3.3.40 \(\int \frac {A+B x}{x^{5/2} (b x+c x^2)^{3/2}} \, dx\)

Optimal. Leaf size=179 \[ -\frac {5 c^2 (6 b B-7 A c) \tanh ^{-1}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right )}{8 b^{9/2}}+\frac {5 c^2 \sqrt {x} (6 b B-7 A c)}{8 b^4 \sqrt {b x+c x^2}}+\frac {5 c (6 b B-7 A c)}{24 b^3 \sqrt {x} \sqrt {b x+c x^2}}-\frac {6 b B-7 A c}{12 b^2 x^{3/2} \sqrt {b x+c x^2}}-\frac {A}{3 b x^{5/2} \sqrt {b x+c x^2}} \]

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Rubi [A]  time = 0.15, antiderivative size = 179, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {792, 672, 666, 660, 207} \begin {gather*} \frac {5 c^2 \sqrt {x} (6 b B-7 A c)}{8 b^4 \sqrt {b x+c x^2}}-\frac {5 c^2 (6 b B-7 A c) \tanh ^{-1}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right )}{8 b^{9/2}}+\frac {5 c (6 b B-7 A c)}{24 b^3 \sqrt {x} \sqrt {b x+c x^2}}-\frac {6 b B-7 A c}{12 b^2 x^{3/2} \sqrt {b x+c x^2}}-\frac {A}{3 b x^{5/2} \sqrt {b x+c x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(A + B*x)/(x^(5/2)*(b*x + c*x^2)^(3/2)),x]

[Out]

-A/(3*b*x^(5/2)*Sqrt[b*x + c*x^2]) - (6*b*B - 7*A*c)/(12*b^2*x^(3/2)*Sqrt[b*x + c*x^2]) + (5*c*(6*b*B - 7*A*c)
)/(24*b^3*Sqrt[x]*Sqrt[b*x + c*x^2]) + (5*c^2*(6*b*B - 7*A*c)*Sqrt[x])/(8*b^4*Sqrt[b*x + c*x^2]) - (5*c^2*(6*b
*B - 7*A*c)*ArcTanh[Sqrt[b*x + c*x^2]/(Sqrt[b]*Sqrt[x])])/(8*b^(9/2))

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 666

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

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^{5/2} \left (b x+c x^2\right )^{3/2}} \, dx &=-\frac {A}{3 b x^{5/2} \sqrt {b x+c x^2}}+\frac {\left (\frac {1}{2} (b B-2 A c)-\frac {5}{2} (-b B+A c)\right ) \int \frac {1}{x^{3/2} \left (b x+c x^2\right )^{3/2}} \, dx}{3 b}\\ &=-\frac {A}{3 b x^{5/2} \sqrt {b x+c x^2}}-\frac {6 b B-7 A c}{12 b^2 x^{3/2} \sqrt {b x+c x^2}}-\frac {(5 c (6 b B-7 A c)) \int \frac {1}{\sqrt {x} \left (b x+c x^2\right )^{3/2}} \, dx}{24 b^2}\\ &=-\frac {A}{3 b x^{5/2} \sqrt {b x+c x^2}}-\frac {6 b B-7 A c}{12 b^2 x^{3/2} \sqrt {b x+c x^2}}+\frac {5 c (6 b B-7 A c)}{24 b^3 \sqrt {x} \sqrt {b x+c x^2}}+\frac {\left (5 c^2 (6 b B-7 A c)\right ) \int \frac {\sqrt {x}}{\left (b x+c x^2\right )^{3/2}} \, dx}{16 b^3}\\ &=-\frac {A}{3 b x^{5/2} \sqrt {b x+c x^2}}-\frac {6 b B-7 A c}{12 b^2 x^{3/2} \sqrt {b x+c x^2}}+\frac {5 c (6 b B-7 A c)}{24 b^3 \sqrt {x} \sqrt {b x+c x^2}}+\frac {5 c^2 (6 b B-7 A c) \sqrt {x}}{8 b^4 \sqrt {b x+c x^2}}+\frac {\left (5 c^2 (6 b B-7 A c)\right ) \int \frac {1}{\sqrt {x} \sqrt {b x+c x^2}} \, dx}{16 b^4}\\ &=-\frac {A}{3 b x^{5/2} \sqrt {b x+c x^2}}-\frac {6 b B-7 A c}{12 b^2 x^{3/2} \sqrt {b x+c x^2}}+\frac {5 c (6 b B-7 A c)}{24 b^3 \sqrt {x} \sqrt {b x+c x^2}}+\frac {5 c^2 (6 b B-7 A c) \sqrt {x}}{8 b^4 \sqrt {b x+c x^2}}+\frac {\left (5 c^2 (6 b B-7 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^4}\\ &=-\frac {A}{3 b x^{5/2} \sqrt {b x+c x^2}}-\frac {6 b B-7 A c}{12 b^2 x^{3/2} \sqrt {b x+c x^2}}+\frac {5 c (6 b B-7 A c)}{24 b^3 \sqrt {x} \sqrt {b x+c x^2}}+\frac {5 c^2 (6 b B-7 A c) \sqrt {x}}{8 b^4 \sqrt {b x+c x^2}}-\frac {5 c^2 (6 b B-7 A c) \tanh ^{-1}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right )}{8 b^{9/2}}\\ \end {align*}

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Mathematica [C]  time = 0.02, size = 62, normalized size = 0.35 \begin {gather*} \frac {c^2 x^3 (6 b B-7 A c) \, _2F_1\left (-\frac {1}{2},3;\frac {1}{2};\frac {c x}{b}+1\right )-A b^3}{3 b^4 x^{5/2} \sqrt {x (b+c x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x)/(x^(5/2)*(b*x + c*x^2)^(3/2)),x]

[Out]

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

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IntegrateAlgebraic [A]  time = 1.93, size = 142, normalized size = 0.79 \begin {gather*} \frac {\sqrt {b x+c x^2} \left (-8 A b^3+14 A b^2 c x-35 A b c^2 x^2-105 A c^3 x^3-12 b^3 B x+30 b^2 B c x^2+90 b B c^2 x^3\right )}{24 b^4 x^{7/2} (b+c x)}-\frac {5 \left (6 b B c^2-7 A c^3\right ) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {b x+c x^2}}\right )}{8 b^{9/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(A + B*x)/(x^(5/2)*(b*x + c*x^2)^(3/2)),x]

[Out]

(Sqrt[b*x + c*x^2]*(-8*A*b^3 - 12*b^3*B*x + 14*A*b^2*c*x + 30*b^2*B*c*x^2 - 35*A*b*c^2*x^2 + 90*b*B*c^2*x^3 -
105*A*c^3*x^3))/(24*b^4*x^(7/2)*(b + c*x)) - (5*(6*b*B*c^2 - 7*A*c^3)*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[b*x + c*x
^2]])/(8*b^(9/2))

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fricas [A]  time = 0.42, size = 359, normalized size = 2.01 \begin {gather*} \left [-\frac {15 \, {\left ({\left (6 \, B b c^{3} - 7 \, A c^{4}\right )} x^{5} + {\left (6 \, B b^{2} c^{2} - 7 \, A b c^{3}\right )} x^{4}\right )} \sqrt {b} \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^{4} - 15 \, {\left (6 \, B b^{2} c^{2} - 7 \, A b c^{3}\right )} x^{3} - 5 \, {\left (6 \, B b^{3} c - 7 \, A b^{2} c^{2}\right )} x^{2} + 2 \, {\left (6 \, B b^{4} - 7 \, A b^{3} c\right )} x\right )} \sqrt {c x^{2} + b x} \sqrt {x}}{48 \, {\left (b^{5} c x^{5} + b^{6} x^{4}\right )}}, \frac {15 \, {\left ({\left (6 \, B b c^{3} - 7 \, A c^{4}\right )} x^{5} + {\left (6 \, B b^{2} c^{2} - 7 \, A b c^{3}\right )} x^{4}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} \sqrt {x}}{\sqrt {c x^{2} + b x}}\right ) - {\left (8 \, A b^{4} - 15 \, {\left (6 \, B b^{2} c^{2} - 7 \, A b c^{3}\right )} x^{3} - 5 \, {\left (6 \, B b^{3} c - 7 \, A b^{2} c^{2}\right )} x^{2} + 2 \, {\left (6 \, B b^{4} - 7 \, A b^{3} c\right )} x\right )} \sqrt {c x^{2} + b x} \sqrt {x}}{24 \, {\left (b^{5} c x^{5} + b^{6} x^{4}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^(5/2)/(c*x^2+b*x)^(3/2),x, algorithm="fricas")

[Out]

[-1/48*(15*((6*B*b*c^3 - 7*A*c^4)*x^5 + (6*B*b^2*c^2 - 7*A*b*c^3)*x^4)*sqrt(b)*log(-(c*x^2 + 2*b*x + 2*sqrt(c*
x^2 + b*x)*sqrt(b)*sqrt(x))/x^2) + 2*(8*A*b^4 - 15*(6*B*b^2*c^2 - 7*A*b*c^3)*x^3 - 5*(6*B*b^3*c - 7*A*b^2*c^2)
*x^2 + 2*(6*B*b^4 - 7*A*b^3*c)*x)*sqrt(c*x^2 + b*x)*sqrt(x))/(b^5*c*x^5 + b^6*x^4), 1/24*(15*((6*B*b*c^3 - 7*A
*c^4)*x^5 + (6*B*b^2*c^2 - 7*A*b*c^3)*x^4)*sqrt(-b)*arctan(sqrt(-b)*sqrt(x)/sqrt(c*x^2 + b*x)) - (8*A*b^4 - 15
*(6*B*b^2*c^2 - 7*A*b*c^3)*x^3 - 5*(6*B*b^3*c - 7*A*b^2*c^2)*x^2 + 2*(6*B*b^4 - 7*A*b^3*c)*x)*sqrt(c*x^2 + b*x
)*sqrt(x))/(b^5*c*x^5 + b^6*x^4)]

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giac [A]  time = 0.33, size = 165, normalized size = 0.92 \begin {gather*} \frac {5 \, {\left (6 \, B b c^{2} - 7 \, A c^{3}\right )} \arctan \left (\frac {\sqrt {c x + b}}{\sqrt {-b}}\right )}{8 \, \sqrt {-b} b^{4}} + \frac {2 \, {\left (B b c^{2} - A c^{3}\right )}}{\sqrt {c x + b} b^{4}} + \frac {42 \, {\left (c x + b\right )}^{\frac {5}{2}} B b c^{2} - 96 \, {\left (c x + b\right )}^{\frac {3}{2}} B b^{2} c^{2} + 54 \, \sqrt {c x + b} B b^{3} c^{2} - 57 \, {\left (c x + b\right )}^{\frac {5}{2}} A c^{3} + 136 \, {\left (c x + b\right )}^{\frac {3}{2}} A b c^{3} - 87 \, \sqrt {c x + b} A b^{2} c^{3}}{24 \, b^{4} c^{3} x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^(5/2)/(c*x^2+b*x)^(3/2),x, algorithm="giac")

[Out]

5/8*(6*B*b*c^2 - 7*A*c^3)*arctan(sqrt(c*x + b)/sqrt(-b))/(sqrt(-b)*b^4) + 2*(B*b*c^2 - A*c^3)/(sqrt(c*x + b)*b
^4) + 1/24*(42*(c*x + b)^(5/2)*B*b*c^2 - 96*(c*x + b)^(3/2)*B*b^2*c^2 + 54*sqrt(c*x + b)*B*b^3*c^2 - 57*(c*x +
b)^(5/2)*A*c^3 + 136*(c*x + b)^(3/2)*A*b*c^3 - 87*sqrt(c*x + b)*A*b^2*c^3)/(b^4*c^3*x^3)

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maple [A]  time = 0.11, size = 150, normalized size = 0.84 \begin {gather*} \frac {\sqrt {\left (c x +b \right ) x}\, \left (105 \sqrt {c x +b}\, A \,c^{3} x^{3} \arctanh \left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right )-90 \sqrt {c x +b}\, B b \,c^{2} x^{3} \arctanh \left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right )-105 A \sqrt {b}\, c^{3} x^{3}+90 B \,b^{\frac {3}{2}} c^{2} x^{3}-35 A \,b^{\frac {3}{2}} c^{2} x^{2}+30 B \,b^{\frac {5}{2}} c \,x^{2}+14 A \,b^{\frac {5}{2}} c x -12 B \,b^{\frac {7}{2}} x -8 A \,b^{\frac {7}{2}}\right )}{24 \left (c x +b \right ) b^{\frac {9}{2}} x^{\frac {7}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)/x^(5/2)/(c*x^2+b*x)^(3/2),x)

[Out]

1/24/x^(7/2)*((c*x+b)*x)^(1/2)*(105*A*(c*x+b)^(1/2)*arctanh((c*x+b)^(1/2)/b^(1/2))*x^3*c^3-90*B*(c*x+b)^(1/2)*
arctanh((c*x+b)^(1/2)/b^(1/2))*x^3*b*c^2-12*B*b^(7/2)*x+30*B*b^(5/2)*x^2*c+90*B*b^(3/2)*x^3*c^2-8*A*b^(7/2)+14
*A*b^(5/2)*x*c-35*A*b^(3/2)*x^2*c^2-105*A*b^(1/2)*x^3*c^3)/(c*x+b)/b^(9/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {B x + A}{{\left (c x^{2} + b x\right )}^{\frac {3}{2}} x^{\frac {5}{2}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^(5/2)/(c*x^2+b*x)^(3/2),x, algorithm="maxima")

[Out]

integrate((B*x + A)/((c*x^2 + b*x)^(3/2)*x^(5/2)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {A+B\,x}{x^{5/2}\,{\left (c\,x^2+b\,x\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*x)/(x^(5/2)*(b*x + c*x^2)^(3/2)),x)

[Out]

int((A + B*x)/(x^(5/2)*(b*x + c*x^2)^(3/2)), 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 {5}{2}} \left (x \left (b + c x\right )\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x**(5/2)/(c*x**2+b*x)**(3/2),x)

[Out]

Integral((A + B*x)/(x**(5/2)*(x*(b + c*x))**(3/2)), x)

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