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

Optimal. Leaf size=111 \[ -\frac {2 A (a+b x)^{7/2}}{7 a x^{7/2}}+2 b^{5/2} B \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )-\frac {2 b^2 B \sqrt {a+b x}}{\sqrt {x}}-\frac {2 B (a+b x)^{5/2}}{5 x^{5/2}}-\frac {2 b B (a+b x)^{3/2}}{3 x^{3/2}} \]

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Rubi [A]  time = 0.04, antiderivative size = 111, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {78, 47, 63, 217, 206} \begin {gather*} -\frac {2 A (a+b x)^{7/2}}{7 a x^{7/2}}-\frac {2 b^2 B \sqrt {a+b x}}{\sqrt {x}}+2 b^{5/2} B \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )-\frac {2 B (a+b x)^{5/2}}{5 x^{5/2}}-\frac {2 b B (a+b x)^{3/2}}{3 x^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*b^2*B*Sqrt[a + b*x])/Sqrt[x] - (2*b*B*(a + b*x)^(3/2))/(3*x^(3/2)) - (2*B*(a + b*x)^(5/2))/(5*x^(5/2)) - (
2*A*(a + b*x)^(7/2))/(7*a*x^(7/2)) + 2*b^(5/2)*B*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[a + b*x]]

Rule 47

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

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

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]))))

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rubi steps

\begin {align*} \int \frac {(a+b x)^{5/2} (A+B x)}{x^{9/2}} \, dx &=-\frac {2 A (a+b x)^{7/2}}{7 a x^{7/2}}+B \int \frac {(a+b x)^{5/2}}{x^{7/2}} \, dx\\ &=-\frac {2 B (a+b x)^{5/2}}{5 x^{5/2}}-\frac {2 A (a+b x)^{7/2}}{7 a x^{7/2}}+(b B) \int \frac {(a+b x)^{3/2}}{x^{5/2}} \, dx\\ &=-\frac {2 b B (a+b x)^{3/2}}{3 x^{3/2}}-\frac {2 B (a+b x)^{5/2}}{5 x^{5/2}}-\frac {2 A (a+b x)^{7/2}}{7 a x^{7/2}}+\left (b^2 B\right ) \int \frac {\sqrt {a+b x}}{x^{3/2}} \, dx\\ &=-\frac {2 b^2 B \sqrt {a+b x}}{\sqrt {x}}-\frac {2 b B (a+b x)^{3/2}}{3 x^{3/2}}-\frac {2 B (a+b x)^{5/2}}{5 x^{5/2}}-\frac {2 A (a+b x)^{7/2}}{7 a x^{7/2}}+\left (b^3 B\right ) \int \frac {1}{\sqrt {x} \sqrt {a+b x}} \, dx\\ &=-\frac {2 b^2 B \sqrt {a+b x}}{\sqrt {x}}-\frac {2 b B (a+b x)^{3/2}}{3 x^{3/2}}-\frac {2 B (a+b x)^{5/2}}{5 x^{5/2}}-\frac {2 A (a+b x)^{7/2}}{7 a x^{7/2}}+\left (2 b^3 B\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+b x^2}} \, dx,x,\sqrt {x}\right )\\ &=-\frac {2 b^2 B \sqrt {a+b x}}{\sqrt {x}}-\frac {2 b B (a+b x)^{3/2}}{3 x^{3/2}}-\frac {2 B (a+b x)^{5/2}}{5 x^{5/2}}-\frac {2 A (a+b x)^{7/2}}{7 a x^{7/2}}+\left (2 b^3 B\right ) \operatorname {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {\sqrt {x}}{\sqrt {a+b x}}\right )\\ &=-\frac {2 b^2 B \sqrt {a+b x}}{\sqrt {x}}-\frac {2 b B (a+b x)^{3/2}}{3 x^{3/2}}-\frac {2 B (a+b x)^{5/2}}{5 x^{5/2}}-\frac {2 A (a+b x)^{7/2}}{7 a x^{7/2}}+2 b^{5/2} B \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )\\ \end {align*}

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Mathematica [C]  time = 0.08, size = 77, normalized size = 0.69 \begin {gather*} \frac {2 \sqrt {a+b x} \left (-\frac {a^4 B \, _2F_1\left (-\frac {7}{2},-\frac {7}{2};-\frac {5}{2};-\frac {b x}{a}\right )}{\sqrt {\frac {b x}{a}+1}}-(a+b x)^3 (A b-a B)\right )}{7 a b x^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[a + b*x]*(-((A*b - a*B)*(a + b*x)^3) - (a^4*B*Hypergeometric2F1[-7/2, -7/2, -5/2, -((b*x)/a)])/Sqrt[1
+ (b*x)/a]))/(7*a*b*x^(7/2))

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IntegrateAlgebraic [A]  time = 0.36, size = 114, normalized size = 1.03 \begin {gather*} -\frac {2 \sqrt {a+b x} \left (15 a^3 A+21 a^3 B x+45 a^2 A b x+77 a^2 b B x^2+45 a A b^2 x^2+161 a b^2 B x^3+15 A b^3 x^3\right )}{105 a x^{7/2}}-2 b^{5/2} B \log \left (\sqrt {a+b x}-\sqrt {b} \sqrt {x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sqrt[a + b*x]*(15*a^3*A + 45*a^2*A*b*x + 21*a^3*B*x + 45*a*A*b^2*x^2 + 77*a^2*b*B*x^2 + 15*A*b^3*x^3 + 161
*a*b^2*B*x^3))/(105*a*x^(7/2)) - 2*b^(5/2)*B*Log[-(Sqrt[b]*Sqrt[x]) + Sqrt[a + b*x]]

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fricas [A]  time = 0.94, size = 231, normalized size = 2.08 \begin {gather*} \left [\frac {105 \, B a b^{\frac {5}{2}} x^{4} \log \left (2 \, b x + 2 \, \sqrt {b x + a} \sqrt {b} \sqrt {x} + a\right ) - 2 \, {\left (15 \, A a^{3} + {\left (161 \, B a b^{2} + 15 \, A b^{3}\right )} x^{3} + {\left (77 \, B a^{2} b + 45 \, A a b^{2}\right )} x^{2} + 3 \, {\left (7 \, B a^{3} + 15 \, A a^{2} b\right )} x\right )} \sqrt {b x + a} \sqrt {x}}{105 \, a x^{4}}, -\frac {2 \, {\left (105 \, B a \sqrt {-b} b^{2} x^{4} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-b}}{b \sqrt {x}}\right ) + {\left (15 \, A a^{3} + {\left (161 \, B a b^{2} + 15 \, A b^{3}\right )} x^{3} + {\left (77 \, B a^{2} b + 45 \, A a b^{2}\right )} x^{2} + 3 \, {\left (7 \, B a^{3} + 15 \, A a^{2} b\right )} x\right )} \sqrt {b x + a} \sqrt {x}\right )}}{105 \, a x^{4}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/105*(105*B*a*b^(5/2)*x^4*log(2*b*x + 2*sqrt(b*x + a)*sqrt(b)*sqrt(x) + a) - 2*(15*A*a^3 + (161*B*a*b^2 + 15
*A*b^3)*x^3 + (77*B*a^2*b + 45*A*a*b^2)*x^2 + 3*(7*B*a^3 + 15*A*a^2*b)*x)*sqrt(b*x + a)*sqrt(x))/(a*x^4), -2/1
05*(105*B*a*sqrt(-b)*b^2*x^4*arctan(sqrt(b*x + a)*sqrt(-b)/(b*sqrt(x))) + (15*A*a^3 + (161*B*a*b^2 + 15*A*b^3)
*x^3 + (77*B*a^2*b + 45*A*a*b^2)*x^2 + 3*(7*B*a^3 + 15*A*a^2*b)*x)*sqrt(b*x + a)*sqrt(x))/(a*x^4)]

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

[Out]

Timed out

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maple [B]  time = 0.02, size = 198, normalized size = 1.78 \begin {gather*} -\frac {\sqrt {b x +a}\, \left (-105 B a \,b^{3} x^{4} \ln \left (\frac {2 b x +a +2 \sqrt {\left (b x +a \right ) x}\, \sqrt {b}}{2 \sqrt {b}}\right )+30 \sqrt {\left (b x +a \right ) x}\, A \,b^{\frac {7}{2}} x^{3}+322 \sqrt {\left (b x +a \right ) x}\, B a \,b^{\frac {5}{2}} x^{3}+90 \sqrt {\left (b x +a \right ) x}\, A a \,b^{\frac {5}{2}} x^{2}+154 \sqrt {\left (b x +a \right ) x}\, B \,a^{2} b^{\frac {3}{2}} x^{2}+90 \sqrt {\left (b x +a \right ) x}\, A \,a^{2} b^{\frac {3}{2}} x +42 \sqrt {\left (b x +a \right ) x}\, B \,a^{3} \sqrt {b}\, x +30 \sqrt {\left (b x +a \right ) x}\, A \,a^{3} \sqrt {b}\right )}{105 \sqrt {\left (b x +a \right ) x}\, a \sqrt {b}\, x^{\frac {7}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/105*(b*x+a)^(1/2)/x^(7/2)*(-105*B*ln(1/2*(2*b*x+a+2*((b*x+a)*x)^(1/2)*b^(1/2))/b^(1/2))*x^4*a*b^3+30*A*((b*
x+a)*x)^(1/2)*b^(7/2)*x^3+322*B*((b*x+a)*x)^(1/2)*b^(5/2)*x^3*a+90*A*((b*x+a)*x)^(1/2)*b^(5/2)*x^2*a+154*B*((b
*x+a)*x)^(1/2)*b^(3/2)*x^2*a^2+90*A*x*a^2*b^(3/2)*((b*x+a)*x)^(1/2)+42*B*x*a^3*((b*x+a)*x)^(1/2)*b^(1/2)+30*A*
a^3*((b*x+a)*x)^(1/2)*b^(1/2))/a/((b*x+a)*x)^(1/2)/b^(1/2)

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maxima [B]  time = 0.93, size = 258, normalized size = 2.32 \begin {gather*} B b^{\frac {5}{2}} \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right ) - \frac {38 \, \sqrt {b x^{2} + a x} B b^{2}}{15 \, x} - \frac {2 \, \sqrt {b x^{2} + a x} A b^{3}}{7 \, a x} - \frac {7 \, \sqrt {b x^{2} + a x} B a b}{30 \, x^{2}} + \frac {\sqrt {b x^{2} + a x} A b^{2}}{7 \, x^{2}} + \frac {3 \, \sqrt {b x^{2} + a x} B a^{2}}{10 \, x^{3}} - \frac {{\left (b x^{2} + a x\right )}^{\frac {3}{2}} B b}{3 \, x^{3}} - \frac {3 \, \sqrt {b x^{2} + a x} A a b}{28 \, x^{3}} - \frac {{\left (b x^{2} + a x\right )}^{\frac {3}{2}} B a}{2 \, x^{4}} - \frac {15 \, \sqrt {b x^{2} + a x} A a^{2}}{28 \, x^{4}} - \frac {{\left (b x^{2} + a x\right )}^{\frac {5}{2}} B}{5 \, x^{5}} + \frac {5 \, {\left (b x^{2} + a x\right )}^{\frac {3}{2}} A a}{4 \, x^{5}} - \frac {{\left (b x^{2} + a x\right )}^{\frac {5}{2}} A}{x^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

B*b^(5/2)*log(2*b*x + a + 2*sqrt(b*x^2 + a*x)*sqrt(b)) - 38/15*sqrt(b*x^2 + a*x)*B*b^2/x - 2/7*sqrt(b*x^2 + a*
x)*A*b^3/(a*x) - 7/30*sqrt(b*x^2 + a*x)*B*a*b/x^2 + 1/7*sqrt(b*x^2 + a*x)*A*b^2/x^2 + 3/10*sqrt(b*x^2 + a*x)*B
*a^2/x^3 - 1/3*(b*x^2 + a*x)^(3/2)*B*b/x^3 - 3/28*sqrt(b*x^2 + a*x)*A*a*b/x^3 - 1/2*(b*x^2 + a*x)^(3/2)*B*a/x^
4 - 15/28*sqrt(b*x^2 + a*x)*A*a^2/x^4 - 1/5*(b*x^2 + a*x)^(5/2)*B/x^5 + 5/4*(b*x^2 + a*x)^(3/2)*A*a/x^5 - (b*x
^2 + a*x)^(5/2)*A/x^6

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\left (A+B\,x\right )\,{\left (a+b\,x\right )}^{5/2}}{x^{9/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 162.69, size = 192, normalized size = 1.73 \begin {gather*} A \left (- \frac {2 a^{2} \sqrt {b} \sqrt {\frac {a}{b x} + 1}}{7 x^{3}} - \frac {6 a b^{\frac {3}{2}} \sqrt {\frac {a}{b x} + 1}}{7 x^{2}} - \frac {6 b^{\frac {5}{2}} \sqrt {\frac {a}{b x} + 1}}{7 x} - \frac {2 b^{\frac {7}{2}} \sqrt {\frac {a}{b x} + 1}}{7 a}\right ) + B \left (- \frac {2 a^{2} \sqrt {b} \sqrt {\frac {a}{b x} + 1}}{5 x^{2}} - \frac {22 a b^{\frac {3}{2}} \sqrt {\frac {a}{b x} + 1}}{15 x} - \frac {46 b^{\frac {5}{2}} \sqrt {\frac {a}{b x} + 1}}{15} - b^{\frac {5}{2}} \log {\left (\frac {a}{b x} \right )} + 2 b^{\frac {5}{2}} \log {\left (\sqrt {\frac {a}{b x} + 1} + 1 \right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

A*(-2*a**2*sqrt(b)*sqrt(a/(b*x) + 1)/(7*x**3) - 6*a*b**(3/2)*sqrt(a/(b*x) + 1)/(7*x**2) - 6*b**(5/2)*sqrt(a/(b
*x) + 1)/(7*x) - 2*b**(7/2)*sqrt(a/(b*x) + 1)/(7*a)) + B*(-2*a**2*sqrt(b)*sqrt(a/(b*x) + 1)/(5*x**2) - 22*a*b*
*(3/2)*sqrt(a/(b*x) + 1)/(15*x) - 46*b**(5/2)*sqrt(a/(b*x) + 1)/15 - b**(5/2)*log(a/(b*x)) + 2*b**(5/2)*log(sq
rt(a/(b*x) + 1) + 1))

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