3.1.83 \(\int \frac {(A+B x) (b x+c x^2)^{3/2}}{x} \, dx\) [83]

Optimal. Leaf size=132 \[ -\frac {b (3 b B-8 A c) (b+2 c x) \sqrt {b x+c x^2}}{64 c^2}-\frac {(3 b B-8 A c) \left (b x+c x^2\right )^{3/2}}{24 c}+\frac {B \left (b x+c x^2\right )^{5/2}}{4 c x}+\frac {b^3 (3 b B-8 A c) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{64 c^{5/2}} \]

[Out]

-1/24*(-8*A*c+3*B*b)*(c*x^2+b*x)^(3/2)/c+1/4*B*(c*x^2+b*x)^(5/2)/c/x+1/64*b^3*(-8*A*c+3*B*b)*arctanh(x*c^(1/2)
/(c*x^2+b*x)^(1/2))/c^(5/2)-1/64*b*(-8*A*c+3*B*b)*(2*c*x+b)*(c*x^2+b*x)^(1/2)/c^2

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Rubi [A]
time = 0.07, antiderivative size = 132, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {808, 678, 626, 634, 212} \begin {gather*} \frac {b^3 (3 b B-8 A c) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{64 c^{5/2}}-\frac {b (b+2 c x) \sqrt {b x+c x^2} (3 b B-8 A c)}{64 c^2}-\frac {\left (b x+c x^2\right )^{3/2} (3 b B-8 A c)}{24 c}+\frac {B \left (b x+c x^2\right )^{5/2}}{4 c x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/64*(b*(3*b*B - 8*A*c)*(b + 2*c*x)*Sqrt[b*x + c*x^2])/c^2 - ((3*b*B - 8*A*c)*(b*x + c*x^2)^(3/2))/(24*c) + (
B*(b*x + c*x^2)^(5/2))/(4*c*x) + (b^3*(3*b*B - 8*A*c)*ArcTanh[(Sqrt[c]*x)/Sqrt[b*x + c*x^2]])/(64*c^(5/2))

Rule 212

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

Rule 626

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(b + 2*c*x)*((a + b*x + c*x^2)^p/(2*c*(2*p + 1
))), x] - Dist[p*((b^2 - 4*a*c)/(2*c*(2*p + 1))), Int[(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c}, x]
&& NeQ[b^2 - 4*a*c, 0] && GtQ[p, 0] && IntegerQ[4*p]

Rule 634

Int[1/Sqrt[(b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(1 - c*x^2), x], x, x/Sqrt[b*x + c*x^2
]], x] /; FreeQ[{b, c}, x]

Rule 678

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

Rule 808

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp
[g*(d + e*x)^m*((a + b*x + c*x^2)^(p + 1)/(c*(m + 2*p + 2))), x] + Dist[(m*(g*(c*d - b*e) + c*e*f) + e*(p + 1)
*(2*c*f - b*g))/(c*e*(m + 2*p + 2)), Int[(d + e*x)^m*(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] && NeQ[m + 2*p + 2, 0] && (NeQ[m, 2] || Eq
Q[d, 0])

Rubi steps

\begin {align*} \int \frac {(A+B x) \left (b x+c x^2\right )^{3/2}}{x} \, dx &=\frac {B \left (b x+c x^2\right )^{5/2}}{4 c x}+\frac {\left (b B-A c+\frac {5}{2} (-b B+2 A c)\right ) \int \frac {\left (b x+c x^2\right )^{3/2}}{x} \, dx}{4 c}\\ &=-\frac {(3 b B-8 A c) \left (b x+c x^2\right )^{3/2}}{24 c}+\frac {B \left (b x+c x^2\right )^{5/2}}{4 c x}-\frac {(b (3 b B-8 A c)) \int \sqrt {b x+c x^2} \, dx}{16 c}\\ &=-\frac {b (3 b B-8 A c) (b+2 c x) \sqrt {b x+c x^2}}{64 c^2}-\frac {(3 b B-8 A c) \left (b x+c x^2\right )^{3/2}}{24 c}+\frac {B \left (b x+c x^2\right )^{5/2}}{4 c x}+\frac {\left (b^3 (3 b B-8 A c)\right ) \int \frac {1}{\sqrt {b x+c x^2}} \, dx}{128 c^2}\\ &=-\frac {b (3 b B-8 A c) (b+2 c x) \sqrt {b x+c x^2}}{64 c^2}-\frac {(3 b B-8 A c) \left (b x+c x^2\right )^{3/2}}{24 c}+\frac {B \left (b x+c x^2\right )^{5/2}}{4 c x}+\frac {\left (b^3 (3 b B-8 A c)\right ) \text {Subst}\left (\int \frac {1}{1-c x^2} \, dx,x,\frac {x}{\sqrt {b x+c x^2}}\right )}{64 c^2}\\ &=-\frac {b (3 b B-8 A c) (b+2 c x) \sqrt {b x+c x^2}}{64 c^2}-\frac {(3 b B-8 A c) \left (b x+c x^2\right )^{3/2}}{24 c}+\frac {B \left (b x+c x^2\right )^{5/2}}{4 c x}+\frac {b^3 (3 b B-8 A c) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{64 c^{5/2}}\\ \end {align*}

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Mathematica [A]
time = 0.22, size = 129, normalized size = 0.98 \begin {gather*} \frac {\sqrt {x (b+c x)} \left (\sqrt {c} \left (-9 b^3 B+6 b^2 c (4 A+B x)+16 c^3 x^2 (4 A+3 B x)+8 b c^2 x (14 A+9 B x)\right )+\frac {3 b^3 (-3 b B+8 A c) \log \left (-\sqrt {c} \sqrt {x}+\sqrt {b+c x}\right )}{\sqrt {x} \sqrt {b+c x}}\right )}{192 c^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[x*(b + c*x)]*(Sqrt[c]*(-9*b^3*B + 6*b^2*c*(4*A + B*x) + 16*c^3*x^2*(4*A + 3*B*x) + 8*b*c^2*x*(14*A + 9*B
*x)) + (3*b^3*(-3*b*B + 8*A*c)*Log[-(Sqrt[c]*Sqrt[x]) + Sqrt[b + c*x]])/(Sqrt[x]*Sqrt[b + c*x])))/(192*c^(5/2)
)

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Maple [A]
time = 0.58, size = 164, normalized size = 1.24

method	result	size
risch	\(\frac {\left (48 B \,c^{3} x^{3}+64 A \,c^{3} x^{2}+72 b B \,x^{2} c^{2}+112 A b \,c^{2} x +6 b^{2} B x c +24 A \,b^{2} c -9 B \,b^{3}\right ) x \left (c x +b \right )}{192 c^{2} \sqrt {x \left (c x +b \right )}}-\frac {b^{3} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right ) A}{16 c^{\frac {3}{2}}}+\frac {3 b^{4} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right ) B}{128 c^{\frac {5}{2}}}\)	\(146\)
default	\(B \left (\frac {\left (2 c x +b \right ) \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{8 c}-\frac {3 b^{2} \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x}}{4 c}-\frac {b^{2} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{8 c^{\frac {3}{2}}}\right )}{16 c}\right )+A \left (\frac {\left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{3}+\frac {b \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x}}{4 c}-\frac {b^{2} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{8 c^{\frac {3}{2}}}\right )}{2}\right )\)	\(164\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(c*x^2+b*x)^(3/2)/x,x,method=_RETURNVERBOSE)

[Out]

B*(1/8*(2*c*x+b)*(c*x^2+b*x)^(3/2)/c-3/16*b^2/c*(1/4*(2*c*x+b)*(c*x^2+b*x)^(1/2)/c-1/8*b^2/c^(3/2)*ln((1/2*b+c
*x)/c^(1/2)+(c*x^2+b*x)^(1/2))))+A*(1/3*(c*x^2+b*x)^(3/2)+1/2*b*(1/4*(2*c*x+b)*(c*x^2+b*x)^(1/2)/c-1/8*b^2/c^(
3/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x)^(1/2))))

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Maxima [A]
time = 0.27, size = 189, normalized size = 1.43 \begin {gather*} \frac {1}{4} \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} B x + \frac {1}{4} \, \sqrt {c x^{2} + b x} A b x - \frac {3 \, \sqrt {c x^{2} + b x} B b^{2} x}{32 \, c} + \frac {3 \, B b^{4} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{128 \, c^{\frac {5}{2}}} - \frac {A b^{3} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{16 \, c^{\frac {3}{2}}} + \frac {1}{3} \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} A - \frac {3 \, \sqrt {c x^{2} + b x} B b^{3}}{64 \, c^{2}} + \frac {{\left (c x^{2} + b x\right )}^{\frac {3}{2}} B b}{8 \, c} + \frac {\sqrt {c x^{2} + b x} A b^{2}}{8 \, c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(c*x^2 + b*x)^(3/2)*B*x + 1/4*sqrt(c*x^2 + b*x)*A*b*x - 3/32*sqrt(c*x^2 + b*x)*B*b^2*x/c + 3/128*B*b^4*log
(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c))/c^(5/2) - 1/16*A*b^3*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c))/c^
(3/2) + 1/3*(c*x^2 + b*x)^(3/2)*A - 3/64*sqrt(c*x^2 + b*x)*B*b^3/c^2 + 1/8*(c*x^2 + b*x)^(3/2)*B*b/c + 1/8*sqr
t(c*x^2 + b*x)*A*b^2/c

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Fricas [A]
time = 3.27, size = 256, normalized size = 1.94 \begin {gather*} \left [-\frac {3 \, {\left (3 \, B b^{4} - 8 \, A b^{3} c\right )} \sqrt {c} \log \left (2 \, c x + b - 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) - 2 \, {\left (48 \, B c^{4} x^{3} - 9 \, B b^{3} c + 24 \, A b^{2} c^{2} + 8 \, {\left (9 \, B b c^{3} + 8 \, A c^{4}\right )} x^{2} + 2 \, {\left (3 \, B b^{2} c^{2} + 56 \, A b c^{3}\right )} x\right )} \sqrt {c x^{2} + b x}}{384 \, c^{3}}, -\frac {3 \, {\left (3 \, B b^{4} - 8 \, A b^{3} c\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x}\right ) - {\left (48 \, B c^{4} x^{3} - 9 \, B b^{3} c + 24 \, A b^{2} c^{2} + 8 \, {\left (9 \, B b c^{3} + 8 \, A c^{4}\right )} x^{2} + 2 \, {\left (3 \, B b^{2} c^{2} + 56 \, A b c^{3}\right )} x\right )} \sqrt {c x^{2} + b x}}{192 \, c^{3}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/384*(3*(3*B*b^4 - 8*A*b^3*c)*sqrt(c)*log(2*c*x + b - 2*sqrt(c*x^2 + b*x)*sqrt(c)) - 2*(48*B*c^4*x^3 - 9*B*
b^3*c + 24*A*b^2*c^2 + 8*(9*B*b*c^3 + 8*A*c^4)*x^2 + 2*(3*B*b^2*c^2 + 56*A*b*c^3)*x)*sqrt(c*x^2 + b*x))/c^3, -
1/192*(3*(3*B*b^4 - 8*A*b^3*c)*sqrt(-c)*arctan(sqrt(c*x^2 + b*x)*sqrt(-c)/(c*x)) - (48*B*c^4*x^3 - 9*B*b^3*c +
24*A*b^2*c^2 + 8*(9*B*b*c^3 + 8*A*c^4)*x^2 + 2*(3*B*b^2*c^2 + 56*A*b*c^3)*x)*sqrt(c*x^2 + b*x))/c^3]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 1.07, size = 138, normalized size = 1.05 \begin {gather*} \frac {1}{192} \, \sqrt {c x^{2} + b x} {\left (2 \, {\left (4 \, {\left (6 \, B c x + \frac {9 \, B b c^{3} + 8 \, A c^{4}}{c^{3}}\right )} x + \frac {3 \, B b^{2} c^{2} + 56 \, A b c^{3}}{c^{3}}\right )} x - \frac {3 \, {\left (3 \, B b^{3} c - 8 \, A b^{2} c^{2}\right )}}{c^{3}}\right )} - \frac {{\left (3 \, B b^{4} - 8 \, A b^{3} c\right )} \log \left ({\left | -2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} \sqrt {c} - b \right |}\right )}{128 \, c^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/192*sqrt(c*x^2 + b*x)*(2*(4*(6*B*c*x + (9*B*b*c^3 + 8*A*c^4)/c^3)*x + (3*B*b^2*c^2 + 56*A*b*c^3)/c^3)*x - 3*
(3*B*b^3*c - 8*A*b^2*c^2)/c^3) - 1/128*(3*B*b^4 - 8*A*b^3*c)*log(abs(-2*(sqrt(c)*x - sqrt(c*x^2 + b*x))*sqrt(c
) - b))/c^(5/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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