(b))/(a^1
8*b^4)) - 210*(3*B*a^16*b^9*abs(b) - 5*A*a^15*b^10*abs(b))/(a^18*b^4))*sqrt(b*x + a)/((b*x + a)*b - a*b)^(7/2)
- 4/3*(9*B*a*(sqrt(b*x + a)*sqrt(b) - sqrt((b*x + a)*b - a*b))^4*b^(9/2) + 24*B*a^2*(sqrt(b*x + a)*sqrt(b) -
sqrt((b*x + a)*b - a*b))^2*b^(11/2) - 12*A*(sqrt(b*x + a)*sqrt(b) - sqrt((b*x + a)*b - a*b))^4*b^(11/2) + 11*B
*a^3*b^(13/2) - 30*A*a*(sqrt(b*x + a)*sqrt(b) - sqrt((b*x + a)*b - a*b))^2*b^(13/2) - 14*A*a^2*b^(15/2))/(((sq
rt(b*x + a)*sqrt(b) - sqrt((b*x + a)*b - a*b))^2 + a*b)^3*a^5*abs(b))

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maple [A]  time = 0.01, size = 125, normalized size = 0.71 \begin {gather*} -\frac {2 \left (-1280 A \,b^{5} x^{5}+896 B a \,b^{4} x^{5}-1920 A a \,b^{4} x^{4}+1344 B \,a^{2} b^{3} x^{4}-480 A \,a^{2} b^{3} x^{3}+336 B \,a^{3} b^{2} x^{3}+80 A \,a^{3} b^{2} x^{2}-56 B \,a^{4} b \,x^{2}-30 A \,a^{4} b x +21 B \,a^{5} x +15 A \,a^{5}\right )}{105 \left (b x +a \right )^{\frac {3}{2}} a^{6} x^{\frac {7}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)/x^(9/2)/(b*x+a)^(5/2),x)

[Out]

-2/105*(-1280*A*b^5*x^5+896*B*a*b^4*x^5-1920*A*a*b^4*x^4+1344*B*a^2*b^3*x^4-480*A*a^2*b^3*x^3+336*B*a^3*b^2*x^
3+80*A*a^3*b^2*x^2-56*B*a^4*b*x^2-30*A*a^4*b*x+21*B*a^5*x+15*A*a^5)/(b*x+a)^(3/2)/x^(7/2)/a^6

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maxima [A]  time = 0.91, size = 224, normalized size = 1.27 \begin {gather*} \frac {32 \, B b^{2} x}{15 \, {\left (b x^{2} + a x\right )}^{\frac {3}{2}} a^{3}} - \frac {256 \, B b^{3} x}{15 \, \sqrt {b x^{2} + a x} a^{5}} - \frac {64 \, A b^{3} x}{21 \, {\left (b x^{2} + a x\right )}^{\frac {3}{2}} a^{4}} + \frac {512 \, A b^{4} x}{21 \, \sqrt {b x^{2} + a x} a^{6}} + \frac {16 \, B b}{15 \, {\left (b x^{2} + a x\right )}^{\frac {3}{2}} a^{2}} - \frac {128 \, B b^{2}}{15 \, \sqrt {b x^{2} + a x} a^{4}} - \frac {32 \, A b^{2}}{21 \, {\left (b x^{2} + a x\right )}^{\frac {3}{2}} a^{3}} + \frac {256 \, A b^{3}}{21 \, \sqrt {b x^{2} + a x} a^{5}} - \frac {2 \, B}{5 \, {\left (b x^{2} + a x\right )}^{\frac {3}{2}} a x} + \frac {4 \, A b}{7 \, {\left (b x^{2} + a x\right )}^{\frac {3}{2}} a^{2} x} - \frac {2 \, A}{7 \, {\left (b x^{2} + a x\right )}^{\frac {3}{2}} a x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^(9/2)/(b*x+a)^(5/2),x, algorithm="maxima")

[Out]

32/15*B*b^2*x/((b*x^2 + a*x)^(3/2)*a^3) - 256/15*B*b^3*x/(sqrt(b*x^2 + a*x)*a^5) - 64/21*A*b^3*x/((b*x^2 + a*x
)^(3/2)*a^4) + 512/21*A*b^4*x/(sqrt(b*x^2 + a*x)*a^6) + 16/15*B*b/((b*x^2 + a*x)^(3/2)*a^2) - 128/15*B*b^2/(sq
rt(b*x^2 + a*x)*a^4) - 32/21*A*b^2/((b*x^2 + a*x)^(3/2)*a^3) + 256/21*A*b^3/(sqrt(b*x^2 + a*x)*a^5) - 2/5*B/((
b*x^2 + a*x)^(3/2)*a*x) + 4/7*A*b/((b*x^2 + a*x)^(3/2)*a^2*x) - 2/7*A/((b*x^2 + a*x)^(3/2)*a*x^2)

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mupad [B]  time = 1.05, size = 147, normalized size = 0.83 \begin {gather*} -\frac {\sqrt {a+b\,x}\,\left (\frac {2\,A}{7\,a\,b^2}-\frac {32\,x^3\,\left (10\,A\,b-7\,B\,a\right )}{35\,a^4}+\frac {16\,x^2\,\left (10\,A\,b-7\,B\,a\right )}{105\,a^3\,b}-\frac {x^5\,\left (2560\,A\,b^5-1792\,B\,a\,b^4\right )}{105\,a^6\,b^2}-\frac {128\,b\,x^4\,\left (10\,A\,b-7\,B\,a\right )}{35\,a^5}+\frac {x\,\left (42\,B\,a^5-60\,A\,a^4\,b\right )}{105\,a^6\,b^2}\right )}{x^{11/2}+\frac {2\,a\,x^{9/2}}{b}+\frac {a^2\,x^{7/2}}{b^2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*x)/(x^(9/2)*(a + b*x)^(5/2)),x)

[Out]

-((a + b*x)^(1/2)*((2*A)/(7*a*b^2) - (32*x^3*(10*A*b - 7*B*a))/(35*a^4) + (16*x^2*(10*A*b - 7*B*a))/(105*a^3*b
) - (x^5*(2560*A*b^5 - 1792*B*a*b^4))/(105*a^6*b^2) - (128*b*x^4*(10*A*b - 7*B*a))/(35*a^5) + (x*(42*B*a^5 - 6
0*A*a^4*b))/(105*a^6*b^2)))/(x^(11/2) + (2*a*x^(9/2))/b + (a^2*x^(7/2))/b^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x**(9/2)/(b*x+a)**(5/2),x)

[Out]

Timed out

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