3.21.14 \(\int (d+e x) (a+b x+c x^2)^{3/2} \, dx\)

Optimal. Leaf size=161 \[ \frac {3 \left (b^2-4 a c\right )^2 (2 c d-b e) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{256 c^{7/2}}-\frac {3 \left (b^2-4 a c\right ) (b+2 c x) \sqrt {a+b x+c x^2} (2 c d-b e)}{128 c^3}+\frac {(b+2 c x) \left (a+b x+c x^2\right )^{3/2} (2 c d-b e)}{16 c^2}+\frac {e \left (a+b x+c x^2\right )^{5/2}}{5 c} \]

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Rubi [A]  time = 0.06, antiderivative size = 161, 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} (2 c d-b e)}{128 c^3}+\frac {3 \left (b^2-4 a c\right )^2 (2 c d-b e) \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} (2 c d-b e)}{16 c^2}+\frac {e \left (a+b x+c x^2\right )^{5/2}}{5 c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*(b^2 - 4*a*c)*(2*c*d - b*e)*(b + 2*c*x)*Sqrt[a + b*x + c*x^2])/(128*c^3) + ((2*c*d - b*e)*(b + 2*c*x)*(a +
b*x + c*x^2)^(3/2))/(16*c^2) + (e*(a + b*x + c*x^2)^(5/2))/(5*c) + (3*(b^2 - 4*a*c)^2*(2*c*d - b*e)*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 (d+e x) \left (a+b x+c x^2\right )^{3/2} \, dx &=\frac {e \left (a+b x+c x^2\right )^{5/2}}{5 c}+\frac {(2 c d-b e) \int \left (a+b x+c x^2\right )^{3/2} \, dx}{2 c}\\ &=\frac {(2 c d-b e) (b+2 c x) \left (a+b x+c x^2\right )^{3/2}}{16 c^2}+\frac {e \left (a+b x+c x^2\right )^{5/2}}{5 c}-\frac {\left (3 \left (b^2-4 a c\right ) (2 c d-b e)\right ) \int \sqrt {a+b x+c x^2} \, dx}{32 c^2}\\ &=-\frac {3 \left (b^2-4 a c\right ) (2 c d-b e) (b+2 c x) \sqrt {a+b x+c x^2}}{128 c^3}+\frac {(2 c d-b e) (b+2 c x) \left (a+b x+c x^2\right )^{3/2}}{16 c^2}+\frac {e \left (a+b x+c x^2\right )^{5/2}}{5 c}+\frac {\left (3 \left (b^2-4 a c\right )^2 (2 c d-b e)\right ) \int \frac {1}{\sqrt {a+b x+c x^2}} \, dx}{256 c^3}\\ &=-\frac {3 \left (b^2-4 a c\right ) (2 c d-b e) (b+2 c x) \sqrt {a+b x+c x^2}}{128 c^3}+\frac {(2 c d-b e) (b+2 c x) \left (a+b x+c x^2\right )^{3/2}}{16 c^2}+\frac {e \left (a+b x+c x^2\right )^{5/2}}{5 c}+\frac {\left (3 \left (b^2-4 a c\right )^2 (2 c d-b e)\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 ) (2 c d-b e) (b+2 c x) \sqrt {a+b x+c x^2}}{128 c^3}+\frac {(2 c d-b e) (b+2 c x) \left (a+b x+c x^2\right )^{3/2}}{16 c^2}+\frac {e \left (a+b x+c x^2\right )^{5/2}}{5 c}+\frac {3 \left (b^2-4 a c\right )^2 (2 c d-b e) \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 = 0.89 \begin {gather*} \frac {5 (2 c d-b e) \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 e (a+x (b+c x))^{5/2}}{10 c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*e*(a + x*(b + c*x))^(5/2) + 5*(2*c*d - b*e)*(((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 = 0.86, size = 243, normalized size = 1.51 \begin {gather*} \frac {\sqrt {a+b x+c x^2} \left (128 a^2 c^2 e-100 a b^2 c e+200 a b c^2 d+56 a b c^2 e x+400 a c^3 d x+256 a c^3 e x^2+15 b^4 e-30 b^3 c d-10 b^3 c e x+20 b^2 c^2 d x+8 b^2 c^2 e x^2+240 b c^3 d x^2+176 b c^3 e x^3+160 c^4 d x^3+128 c^4 e x^4\right )}{640 c^3}+\frac {3 \left (16 a^2 b c^2 e-32 a^2 c^3 d-8 a b^3 c e+16 a b^2 c^2 d+b^5 e-2 b^4 c d\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[(d + e*x)*(a + b*x + c*x^2)^(3/2),x]

[Out]

(Sqrt[a + b*x + c*x^2]*(-30*b^3*c*d + 200*a*b*c^2*d + 15*b^4*e - 100*a*b^2*c*e + 128*a^2*c^2*e + 20*b^2*c^2*d*
x + 400*a*c^3*d*x - 10*b^3*c*e*x + 56*a*b*c^2*e*x + 240*b*c^3*d*x^2 + 8*b^2*c^2*e*x^2 + 256*a*c^3*e*x^2 + 160*
c^4*d*x^3 + 176*b*c^3*e*x^3 + 128*c^4*e*x^4))/(640*c^3) + (3*(-2*b^4*c*d + 16*a*b^2*c^2*d - 32*a^2*c^3*d + b^5
*e - 8*a*b^3*c*e + 16*a^2*b*c^2*e)*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 = 0.46, size = 529, normalized size = 3.29 \begin {gather*} \left [-\frac {15 \, {\left (2 \, {\left (b^{4} c - 8 \, a b^{2} c^{2} + 16 \, a^{2} c^{3}\right )} d - {\left (b^{5} - 8 \, a b^{3} c + 16 \, a^{2} b c^{2}\right )} e\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 \, c^{5} e x^{4} + 16 \, {\left (10 \, c^{5} d + 11 \, b c^{4} e\right )} x^{3} + 8 \, {\left (30 \, b c^{4} d + {\left (b^{2} c^{3} + 32 \, a c^{4}\right )} e\right )} x^{2} - 10 \, {\left (3 \, b^{3} c^{2} - 20 \, a b c^{3}\right )} d + {\left (15 \, b^{4} c - 100 \, a b^{2} c^{2} + 128 \, a^{2} c^{3}\right )} e + 2 \, {\left (10 \, {\left (b^{2} c^{3} + 20 \, a c^{4}\right )} d - {\left (5 \, b^{3} c^{2} - 28 \, a b c^{3}\right )} e\right )} x\right )} \sqrt {c x^{2} + b x + a}}{2560 \, c^{4}}, -\frac {15 \, {\left (2 \, {\left (b^{4} c - 8 \, a b^{2} c^{2} + 16 \, a^{2} c^{3}\right )} d - {\left (b^{5} - 8 \, a b^{3} c + 16 \, a^{2} b c^{2}\right )} e\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 \, c^{5} e x^{4} + 16 \, {\left (10 \, c^{5} d + 11 \, b c^{4} e\right )} x^{3} + 8 \, {\left (30 \, b c^{4} d + {\left (b^{2} c^{3} + 32 \, a c^{4}\right )} e\right )} x^{2} - 10 \, {\left (3 \, b^{3} c^{2} - 20 \, a b c^{3}\right )} d + {\left (15 \, b^{4} c - 100 \, a b^{2} c^{2} + 128 \, a^{2} c^{3}\right )} e + 2 \, {\left (10 \, {\left (b^{2} c^{3} + 20 \, a c^{4}\right )} d - {\left (5 \, b^{3} c^{2} - 28 \, a b c^{3}\right )} e\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((e*x+d)*(c*x^2+b*x+a)^(3/2),x, algorithm="fricas")

[Out]

[-1/2560*(15*(2*(b^4*c - 8*a*b^2*c^2 + 16*a^2*c^3)*d - (b^5 - 8*a*b^3*c + 16*a^2*b*c^2)*e)*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*c^5*e*x^4 + 16*(10*c^5*d +
11*b*c^4*e)*x^3 + 8*(30*b*c^4*d + (b^2*c^3 + 32*a*c^4)*e)*x^2 - 10*(3*b^3*c^2 - 20*a*b*c^3)*d + (15*b^4*c - 1
00*a*b^2*c^2 + 128*a^2*c^3)*e + 2*(10*(b^2*c^3 + 20*a*c^4)*d - (5*b^3*c^2 - 28*a*b*c^3)*e)*x)*sqrt(c*x^2 + b*x
+ a))/c^4, -1/1280*(15*(2*(b^4*c - 8*a*b^2*c^2 + 16*a^2*c^3)*d - (b^5 - 8*a*b^3*c + 16*a^2*b*c^2)*e)*sqrt(-c)
*arctan(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*c^5*e*x^4 + 16*(10*c^
5*d + 11*b*c^4*e)*x^3 + 8*(30*b*c^4*d + (b^2*c^3 + 32*a*c^4)*e)*x^2 - 10*(3*b^3*c^2 - 20*a*b*c^3)*d + (15*b^4*
c - 100*a*b^2*c^2 + 128*a^2*c^3)*e + 2*(10*(b^2*c^3 + 20*a*c^4)*d - (5*b^3*c^2 - 28*a*b*c^3)*e)*x)*sqrt(c*x^2
+ b*x + a))/c^4]

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giac [A]  time = 0.29, size = 264, normalized size = 1.64 \begin {gather*} \frac {1}{640} \, \sqrt {c x^{2} + b x + a} {\left (2 \, {\left (4 \, {\left (2 \, {\left (8 \, c x e + \frac {10 \, c^{5} d + 11 \, b c^{4} e}{c^{4}}\right )} x + \frac {30 \, b c^{4} d + b^{2} c^{3} e + 32 \, a c^{4} e}{c^{4}}\right )} x + \frac {10 \, b^{2} c^{3} d + 200 \, a c^{4} d - 5 \, b^{3} c^{2} e + 28 \, a b c^{3} e}{c^{4}}\right )} x - \frac {30 \, b^{3} c^{2} d - 200 \, a b c^{3} d - 15 \, b^{4} c e + 100 \, a b^{2} c^{2} e - 128 \, a^{2} c^{3} e}{c^{4}}\right )} - \frac {3 \, {\left (2 \, b^{4} c d - 16 \, a b^{2} c^{2} d + 32 \, a^{2} c^{3} d - b^{5} e + 8 \, a b^{3} c e - 16 \, a^{2} b c^{2} e\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((e*x+d)*(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*c*x*e + (10*c^5*d + 11*b*c^4*e)/c^4)*x + (30*b*c^4*d + b^2*c^3*e + 32*
a*c^4*e)/c^4)*x + (10*b^2*c^3*d + 200*a*c^4*d - 5*b^3*c^2*e + 28*a*b*c^3*e)/c^4)*x - (30*b^3*c^2*d - 200*a*b*c
^3*d - 15*b^4*c*e + 100*a*b^2*c^2*e - 128*a^2*c^3*e)/c^4) - 3/256*(2*b^4*c*d - 16*a*b^2*c^2*d + 32*a^2*c^3*d -
b^5*e + 8*a*b^3*c*e - 16*a^2*b*c^2*e)*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 = 0.06, size = 469, normalized size = 2.91 \begin {gather*} -\frac {3 a^{2} b e \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^{2} d \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{8 \sqrt {c}}+\frac {3 a \,b^{3} e \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 a \,b^{2} d \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^{5} e \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 b^{4} d \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 \sqrt {c \,x^{2}+b x +a}\, a b e x}{16 c}+\frac {3 \sqrt {c \,x^{2}+b x +a}\, a d x}{8}+\frac {3 \sqrt {c \,x^{2}+b x +a}\, b^{3} e x}{64 c^{2}}-\frac {3 \sqrt {c \,x^{2}+b x +a}\, b^{2} d x}{32 c}-\frac {3 \sqrt {c \,x^{2}+b x +a}\, a \,b^{2} e}{32 c^{2}}+\frac {3 \sqrt {c \,x^{2}+b x +a}\, a b d}{16 c}+\frac {3 \sqrt {c \,x^{2}+b x +a}\, b^{4} e}{128 c^{3}}-\frac {3 \sqrt {c \,x^{2}+b x +a}\, b^{3} d}{64 c^{2}}-\frac {\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} b e x}{8 c}+\frac {\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} d x}{4}-\frac {\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} b^{2} e}{16 c^{2}}+\frac {\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} b d}{8 c}+\frac {\left (c \,x^{2}+b x +a \right )^{\frac {5}{2}} e}{5 c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)*(c*x^2+b*x+a)^(3/2),x)

[Out]

1/5*e*(c*x^2+b*x+a)^(5/2)/c-1/8*e*b/c*x*(c*x^2+b*x+a)^(3/2)-1/16*e*b^2/c^2*(c*x^2+b*x+a)^(3/2)-3/16*e*b/c*(c*x
^2+b*x+a)^(1/2)*x*a+3/64*e*b^3/c^2*(c*x^2+b*x+a)^(1/2)*x-3/32*e*b^2/c^2*(c*x^2+b*x+a)^(1/2)*a+3/128*e*b^4/c^3*
(c*x^2+b*x+a)^(1/2)-3/16*e*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*e*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*e*b^5/c^(7/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))+1/4*d*
x*(c*x^2+b*x+a)^(3/2)+1/8*d/c*(c*x^2+b*x+a)^(3/2)*b+3/8*d*(c*x^2+b*x+a)^(1/2)*x*a-3/32*d/c*(c*x^2+b*x+a)^(1/2)
*x*b^2+3/16*d/c*(c*x^2+b*x+a)^(1/2)*b*a-3/64*d/c^2*(c*x^2+b*x+a)^(1/2)*b^3+3/8*d/c^(1/2)*ln((c*x+1/2*b)/c^(1/2
)+(c*x^2+b*x+a)^(1/2))*a^2-3/16*d/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*d/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((e*x+d)*(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 = 1.51, size = 305, normalized size = 1.89 \begin {gather*} \frac {e\,{\left (c\,x^2+b\,x+a\right )}^{5/2}}{5\,c}+\frac {d\,\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\,e\,\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 {d\,\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((d + e*x)*(a + b*x + c*x^2)^(3/2),x)

[Out]

(e*(a + b*x + c*x^2)^(5/2))/(5*c) + (d*((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*e*((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) + (d*(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 (d + e 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((e*x+d)*(c*x**2+b*x+a)**(3/2),x)

[Out]

Integral((d + e*x)*(a + b*x + c*x**2)**(3/2), x)

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