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

Optimal. Leaf size=212 \[ \frac {\left (b^2-4 a c\right )^2 \left (-4 a c C+24 A c^2+7 b^2 C\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{1024 c^{9/2}}-\frac {\left (b^2-4 a c\right ) (b+2 c x) \sqrt {a+b x+c x^2} \left (-4 a c C+24 A c^2+7 b^2 C\right )}{512 c^4}+\frac {(b+2 c x) \left (a+b x+c x^2\right )^{3/2} \left (-4 a c C+24 A c^2+7 b^2 C\right )}{192 c^3}-\frac {7 b C \left (a+b x+c x^2\right )^{5/2}}{60 c^2}+\frac {C x \left (a+b x+c x^2\right )^{5/2}}{6 c} \]

[Out]

1/192*(24*A*c^2-4*C*a*c+7*C*b^2)*(2*c*x+b)*(c*x^2+b*x+a)^(3/2)/c^3-7/60*b*C*(c*x^2+b*x+a)^(5/2)/c^2+1/6*C*x*(c
*x^2+b*x+a)^(5/2)/c+1/1024*(-4*a*c+b^2)^2*(24*A*c^2-4*C*a*c+7*C*b^2)*arctanh(1/2*(2*c*x+b)/c^(1/2)/(c*x^2+b*x+
a)^(1/2))/c^(9/2)-1/512*(-4*a*c+b^2)*(24*A*c^2-4*C*a*c+7*C*b^2)*(2*c*x+b)*(c*x^2+b*x+a)^(1/2)/c^4

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Rubi [A]  time = 0.18, antiderivative size = 212, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {1661, 640, 612, 621, 206} \[ \frac {(b+2 c x) \left (a+b x+c x^2\right )^{3/2} \left (-4 a c C+24 A c^2+7 b^2 C\right )}{192 c^3}-\frac {\left (b^2-4 a c\right ) (b+2 c x) \sqrt {a+b x+c x^2} \left (-4 a c C+24 A c^2+7 b^2 C\right )}{512 c^4}+\frac {\left (b^2-4 a c\right )^2 \left (-4 a c C+24 A c^2+7 b^2 C\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{1024 c^{9/2}}-\frac {7 b C \left (a+b x+c x^2\right )^{5/2}}{60 c^2}+\frac {C x \left (a+b x+c x^2\right )^{5/2}}{6 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((b^2 - 4*a*c)*(24*A*c^2 + 7*b^2*C - 4*a*c*C)*(b + 2*c*x)*Sqrt[a + b*x + c*x^2])/(512*c^4) + ((24*A*c^2 + 7*b
^2*C - 4*a*c*C)*(b + 2*c*x)*(a + b*x + c*x^2)^(3/2))/(192*c^3) - (7*b*C*(a + b*x + c*x^2)^(5/2))/(60*c^2) + (C
*x*(a + b*x + c*x^2)^(5/2))/(6*c) + ((b^2 - 4*a*c)^2*(24*A*c^2 + 7*b^2*C - 4*a*c*C)*ArcTanh[(b + 2*c*x)/(2*Sqr
t[c]*Sqrt[a + b*x + c*x^2])])/(1024*c^(9/2))

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]

Rule 1661

Int[(Pq_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], e = Coeff[Pq, x, Expo
n[Pq, x]]}, Simp[(e*x^(q - 1)*(a + b*x + c*x^2)^(p + 1))/(c*(q + 2*p + 1)), x] + Dist[1/(c*(q + 2*p + 1)), Int
[(a + b*x + c*x^2)^p*ExpandToSum[c*(q + 2*p + 1)*Pq - a*e*(q - 1)*x^(q - 2) - b*e*(q + p)*x^(q - 1) - c*e*(q +
2*p + 1)*x^q, x], x], x]] /; FreeQ[{a, b, c, p}, x] && PolyQ[Pq, x] && NeQ[b^2 - 4*a*c, 0] &&  !LeQ[p, -1]

Rubi steps

\begin {align*} \int \left (a+b x+c x^2\right )^{3/2} \left (A+C x^2\right ) \, dx &=\frac {C x \left (a+b x+c x^2\right )^{5/2}}{6 c}+\frac {\int \left (6 A c-a C-\frac {7 b C x}{2}\right ) \left (a+b x+c x^2\right )^{3/2} \, dx}{6 c}\\ &=-\frac {7 b C \left (a+b x+c x^2\right )^{5/2}}{60 c^2}+\frac {C x \left (a+b x+c x^2\right )^{5/2}}{6 c}+\frac {\left (\frac {7 b^2 C}{2}+2 c (6 A c-a C)\right ) \int \left (a+b x+c x^2\right )^{3/2} \, dx}{12 c^2}\\ &=\frac {\left (24 A c^2+7 b^2 C-4 a c C\right ) (b+2 c x) \left (a+b x+c x^2\right )^{3/2}}{192 c^3}-\frac {7 b C \left (a+b x+c x^2\right )^{5/2}}{60 c^2}+\frac {C x \left (a+b x+c x^2\right )^{5/2}}{6 c}-\frac {\left (\left (b^2-4 a c\right ) \left (24 A c^2+7 b^2 C-4 a c C\right )\right ) \int \sqrt {a+b x+c x^2} \, dx}{128 c^3}\\ &=-\frac {\left (b^2-4 a c\right ) \left (24 A c^2+7 b^2 C-4 a c C\right ) (b+2 c x) \sqrt {a+b x+c x^2}}{512 c^4}+\frac {\left (24 A c^2+7 b^2 C-4 a c C\right ) (b+2 c x) \left (a+b x+c x^2\right )^{3/2}}{192 c^3}-\frac {7 b C \left (a+b x+c x^2\right )^{5/2}}{60 c^2}+\frac {C x \left (a+b x+c x^2\right )^{5/2}}{6 c}+\frac {\left (\left (b^2-4 a c\right )^2 \left (24 A c^2+7 b^2 C-4 a c C\right )\right ) \int \frac {1}{\sqrt {a+b x+c x^2}} \, dx}{1024 c^4}\\ &=-\frac {\left (b^2-4 a c\right ) \left (24 A c^2+7 b^2 C-4 a c C\right ) (b+2 c x) \sqrt {a+b x+c x^2}}{512 c^4}+\frac {\left (24 A c^2+7 b^2 C-4 a c C\right ) (b+2 c x) \left (a+b x+c x^2\right )^{3/2}}{192 c^3}-\frac {7 b C \left (a+b x+c x^2\right )^{5/2}}{60 c^2}+\frac {C x \left (a+b x+c x^2\right )^{5/2}}{6 c}+\frac {\left (\left (b^2-4 a c\right )^2 \left (24 A c^2+7 b^2 C-4 a c C\right )\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 )}{512 c^4}\\ &=-\frac {\left (b^2-4 a c\right ) \left (24 A c^2+7 b^2 C-4 a c C\right ) (b+2 c x) \sqrt {a+b x+c x^2}}{512 c^4}+\frac {\left (24 A c^2+7 b^2 C-4 a c C\right ) (b+2 c x) \left (a+b x+c x^2\right )^{3/2}}{192 c^3}-\frac {7 b C \left (a+b x+c x^2\right )^{5/2}}{60 c^2}+\frac {C x \left (a+b x+c x^2\right )^{5/2}}{6 c}+\frac {\left (b^2-4 a c\right )^2 \left (24 A c^2+7 b^2 C-4 a c C\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{1024 c^{9/2}}\\ \end {align*}

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Mathematica [A]  time = 0.58, size = 267, normalized size = 1.26 \[ \frac {\frac {360 A \left (b^2-4 a c\right ) \left (\left (b^2-4 a c\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+x (b+c x)}}\right )-2 \sqrt {c} (b+2 c x) \sqrt {a+x (b+c x)}\right )}{c^{3/2}}+1920 A (b+2 c x) (a+x (b+c x))^{3/2}+\frac {C \left (5 \left (7 b^2-4 a c\right ) \left (\frac {3 \left (b^2-4 a c\right ) \left (\left (b^2-4 a c\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+x (b+c x)}}\right )-2 \sqrt {c} (b+2 c x) \sqrt {a+x (b+c x)}\right )}{c^{5/2}}+\frac {16 (b+2 c x) (a+x (b+c x))^{3/2}}{c}\right )-1792 b (a+x (b+c x))^{5/2}\right )}{c}+2560 C x (a+x (b+c x))^{5/2}}{15360 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(1920*A*(b + 2*c*x)*(a + x*(b + c*x))^(3/2) + 2560*C*x*(a + x*(b + c*x))^(5/2) + (360*A*(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)])]))
/c^(3/2) + (C*(-1792*b*(a + x*(b + c*x))^(5/2) + 5*(7*b^2 - 4*a*c)*((16*(b + 2*c*x)*(a + x*(b + c*x))^(3/2))/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)])]))/c^(5/2))))/c)/(15360*c)

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fricas [A]  time = 1.09, size = 605, normalized size = 2.85 \[ \left [\frac {15 \, {\left (7 \, C b^{6} - 60 \, C a b^{4} c + 384 \, A a^{2} c^{4} - 64 \, {\left (C a^{3} + 3 \, A a b^{2}\right )} c^{3} + 24 \, {\left (6 \, C a^{2} b^{2} + A b^{4}\right )} c^{2}\right )} \sqrt {c} \log \left (-8 \, c^{2} x^{2} - 8 \, b c x - b^{2} - 4 \, \sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )} \sqrt {c} - 4 \, a c\right ) + 4 \, {\left (1280 \, C c^{6} x^{5} + 1664 \, C b c^{5} x^{4} - 105 \, C b^{5} c + 760 \, C a b^{3} c^{2} + 2400 \, A a b c^{4} - 72 \, {\left (18 \, C a^{2} b + 5 \, A b^{3}\right )} c^{3} + 16 \, {\left (3 \, C b^{2} c^{4} + 140 \, C a c^{5} + 120 \, A c^{6}\right )} x^{3} - 8 \, {\left (7 \, C b^{3} c^{3} - 36 \, C a b c^{4} - 360 \, A b c^{5}\right )} x^{2} + 2 \, {\left (35 \, C b^{4} c^{2} - 216 \, C a b^{2} c^{3} + 2400 \, A a c^{5} + 120 \, {\left (2 \, C a^{2} + A b^{2}\right )} c^{4}\right )} x\right )} \sqrt {c x^{2} + b x + a}}{30720 \, c^{5}}, -\frac {15 \, {\left (7 \, C b^{6} - 60 \, C a b^{4} c + 384 \, A a^{2} c^{4} - 64 \, {\left (C a^{3} + 3 \, A a b^{2}\right )} c^{3} + 24 \, {\left (6 \, C a^{2} b^{2} + A b^{4}\right )} c^{2}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )} \sqrt {-c}}{2 \, {\left (c^{2} x^{2} + b c x + a c\right )}}\right ) - 2 \, {\left (1280 \, C c^{6} x^{5} + 1664 \, C b c^{5} x^{4} - 105 \, C b^{5} c + 760 \, C a b^{3} c^{2} + 2400 \, A a b c^{4} - 72 \, {\left (18 \, C a^{2} b + 5 \, A b^{3}\right )} c^{3} + 16 \, {\left (3 \, C b^{2} c^{4} + 140 \, C a c^{5} + 120 \, A c^{6}\right )} x^{3} - 8 \, {\left (7 \, C b^{3} c^{3} - 36 \, C a b c^{4} - 360 \, A b c^{5}\right )} x^{2} + 2 \, {\left (35 \, C b^{4} c^{2} - 216 \, C a b^{2} c^{3} + 2400 \, A a c^{5} + 120 \, {\left (2 \, C a^{2} + A b^{2}\right )} c^{4}\right )} x\right )} \sqrt {c x^{2} + b x + a}}{15360 \, c^{5}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/30720*(15*(7*C*b^6 - 60*C*a*b^4*c + 384*A*a^2*c^4 - 64*(C*a^3 + 3*A*a*b^2)*c^3 + 24*(6*C*a^2*b^2 + A*b^4)*c
^2)*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*(1280*C*
c^6*x^5 + 1664*C*b*c^5*x^4 - 105*C*b^5*c + 760*C*a*b^3*c^2 + 2400*A*a*b*c^4 - 72*(18*C*a^2*b + 5*A*b^3)*c^3 +
16*(3*C*b^2*c^4 + 140*C*a*c^5 + 120*A*c^6)*x^3 - 8*(7*C*b^3*c^3 - 36*C*a*b*c^4 - 360*A*b*c^5)*x^2 + 2*(35*C*b^
4*c^2 - 216*C*a*b^2*c^3 + 2400*A*a*c^5 + 120*(2*C*a^2 + A*b^2)*c^4)*x)*sqrt(c*x^2 + b*x + a))/c^5, -1/15360*(1
5*(7*C*b^6 - 60*C*a*b^4*c + 384*A*a^2*c^4 - 64*(C*a^3 + 3*A*a*b^2)*c^3 + 24*(6*C*a^2*b^2 + A*b^4)*c^2)*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*(1280*C*c^6*x^5 + 1664*C*
b*c^5*x^4 - 105*C*b^5*c + 760*C*a*b^3*c^2 + 2400*A*a*b*c^4 - 72*(18*C*a^2*b + 5*A*b^3)*c^3 + 16*(3*C*b^2*c^4 +
140*C*a*c^5 + 120*A*c^6)*x^3 - 8*(7*C*b^3*c^3 - 36*C*a*b*c^4 - 360*A*b*c^5)*x^2 + 2*(35*C*b^4*c^2 - 216*C*a*b
^2*c^3 + 2400*A*a*c^5 + 120*(2*C*a^2 + A*b^2)*c^4)*x)*sqrt(c*x^2 + b*x + a))/c^5]

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giac [A]  time = 0.28, size = 297, normalized size = 1.40 \[ \frac {1}{7680} \, \sqrt {c x^{2} + b x + a} {\left (2 \, {\left (4 \, {\left (2 \, {\left (8 \, {\left (10 \, C c x + 13 \, C b\right )} x + \frac {3 \, C b^{2} c^{4} + 140 \, C a c^{5} + 120 \, A c^{6}}{c^{5}}\right )} x - \frac {7 \, C b^{3} c^{3} - 36 \, C a b c^{4} - 360 \, A b c^{5}}{c^{5}}\right )} x + \frac {35 \, C b^{4} c^{2} - 216 \, C a b^{2} c^{3} + 240 \, C a^{2} c^{4} + 120 \, A b^{2} c^{4} + 2400 \, A a c^{5}}{c^{5}}\right )} x - \frac {105 \, C b^{5} c - 760 \, C a b^{3} c^{2} + 1296 \, C a^{2} b c^{3} + 360 \, A b^{3} c^{3} - 2400 \, A a b c^{4}}{c^{5}}\right )} - \frac {{\left (7 \, C b^{6} - 60 \, C a b^{4} c + 144 \, C a^{2} b^{2} c^{2} + 24 \, A b^{4} c^{2} - 64 \, C a^{3} c^{3} - 192 \, A a b^{2} c^{3} + 384 \, A a^{2} c^{4}\right )} \log \left ({\left | -2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} \sqrt {c} - b \right |}\right )}{1024 \, c^{\frac {9}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/7680*sqrt(c*x^2 + b*x + a)*(2*(4*(2*(8*(10*C*c*x + 13*C*b)*x + (3*C*b^2*c^4 + 140*C*a*c^5 + 120*A*c^6)/c^5)*
x - (7*C*b^3*c^3 - 36*C*a*b*c^4 - 360*A*b*c^5)/c^5)*x + (35*C*b^4*c^2 - 216*C*a*b^2*c^3 + 240*C*a^2*c^4 + 120*
A*b^2*c^4 + 2400*A*a*c^5)/c^5)*x - (105*C*b^5*c - 760*C*a*b^3*c^2 + 1296*C*a^2*b*c^3 + 360*A*b^3*c^3 - 2400*A*
a*b*c^4)/c^5) - 1/1024*(7*C*b^6 - 60*C*a*b^4*c + 144*C*a^2*b^2*c^2 + 24*A*b^4*c^2 - 64*C*a^3*c^3 - 192*A*a*b^2
*c^3 + 384*A*a^2*c^4)*log(abs(-2*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*sqrt(c) - b))/c^(9/2)

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maple [B]  time = 0.01, size = 613, normalized size = 2.89 \[ \frac {3 A \,a^{2} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{8 \sqrt {c}}-\frac {3 A a \,b^{2} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{16 c^{\frac {3}{2}}}+\frac {3 A \,b^{4} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{128 c^{\frac {5}{2}}}-\frac {C \,a^{3} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{16 c^{\frac {3}{2}}}+\frac {9 C \,a^{2} b^{2} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{64 c^{\frac {5}{2}}}-\frac {15 C a \,b^{4} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{256 c^{\frac {7}{2}}}+\frac {7 C \,b^{6} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{1024 c^{\frac {9}{2}}}+\frac {3 \sqrt {c \,x^{2}+b x +a}\, A a x}{8}-\frac {3 \sqrt {c \,x^{2}+b x +a}\, A \,b^{2} x}{32 c}-\frac {\sqrt {c \,x^{2}+b x +a}\, C \,a^{2} x}{16 c}+\frac {\sqrt {c \,x^{2}+b x +a}\, C a \,b^{2} x}{8 c^{2}}-\frac {7 \sqrt {c \,x^{2}+b x +a}\, C \,b^{4} x}{256 c^{3}}+\frac {3 \sqrt {c \,x^{2}+b x +a}\, A a b}{16 c}-\frac {3 \sqrt {c \,x^{2}+b x +a}\, A \,b^{3}}{64 c^{2}}+\frac {\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} A x}{4}-\frac {\sqrt {c \,x^{2}+b x +a}\, C \,a^{2} b}{32 c^{2}}+\frac {\sqrt {c \,x^{2}+b x +a}\, C a \,b^{3}}{16 c^{3}}-\frac {\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} C a x}{24 c}-\frac {7 \sqrt {c \,x^{2}+b x +a}\, C \,b^{5}}{512 c^{4}}+\frac {7 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} C \,b^{2} x}{96 c^{2}}+\frac {\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} A b}{8 c}-\frac {\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} C a b}{48 c^{2}}+\frac {7 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} C \,b^{3}}{192 c^{3}}+\frac {\left (c \,x^{2}+b x +a \right )^{\frac {5}{2}} C x}{6 c}-\frac {7 \left (c \,x^{2}+b x +a \right )^{\frac {5}{2}} C b}{60 c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*C/c^2*b^2*(c*x^2+b*x+a)^(1/2)*x*a-7/60*b*C*(c*x^2+b*x+a)^(5/2)/c^2+1/6*C*x*(c*x^2+b*x+a)^(5/2)/c+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+3/128*A/c^(5/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2
))*b^4+7/192*C/c^3*b^3*(c*x^2+b*x+a)^(3/2)-7/512*C/c^4*b^5*(c*x^2+b*x+a)^(1/2)+7/1024*C/c^(9/2)*b^6*ln((c*x+1/
2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))-1/16*C*a^3/c^(3/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))+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-3/64*A/c^2*(c*x^2+b*x+a)^(1/2)*b^3-3/32*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+1/4*A*(c*x^2+b*x+a)^(3/2)*x-1/32*C*a^2/c^2*(c*x^2+b*x+a)^(1/2)*b+7/96*
C/c^2*b^2*(c*x^2+b*x+a)^(3/2)*x-7/256*C/c^3*b^4*(c*x^2+b*x+a)^(1/2)*x+1/16*C/c^3*b^3*(c*x^2+b*x+a)^(1/2)*a-3/1
6*A/c^(3/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*b^2*a+9/64*C/c^(5/2)*b^2*ln((c*x+1/2*b)/c^(1/2)+(c*x^2
+b*x+a)^(1/2))*a^2-15/256*C/c^(7/2)*b^4*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*a-1/24*C*a/c*(c*x^2+b*x+a)
^(3/2)*x-1/48*C*a/c^2*(c*x^2+b*x+a)^(3/2)*b-1/16*C*a^2/c*(c*x^2+b*x+a)^(1/2)*x

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x+a)^(3/2)*(C*x^2+A),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 = -0.00 \[ \int \left (C\,x^2+A\right )\,{\left (c\,x^2+b\,x+a\right )}^{3/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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