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

Optimal. Leaf size=89 \[ -\frac {2 A (a+b x)^{5/2}}{5 a x^{5/2}}+2 b^{3/2} B \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )-\frac {2 B (a+b x)^{3/2}}{3 x^{3/2}}-\frac {2 b B \sqrt {a+b x}}{\sqrt {x}} \]

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Rubi [A]  time = 0.03, antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 6, 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)^{5/2}}{5 a x^{5/2}}+2 b^{3/2} B \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )-\frac {2 B (a+b x)^{3/2}}{3 x^{3/2}}-\frac {2 b B \sqrt {a+b x}}{\sqrt {x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*b*B*Sqrt[a + b*x])/Sqrt[x] - (2*B*(a + b*x)^(3/2))/(3*x^(3/2)) - (2*A*(a + b*x)^(5/2))/(5*a*x^(5/2)) + 2*b
^(3/2)*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)^{3/2} (A+B x)}{x^{7/2}} \, dx &=-\frac {2 A (a+b x)^{5/2}}{5 a x^{5/2}}+B \int \frac {(a+b x)^{3/2}}{x^{5/2}} \, dx\\ &=-\frac {2 B (a+b x)^{3/2}}{3 x^{3/2}}-\frac {2 A (a+b x)^{5/2}}{5 a x^{5/2}}+(b B) \int \frac {\sqrt {a+b x}}{x^{3/2}} \, dx\\ &=-\frac {2 b B \sqrt {a+b x}}{\sqrt {x}}-\frac {2 B (a+b x)^{3/2}}{3 x^{3/2}}-\frac {2 A (a+b x)^{5/2}}{5 a x^{5/2}}+\left (b^2 B\right ) \int \frac {1}{\sqrt {x} \sqrt {a+b x}} \, dx\\ &=-\frac {2 b B \sqrt {a+b x}}{\sqrt {x}}-\frac {2 B (a+b x)^{3/2}}{3 x^{3/2}}-\frac {2 A (a+b x)^{5/2}}{5 a x^{5/2}}+\left (2 b^2 B\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+b x^2}} \, dx,x,\sqrt {x}\right )\\ &=-\frac {2 b B \sqrt {a+b x}}{\sqrt {x}}-\frac {2 B (a+b x)^{3/2}}{3 x^{3/2}}-\frac {2 A (a+b x)^{5/2}}{5 a x^{5/2}}+\left (2 b^2 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 B \sqrt {a+b x}}{\sqrt {x}}-\frac {2 B (a+b x)^{3/2}}{3 x^{3/2}}-\frac {2 A (a+b x)^{5/2}}{5 a x^{5/2}}+2 b^{3/2} 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 = 0.87 \begin {gather*} \frac {2 \sqrt {a+b x} \left (-\frac {a^3 B \, _2F_1\left (-\frac {5}{2},-\frac {5}{2};-\frac {3}{2};-\frac {b x}{a}\right )}{\sqrt {\frac {b x}{a}+1}}-(a+b x)^2 (A b-a B)\right )}{5 a b x^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.25, size = 90, normalized size = 1.01 \begin {gather*} -\frac {2 \sqrt {a+b x} \left (3 a^2 A+5 a^2 B x+6 a A b x+20 a b B x^2+3 A b^2 x^2\right )}{15 a x^{5/2}}-2 b^{3/2} B \log \left (\sqrt {a+b x}-\sqrt {b} \sqrt {x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.17, size = 181, normalized size = 2.03 \begin {gather*} \left [\frac {15 \, B a b^{\frac {3}{2}} x^{3} \log \left (2 \, b x + 2 \, \sqrt {b x + a} \sqrt {b} \sqrt {x} + a\right ) - 2 \, {\left (3 \, A a^{2} + {\left (20 \, B a b + 3 \, A b^{2}\right )} x^{2} + {\left (5 \, B a^{2} + 6 \, A a b\right )} x\right )} \sqrt {b x + a} \sqrt {x}}{15 \, a x^{3}}, -\frac {2 \, {\left (15 \, B a \sqrt {-b} b x^{3} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-b}}{b \sqrt {x}}\right ) + {\left (3 \, A a^{2} + {\left (20 \, B a b + 3 \, A b^{2}\right )} x^{2} + {\left (5 \, B a^{2} + 6 \, A a b\right )} x\right )} \sqrt {b x + a} \sqrt {x}\right )}}{15 \, a x^{3}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/15*(15*B*a*b^(3/2)*x^3*log(2*b*x + 2*sqrt(b*x + a)*sqrt(b)*sqrt(x) + a) - 2*(3*A*a^2 + (20*B*a*b + 3*A*b^2)
*x^2 + (5*B*a^2 + 6*A*a*b)*x)*sqrt(b*x + a)*sqrt(x))/(a*x^3), -2/15*(15*B*a*sqrt(-b)*b*x^3*arctan(sqrt(b*x + a
)*sqrt(-b)/(b*sqrt(x))) + (3*A*a^2 + (20*B*a*b + 3*A*b^2)*x^2 + (5*B*a^2 + 6*A*a*b)*x)*sqrt(b*x + a)*sqrt(x))/
(a*x^3)]

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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)^(3/2)*(B*x+A)/x^(7/2),x, algorithm="giac")

[Out]

Timed out

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maple [B]  time = 0.02, size = 156, normalized size = 1.75 \begin {gather*} -\frac {\sqrt {b x +a}\, \left (-15 B a \,b^{2} x^{3} \ln \left (\frac {2 b x +a +2 \sqrt {\left (b x +a \right ) x}\, \sqrt {b}}{2 \sqrt {b}}\right )+6 \sqrt {\left (b x +a \right ) x}\, A \,b^{\frac {5}{2}} x^{2}+40 \sqrt {\left (b x +a \right ) x}\, B a \,b^{\frac {3}{2}} x^{2}+12 \sqrt {\left (b x +a \right ) x}\, A a \,b^{\frac {3}{2}} x +10 \sqrt {\left (b x +a \right ) x}\, B \,a^{2} \sqrt {b}\, x +6 \sqrt {\left (b x +a \right ) x}\, A \,a^{2} \sqrt {b}\right )}{15 \sqrt {\left (b x +a \right ) x}\, a \sqrt {b}\, x^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/15*(b*x+a)^(1/2)/x^(5/2)*(-15*B*ln(1/2*(2*b*x+a+2*((b*x+a)*x)^(1/2)*b^(1/2))/b^(1/2))*x^3*a*b^2+6*A*b^(5/2)
*((b*x+a)*x)^(1/2)*x^2+40*B*b^(3/2)*((b*x+a)*x)^(1/2)*x^2*a+12*A*x*a*b^(3/2)*((b*x+a)*x)^(1/2)+10*B*x*a^2*((b*
x+a)*x)^(1/2)*b^(1/2)+6*A*a^2*((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 = 0.94, size = 158, normalized size = 1.78 \begin {gather*} B b^{\frac {3}{2}} \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right ) - \frac {7 \, \sqrt {b x^{2} + a x} B b}{3 \, x} - \frac {2 \, \sqrt {b x^{2} + a x} A b^{2}}{5 \, a x} - \frac {\sqrt {b x^{2} + a x} B a}{3 \, x^{2}} + \frac {\sqrt {b x^{2} + a x} A b}{5 \, x^{2}} - \frac {{\left (b x^{2} + a x\right )}^{\frac {3}{2}} B}{3 \, x^{3}} + \frac {3 \, \sqrt {b x^{2} + a x} A a}{5 \, x^{3}} - \frac {{\left (b x^{2} + a x\right )}^{\frac {3}{2}} A}{x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

B*b^(3/2)*log(2*b*x + a + 2*sqrt(b*x^2 + a*x)*sqrt(b)) - 7/3*sqrt(b*x^2 + a*x)*B*b/x - 2/5*sqrt(b*x^2 + a*x)*A
*b^2/(a*x) - 1/3*sqrt(b*x^2 + a*x)*B*a/x^2 + 1/5*sqrt(b*x^2 + a*x)*A*b/x^2 - 1/3*(b*x^2 + a*x)^(3/2)*B/x^3 + 3
/5*sqrt(b*x^2 + a*x)*A*a/x^3 - (b*x^2 + a*x)^(3/2)*A/x^4

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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 )}^{3/2}}{x^{7/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 70.27, size = 141, normalized size = 1.58 \begin {gather*} A \left (- \frac {2 a \sqrt {b} \sqrt {\frac {a}{b x} + 1}}{5 x^{2}} - \frac {4 b^{\frac {3}{2}} \sqrt {\frac {a}{b x} + 1}}{5 x} - \frac {2 b^{\frac {5}{2}} \sqrt {\frac {a}{b x} + 1}}{5 a}\right ) + B \left (- \frac {2 a \sqrt {b} \sqrt {\frac {a}{b x} + 1}}{3 x} - \frac {8 b^{\frac {3}{2}} \sqrt {\frac {a}{b x} + 1}}{3} - b^{\frac {3}{2}} \log {\left (\frac {a}{b x} \right )} + 2 b^{\frac {3}{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)**(3/2)*(B*x+A)/x**(7/2),x)

[Out]

A*(-2*a*sqrt(b)*sqrt(a/(b*x) + 1)/(5*x**2) - 4*b**(3/2)*sqrt(a/(b*x) + 1)/(5*x) - 2*b**(5/2)*sqrt(a/(b*x) + 1)
/(5*a)) + B*(-2*a*sqrt(b)*sqrt(a/(b*x) + 1)/(3*x) - 8*b**(3/2)*sqrt(a/(b*x) + 1)/3 - b**(3/2)*log(a/(b*x)) + 2
*b**(3/2)*log(sqrt(a/(b*x) + 1) + 1))

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