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

Optimal. Leaf size=134 \[ -\frac {3 b^4 (b B-2 A c) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{128 c^{7/2}}+\frac {3 b^2 (b+2 c x) \sqrt {b x+c x^2} (b B-2 A c)}{128 c^3}-\frac {(b+2 c x) \left (b x+c x^2\right )^{3/2} (b B-2 A c)}{16 c^2}+\frac {B \left (b x+c x^2\right )^{5/2}}{5 c} \]

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Rubi [A]  time = 0.05, antiderivative size = 134, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {640, 612, 620, 206} \begin {gather*} \frac {3 b^2 (b+2 c x) \sqrt {b x+c x^2} (b B-2 A c)}{128 c^3}-\frac {3 b^4 (b B-2 A c) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{128 c^{7/2}}-\frac {(b+2 c x) \left (b x+c x^2\right )^{3/2} (b B-2 A c)}{16 c^2}+\frac {B \left (b x+c x^2\right )^{5/2}}{5 c} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(A + B*x)*(b*x + c*x^2)^(3/2),x]

[Out]

(3*b^2*(b*B - 2*A*c)*(b + 2*c*x)*Sqrt[b*x + c*x^2])/(128*c^3) - ((b*B - 2*A*c)*(b + 2*c*x)*(b*x + c*x^2)^(3/2)
)/(16*c^2) + (B*(b*x + c*x^2)^(5/2))/(5*c) - (3*b^4*(b*B - 2*A*c)*ArcTanh[(Sqrt[c]*x)/Sqrt[b*x + c*x^2]])/(128
*c^(7/2))

Rule 206

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

Rule 612

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 620

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 640

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

Rubi steps

\begin {align*} \int (A+B x) \left (b x+c x^2\right )^{3/2} \, dx &=\frac {B \left (b x+c x^2\right )^{5/2}}{5 c}+\frac {(-b B+2 A c) \int \left (b x+c x^2\right )^{3/2} \, dx}{2 c}\\ &=-\frac {(b B-2 A c) (b+2 c x) \left (b x+c x^2\right )^{3/2}}{16 c^2}+\frac {B \left (b x+c x^2\right )^{5/2}}{5 c}+\frac {\left (3 b^2 (b B-2 A c)\right ) \int \sqrt {b x+c x^2} \, dx}{32 c^2}\\ &=\frac {3 b^2 (b B-2 A c) (b+2 c x) \sqrt {b x+c x^2}}{128 c^3}-\frac {(b B-2 A c) (b+2 c x) \left (b x+c x^2\right )^{3/2}}{16 c^2}+\frac {B \left (b x+c x^2\right )^{5/2}}{5 c}-\frac {\left (3 b^4 (b B-2 A c)\right ) \int \frac {1}{\sqrt {b x+c x^2}} \, dx}{256 c^3}\\ &=\frac {3 b^2 (b B-2 A c) (b+2 c x) \sqrt {b x+c x^2}}{128 c^3}-\frac {(b B-2 A c) (b+2 c x) \left (b x+c x^2\right )^{3/2}}{16 c^2}+\frac {B \left (b x+c x^2\right )^{5/2}}{5 c}-\frac {\left (3 b^4 (b B-2 A c)\right ) \operatorname {Subst}\left (\int \frac {1}{1-c x^2} \, dx,x,\frac {x}{\sqrt {b x+c x^2}}\right )}{128 c^3}\\ &=\frac {3 b^2 (b B-2 A c) (b+2 c x) \sqrt {b x+c x^2}}{128 c^3}-\frac {(b B-2 A c) (b+2 c x) \left (b x+c x^2\right )^{3/2}}{16 c^2}+\frac {B \left (b x+c x^2\right )^{5/2}}{5 c}-\frac {3 b^4 (b B-2 A c) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{128 c^{7/2}}\\ \end {align*}

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Mathematica [A]  time = 0.25, size = 146, normalized size = 1.09 \begin {gather*} \frac {\sqrt {x (b+c x)} \left (\sqrt {c} \left (-10 b^3 c (3 A+B x)+4 b^2 c^2 x (5 A+2 B x)+16 b c^3 x^2 (15 A+11 B x)+32 c^4 x^3 (5 A+4 B x)+15 b^4 B\right )-\frac {15 b^{7/2} (b B-2 A c) \sinh ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{\sqrt {x} \sqrt {\frac {c x}{b}+1}}\right )}{640 c^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[x*(b + c*x)]*(Sqrt[c]*(15*b^4*B - 10*b^3*c*(3*A + B*x) + 4*b^2*c^2*x*(5*A + 2*B*x) + 32*c^4*x^3*(5*A + 4
*B*x) + 16*b*c^3*x^2*(15*A + 11*B*x)) - (15*b^(7/2)*(b*B - 2*A*c)*ArcSinh[(Sqrt[c]*Sqrt[x])/Sqrt[b]])/(Sqrt[x]
*Sqrt[1 + (c*x)/b])))/(640*c^(7/2))

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IntegrateAlgebraic [A]  time = 0.66, size = 152, normalized size = 1.13 \begin {gather*} \frac {3 \left (b^5 B-2 A b^4 c\right ) \log \left (-2 \sqrt {c} \sqrt {b x+c x^2}+b+2 c x\right )}{256 c^{7/2}}+\frac {\sqrt {b x+c x^2} \left (-30 A b^3 c+20 A b^2 c^2 x+240 A b c^3 x^2+160 A c^4 x^3+15 b^4 B-10 b^3 B c x+8 b^2 B c^2 x^2+176 b B c^3 x^3+128 B c^4 x^4\right )}{640 c^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[b*x + c*x^2]*(15*b^4*B - 30*A*b^3*c - 10*b^3*B*c*x + 20*A*b^2*c^2*x + 8*b^2*B*c^2*x^2 + 240*A*b*c^3*x^2
+ 176*b*B*c^3*x^3 + 160*A*c^4*x^3 + 128*B*c^4*x^4))/(640*c^3) + (3*(b^5*B - 2*A*b^4*c)*Log[b + 2*c*x - 2*Sqrt[
c]*Sqrt[b*x + c*x^2]])/(256*c^(7/2))

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fricas [A]  time = 0.42, size = 297, normalized size = 2.22 \begin {gather*} \left [-\frac {15 \, {\left (B b^{5} - 2 \, A b^{4} c\right )} \sqrt {c} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) - 2 \, {\left (128 \, B c^{5} x^{4} + 15 \, B b^{4} c - 30 \, A b^{3} c^{2} + 16 \, {\left (11 \, B b c^{4} + 10 \, A c^{5}\right )} x^{3} + 8 \, {\left (B b^{2} c^{3} + 30 \, A b c^{4}\right )} x^{2} - 10 \, {\left (B b^{3} c^{2} - 2 \, A b^{2} c^{3}\right )} x\right )} \sqrt {c x^{2} + b x}}{1280 \, c^{4}}, \frac {15 \, {\left (B b^{5} - 2 \, A b^{4} c\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x}\right ) + {\left (128 \, B c^{5} x^{4} + 15 \, B b^{4} c - 30 \, A b^{3} c^{2} + 16 \, {\left (11 \, B b c^{4} + 10 \, A c^{5}\right )} x^{3} + 8 \, {\left (B b^{2} c^{3} + 30 \, A b c^{4}\right )} x^{2} - 10 \, {\left (B b^{3} c^{2} - 2 \, A b^{2} c^{3}\right )} x\right )} \sqrt {c x^{2} + b x}}{640 \, c^{4}}\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, algorithm="fricas")

[Out]

[-1/1280*(15*(B*b^5 - 2*A*b^4*c)*sqrt(c)*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c)) - 2*(128*B*c^5*x^4 + 15*
B*b^4*c - 30*A*b^3*c^2 + 16*(11*B*b*c^4 + 10*A*c^5)*x^3 + 8*(B*b^2*c^3 + 30*A*b*c^4)*x^2 - 10*(B*b^3*c^2 - 2*A
*b^2*c^3)*x)*sqrt(c*x^2 + b*x))/c^4, 1/640*(15*(B*b^5 - 2*A*b^4*c)*sqrt(-c)*arctan(sqrt(c*x^2 + b*x)*sqrt(-c)/
(c*x)) + (128*B*c^5*x^4 + 15*B*b^4*c - 30*A*b^3*c^2 + 16*(11*B*b*c^4 + 10*A*c^5)*x^3 + 8*(B*b^2*c^3 + 30*A*b*c
^4)*x^2 - 10*(B*b^3*c^2 - 2*A*b^2*c^3)*x)*sqrt(c*x^2 + b*x))/c^4]

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giac [A]  time = 0.22, size = 162, normalized size = 1.21 \begin {gather*} \frac {1}{640} \, \sqrt {c x^{2} + b x} {\left (2 \, {\left (4 \, {\left (2 \, {\left (8 \, B c x + \frac {11 \, B b c^{4} + 10 \, A c^{5}}{c^{4}}\right )} x + \frac {B b^{2} c^{3} + 30 \, A b c^{4}}{c^{4}}\right )} x - \frac {5 \, {\left (B b^{3} c^{2} - 2 \, A b^{2} c^{3}\right )}}{c^{4}}\right )} x + \frac {15 \, {\left (B b^{4} c - 2 \, A b^{3} c^{2}\right )}}{c^{4}}\right )} + \frac {3 \, {\left (B b^{5} - 2 \, A b^{4} c\right )} \log \left ({\left | -2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} \sqrt {c} - b \right |}\right )}{256 \, c^{\frac {7}{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, algorithm="giac")

[Out]

1/640*sqrt(c*x^2 + b*x)*(2*(4*(2*(8*B*c*x + (11*B*b*c^4 + 10*A*c^5)/c^4)*x + (B*b^2*c^3 + 30*A*b*c^4)/c^4)*x -
5*(B*b^3*c^2 - 2*A*b^2*c^3)/c^4)*x + 15*(B*b^4*c - 2*A*b^3*c^2)/c^4) + 3/256*(B*b^5 - 2*A*b^4*c)*log(abs(-2*(
sqrt(c)*x - sqrt(c*x^2 + b*x))*sqrt(c) - b))/c^(7/2)

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maple [B]  time = 0.05, size = 239, normalized size = 1.78 \begin {gather*} \frac {3 A \,b^{4} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{128 c^{\frac {5}{2}}}-\frac {3 B \,b^{5} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{256 c^{\frac {7}{2}}}-\frac {3 \sqrt {c \,x^{2}+b x}\, A \,b^{2} x}{32 c}+\frac {3 \sqrt {c \,x^{2}+b x}\, B \,b^{3} x}{64 c^{2}}-\frac {3 \sqrt {c \,x^{2}+b x}\, A \,b^{3}}{64 c^{2}}+\frac {\left (c \,x^{2}+b x \right )^{\frac {3}{2}} A x}{4}+\frac {3 \sqrt {c \,x^{2}+b x}\, B \,b^{4}}{128 c^{3}}-\frac {\left (c \,x^{2}+b x \right )^{\frac {3}{2}} B b x}{8 c}+\frac {\left (c \,x^{2}+b x \right )^{\frac {3}{2}} A b}{8 c}-\frac {\left (c \,x^{2}+b x \right )^{\frac {3}{2}} B \,b^{2}}{16 c^{2}}+\frac {\left (c \,x^{2}+b x \right )^{\frac {5}{2}} B}{5 c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*B*(c*x^2+b*x)^(5/2)/c-1/8*B*b/c*x*(c*x^2+b*x)^(3/2)-1/16*B*b^2/c^2*(c*x^2+b*x)^(3/2)+3/64*B*b^3/c^2*(c*x^2
+b*x)^(1/2)*x+3/128*B*b^4/c^3*(c*x^2+b*x)^(1/2)-3/256*B*b^5/c^(7/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x)^(1/2))+
1/4*A*x*(c*x^2+b*x)^(3/2)+1/8*A/c*(c*x^2+b*x)^(3/2)*b-3/32*A*b^2/c*(c*x^2+b*x)^(1/2)*x-3/64*A*b^3/c^2*(c*x^2+b
*x)^(1/2)+3/128*A*b^4/c^(5/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x)^(1/2))

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maxima [B]  time = 0.89, size = 236, normalized size = 1.76 \begin {gather*} \frac {1}{4} \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} A x + \frac {3 \, \sqrt {c x^{2} + b x} B b^{3} x}{64 \, c^{2}} - \frac {{\left (c x^{2} + b x\right )}^{\frac {3}{2}} B b x}{8 \, c} - \frac {3 \, \sqrt {c x^{2} + b x} A b^{2} x}{32 \, c} - \frac {3 \, B b^{5} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{256 \, c^{\frac {7}{2}}} + \frac {3 \, A b^{4} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{128 \, c^{\frac {5}{2}}} + \frac {3 \, \sqrt {c x^{2} + b x} B b^{4}}{128 \, c^{3}} - \frac {{\left (c x^{2} + b x\right )}^{\frac {3}{2}} B b^{2}}{16 \, c^{2}} - \frac {3 \, \sqrt {c x^{2} + b x} A b^{3}}{64 \, c^{2}} + \frac {{\left (c x^{2} + b x\right )}^{\frac {5}{2}} B}{5 \, c} + \frac {{\left (c x^{2} + b x\right )}^{\frac {3}{2}} A b}{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, algorithm="maxima")

[Out]

1/4*(c*x^2 + b*x)^(3/2)*A*x + 3/64*sqrt(c*x^2 + b*x)*B*b^3*x/c^2 - 1/8*(c*x^2 + b*x)^(3/2)*B*b*x/c - 3/32*sqrt
(c*x^2 + b*x)*A*b^2*x/c - 3/256*B*b^5*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c))/c^(7/2) + 3/128*A*b^4*log(2
*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c))/c^(5/2) + 3/128*sqrt(c*x^2 + b*x)*B*b^4/c^3 - 1/16*(c*x^2 + b*x)^(3/2)
*B*b^2/c^2 - 3/64*sqrt(c*x^2 + b*x)*A*b^3/c^2 + 1/5*(c*x^2 + b*x)^(5/2)*B/c + 1/8*(c*x^2 + b*x)^(3/2)*A*b/c

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mupad [B]  time = 1.45, size = 208, normalized size = 1.55 \begin {gather*} \frac {B\,{\left (c\,x^2+b\,x\right )}^{5/2}}{5\,c}+\frac {A\,{\left (c\,x^2+b\,x\right )}^{3/2}\,\left (\frac {b}{2}+c\,x\right )}{4\,c}-\frac {B\,b\,\left (\frac {x\,{\left (c\,x^2+b\,x\right )}^{3/2}}{4}+\frac {b\,{\left (c\,x^2+b\,x\right )}^{3/2}}{8\,c}-\frac {3\,b^2\,\left (\frac {\sqrt {c\,x^2+b\,x}\,\left (b+2\,c\,x\right )}{4\,c}-\frac {b^2\,\ln \left (\frac {\frac {b}{2}+c\,x}{\sqrt {c}}+\sqrt {c\,x^2+b\,x}\right )}{8\,c^{3/2}}\right )}{16\,c}\right )}{2\,c}-\frac {3\,A\,b^2\,\left (\sqrt {c\,x^2+b\,x}\,\left (\frac {x}{2}+\frac {b}{4\,c}\right )-\frac {b^2\,\ln \left (\frac {\frac {b}{2}+c\,x}{\sqrt {c}}+\sqrt {c\,x^2+b\,x}\right )}{8\,c^{3/2}}\right )}{16\,c} \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)

[Out]

(B*(b*x + c*x^2)^(5/2))/(5*c) + (A*(b*x + c*x^2)^(3/2)*(b/2 + c*x))/(4*c) - (B*b*((x*(b*x + c*x^2)^(3/2))/4 +
(b*(b*x + c*x^2)^(3/2))/(8*c) - (3*b^2*(((b*x + c*x^2)^(1/2)*(b + 2*c*x))/(4*c) - (b^2*log((b/2 + c*x)/c^(1/2)
+ (b*x + c*x^2)^(1/2)))/(8*c^(3/2))))/(16*c)))/(2*c) - (3*A*b^2*((b*x + c*x^2)^(1/2)*(x/2 + b/(4*c)) - (b^2*l
og((b/2 + c*x)/c^(1/2) + (b*x + c*x^2)^(1/2)))/(8*c^(3/2))))/(16*c)

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

[Out]

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

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