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

Optimal. Leaf size=146 \[ -\frac {(2 b B-5 A c) \tanh ^{-1}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right )}{b^{7/2}}+\frac {\sqrt {x} (2 b B-5 A c)}{b^3 \sqrt {b x+c x^2}}+\frac {2 b B-5 A c}{3 b^2 c \sqrt {x} \sqrt {b x+c x^2}}-\frac {2 \sqrt {x} (b B-A c)}{3 b c \left (b x+c x^2\right )^{3/2}} \]

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Rubi [A]  time = 0.12, antiderivative size = 146, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {788, 672, 666, 660, 207} \begin {gather*} \frac {\sqrt {x} (2 b B-5 A c)}{b^3 \sqrt {b x+c x^2}}+\frac {2 b B-5 A c}{3 b^2 c \sqrt {x} \sqrt {b x+c x^2}}-\frac {(2 b B-5 A c) \tanh ^{-1}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right )}{b^{7/2}}-\frac {2 \sqrt {x} (b B-A c)}{3 b c \left (b x+c x^2\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(b*B - A*c)*Sqrt[x])/(3*b*c*(b*x + c*x^2)^(3/2)) + (2*b*B - 5*A*c)/(3*b^2*c*Sqrt[x]*Sqrt[b*x + c*x^2]) + (
(2*b*B - 5*A*c)*Sqrt[x])/(b^3*Sqrt[b*x + c*x^2]) - ((2*b*B - 5*A*c)*ArcTanh[Sqrt[b*x + c*x^2]/(Sqrt[b]*Sqrt[x]
)])/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 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 788

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

Rubi steps

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

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Mathematica [C]  time = 0.03, size = 55, normalized size = 0.38 \begin {gather*} \frac {\sqrt {x} \left (x (2 b B-5 A c) \, _2F_1\left (-\frac {3}{2},1;-\frac {1}{2};\frac {c x}{b}+1\right )-3 A b\right )}{3 b^2 (x (b+c x))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 1.73, size = 110, normalized size = 0.75 \begin {gather*} \frac {(5 A c-2 b B) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {b x+c x^2}}\right )}{b^{7/2}}+\frac {\sqrt {b x+c x^2} \left (-3 A b^2-20 A b c x-15 A c^2 x^2+8 b^2 B x+6 b B c x^2\right )}{3 b^3 x^{3/2} (b+c x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[b*x + c*x^2]*(-3*A*b^2 + 8*b^2*B*x - 20*A*b*c*x + 6*b*B*c*x^2 - 15*A*c^2*x^2))/(3*b^3*x^(3/2)*(b + c*x)^
2) + ((-2*b*B + 5*A*c)*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[b*x + c*x^2]])/b^(7/2)

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fricas [A]  time = 0.44, size = 367, normalized size = 2.51 \begin {gather*} \left [-\frac {3 \, {\left ({\left (2 \, B b c^{2} - 5 \, A c^{3}\right )} x^{4} + 2 \, {\left (2 \, B b^{2} c - 5 \, A b c^{2}\right )} x^{3} + {\left (2 \, B b^{3} - 5 \, A b^{2} c\right )} x^{2}\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 (3 \, A b^{3} - 3 \, {\left (2 \, B b^{2} c - 5 \, A b c^{2}\right )} x^{2} - 4 \, {\left (2 \, B b^{3} - 5 \, A b^{2} c\right )} x\right )} \sqrt {c x^{2} + b x} \sqrt {x}}{6 \, {\left (b^{4} c^{2} x^{4} + 2 \, b^{5} c x^{3} + b^{6} x^{2}\right )}}, \frac {3 \, {\left ({\left (2 \, B b c^{2} - 5 \, A c^{3}\right )} x^{4} + 2 \, {\left (2 \, B b^{2} c - 5 \, A b c^{2}\right )} x^{3} + {\left (2 \, B b^{3} - 5 \, A b^{2} c\right )} x^{2}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} \sqrt {x}}{\sqrt {c x^{2} + b x}}\right ) - {\left (3 \, A b^{3} - 3 \, {\left (2 \, B b^{2} c - 5 \, A b c^{2}\right )} x^{2} - 4 \, {\left (2 \, B b^{3} - 5 \, A b^{2} c\right )} x\right )} \sqrt {c x^{2} + b x} \sqrt {x}}{3 \, {\left (b^{4} c^{2} x^{4} + 2 \, b^{5} c x^{3} + b^{6} x^{2}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.30, size = 90, normalized size = 0.62 \begin {gather*} \frac {{\left (2 \, B b - 5 \, A c\right )} \arctan \left (\frac {\sqrt {c x + b}}{\sqrt {-b}}\right )}{\sqrt {-b} b^{3}} - \frac {\sqrt {c x + b} A}{b^{3} x} + \frac {2 \, {\left (3 \, {\left (c x + b\right )} B b + B b^{2} - 6 \, {\left (c x + b\right )} A c - A b c\right )}}{3 \, {\left (c x + b\right )}^{\frac {3}{2}} b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*B*b - 5*A*c)*arctan(sqrt(c*x + b)/sqrt(-b))/(sqrt(-b)*b^3) - sqrt(c*x + b)*A/(b^3*x) + 2/3*(3*(c*x + b)*B*b
+ B*b^2 - 6*(c*x + b)*A*c - A*b*c)/((c*x + b)^(3/2)*b^3)

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maple [A]  time = 0.08, size = 175, normalized size = 1.20 \begin {gather*} \frac {\sqrt {\left (c x +b \right ) x}\, \left (15 \sqrt {c x +b}\, A \,c^{2} x^{2} \arctanh \left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right )-6 \sqrt {c x +b}\, B b c \,x^{2} \arctanh \left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right )-15 A \sqrt {b}\, c^{2} x^{2}+6 B \,b^{\frac {3}{2}} c \,x^{2}+15 \sqrt {c x +b}\, A b c x \arctanh \left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right )-6 \sqrt {c x +b}\, B \,b^{2} x \arctanh \left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right )-20 A \,b^{\frac {3}{2}} c x +8 B \,b^{\frac {5}{2}} x -3 A \,b^{\frac {5}{2}}\right )}{3 \left (c x +b \right )^{2} b^{\frac {7}{2}} x^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*((c*x+b)*x)^(1/2)*(15*A*(c*x+b)^(1/2)*arctanh((c*x+b)^(1/2)/b^(1/2))*x^2*c^2-6*B*(c*x+b)^(1/2)*arctanh((c*
x+b)^(1/2)/b^(1/2))*x^2*b*c+15*A*arctanh((c*x+b)^(1/2)/b^(1/2))*x*b*c*(c*x+b)^(1/2)-15*A*b^(1/2)*c^2*x^2-6*B*a
rctanh((c*x+b)^(1/2)/b^(1/2))*x*b^2*(c*x+b)^(1/2)+6*B*b^(3/2)*c*x^2-20*A*b^(3/2)*c*x+8*B*b^(5/2)*x-3*A*b^(5/2)
)/x^(3/2)/(c*x+b)^2/b^(7/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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