### 3.1203 $$\int (b d+2 c d x)^4 (a+b x+c x^2)^{3/2} \, dx$$

Optimal. Leaf size=207 $-\frac{d^4 \left (b^2-4 a c\right ) (b+2 c x)^5 \sqrt{a+b x+c x^2}}{128 c^2}+\frac{d^4 \left (b^2-4 a c\right )^2 (b+2 c x)^3 \sqrt{a+b x+c x^2}}{512 c^2}+\frac{3 d^4 \left (b^2-4 a c\right )^3 (b+2 c x) \sqrt{a+b x+c x^2}}{1024 c^2}+\frac{3 d^4 \left (b^2-4 a c\right )^4 \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{2048 c^{5/2}}+\frac{d^4 (b+2 c x)^5 \left (a+b x+c x^2\right )^{3/2}}{16 c}$

[Out]

(3*(b^2 - 4*a*c)^3*d^4*(b + 2*c*x)*Sqrt[a + b*x + c*x^2])/(1024*c^2) + ((b^2 - 4*a*c)^2*d^4*(b + 2*c*x)^3*Sqrt
[a + b*x + c*x^2])/(512*c^2) - ((b^2 - 4*a*c)*d^4*(b + 2*c*x)^5*Sqrt[a + b*x + c*x^2])/(128*c^2) + (d^4*(b + 2
*c*x)^5*(a + b*x + c*x^2)^(3/2))/(16*c) + (3*(b^2 - 4*a*c)^4*d^4*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x +
c*x^2])])/(2048*c^(5/2))

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Rubi [A]  time = 0.128642, antiderivative size = 207, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 26, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.154, Rules used = {685, 692, 621, 206} $-\frac{d^4 \left (b^2-4 a c\right ) (b+2 c x)^5 \sqrt{a+b x+c x^2}}{128 c^2}+\frac{d^4 \left (b^2-4 a c\right )^2 (b+2 c x)^3 \sqrt{a+b x+c x^2}}{512 c^2}+\frac{3 d^4 \left (b^2-4 a c\right )^3 (b+2 c x) \sqrt{a+b x+c x^2}}{1024 c^2}+\frac{3 d^4 \left (b^2-4 a c\right )^4 \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{2048 c^{5/2}}+\frac{d^4 (b+2 c x)^5 \left (a+b x+c x^2\right )^{3/2}}{16 c}$

Antiderivative was successfully veriﬁed.

[In]

Int[(b*d + 2*c*d*x)^4*(a + b*x + c*x^2)^(3/2),x]

[Out]

(3*(b^2 - 4*a*c)^3*d^4*(b + 2*c*x)*Sqrt[a + b*x + c*x^2])/(1024*c^2) + ((b^2 - 4*a*c)^2*d^4*(b + 2*c*x)^3*Sqrt
[a + b*x + c*x^2])/(512*c^2) - ((b^2 - 4*a*c)*d^4*(b + 2*c*x)^5*Sqrt[a + b*x + c*x^2])/(128*c^2) + (d^4*(b + 2
*c*x)^5*(a + b*x + c*x^2)^(3/2))/(16*c) + (3*(b^2 - 4*a*c)^4*d^4*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x +
c*x^2])])/(2048*c^(5/2))

Rule 685

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x)^(m + 1)*(
a + b*x + c*x^2)^p)/(e*(m + 2*p + 1)), x] - Dist[(d*p*(b^2 - 4*a*c))/(b*e*(m + 2*p + 1)), Int[(d + e*x)^m*(a +
b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[2*c*d - b*e, 0] &&
NeQ[m + 2*p + 3, 0] && GtQ[p, 0] &&  !LtQ[m, -1] &&  !(IGtQ[(m - 1)/2, 0] && ( !IntegerQ[p] || LtQ[m, 2*p]))
&& RationalQ[m] && IntegerQ[2*p]

Rule 692

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(2*d*(d + e*x)^(m -
1)*(a + b*x + c*x^2)^(p + 1))/(b*(m + 2*p + 1)), x] + Dist[(d^2*(m - 1)*(b^2 - 4*a*c))/(b^2*(m + 2*p + 1)), In
t[(d + e*x)^(m - 2)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[
2*c*d - b*e, 0] && NeQ[m + 2*p + 3, 0] && GtQ[m, 1] && NeQ[m + 2*p + 1, 0] && (IntegerQ[2*p] || (IntegerQ[m] &
& RationalQ[p]) || OddQ[m])

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 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])

Rubi steps

\begin{align*} \int (b d+2 c d x)^4 \left (a+b x+c x^2\right )^{3/2} \, dx &=\frac{d^4 (b+2 c x)^5 \left (a+b x+c x^2\right )^{3/2}}{16 c}-\frac{\left (3 \left (b^2-4 a c\right )\right ) \int (b d+2 c d x)^4 \sqrt{a+b x+c x^2} \, dx}{32 c}\\ &=-\frac{\left (b^2-4 a c\right ) d^4 (b+2 c x)^5 \sqrt{a+b x+c x^2}}{128 c^2}+\frac{d^4 (b+2 c x)^5 \left (a+b x+c x^2\right )^{3/2}}{16 c}+\frac{\left (b^2-4 a c\right )^2 \int \frac{(b d+2 c d x)^4}{\sqrt{a+b x+c x^2}} \, dx}{256 c^2}\\ &=\frac{\left (b^2-4 a c\right )^2 d^4 (b+2 c x)^3 \sqrt{a+b x+c x^2}}{512 c^2}-\frac{\left (b^2-4 a c\right ) d^4 (b+2 c x)^5 \sqrt{a+b x+c x^2}}{128 c^2}+\frac{d^4 (b+2 c x)^5 \left (a+b x+c x^2\right )^{3/2}}{16 c}+\frac{\left (3 \left (b^2-4 a c\right )^3 d^2\right ) \int \frac{(b d+2 c d x)^2}{\sqrt{a+b x+c x^2}} \, dx}{1024 c^2}\\ &=\frac{3 \left (b^2-4 a c\right )^3 d^4 (b+2 c x) \sqrt{a+b x+c x^2}}{1024 c^2}+\frac{\left (b^2-4 a c\right )^2 d^4 (b+2 c x)^3 \sqrt{a+b x+c x^2}}{512 c^2}-\frac{\left (b^2-4 a c\right ) d^4 (b+2 c x)^5 \sqrt{a+b x+c x^2}}{128 c^2}+\frac{d^4 (b+2 c x)^5 \left (a+b x+c x^2\right )^{3/2}}{16 c}+\frac{\left (3 \left (b^2-4 a c\right )^4 d^4\right ) \int \frac{1}{\sqrt{a+b x+c x^2}} \, dx}{2048 c^2}\\ &=\frac{3 \left (b^2-4 a c\right )^3 d^4 (b+2 c x) \sqrt{a+b x+c x^2}}{1024 c^2}+\frac{\left (b^2-4 a c\right )^2 d^4 (b+2 c x)^3 \sqrt{a+b x+c x^2}}{512 c^2}-\frac{\left (b^2-4 a c\right ) d^4 (b+2 c x)^5 \sqrt{a+b x+c x^2}}{128 c^2}+\frac{d^4 (b+2 c x)^5 \left (a+b x+c x^2\right )^{3/2}}{16 c}+\frac{\left (3 \left (b^2-4 a c\right )^4 d^4\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 )}{1024 c^2}\\ &=\frac{3 \left (b^2-4 a c\right )^3 d^4 (b+2 c x) \sqrt{a+b x+c x^2}}{1024 c^2}+\frac{\left (b^2-4 a c\right )^2 d^4 (b+2 c x)^3 \sqrt{a+b x+c x^2}}{512 c^2}-\frac{\left (b^2-4 a c\right ) d^4 (b+2 c x)^5 \sqrt{a+b x+c x^2}}{128 c^2}+\frac{d^4 (b+2 c x)^5 \left (a+b x+c x^2\right )^{3/2}}{16 c}+\frac{3 \left (b^2-4 a c\right )^4 d^4 \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{2048 c^{5/2}}\\ \end{align*}

Mathematica [A]  time = 2.10874, size = 249, normalized size = 1.2 $\frac{1}{4} d^4 \left ((b+2 c x)^3 (a+x (b+c x))^{5/2}-2 c \left (a-\frac{b^2}{4 c}\right ) (b+2 c x) \sqrt{a+x (b+c x)} \left ((a+x (b+c x))^2-\frac{(a+x (b+c x)) \left (2 (b+2 c x) \sqrt{\frac{c (a+x (b+c x))}{4 a c-b^2}} \left (4 c \left (5 a+2 c x^2\right )-3 b^2+8 b c x\right )+3 \sqrt{c} \sqrt{4 a-\frac{b^2}{c}} \left (4 a c-b^2\right ) \sinh ^{-1}\left (\frac{b+2 c x}{\sqrt{c} \sqrt{4 a-\frac{b^2}{c}}}\right )\right )}{256 c (b+2 c x) \left (\frac{c (a+x (b+c x))}{4 a c-b^2}\right )^{3/2}}\right )\right )$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(b*d + 2*c*d*x)^4*(a + b*x + c*x^2)^(3/2),x]

[Out]

(d^4*((b + 2*c*x)^3*(a + x*(b + c*x))^(5/2) - 2*(a - b^2/(4*c))*c*(b + 2*c*x)*Sqrt[a + x*(b + c*x)]*((a + x*(b
+ c*x))^2 - ((a + x*(b + c*x))*(2*(b + 2*c*x)*Sqrt[(c*(a + x*(b + c*x)))/(-b^2 + 4*a*c)]*(-3*b^2 + 8*b*c*x +
4*c*(5*a + 2*c*x^2)) + 3*Sqrt[4*a - b^2/c]*Sqrt[c]*(-b^2 + 4*a*c)*ArcSinh[(b + 2*c*x)/(Sqrt[4*a - b^2/c]*Sqrt[
c])]))/(256*c*(b + 2*c*x)*((c*(a + x*(b + c*x)))/(-b^2 + 4*a*c))^(3/2)))))/4

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Maple [B]  time = 0.056, size = 641, normalized size = 3.1 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((2*c*d*x+b*d)^4*(c*x^2+b*x+a)^(3/2),x)

[Out]

3*d^4*c^2*b*x^2*(c*x^2+b*x+a)^(5/2)-1/2*d^4*c*b*a*(c*x^2+b*x+a)^(5/2)+7/4*d^4*c*b^2*x*(c*x^2+b*x+a)^(5/2)-3/51
2*d^4/c*b^6*(c*x^2+b*x+a)^(1/2)*x+9/128*d^4*b^4*(c*x^2+b*x+a)^(1/2)*x*a+9/64*d^4*b^4/c^(1/2)*ln((1/2*b+c*x)/c^
(1/2)+(c*x^2+b*x+a)^(1/2))*a^2-3/128*d^4*b^6/c^(3/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*a-3/8*d^4*c^(
1/2)*b^2*a^3*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))+1/4*d^4*c^2*a^2*x*(c*x^2+b*x+a)^(3/2)+1/8*d^4*c*a^2*(
c*x^2+b*x+a)^(3/2)*b+3/8*d^4*c^2*a^3*(c*x^2+b*x+a)^(1/2)*x+3/16*d^4*c*a^3*(c*x^2+b*x+a)^(1/2)*b-d^4*c^2*a*x*(c
*x^2+b*x+a)^(5/2)+2*d^4*c^3*x^3*(c*x^2+b*x+a)^(5/2)+1/128*d^4/c*b^5*(c*x^2+b*x+a)^(3/2)-3/1024*d^4/c^2*b^7*(c*
x^2+b*x+a)^(1/2)+3/8*d^4*c^(3/2)*a^4*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))-1/16*d^4*b^3*a*(c*x^2+b*x+a)^
(3/2)-9/64*d^4*b^3*a^2*(c*x^2+b*x+a)^(1/2)+1/64*d^4*b^4*x*(c*x^2+b*x+a)^(3/2)+3/2048*d^4*b^8/c^(5/2)*ln((1/2*b
+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))+9/256*d^4/c*b^5*(c*x^2+b*x+a)^(1/2)*a+3/8*d^4*b^3*(c*x^2+b*x+a)^(5/2)-9/32*
d^4*c*b^2*a^2*(c*x^2+b*x+a)^(1/2)*x-1/8*d^4*c*b^2*a*x*(c*x^2+b*x+a)^(3/2)

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*d*x+b*d)^4*(c*x^2+b*x+a)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.80686, size = 1540, normalized size = 7.44 \begin{align*} \left [\frac{3 \,{\left (b^{8} - 16 \, a b^{6} c + 96 \, a^{2} b^{4} c^{2} - 256 \, a^{3} b^{2} c^{3} + 256 \, a^{4} c^{4}\right )} \sqrt{c} d^{4} \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 (2048 \, c^{8} d^{4} x^{7} + 7168 \, b c^{7} d^{4} x^{6} + 768 \,{\left (13 \, b^{2} c^{6} + 4 \, a c^{7}\right )} d^{4} x^{5} + 640 \,{\left (11 \, b^{3} c^{5} + 12 \, a b c^{6}\right )} d^{4} x^{4} + 16 \,{\left (161 \, b^{4} c^{4} + 472 \, a b^{2} c^{5} + 16 \, a^{2} c^{6}\right )} d^{4} x^{3} + 24 \,{\left (17 \, b^{5} c^{3} + 152 \, a b^{3} c^{4} + 16 \, a^{2} b c^{5}\right )} d^{4} x^{2} + 2 \,{\left (b^{6} c^{2} + 396 \, a b^{4} c^{3} + 240 \, a^{2} b^{2} c^{4} - 192 \, a^{3} c^{5}\right )} d^{4} x -{\left (3 \, b^{7} c - 44 \, a b^{5} c^{2} - 176 \, a^{2} b^{3} c^{3} + 192 \, a^{3} b c^{4}\right )} d^{4}\right )} \sqrt{c x^{2} + b x + a}}{4096 \, c^{3}}, -\frac{3 \,{\left (b^{8} - 16 \, a b^{6} c + 96 \, a^{2} b^{4} c^{2} - 256 \, a^{3} b^{2} c^{3} + 256 \, a^{4} c^{4}\right )} \sqrt{-c} d^{4} \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 (2048 \, c^{8} d^{4} x^{7} + 7168 \, b c^{7} d^{4} x^{6} + 768 \,{\left (13 \, b^{2} c^{6} + 4 \, a c^{7}\right )} d^{4} x^{5} + 640 \,{\left (11 \, b^{3} c^{5} + 12 \, a b c^{6}\right )} d^{4} x^{4} + 16 \,{\left (161 \, b^{4} c^{4} + 472 \, a b^{2} c^{5} + 16 \, a^{2} c^{6}\right )} d^{4} x^{3} + 24 \,{\left (17 \, b^{5} c^{3} + 152 \, a b^{3} c^{4} + 16 \, a^{2} b c^{5}\right )} d^{4} x^{2} + 2 \,{\left (b^{6} c^{2} + 396 \, a b^{4} c^{3} + 240 \, a^{2} b^{2} c^{4} - 192 \, a^{3} c^{5}\right )} d^{4} x -{\left (3 \, b^{7} c - 44 \, a b^{5} c^{2} - 176 \, a^{2} b^{3} c^{3} + 192 \, a^{3} b c^{4}\right )} d^{4}\right )} \sqrt{c x^{2} + b x + a}}{2048 \, c^{3}}\right ] \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*d*x+b*d)^4*(c*x^2+b*x+a)^(3/2),x, algorithm="fricas")

[Out]

[1/4096*(3*(b^8 - 16*a*b^6*c + 96*a^2*b^4*c^2 - 256*a^3*b^2*c^3 + 256*a^4*c^4)*sqrt(c)*d^4*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*(2048*c^8*d^4*x^7 + 7168*b*c^7*d^4*x^6
+ 768*(13*b^2*c^6 + 4*a*c^7)*d^4*x^5 + 640*(11*b^3*c^5 + 12*a*b*c^6)*d^4*x^4 + 16*(161*b^4*c^4 + 472*a*b^2*c^5
+ 16*a^2*c^6)*d^4*x^3 + 24*(17*b^5*c^3 + 152*a*b^3*c^4 + 16*a^2*b*c^5)*d^4*x^2 + 2*(b^6*c^2 + 396*a*b^4*c^3 +
240*a^2*b^2*c^4 - 192*a^3*c^5)*d^4*x - (3*b^7*c - 44*a*b^5*c^2 - 176*a^2*b^3*c^3 + 192*a^3*b*c^4)*d^4)*sqrt(c
*x^2 + b*x + a))/c^3, -1/2048*(3*(b^8 - 16*a*b^6*c + 96*a^2*b^4*c^2 - 256*a^3*b^2*c^3 + 256*a^4*c^4)*sqrt(-c)*
d^4*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*(2048*c^8*d^4*x^7 + 716
8*b*c^7*d^4*x^6 + 768*(13*b^2*c^6 + 4*a*c^7)*d^4*x^5 + 640*(11*b^3*c^5 + 12*a*b*c^6)*d^4*x^4 + 16*(161*b^4*c^4
+ 472*a*b^2*c^5 + 16*a^2*c^6)*d^4*x^3 + 24*(17*b^5*c^3 + 152*a*b^3*c^4 + 16*a^2*b*c^5)*d^4*x^2 + 2*(b^6*c^2 +
396*a*b^4*c^3 + 240*a^2*b^2*c^4 - 192*a^3*c^5)*d^4*x - (3*b^7*c - 44*a*b^5*c^2 - 176*a^2*b^3*c^3 + 192*a^3*b*
c^4)*d^4)*sqrt(c*x^2 + b*x + a))/c^3]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} d^{4} \left (\int a b^{4} \sqrt{a + b x + c x^{2}}\, dx + \int b^{5} x \sqrt{a + b x + c x^{2}}\, dx + \int 16 c^{5} x^{6} \sqrt{a + b x + c x^{2}}\, dx + \int 16 a c^{4} x^{4} \sqrt{a + b x + c x^{2}}\, dx + \int 48 b c^{4} x^{5} \sqrt{a + b x + c x^{2}}\, dx + \int 56 b^{2} c^{3} x^{4} \sqrt{a + b x + c x^{2}}\, dx + \int 32 b^{3} c^{2} x^{3} \sqrt{a + b x + c x^{2}}\, dx + \int 9 b^{4} c x^{2} \sqrt{a + b x + c x^{2}}\, dx + \int 32 a b c^{3} x^{3} \sqrt{a + b x + c x^{2}}\, dx + \int 24 a b^{2} c^{2} x^{2} \sqrt{a + b x + c x^{2}}\, dx + \int 8 a b^{3} c x \sqrt{a + b x + c x^{2}}\, dx\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*d*x+b*d)**4*(c*x**2+b*x+a)**(3/2),x)

[Out]

d**4*(Integral(a*b**4*sqrt(a + b*x + c*x**2), x) + Integral(b**5*x*sqrt(a + b*x + c*x**2), x) + Integral(16*c*
*5*x**6*sqrt(a + b*x + c*x**2), x) + Integral(16*a*c**4*x**4*sqrt(a + b*x + c*x**2), x) + Integral(48*b*c**4*x
**5*sqrt(a + b*x + c*x**2), x) + Integral(56*b**2*c**3*x**4*sqrt(a + b*x + c*x**2), x) + Integral(32*b**3*c**2
*x**3*sqrt(a + b*x + c*x**2), x) + Integral(9*b**4*c*x**2*sqrt(a + b*x + c*x**2), x) + Integral(32*a*b*c**3*x*
*3*sqrt(a + b*x + c*x**2), x) + Integral(24*a*b**2*c**2*x**2*sqrt(a + b*x + c*x**2), x) + Integral(8*a*b**3*c*
x*sqrt(a + b*x + c*x**2), x))

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Giac [B]  time = 1.19455, size = 528, normalized size = 2.55 \begin{align*} \frac{1}{1024} \, \sqrt{c x^{2} + b x + a}{\left (2 \,{\left (4 \,{\left (2 \,{\left (8 \,{\left (2 \,{\left (4 \,{\left (2 \, c^{5} d^{4} x + 7 \, b c^{4} d^{4}\right )} x + \frac{3 \,{\left (13 \, b^{2} c^{10} d^{4} + 4 \, a c^{11} d^{4}\right )}}{c^{7}}\right )} x + \frac{5 \,{\left (11 \, b^{3} c^{9} d^{4} + 12 \, a b c^{10} d^{4}\right )}}{c^{7}}\right )} x + \frac{161 \, b^{4} c^{8} d^{4} + 472 \, a b^{2} c^{9} d^{4} + 16 \, a^{2} c^{10} d^{4}}{c^{7}}\right )} x + \frac{3 \,{\left (17 \, b^{5} c^{7} d^{4} + 152 \, a b^{3} c^{8} d^{4} + 16 \, a^{2} b c^{9} d^{4}\right )}}{c^{7}}\right )} x + \frac{b^{6} c^{6} d^{4} + 396 \, a b^{4} c^{7} d^{4} + 240 \, a^{2} b^{2} c^{8} d^{4} - 192 \, a^{3} c^{9} d^{4}}{c^{7}}\right )} x - \frac{3 \, b^{7} c^{5} d^{4} - 44 \, a b^{5} c^{6} d^{4} - 176 \, a^{2} b^{3} c^{7} d^{4} + 192 \, a^{3} b c^{8} d^{4}}{c^{7}}\right )} - \frac{3 \,{\left (b^{8} d^{4} - 16 \, a b^{6} c d^{4} + 96 \, a^{2} b^{4} c^{2} d^{4} - 256 \, a^{3} b^{2} c^{3} d^{4} + 256 \, a^{4} c^{4} d^{4}\right )} \log \left ({\left | -2 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )} \sqrt{c} - b \right |}\right )}{2048 \, c^{\frac{5}{2}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*d*x+b*d)^4*(c*x^2+b*x+a)^(3/2),x, algorithm="giac")

[Out]

1/1024*sqrt(c*x^2 + b*x + a)*(2*(4*(2*(8*(2*(4*(2*c^5*d^4*x + 7*b*c^4*d^4)*x + 3*(13*b^2*c^10*d^4 + 4*a*c^11*d
^4)/c^7)*x + 5*(11*b^3*c^9*d^4 + 12*a*b*c^10*d^4)/c^7)*x + (161*b^4*c^8*d^4 + 472*a*b^2*c^9*d^4 + 16*a^2*c^10*
d^4)/c^7)*x + 3*(17*b^5*c^7*d^4 + 152*a*b^3*c^8*d^4 + 16*a^2*b*c^9*d^4)/c^7)*x + (b^6*c^6*d^4 + 396*a*b^4*c^7*
d^4 + 240*a^2*b^2*c^8*d^4 - 192*a^3*c^9*d^4)/c^7)*x - (3*b^7*c^5*d^4 - 44*a*b^5*c^6*d^4 - 176*a^2*b^3*c^7*d^4
+ 192*a^3*b*c^8*d^4)/c^7) - 3/2048*(b^8*d^4 - 16*a*b^6*c*d^4 + 96*a^2*b^4*c^2*d^4 - 256*a^3*b^2*c^3*d^4 + 256*
a^4*c^4*d^4)*log(abs(-2*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*sqrt(c) - b))/c^(5/2)