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

Optimal. Leaf size=174 \[ \frac {5 c (4 b B-7 A c) \tanh ^{-1}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right )}{4 b^{9/2}}-\frac {5 c \sqrt {x} (4 b B-7 A c)}{4 b^4 \sqrt {b x+c x^2}}-\frac {5 (4 b B-7 A c)}{12 b^3 \sqrt {x} \sqrt {b x+c x^2}}+\frac {\sqrt {x} (4 b B-7 A c)}{6 b^2 \left (b x+c x^2\right )^{3/2}}-\frac {A}{2 b \sqrt {x} \left (b x+c x^2\right )^{3/2}} \]

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Rubi [A]  time = 0.15, antiderivative size = 174, 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, 666, 672, 660, 207} \begin {gather*} -\frac {5 c \sqrt {x} (4 b B-7 A c)}{4 b^4 \sqrt {b x+c x^2}}-\frac {5 (4 b B-7 A c)}{12 b^3 \sqrt {x} \sqrt {b x+c x^2}}+\frac {\sqrt {x} (4 b B-7 A c)}{6 b^2 \left (b x+c x^2\right )^{3/2}}+\frac {5 c (4 b B-7 A c) \tanh ^{-1}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right )}{4 b^{9/2}}-\frac {A}{2 b \sqrt {x} \left (b x+c x^2\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [C]  time = 0.03, size = 60, normalized size = 0.34 \begin {gather*} \frac {c x^2 (7 A c-4 b B) \, _2F_1\left (-\frac {3}{2},2;-\frac {1}{2};\frac {c x}{b}+1\right )-3 A b^2}{6 b^3 \sqrt {x} (x (b+c x))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 2.44, size = 140, normalized size = 0.80 \begin {gather*} \frac {5 \left (4 b B c-7 A c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {b x+c x^2}}\right )}{4 b^{9/2}}+\frac {\sqrt {b x+c x^2} \left (-6 A b^3+21 A b^2 c x+140 A b c^2 x^2+105 A c^3 x^3-12 b^3 B x-80 b^2 B c x^2-60 b B c^2 x^3\right )}{12 b^4 x^{5/2} (b+c x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[b*x + c*x^2]*(-6*A*b^3 - 12*b^3*B*x + 21*A*b^2*c*x - 80*b^2*B*c*x^2 + 140*A*b*c^2*x^2 - 60*b*B*c^2*x^3 +
105*A*c^3*x^3))/(12*b^4*x^(5/2)*(b + c*x)^2) + (5*(4*b*B*c - 7*A*c^2)*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[b*x + c*
x^2]])/(4*b^(9/2))

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fricas [A]  time = 0.42, size = 424, normalized size = 2.44 \begin {gather*} \left [-\frac {15 \, {\left ({\left (4 \, B b c^{3} - 7 \, A c^{4}\right )} x^{5} + 2 \, {\left (4 \, B b^{2} c^{2} - 7 \, A b c^{3}\right )} x^{4} + {\left (4 \, B b^{3} c - 7 \, A b^{2} c^{2}\right )} x^{3}\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 (6 \, A b^{4} + 15 \, {\left (4 \, B b^{2} c^{2} - 7 \, A b c^{3}\right )} x^{3} + 20 \, {\left (4 \, B b^{3} c - 7 \, A b^{2} c^{2}\right )} x^{2} + 3 \, {\left (4 \, B b^{4} - 7 \, A b^{3} c\right )} x\right )} \sqrt {c x^{2} + b x} \sqrt {x}}{24 \, {\left (b^{5} c^{2} x^{5} + 2 \, b^{6} c x^{4} + b^{7} x^{3}\right )}}, -\frac {15 \, {\left ({\left (4 \, B b c^{3} - 7 \, A c^{4}\right )} x^{5} + 2 \, {\left (4 \, B b^{2} c^{2} - 7 \, A b c^{3}\right )} x^{4} + {\left (4 \, B b^{3} c - 7 \, A b^{2} c^{2}\right )} x^{3}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} \sqrt {x}}{\sqrt {c x^{2} + b x}}\right ) + {\left (6 \, A b^{4} + 15 \, {\left (4 \, B b^{2} c^{2} - 7 \, A b c^{3}\right )} x^{3} + 20 \, {\left (4 \, B b^{3} c - 7 \, A b^{2} c^{2}\right )} x^{2} + 3 \, {\left (4 \, B b^{4} - 7 \, A b^{3} c\right )} x\right )} \sqrt {c x^{2} + b x} \sqrt {x}}{12 \, {\left (b^{5} c^{2} x^{5} + 2 \, b^{6} c x^{4} + b^{7} x^{3}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/(c*x^2+b*x)^(5/2)/x^(1/2),x, algorithm="fricas")

[Out]

[-1/24*(15*((4*B*b*c^3 - 7*A*c^4)*x^5 + 2*(4*B*b^2*c^2 - 7*A*b*c^3)*x^4 + (4*B*b^3*c - 7*A*b^2*c^2)*x^3)*sqrt(
b)*log(-(c*x^2 + 2*b*x - 2*sqrt(c*x^2 + b*x)*sqrt(b)*sqrt(x))/x^2) + 2*(6*A*b^4 + 15*(4*B*b^2*c^2 - 7*A*b*c^3)
*x^3 + 20*(4*B*b^3*c - 7*A*b^2*c^2)*x^2 + 3*(4*B*b^4 - 7*A*b^3*c)*x)*sqrt(c*x^2 + b*x)*sqrt(x))/(b^5*c^2*x^5 +
2*b^6*c*x^4 + b^7*x^3), -1/12*(15*((4*B*b*c^3 - 7*A*c^4)*x^5 + 2*(4*B*b^2*c^2 - 7*A*b*c^3)*x^4 + (4*B*b^3*c -
7*A*b^2*c^2)*x^3)*sqrt(-b)*arctan(sqrt(-b)*sqrt(x)/sqrt(c*x^2 + b*x)) + (6*A*b^4 + 15*(4*B*b^2*c^2 - 7*A*b*c^
3)*x^3 + 20*(4*B*b^3*c - 7*A*b^2*c^2)*x^2 + 3*(4*B*b^4 - 7*A*b^3*c)*x)*sqrt(c*x^2 + b*x)*sqrt(x))/(b^5*c^2*x^5
+ 2*b^6*c*x^4 + b^7*x^3)]

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giac [A]  time = 0.30, size = 149, normalized size = 0.86 \begin {gather*} -\frac {5 \, {\left (4 \, B b c - 7 \, A c^{2}\right )} \arctan \left (\frac {\sqrt {c x + b}}{\sqrt {-b}}\right )}{4 \, \sqrt {-b} b^{4}} - \frac {2 \, {\left (6 \, {\left (c x + b\right )} B b c + B b^{2} c - 9 \, {\left (c x + b\right )} A c^{2} - A b c^{2}\right )}}{3 \, {\left (c x + b\right )}^{\frac {3}{2}} b^{4}} - \frac {4 \, {\left (c x + b\right )}^{\frac {3}{2}} B b c - 4 \, \sqrt {c x + b} B b^{2} c - 11 \, {\left (c x + b\right )}^{\frac {3}{2}} A c^{2} + 13 \, \sqrt {c x + b} A b c^{2}}{4 \, b^{4} c^{2} x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/(c*x^2+b*x)^(5/2)/x^(1/2),x, algorithm="giac")

[Out]

-5/4*(4*B*b*c - 7*A*c^2)*arctan(sqrt(c*x + b)/sqrt(-b))/(sqrt(-b)*b^4) - 2/3*(6*(c*x + b)*B*b*c + B*b^2*c - 9*
(c*x + b)*A*c^2 - A*b*c^2)/((c*x + b)^(3/2)*b^4) - 1/4*(4*(c*x + b)^(3/2)*B*b*c - 4*sqrt(c*x + b)*B*b^2*c - 11
*(c*x + b)^(3/2)*A*c^2 + 13*sqrt(c*x + b)*A*b*c^2)/(b^4*c^2*x^2)

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maple [A]  time = 0.08, size = 208, normalized size = 1.20 \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 )-60 \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}+60 B \,b^{\frac {3}{2}} c^{2} x^{3}+105 \sqrt {c x +b}\, A b \,c^{2} x^{2} \arctanh \left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right )-60 \sqrt {c x +b}\, B \,b^{2} c \,x^{2} \arctanh \left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right )-140 A \,b^{\frac {3}{2}} c^{2} x^{2}+80 B \,b^{\frac {5}{2}} c \,x^{2}-21 A \,b^{\frac {5}{2}} c x +12 B \,b^{\frac {7}{2}} x +6 A \,b^{\frac {7}{2}}\right )}{12 \left (c x +b \right )^{2} b^{\frac {9}{2}} x^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)/(c*x^2+b*x)^(5/2)/x^(1/2),x)

[Out]

-1/12*((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^3-60*B*(c*x+b)^(1/2)*arctanh
((c*x+b)^(1/2)/b^(1/2))*x^3*b*c^2+105*A*arctanh((c*x+b)^(1/2)/b^(1/2))*x^2*b*c^2*(c*x+b)^(1/2)-105*A*b^(1/2)*c
^3*x^3-60*B*arctanh((c*x+b)^(1/2)/b^(1/2))*x^2*b^2*c*(c*x+b)^(1/2)+60*B*b^(3/2)*c^2*x^3-140*A*b^(3/2)*c^2*x^2+
80*B*b^(5/2)*c*x^2-21*A*b^(5/2)*c*x+12*B*b^(7/2)*x+6*A*b^(7/2))/x^(5/2)/(c*x+b)^2/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 {5}{2}} \sqrt {x}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/(c*x^2+b*x)^(5/2)/x^(1/2),x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*x)/(x^(1/2)*(b*x + c*x^2)^(5/2)),x)

[Out]

int((A + B*x)/(x^(1/2)*(b*x + c*x^2)^(5/2)), x)

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sympy [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)/(c*x**2+b*x)**(5/2)/x**(1/2),x)

[Out]

Timed out

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