3.3.40 \(\int \frac {A+B x}{x^{5/2} (b x+c x^2)^{3/2}} \, dx\) [240]

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

[Out]

-5/8*c^2*(-7*A*c+6*B*b)*arctanh((c*x^2+b*x)^(1/2)/b^(1/2)/x^(1/2))/b^(9/2)-1/3*A/b/x^(5/2)/(c*x^2+b*x)^(1/2)+1
/12*(7*A*c-6*B*b)/b^2/x^(3/2)/(c*x^2+b*x)^(1/2)+5/24*c*(-7*A*c+6*B*b)/b^3/x^(1/2)/(c*x^2+b*x)^(1/2)+5/8*c^2*(-
7*A*c+6*B*b)*x^(1/2)/b^4/(c*x^2+b*x)^(1/2)

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Rubi [A]
time = 0.10, 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 = {806, 686, 680, 674, 213} \begin {gather*} -\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}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/3*A/(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 213

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

Rule 674

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 680

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 686

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))),
Int[(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] && E
qQ[c*d^2 - b*d*e + a*e^2, 0] && LtQ[m, 0] && NeQ[m + p + 1, 0] && IntegerQ[2*p]

Rule 806

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 ) \text {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 [A]
time = 0.20, size = 132, normalized size = 0.74 \begin {gather*} \frac {\sqrt {b} \left (6 b B x \left (-2 b^2+5 b c x+15 c^2 x^2\right )-A \left (8 b^3-14 b^2 c x+35 b c^2 x^2+105 c^3 x^3\right )\right )+15 c^2 (-6 b B+7 A c) x^3 \sqrt {b+c x} \tanh ^{-1}\left (\frac {\sqrt {b+c x}}{\sqrt {b}}\right )}{24 b^{9/2} 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]

(Sqrt[b]*(6*b*B*x*(-2*b^2 + 5*b*c*x + 15*c^2*x^2) - A*(8*b^3 - 14*b^2*c*x + 35*b*c^2*x^2 + 105*c^3*x^3)) + 15*
c^2*(-6*b*B + 7*A*c)*x^3*Sqrt[b + c*x]*ArcTanh[Sqrt[b + c*x]/Sqrt[b]])/(24*b^(9/2)*x^(5/2)*Sqrt[x*(b + c*x)])

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Maple [A]
time = 0.55, size = 150, normalized size = 0.84

method	result	size
risch	\(-\frac {\left (c x +b \right ) \left (57 A \,c^{2} x^{2}-42 b B \,x^{2} c -22 A b c x +12 b^{2} B x +8 b^{2} A \right )}{24 b^{4} x^{\frac {5}{2}} \sqrt {x \left (c x +b \right )}}-\frac {c^{2} \left (-\frac {2 \left (35 A c -30 B b \right ) \arctanh \left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right )}{\sqrt {b}}-\frac {2 \left (-16 A c +16 B b \right )}{\sqrt {c x +b}}\right ) \sqrt {c x +b}\, \sqrt {x}}{16 b^{4} \sqrt {x \left (c x +b \right )}}\)	\(133\)
default	\(\frac {\sqrt {x \left (c x +b \right )}\, \left (105 A \arctanh \left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right ) \sqrt {c x +b}\, c^{3} x^{3}-90 B \arctanh \left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right ) \sqrt {c x +b}\, b \,c^{2} x^{3}-105 A \sqrt {b}\, c^{3} x^{3}+90 B \,b^{\frac {3}{2}} c^{2} x^{3}-35 A \,b^{\frac {3}{2}} c^{2} x^{2}+30 B \,b^{\frac {5}{2}} c \,x^{2}+14 A \,b^{\frac {5}{2}} c x -12 B \,b^{\frac {7}{2}} x -8 A \,b^{\frac {7}{2}}\right )}{24 x^{\frac {7}{2}} \left (c x +b \right ) b^{\frac {9}{2}}}\)	\(150\)

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,method=_RETURNVERBOSE)

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \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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Fricas [A]
time = 3.38, 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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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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Giac [A]
time = 0.99, 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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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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