3.3.32 \(\int \frac {A+B x}{x^{7/2} \sqrt {b x+c x^2}} \, dx\)

Optimal. Leaf size=142 \[ -\frac {c^2 (6 b B-5 A c) \tanh ^{-1}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right )}{8 b^{7/2}}+\frac {c \sqrt {b x+c x^2} (6 b B-5 A c)}{8 b^3 x^{3/2}}-\frac {\sqrt {b x+c x^2} (6 b B-5 A c)}{12 b^2 x^{5/2}}-\frac {A \sqrt {b x+c x^2}}{3 b x^{7/2}} \]

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Rubi [A]  time = 0.12, antiderivative size = 142, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {792, 672, 660, 207} \begin {gather*} -\frac {c^2 (6 b B-5 A c) \tanh ^{-1}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right )}{8 b^{7/2}}+\frac {c \sqrt {b x+c x^2} (6 b B-5 A c)}{8 b^3 x^{3/2}}-\frac {\sqrt {b x+c x^2} (6 b B-5 A c)}{12 b^2 x^{5/2}}-\frac {A \sqrt {b x+c x^2}}{3 b x^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(A*Sqrt[b*x + c*x^2])/(3*b*x^(7/2)) - ((6*b*B - 5*A*c)*Sqrt[b*x + c*x^2])/(12*b^2*x^(5/2)) + (c*(6*b*B - 5*A*
c)*Sqrt[b*x + c*x^2])/(8*b^3*x^(3/2)) - (c^2*(6*b*B - 5*A*c)*ArcTanh[Sqrt[b*x + c*x^2]/(Sqrt[b]*Sqrt[x])])/(8*
b^(7/2))

Rule 207

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

Rule 660

Int[1/(Sqrt[(d_.) + (e_.)*(x_)]*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[2*e, Subst[Int[1/(
2*c*d - b*e + e^2*x^2), x], x, Sqrt[a + b*x + c*x^2]/Sqrt[d + e*x]], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^
2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0]

Rule 672

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[(e*(d + e*x)^m*(a +
b*x + c*x^2)^(p + 1))/((m + p + 1)*(2*c*d - b*e)), x] + Dist[(c*(m + 2*p + 2))/((m + p + 1)*(2*c*d - b*e)), I
nt[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ
[c*d^2 - b*d*e + a*e^2, 0] && LtQ[m, 0] && NeQ[m + p + 1, 0] && IntegerQ[2*p]

Rule 792

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp
[((d*g - e*f)*(d + e*x)^m*(a + b*x + c*x^2)^(p + 1))/((2*c*d - b*e)*(m + p + 1)), x] + Dist[(m*(g*(c*d - b*e)
+ c*e*f) + e*(p + 1)*(2*c*f - b*g))/(e*(2*c*d - b*e)*(m + p + 1)), Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p,
x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && ((L
tQ[m, -1] &&  !IGtQ[m + p + 1, 0]) || (LtQ[m, 0] && LtQ[p, -1]) || EqQ[m + 2*p + 2, 0]) && NeQ[m + p + 1, 0]

Rubi steps

\begin {align*} \int \frac {A+B x}{x^{7/2} \sqrt {b x+c x^2}} \, dx &=-\frac {A \sqrt {b x+c x^2}}{3 b x^{7/2}}+\frac {\left (-\frac {7}{2} (-b B+A c)+\frac {1}{2} (-b B+2 A c)\right ) \int \frac {1}{x^{5/2} \sqrt {b x+c x^2}} \, dx}{3 b}\\ &=-\frac {A \sqrt {b x+c x^2}}{3 b x^{7/2}}-\frac {(6 b B-5 A c) \sqrt {b x+c x^2}}{12 b^2 x^{5/2}}-\frac {(c (6 b B-5 A c)) \int \frac {1}{x^{3/2} \sqrt {b x+c x^2}} \, dx}{8 b^2}\\ &=-\frac {A \sqrt {b x+c x^2}}{3 b x^{7/2}}-\frac {(6 b B-5 A c) \sqrt {b x+c x^2}}{12 b^2 x^{5/2}}+\frac {c (6 b B-5 A c) \sqrt {b x+c x^2}}{8 b^3 x^{3/2}}+\frac {\left (c^2 (6 b B-5 A c)\right ) \int \frac {1}{\sqrt {x} \sqrt {b x+c x^2}} \, dx}{16 b^3}\\ &=-\frac {A \sqrt {b x+c x^2}}{3 b x^{7/2}}-\frac {(6 b B-5 A c) \sqrt {b x+c x^2}}{12 b^2 x^{5/2}}+\frac {c (6 b B-5 A c) \sqrt {b x+c x^2}}{8 b^3 x^{3/2}}+\frac {\left (c^2 (6 b B-5 A c)\right ) \operatorname {Subst}\left (\int \frac {1}{-b+x^2} \, dx,x,\frac {\sqrt {b x+c x^2}}{\sqrt {x}}\right )}{8 b^3}\\ &=-\frac {A \sqrt {b x+c x^2}}{3 b x^{7/2}}-\frac {(6 b B-5 A c) \sqrt {b x+c x^2}}{12 b^2 x^{5/2}}+\frac {c (6 b B-5 A c) \sqrt {b x+c x^2}}{8 b^3 x^{3/2}}-\frac {c^2 (6 b B-5 A c) \tanh ^{-1}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right )}{8 b^{7/2}}\\ \end {align*}

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Mathematica [C]  time = 0.03, size = 61, normalized size = 0.43 \begin {gather*} -\frac {\sqrt {x (b+c x)} \left (A b^3+c^2 x^3 (6 b B-5 A c) \, _2F_1\left (\frac {1}{2},3;\frac {3}{2};\frac {c x}{b}+1\right )\right )}{3 b^4 x^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/3*(Sqrt[x*(b + c*x)]*(A*b^3 + c^2*(6*b*B - 5*A*c)*x^3*Hypergeometric2F1[1/2, 3, 3/2, 1 + (c*x)/b]))/(b^4*x^
(7/2))

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IntegrateAlgebraic [A]  time = 0.35, size = 111, normalized size = 0.78 \begin {gather*} \frac {\left (5 A c^3-6 b B c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {b x+c x^2}}\right )}{8 b^{7/2}}+\frac {\sqrt {b x+c x^2} \left (-8 A b^2+10 A b c x-15 A c^2 x^2-12 b^2 B x+18 b B c x^2\right )}{24 b^3 x^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[b*x + c*x^2]*(-8*A*b^2 - 12*b^2*B*x + 10*A*b*c*x + 18*b*B*c*x^2 - 15*A*c^2*x^2))/(24*b^3*x^(7/2)) + ((-6
*b*B*c^2 + 5*A*c^3)*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[b*x + c*x^2]])/(8*b^(7/2))

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fricas [A]  time = 0.41, size = 241, normalized size = 1.70 \begin {gather*} \left [-\frac {3 \, {\left (6 \, B b c^{2} - 5 \, A c^{3}\right )} \sqrt {b} x^{4} \log \left (-\frac {c x^{2} + 2 \, b x + 2 \, \sqrt {c x^{2} + b x} \sqrt {b} \sqrt {x}}{x^{2}}\right ) + 2 \, {\left (8 \, A b^{3} - 3 \, {\left (6 \, B b^{2} c - 5 \, A b c^{2}\right )} x^{2} + 2 \, {\left (6 \, B b^{3} - 5 \, A b^{2} c\right )} x\right )} \sqrt {c x^{2} + b x} \sqrt {x}}{48 \, b^{4} x^{4}}, \frac {3 \, {\left (6 \, B b c^{2} - 5 \, A c^{3}\right )} \sqrt {-b} x^{4} \arctan \left (\frac {\sqrt {-b} \sqrt {x}}{\sqrt {c x^{2} + b x}}\right ) - {\left (8 \, A b^{3} - 3 \, {\left (6 \, B b^{2} c - 5 \, A b c^{2}\right )} x^{2} + 2 \, {\left (6 \, B b^{3} - 5 \, A b^{2} c\right )} x\right )} \sqrt {c x^{2} + b x} \sqrt {x}}{24 \, b^{4} x^{4}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^(7/2)/(c*x^2+b*x)^(1/2),x, algorithm="fricas")

[Out]

[-1/48*(3*(6*B*b*c^2 - 5*A*c^3)*sqrt(b)*x^4*log(-(c*x^2 + 2*b*x + 2*sqrt(c*x^2 + b*x)*sqrt(b)*sqrt(x))/x^2) +
2*(8*A*b^3 - 3*(6*B*b^2*c - 5*A*b*c^2)*x^2 + 2*(6*B*b^3 - 5*A*b^2*c)*x)*sqrt(c*x^2 + b*x)*sqrt(x))/(b^4*x^4),
1/24*(3*(6*B*b*c^2 - 5*A*c^3)*sqrt(-b)*x^4*arctan(sqrt(-b)*sqrt(x)/sqrt(c*x^2 + b*x)) - (8*A*b^3 - 3*(6*B*b^2*
c - 5*A*b*c^2)*x^2 + 2*(6*B*b^3 - 5*A*b^2*c)*x)*sqrt(c*x^2 + b*x)*sqrt(x))/(b^4*x^4)]

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giac [A]  time = 0.28, size = 144, normalized size = 1.01 \begin {gather*} \frac {\frac {3 \, {\left (6 \, B b c^{3} - 5 \, A c^{4}\right )} \arctan \left (\frac {\sqrt {c x + b}}{\sqrt {-b}}\right )}{\sqrt {-b} b^{3}} + \frac {18 \, {\left (c x + b\right )}^{\frac {5}{2}} B b c^{3} - 48 \, {\left (c x + b\right )}^{\frac {3}{2}} B b^{2} c^{3} + 30 \, \sqrt {c x + b} B b^{3} c^{3} - 15 \, {\left (c x + b\right )}^{\frac {5}{2}} A c^{4} + 40 \, {\left (c x + b\right )}^{\frac {3}{2}} A b c^{4} - 33 \, \sqrt {c x + b} A b^{2} c^{4}}{b^{3} c^{3} x^{3}}}{24 \, c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^(7/2)/(c*x^2+b*x)^(1/2),x, algorithm="giac")

[Out]

1/24*(3*(6*B*b*c^3 - 5*A*c^4)*arctan(sqrt(c*x + b)/sqrt(-b))/(sqrt(-b)*b^3) + (18*(c*x + b)^(5/2)*B*b*c^3 - 48
*(c*x + b)^(3/2)*B*b^2*c^3 + 30*sqrt(c*x + b)*B*b^3*c^3 - 15*(c*x + b)^(5/2)*A*c^4 + 40*(c*x + b)^(3/2)*A*b*c^
4 - 33*sqrt(c*x + b)*A*b^2*c^4)/(b^3*c^3*x^3))/c

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maple [A]  time = 0.06, size = 147, normalized size = 1.04 \begin {gather*} \frac {\sqrt {\left (c x +b \right ) x}\, \left (15 A \,c^{3} x^{3} \arctanh \left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right )-18 B b \,c^{2} x^{3} \arctanh \left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right )-15 \sqrt {c x +b}\, A \sqrt {b}\, c^{2} x^{2}+18 \sqrt {c x +b}\, B \,b^{\frac {3}{2}} c \,x^{2}+10 \sqrt {c x +b}\, A \,b^{\frac {3}{2}} c x -12 \sqrt {c x +b}\, B \,b^{\frac {5}{2}} x -8 \sqrt {c x +b}\, A \,b^{\frac {5}{2}}\right )}{24 \sqrt {c x +b}\, b^{\frac {7}{2}} x^{\frac {7}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)/x^(7/2)/(c*x^2+b*x)^(1/2),x)

[Out]

1/24*((c*x+b)*x)^(1/2)/b^(7/2)*(15*A*arctanh((c*x+b)^(1/2)/b^(1/2))*x^3*c^3-18*B*arctanh((c*x+b)^(1/2)/b^(1/2)
)*x^3*b*c^2-15*(c*x+b)^(1/2)*A*b^(1/2)*c^2*x^2+18*(c*x+b)^(1/2)*B*b^(3/2)*c*x^2+10*(c*x+b)^(1/2)*A*b^(3/2)*c*x
-12*(c*x+b)^(1/2)*B*b^(5/2)*x-8*(c*x+b)^(1/2)*A*b^(5/2))/x^(7/2)/(c*x+b)^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {B x + A}{\sqrt {c x^{2} + b x} x^{\frac {7}{2}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^(7/2)/(c*x^2+b*x)^(1/2),x, algorithm="maxima")

[Out]

integrate((B*x + A)/(sqrt(c*x^2 + b*x)*x^(7/2)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {A+B\,x}{x^{7/2}\,\sqrt {c\,x^2+b\,x}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*x)/(x^(7/2)*(b*x + c*x^2)^(1/2)),x)

[Out]

int((A + B*x)/(x^(7/2)*(b*x + c*x^2)^(1/2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {A + B x}{x^{\frac {7}{2}} \sqrt {x \left (b + c x\right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x**(7/2)/(c*x**2+b*x)**(1/2),x)

[Out]

Integral((A + B*x)/(x**(7/2)*sqrt(x*(b + c*x))), x)

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