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

Optimal. Leaf size=160 \[ \frac {e \left (b^2-4 a c\right )^3 \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{512 c^{7/2}}-\frac {e \left (b^2-4 a c\right )^2 (b+2 c x) \sqrt {a+b x+c x^2}}{256 c^3}+\frac {e \left (b^2-4 a c\right ) (b+2 c x) \left (a+b x+c x^2\right )^{3/2}}{96 c^2}+\frac {\left (a+b x+c x^2\right )^{5/2} (-b e+12 c d+10 c e x)}{30 c} \]

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Rubi [A]  time = 0.12, antiderivative size = 160, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {779, 612, 621, 206} \begin {gather*} -\frac {e \left (b^2-4 a c\right )^2 (b+2 c x) \sqrt {a+b x+c x^2}}{256 c^3}+\frac {e \left (b^2-4 a c\right ) (b+2 c x) \left (a+b x+c x^2\right )^{3/2}}{96 c^2}+\frac {e \left (b^2-4 a c\right )^3 \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{512 c^{7/2}}+\frac {\left (a+b x+c x^2\right )^{5/2} (-b e+12 c d+10 c e x)}{30 c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((b^2 - 4*a*c)^2*e*(b + 2*c*x)*Sqrt[a + b*x + c*x^2])/(256*c^3) + ((b^2 - 4*a*c)*e*(b + 2*c*x)*(a + b*x + c*x
^2)^(3/2))/(96*c^2) + ((12*c*d - b*e + 10*c*e*x)*(a + b*x + c*x^2)^(5/2))/(30*c) + ((b^2 - 4*a*c)^3*e*ArcTanh[
(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(512*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 779

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((b
*e*g*(p + 2) - c*(e*f + d*g)*(2*p + 3) - 2*c*e*g*(p + 1)*x)*(a + b*x + c*x^2)^(p + 1))/(2*c^2*(p + 1)*(2*p + 3
)), x] + Dist[(b^2*e*g*(p + 2) - 2*a*c*e*g + c*(2*c*d*f - b*(e*f + d*g))*(2*p + 3))/(2*c^2*(2*p + 3)), Int[(a
+ b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[b^2 - 4*a*c, 0] &&  !LeQ[p, -1]

Rubi steps

\begin {align*} \int (b+2 c x) (d+e x) \left (a+b x+c x^2\right )^{3/2} \, dx &=\frac {(12 c d-b e+10 c e x) \left (a+b x+c x^2\right )^{5/2}}{30 c}+\frac {\left (\left (b^2-4 a c\right ) e\right ) \int \left (a+b x+c x^2\right )^{3/2} \, dx}{12 c}\\ &=\frac {\left (b^2-4 a c\right ) e (b+2 c x) \left (a+b x+c x^2\right )^{3/2}}{96 c^2}+\frac {(12 c d-b e+10 c e x) \left (a+b x+c x^2\right )^{5/2}}{30 c}-\frac {\left (\left (b^2-4 a c\right )^2 e\right ) \int \sqrt {a+b x+c x^2} \, dx}{64 c^2}\\ &=-\frac {\left (b^2-4 a c\right )^2 e (b+2 c x) \sqrt {a+b x+c x^2}}{256 c^3}+\frac {\left (b^2-4 a c\right ) e (b+2 c x) \left (a+b x+c x^2\right )^{3/2}}{96 c^2}+\frac {(12 c d-b e+10 c e x) \left (a+b x+c x^2\right )^{5/2}}{30 c}+\frac {\left (\left (b^2-4 a c\right )^3 e\right ) \int \frac {1}{\sqrt {a+b x+c x^2}} \, dx}{512 c^3}\\ &=-\frac {\left (b^2-4 a c\right )^2 e (b+2 c x) \sqrt {a+b x+c x^2}}{256 c^3}+\frac {\left (b^2-4 a c\right ) e (b+2 c x) \left (a+b x+c x^2\right )^{3/2}}{96 c^2}+\frac {(12 c d-b e+10 c e x) \left (a+b x+c x^2\right )^{5/2}}{30 c}+\frac {\left (\left (b^2-4 a c\right )^3 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 )}{256 c^3}\\ &=-\frac {\left (b^2-4 a c\right )^2 e (b+2 c x) \sqrt {a+b x+c x^2}}{256 c^3}+\frac {\left (b^2-4 a c\right ) e (b+2 c x) \left (a+b x+c x^2\right )^{3/2}}{96 c^2}+\frac {(12 c d-b e+10 c e x) \left (a+b x+c x^2\right )^{5/2}}{30 c}+\frac {\left (b^2-4 a c\right )^3 e \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{512 c^{7/2}}\\ \end {align*}

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Mathematica [A]  time = 0.19, size = 147, normalized size = 0.92 \begin {gather*} \frac {e \left (b^2-4 a c\right ) \left (2 \sqrt {c} (b+2 c x) \sqrt {a+x (b+c x)} \left (4 c \left (5 a+2 c x^2\right )-3 b^2+8 b c x\right )+3 \left (b^2-4 a c\right )^2 \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+x (b+c x)}}\right )\right )}{1536 c^{7/2}}+\frac {(a+x (b+c x))^{5/2} (2 c (6 d+5 e x)-b e)}{30 c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((a + x*(b + c*x))^(5/2)*(-(b*e) + 2*c*(6*d + 5*e*x)))/(30*c) + ((b^2 - 4*a*c)*e*(2*Sqrt[c]*(b + 2*c*x)*Sqrt[a
+ x*(b + c*x)]*(-3*b^2 + 8*b*c*x + 4*c*(5*a + 2*c*x^2)) + 3*(b^2 - 4*a*c)^2*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sq
rt[a + x*(b + c*x)])]))/(1536*c^(7/2))

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IntegrateAlgebraic [A]  time = 1.16, size = 267, normalized size = 1.67 \begin {gather*} \frac {\sqrt {a+b x+c x^2} \left (-528 a^2 b c^2 e+1536 a^2 c^3 d+480 a^2 c^3 e x+160 a b^3 c e-96 a b^2 c^2 e x+3072 a b c^3 d x+1824 a b c^3 e x^2+3072 a c^4 d x^2+2240 a c^4 e x^3-15 b^5 e+10 b^4 c e x-8 b^3 c^2 e x^2+1536 b^2 c^3 d x^2+1104 b^2 c^3 e x^3+3072 b c^4 d x^3+2432 b c^4 e x^4+1536 c^5 d x^4+1280 c^5 e x^5\right )}{3840 c^3}-\frac {e \left (-64 a^3 c^3+48 a^2 b^2 c^2-12 a b^4 c+b^6\right ) \log \left (-2 \sqrt {c} \sqrt {a+b x+c x^2}+b+2 c x\right )}{512 c^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + b*x + c*x^2]*(1536*a^2*c^3*d - 15*b^5*e + 160*a*b^3*c*e - 528*a^2*b*c^2*e + 3072*a*b*c^3*d*x + 10*b^
4*c*e*x - 96*a*b^2*c^2*e*x + 480*a^2*c^3*e*x + 1536*b^2*c^3*d*x^2 + 3072*a*c^4*d*x^2 - 8*b^3*c^2*e*x^2 + 1824*
a*b*c^3*e*x^2 + 3072*b*c^4*d*x^3 + 1104*b^2*c^3*e*x^3 + 2240*a*c^4*e*x^3 + 1536*c^5*d*x^4 + 2432*b*c^4*e*x^4 +
1280*c^5*e*x^5))/(3840*c^3) - ((b^6 - 12*a*b^4*c + 48*a^2*b^2*c^2 - 64*a^3*c^3)*e*Log[b + 2*c*x - 2*Sqrt[c]*S
qrt[a + b*x + c*x^2]])/(512*c^(7/2))

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fricas [B]  time = 0.48, size = 559, normalized size = 3.49 \begin {gather*} \left [-\frac {15 \, {\left (b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}\right )} \sqrt {c} e \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 (1280 \, c^{6} e x^{5} + 1536 \, a^{2} c^{4} d + 128 \, {\left (12 \, c^{6} d + 19 \, b c^{5} e\right )} x^{4} + 16 \, {\left (192 \, b c^{5} d + {\left (69 \, b^{2} c^{4} + 140 \, a c^{5}\right )} e\right )} x^{3} + 8 \, {\left (192 \, {\left (b^{2} c^{4} + 2 \, a c^{5}\right )} d - {\left (b^{3} c^{3} - 228 \, a b c^{4}\right )} e\right )} x^{2} - {\left (15 \, b^{5} c - 160 \, a b^{3} c^{2} + 528 \, a^{2} b c^{3}\right )} e + 2 \, {\left (1536 \, a b c^{4} d + {\left (5 \, b^{4} c^{2} - 48 \, a b^{2} c^{3} + 240 \, a^{2} c^{4}\right )} e\right )} x\right )} \sqrt {c x^{2} + b x + a}}{15360 \, c^{4}}, -\frac {15 \, {\left (b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}\right )} \sqrt {-c} e \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 (1280 \, c^{6} e x^{5} + 1536 \, a^{2} c^{4} d + 128 \, {\left (12 \, c^{6} d + 19 \, b c^{5} e\right )} x^{4} + 16 \, {\left (192 \, b c^{5} d + {\left (69 \, b^{2} c^{4} + 140 \, a c^{5}\right )} e\right )} x^{3} + 8 \, {\left (192 \, {\left (b^{2} c^{4} + 2 \, a c^{5}\right )} d - {\left (b^{3} c^{3} - 228 \, a b c^{4}\right )} e\right )} x^{2} - {\left (15 \, b^{5} c - 160 \, a b^{3} c^{2} + 528 \, a^{2} b c^{3}\right )} e + 2 \, {\left (1536 \, a b c^{4} d + {\left (5 \, b^{4} c^{2} - 48 \, a b^{2} c^{3} + 240 \, a^{2} c^{4}\right )} e\right )} x\right )} \sqrt {c x^{2} + b x + a}}{7680 \, c^{4}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*x+b)*(e*x+d)*(c*x^2+b*x+a)^(3/2),x, algorithm="fricas")

[Out]

[-1/15360*(15*(b^6 - 12*a*b^4*c + 48*a^2*b^2*c^2 - 64*a^3*c^3)*sqrt(c)*e*log(-8*c^2*x^2 - 8*b*c*x - b^2 + 4*sq
rt(c*x^2 + b*x + a)*(2*c*x + b)*sqrt(c) - 4*a*c) - 4*(1280*c^6*e*x^5 + 1536*a^2*c^4*d + 128*(12*c^6*d + 19*b*c
^5*e)*x^4 + 16*(192*b*c^5*d + (69*b^2*c^4 + 140*a*c^5)*e)*x^3 + 8*(192*(b^2*c^4 + 2*a*c^5)*d - (b^3*c^3 - 228*
a*b*c^4)*e)*x^2 - (15*b^5*c - 160*a*b^3*c^2 + 528*a^2*b*c^3)*e + 2*(1536*a*b*c^4*d + (5*b^4*c^2 - 48*a*b^2*c^3
+ 240*a^2*c^4)*e)*x)*sqrt(c*x^2 + b*x + a))/c^4, -1/7680*(15*(b^6 - 12*a*b^4*c + 48*a^2*b^2*c^2 - 64*a^3*c^3)
*sqrt(-c)*e*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*(1280*c^6*e*x^5
+ 1536*a^2*c^4*d + 128*(12*c^6*d + 19*b*c^5*e)*x^4 + 16*(192*b*c^5*d + (69*b^2*c^4 + 140*a*c^5)*e)*x^3 + 8*(1
92*(b^2*c^4 + 2*a*c^5)*d - (b^3*c^3 - 228*a*b*c^4)*e)*x^2 - (15*b^5*c - 160*a*b^3*c^2 + 528*a^2*b*c^3)*e + 2*(
1536*a*b*c^4*d + (5*b^4*c^2 - 48*a*b^2*c^3 + 240*a^2*c^4)*e)*x)*sqrt(c*x^2 + b*x + a))/c^4]

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giac [B]  time = 0.26, size = 294, normalized size = 1.84 \begin {gather*} \frac {1}{3840} \, \sqrt {c x^{2} + b x + a} {\left (2 \, {\left (4 \, {\left (2 \, {\left (8 \, {\left (10 \, c^{2} x e + \frac {12 \, c^{7} d + 19 \, b c^{6} e}{c^{5}}\right )} x + \frac {192 \, b c^{6} d + 69 \, b^{2} c^{5} e + 140 \, a c^{6} e}{c^{5}}\right )} x + \frac {192 \, b^{2} c^{5} d + 384 \, a c^{6} d - b^{3} c^{4} e + 228 \, a b c^{5} e}{c^{5}}\right )} x + \frac {1536 \, a b c^{5} d + 5 \, b^{4} c^{3} e - 48 \, a b^{2} c^{4} e + 240 \, a^{2} c^{5} e}{c^{5}}\right )} x + \frac {1536 \, a^{2} c^{5} d - 15 \, b^{5} c^{2} e + 160 \, a b^{3} c^{3} e - 528 \, a^{2} b c^{4} e}{c^{5}}\right )} - \frac {{\left (b^{6} e - 12 \, a b^{4} c e + 48 \, a^{2} b^{2} c^{2} e - 64 \, a^{3} c^{3} e\right )} \log \left ({\left | -2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} \sqrt {c} - b \right |}\right )}{512 \, c^{\frac {7}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*x+b)*(e*x+d)*(c*x^2+b*x+a)^(3/2),x, algorithm="giac")

[Out]

1/3840*sqrt(c*x^2 + b*x + a)*(2*(4*(2*(8*(10*c^2*x*e + (12*c^7*d + 19*b*c^6*e)/c^5)*x + (192*b*c^6*d + 69*b^2*
c^5*e + 140*a*c^6*e)/c^5)*x + (192*b^2*c^5*d + 384*a*c^6*d - b^3*c^4*e + 228*a*b*c^5*e)/c^5)*x + (1536*a*b*c^5
*d + 5*b^4*c^3*e - 48*a*b^2*c^4*e + 240*a^2*c^5*e)/c^5)*x + (1536*a^2*c^5*d - 15*b^5*c^2*e + 160*a*b^3*c^3*e -
528*a^2*b*c^4*e)/c^5) - 1/512*(b^6*e - 12*a*b^4*c*e + 48*a^2*b^2*c^2*e - 64*a^3*c^3*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 = 401, normalized size = 2.51 \begin {gather*} -\frac {a^{3} e \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{8 \sqrt {c}}+\frac {3 a^{2} b^{2} e \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{32 c^{\frac {3}{2}}}-\frac {3 a \,b^{4} e \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{128 c^{\frac {5}{2}}}+\frac {b^{6} e \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{512 c^{\frac {7}{2}}}-\frac {\sqrt {c \,x^{2}+b x +a}\, a^{2} e x}{8}+\frac {\sqrt {c \,x^{2}+b x +a}\, a \,b^{2} e x}{16 c}-\frac {\sqrt {c \,x^{2}+b x +a}\, b^{4} e x}{128 c^{2}}-\frac {\sqrt {c \,x^{2}+b x +a}\, a^{2} b e}{16 c}+\frac {\sqrt {c \,x^{2}+b x +a}\, a \,b^{3} e}{32 c^{2}}-\frac {\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} a e x}{12}-\frac {\sqrt {c \,x^{2}+b x +a}\, b^{5} e}{256 c^{3}}+\frac {\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} b^{2} e x}{48 c}-\frac {\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} a b e}{24 c}+\frac {\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} b^{3} e}{96 c^{2}}+\frac {\left (c \,x^{2}+b x +a \right )^{\frac {5}{2}} e x}{3}-\frac {\left (c \,x^{2}+b x +a \right )^{\frac {5}{2}} b e}{30 c}+\frac {2 \left (c \,x^{2}+b x +a \right )^{\frac {5}{2}} d}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*e*a^2*(c*x^2+b*x+a)^(1/2)*x-1/12*e*a*x*(c*x^2+b*x+a)^(3/2)+1/96/c^2*e*b^3*(c*x^2+b*x+a)^(3/2)-1/256/c^3*e
*b^5*(c*x^2+b*x+a)^(1/2)-1/30/c*e*b*(c*x^2+b*x+a)^(5/2)-3/128/c^(5/2)*e*b^4*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+
a)^(1/2))*a+1/16/c*e*b^2*(c*x^2+b*x+a)^(1/2)*x*a+3/32/c^(3/2)*e*b^2*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2)
)*a^2-1/16/c*e*a^2*(c*x^2+b*x+a)^(1/2)*b-1/8/c^(1/2)*e*a^3*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))-1/24/c*
e*a*(c*x^2+b*x+a)^(3/2)*b+1/512/c^(7/2)*e*b^6*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))-1/128/c^2*e*b^4*(c*x
^2+b*x+a)^(1/2)*x+1/32/c^2*e*b^3*(c*x^2+b*x+a)^(1/2)*a+1/48/c*e*b^2*x*(c*x^2+b*x+a)^(3/2)+2/5*(c*x^2+b*x+a)^(5
/2)*d+1/3*e*x*(c*x^2+b*x+a)^(5/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (b + 2 c x\right ) \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((2*c*x+b)*(e*x+d)*(c*x**2+b*x+a)**(3/2),x)

[Out]

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

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