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

Optimal. Leaf size=158 \[ -\frac {3 \left (b^2-4 a c\right )^2 (b B-2 A c) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{256 c^{7/2}}+\frac {3 \left (b^2-4 a c\right ) (b+2 c x) \sqrt {a+b x+c x^2} (b B-2 A c)}{128 c^3}-\frac {(b+2 c x) \left (a+b x+c x^2\right )^{3/2} (b B-2 A c)}{16 c^2}+\frac {B \left (a+b x+c x^2\right )^{5/2}}{5 c} \]

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Rubi [A]  time = 0.07, antiderivative size = 158, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {640, 612, 621, 206} \begin {gather*} \frac {3 \left (b^2-4 a c\right ) (b+2 c x) \sqrt {a+b x+c x^2} (b B-2 A c)}{128 c^3}-\frac {3 \left (b^2-4 a c\right )^2 (b B-2 A c) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{256 c^{7/2}}-\frac {(b+2 c x) \left (a+b x+c x^2\right )^{3/2} (b B-2 A c)}{16 c^2}+\frac {B \left (a+b x+c x^2\right )^{5/2}}{5 c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(3*(b^2 - 4*a*c)*(b*B - 2*A*c)*(b + 2*c*x)*Sqrt[a + b*x + c*x^2])/(128*c^3) - ((b*B - 2*A*c)*(b + 2*c*x)*(a +
b*x + c*x^2)^(3/2))/(16*c^2) + (B*(a + b*x + c*x^2)^(5/2))/(5*c) - (3*(b^2 - 4*a*c)^2*(b*B - 2*A*c)*ArcTanh[(b
+ 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(256*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 621

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

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

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Mathematica [A]  time = 0.28, size = 144, normalized size = 0.91 \begin {gather*} \frac {5 (2 A c-b B) \left (\frac {3 \left (b^2-4 a c\right ) \left (\left (b^2-4 a c\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+x (b+c x)}}\right )-2 \sqrt {c} (b+2 c x) \sqrt {a+x (b+c x)}\right )}{128 c^{5/2}}+\frac {(b+2 c x) (a+x (b+c x))^{3/2}}{8 c}\right )+2 B (a+x (b+c x))^{5/2}}{10 c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*B*(a + x*(b + c*x))^(5/2) + 5*(-(b*B) + 2*A*c)*(((b + 2*c*x)*(a + x*(b + c*x))^(3/2))/(8*c) + (3*(b^2 - 4*a
*c)*(-2*Sqrt[c]*(b + 2*c*x)*Sqrt[a + x*(b + c*x)] + (b^2 - 4*a*c)*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + x*(b
+ c*x)])]))/(128*c^(5/2))))/(10*c)

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IntegrateAlgebraic [A]  time = 0.85, size = 243, normalized size = 1.54 \begin {gather*} \frac {\sqrt {a+b x+c x^2} \left (128 a^2 B c^2+200 a A b c^2+400 a A c^3 x-100 a b^2 B c+56 a b B c^2 x+256 a B c^3 x^2-30 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}+\frac {3 \left (-32 a^2 A c^3+16 a^2 b B c^2+16 a A b^2 c^2-8 a b^3 B c-2 A b^4 c+b^5 B\right ) \log \left (-2 \sqrt {c} \sqrt {a+b x+c x^2}+b+2 c x\right )}{256 c^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + b*x + c*x^2]*(15*b^4*B - 30*A*b^3*c - 100*a*b^2*B*c + 200*a*A*b*c^2 + 128*a^2*B*c^2 - 10*b^3*B*c*x +
20*A*b^2*c^2*x + 56*a*b*B*c^2*x + 400*a*A*c^3*x + 8*b^2*B*c^2*x^2 + 240*A*b*c^3*x^2 + 256*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 - 8*a*b^3*B*c + 16*a*A*b^2*c^2
+ 16*a^2*b*B*c^2 - 32*a^2*A*c^3)*Log[b + 2*c*x - 2*Sqrt[c]*Sqrt[a + b*x + c*x^2]])/(256*c^(7/2))

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fricas [A]  time = 0.47, size = 515, normalized size = 3.26 \begin {gather*} \left [\frac {15 \, {\left (B b^{5} - 32 \, A a^{2} c^{3} + 16 \, {\left (B a^{2} b + A a b^{2}\right )} c^{2} - 2 \, {\left (4 \, B a b^{3} + A b^{4}\right )} c\right )} \sqrt {c} \log \left (-8 \, c^{2} x^{2} - 8 \, b c x - b^{2} + 4 \, \sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )} \sqrt {c} - 4 \, a c\right ) + 4 \, {\left (128 \, B c^{5} x^{4} + 15 \, B b^{4} c + 8 \, {\left (16 \, B a^{2} + 25 \, A a b\right )} c^{3} + 16 \, {\left (11 \, B b c^{4} + 10 \, A c^{5}\right )} x^{3} - 10 \, {\left (10 \, B a b^{2} + 3 \, A b^{3}\right )} c^{2} + 8 \, {\left (B b^{2} c^{3} + 2 \, {\left (16 \, B a + 15 \, A b\right )} c^{4}\right )} x^{2} - 2 \, {\left (5 \, B b^{3} c^{2} - 200 \, A a c^{4} - 2 \, {\left (14 \, B a b + 5 \, A b^{2}\right )} c^{3}\right )} x\right )} \sqrt {c x^{2} + b x + a}}{2560 \, c^{4}}, \frac {15 \, {\left (B b^{5} - 32 \, A a^{2} c^{3} + 16 \, {\left (B a^{2} b + A a b^{2}\right )} c^{2} - 2 \, {\left (4 \, B a b^{3} + A b^{4}\right )} c\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )} \sqrt {-c}}{2 \, {\left (c^{2} x^{2} + b c x + a c\right )}}\right ) + 2 \, {\left (128 \, B c^{5} x^{4} + 15 \, B b^{4} c + 8 \, {\left (16 \, B a^{2} + 25 \, A a b\right )} c^{3} + 16 \, {\left (11 \, B b c^{4} + 10 \, A c^{5}\right )} x^{3} - 10 \, {\left (10 \, B a b^{2} + 3 \, A b^{3}\right )} c^{2} + 8 \, {\left (B b^{2} c^{3} + 2 \, {\left (16 \, B a + 15 \, A b\right )} c^{4}\right )} x^{2} - 2 \, {\left (5 \, B b^{3} c^{2} - 200 \, A a c^{4} - 2 \, {\left (14 \, B a b + 5 \, A b^{2}\right )} c^{3}\right )} x\right )} \sqrt {c x^{2} + b x + a}}{1280 \, 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+a)^(3/2),x, algorithm="fricas")

[Out]

[1/2560*(15*(B*b^5 - 32*A*a^2*c^3 + 16*(B*a^2*b + A*a*b^2)*c^2 - 2*(4*B*a*b^3 + A*b^4)*c)*sqrt(c)*log(-8*c^2*x
^2 - 8*b*c*x - b^2 + 4*sqrt(c*x^2 + b*x + a)*(2*c*x + b)*sqrt(c) - 4*a*c) + 4*(128*B*c^5*x^4 + 15*B*b^4*c + 8*
(16*B*a^2 + 25*A*a*b)*c^3 + 16*(11*B*b*c^4 + 10*A*c^5)*x^3 - 10*(10*B*a*b^2 + 3*A*b^3)*c^2 + 8*(B*b^2*c^3 + 2*
(16*B*a + 15*A*b)*c^4)*x^2 - 2*(5*B*b^3*c^2 - 200*A*a*c^4 - 2*(14*B*a*b + 5*A*b^2)*c^3)*x)*sqrt(c*x^2 + b*x +
a))/c^4, 1/1280*(15*(B*b^5 - 32*A*a^2*c^3 + 16*(B*a^2*b + A*a*b^2)*c^2 - 2*(4*B*a*b^3 + A*b^4)*c)*sqrt(-c)*arc
tan(1/2*sqrt(c*x^2 + b*x + a)*(2*c*x + b)*sqrt(-c)/(c^2*x^2 + b*c*x + a*c)) + 2*(128*B*c^5*x^4 + 15*B*b^4*c +
8*(16*B*a^2 + 25*A*a*b)*c^3 + 16*(11*B*b*c^4 + 10*A*c^5)*x^3 - 10*(10*B*a*b^2 + 3*A*b^3)*c^2 + 8*(B*b^2*c^3 +
2*(16*B*a + 15*A*b)*c^4)*x^2 - 2*(5*B*b^3*c^2 - 200*A*a*c^4 - 2*(14*B*a*b + 5*A*b^2)*c^3)*x)*sqrt(c*x^2 + b*x
+ a))/c^4]

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giac [A]  time = 0.23, size = 251, normalized size = 1.59 \begin {gather*} \frac {1}{640} \, \sqrt {c x^{2} + b x + a} {\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} + 32 \, B a c^{4} + 30 \, A b c^{4}}{c^{4}}\right )} x - \frac {5 \, B b^{3} c^{2} - 28 \, B a b c^{3} - 10 \, A b^{2} c^{3} - 200 \, A a c^{4}}{c^{4}}\right )} x + \frac {15 \, B b^{4} c - 100 \, B a b^{2} c^{2} - 30 \, A b^{3} c^{2} + 128 \, B a^{2} c^{3} + 200 \, A a b c^{3}}{c^{4}}\right )} + \frac {3 \, {\left (B b^{5} - 8 \, B a b^{3} c - 2 \, A b^{4} c + 16 \, B a^{2} b c^{2} + 16 \, A a b^{2} c^{2} - 32 \, A a^{2} c^{3}\right )} \log \left ({\left | -2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\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+a)^(3/2),x, algorithm="giac")

[Out]

1/640*sqrt(c*x^2 + b*x + a)*(2*(4*(2*(8*B*c*x + (11*B*b*c^4 + 10*A*c^5)/c^4)*x + (B*b^2*c^3 + 32*B*a*c^4 + 30*
A*b*c^4)/c^4)*x - (5*B*b^3*c^2 - 28*B*a*b*c^3 - 10*A*b^2*c^3 - 200*A*a*c^4)/c^4)*x + (15*B*b^4*c - 100*B*a*b^2
*c^2 - 30*A*b^3*c^2 + 128*B*a^2*c^3 + 200*A*a*b*c^3)/c^4) + 3/256*(B*b^5 - 8*B*a*b^3*c - 2*A*b^4*c + 16*B*a^2*
b*c^2 + 16*A*a*b^2*c^2 - 32*A*a^2*c^3)*log(abs(-2*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*sqrt(c) - b))/c^(7/2)

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maple [B]  time = 0.05, size = 469, normalized size = 2.97 \begin {gather*} \frac {3 A \,a^{2} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{8 \sqrt {c}}-\frac {3 A a \,b^{2} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{16 c^{\frac {3}{2}}}+\frac {3 A \,b^{4} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{128 c^{\frac {5}{2}}}-\frac {3 B \,a^{2} b \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{16 c^{\frac {3}{2}}}+\frac {3 B a \,b^{3} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{32 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 +a}\right )}{256 c^{\frac {7}{2}}}+\frac {3 \sqrt {c \,x^{2}+b x +a}\, A a x}{8}-\frac {3 \sqrt {c \,x^{2}+b x +a}\, A \,b^{2} x}{32 c}-\frac {3 \sqrt {c \,x^{2}+b x +a}\, B a b x}{16 c}+\frac {3 \sqrt {c \,x^{2}+b x +a}\, B \,b^{3} x}{64 c^{2}}+\frac {3 \sqrt {c \,x^{2}+b x +a}\, A a b}{16 c}-\frac {3 \sqrt {c \,x^{2}+b x +a}\, A \,b^{3}}{64 c^{2}}+\frac {\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} A x}{4}-\frac {3 \sqrt {c \,x^{2}+b x +a}\, B a \,b^{2}}{32 c^{2}}+\frac {3 \sqrt {c \,x^{2}+b x +a}\, B \,b^{4}}{128 c^{3}}-\frac {\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} B b x}{8 c}+\frac {\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} A b}{8 c}-\frac {\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} B \,b^{2}}{16 c^{2}}+\frac {\left (c \,x^{2}+b x +a \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+a)^(3/2),x)

[Out]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a*c-b^2>0)', see `assume?` f
or more details)Is 4*a*c-b^2 positive, negative or zero?

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mupad [B]  time = 1.61, size = 305, normalized size = 1.93 \begin {gather*} \frac {B\,{\left (c\,x^2+b\,x+a\right )}^{5/2}}{5\,c}+\frac {A\,\left (\left (\frac {x}{2}+\frac {b}{4\,c}\right )\,\sqrt {c\,x^2+b\,x+a}+\frac {\ln \left (\frac {\frac {b}{2}+c\,x}{\sqrt {c}}+\sqrt {c\,x^2+b\,x+a}\right )\,\left (a\,c-\frac {b^2}{4}\right )}{2\,c^{3/2}}\right )\,\left (3\,a\,c-\frac {3\,b^2}{4}\right )}{4\,c}-\frac {B\,b\,\left (\frac {3\,a\,\left (\ln \left (\frac {\frac {b}{2}+c\,x}{\sqrt {c}}+\sqrt {c\,x^2+b\,x+a}\right )\,\left (\frac {a}{2\,\sqrt {c}}-\frac {b^2}{8\,c^{3/2}}\right )+\frac {\left (b+2\,c\,x\right )\,\sqrt {c\,x^2+b\,x+a}}{4\,c}\right )}{4}+\frac {x\,{\left (c\,x^2+b\,x+a\right )}^{3/2}}{4}+\frac {b\,{\left (c\,x^2+b\,x+a\right )}^{3/2}}{8\,c}-\frac {3\,b^2\,\left (\ln \left (\frac {\frac {b}{2}+c\,x}{\sqrt {c}}+\sqrt {c\,x^2+b\,x+a}\right )\,\left (\frac {a}{2\,\sqrt {c}}-\frac {b^2}{8\,c^{3/2}}\right )+\frac {\left (b+2\,c\,x\right )\,\sqrt {c\,x^2+b\,x+a}}{4\,c}\right )}{16\,c}\right )}{2\,c}+\frac {A\,\left (\frac {b}{2}+c\,x\right )\,{\left (c\,x^2+b\,x+a\right )}^{3/2}}{4\,c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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