3.11.32 \(\int (b d+2 c d x)^2 (a+b x+c x^2)^{5/2} \, dx\)

Optimal. Leaf size=207 \[ -\frac {5 d^2 \left (b^2-4 a c\right )^4 \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{8192 c^{7/2}}-\frac {5 d^2 \left (b^2-4 a c\right )^3 (b+2 c x) \sqrt {a+b x+c x^2}}{4096 c^3}+\frac {5 d^2 \left (b^2-4 a c\right )^2 (b+2 c x)^3 \sqrt {a+b x+c x^2}}{2048 c^3}-\frac {5 d^2 \left (b^2-4 a c\right ) (b+2 c x)^3 \left (a+b x+c x^2\right )^{3/2}}{384 c^2}+\frac {d^2 (b+2 c x)^3 \left (a+b x+c x^2\right )^{5/2}}{16 c} \]

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Rubi [A]  time = 0.12, antiderivative size = 207, normalized size of antiderivative = 1.00, 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} \begin {gather*} -\frac {5 d^2 \left (b^2-4 a c\right )^3 (b+2 c x) \sqrt {a+b x+c x^2}}{4096 c^3}+\frac {5 d^2 \left (b^2-4 a c\right )^2 (b+2 c x)^3 \sqrt {a+b x+c x^2}}{2048 c^3}-\frac {5 d^2 \left (b^2-4 a c\right ) (b+2 c x)^3 \left (a+b x+c x^2\right )^{3/2}}{384 c^2}-\frac {5 d^2 \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 )}{8192 c^{7/2}}+\frac {d^2 (b+2 c x)^3 \left (a+b x+c x^2\right )^{5/2}}{16 c} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(b*d + 2*c*d*x)^2*(a + b*x + c*x^2)^(5/2),x]

[Out]

(-5*(b^2 - 4*a*c)^3*d^2*(b + 2*c*x)*Sqrt[a + b*x + c*x^2])/(4096*c^3) + (5*(b^2 - 4*a*c)^2*d^2*(b + 2*c*x)^3*S
qrt[a + b*x + c*x^2])/(2048*c^3) - (5*(b^2 - 4*a*c)*d^2*(b + 2*c*x)^3*(a + b*x + c*x^2)^(3/2))/(384*c^2) + (d^
2*(b + 2*c*x)^3*(a + b*x + c*x^2)^(5/2))/(16*c) - (5*(b^2 - 4*a*c)^4*d^2*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a
+ b*x + c*x^2])])/(8192*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 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 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])

Rubi steps

\begin {align*} \int (b d+2 c d x)^2 \left (a+b x+c x^2\right )^{5/2} \, dx &=\frac {d^2 (b+2 c x)^3 \left (a+b x+c x^2\right )^{5/2}}{16 c}-\frac {\left (5 \left (b^2-4 a c\right )\right ) \int (b d+2 c d x)^2 \left (a+b x+c x^2\right )^{3/2} \, dx}{32 c}\\ &=-\frac {5 \left (b^2-4 a c\right ) d^2 (b+2 c x)^3 \left (a+b x+c x^2\right )^{3/2}}{384 c^2}+\frac {d^2 (b+2 c x)^3 \left (a+b x+c x^2\right )^{5/2}}{16 c}+\frac {\left (5 \left (b^2-4 a c\right )^2\right ) \int (b d+2 c d x)^2 \sqrt {a+b x+c x^2} \, dx}{256 c^2}\\ &=\frac {5 \left (b^2-4 a c\right )^2 d^2 (b+2 c x)^3 \sqrt {a+b x+c x^2}}{2048 c^3}-\frac {5 \left (b^2-4 a c\right ) d^2 (b+2 c x)^3 \left (a+b x+c x^2\right )^{3/2}}{384 c^2}+\frac {d^2 (b+2 c x)^3 \left (a+b x+c x^2\right )^{5/2}}{16 c}-\frac {\left (5 \left (b^2-4 a c\right )^3\right ) \int \frac {(b d+2 c d x)^2}{\sqrt {a+b x+c x^2}} \, dx}{4096 c^3}\\ &=-\frac {5 \left (b^2-4 a c\right )^3 d^2 (b+2 c x) \sqrt {a+b x+c x^2}}{4096 c^3}+\frac {5 \left (b^2-4 a c\right )^2 d^2 (b+2 c x)^3 \sqrt {a+b x+c x^2}}{2048 c^3}-\frac {5 \left (b^2-4 a c\right ) d^2 (b+2 c x)^3 \left (a+b x+c x^2\right )^{3/2}}{384 c^2}+\frac {d^2 (b+2 c x)^3 \left (a+b x+c x^2\right )^{5/2}}{16 c}-\frac {\left (5 \left (b^2-4 a c\right )^4 d^2\right ) \int \frac {1}{\sqrt {a+b x+c x^2}} \, dx}{8192 c^3}\\ &=-\frac {5 \left (b^2-4 a c\right )^3 d^2 (b+2 c x) \sqrt {a+b x+c x^2}}{4096 c^3}+\frac {5 \left (b^2-4 a c\right )^2 d^2 (b+2 c x)^3 \sqrt {a+b x+c x^2}}{2048 c^3}-\frac {5 \left (b^2-4 a c\right ) d^2 (b+2 c x)^3 \left (a+b x+c x^2\right )^{3/2}}{384 c^2}+\frac {d^2 (b+2 c x)^3 \left (a+b x+c x^2\right )^{5/2}}{16 c}-\frac {\left (5 \left (b^2-4 a c\right )^4 d^2\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 )}{4096 c^3}\\ &=-\frac {5 \left (b^2-4 a c\right )^3 d^2 (b+2 c x) \sqrt {a+b x+c x^2}}{4096 c^3}+\frac {5 \left (b^2-4 a c\right )^2 d^2 (b+2 c x)^3 \sqrt {a+b x+c x^2}}{2048 c^3}-\frac {5 \left (b^2-4 a c\right ) d^2 (b+2 c x)^3 \left (a+b x+c x^2\right )^{3/2}}{384 c^2}+\frac {d^2 (b+2 c x)^3 \left (a+b x+c x^2\right )^{5/2}}{16 c}-\frac {5 \left (b^2-4 a c\right )^4 d^2 \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{8192 c^{7/2}}\\ \end {align*}

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Mathematica [A]  time = 0.93, size = 225, normalized size = 1.09 \begin {gather*} \frac {1}{4} d^2 (b+2 c x) \sqrt {a+x (b+c x)} \left (\frac {\left (b^2-4 a c\right ) \left (16 c^2 \left (33 a^2+26 a c x^2+8 c^2 x^4\right )+8 b^2 c \left (11 c x^2-20 a\right )+32 b c^2 x \left (13 a+8 c x^2\right )+15 b^4-40 b^3 c x\right )}{3072 c^3}-\frac {5 \sqrt {c} \sqrt {4 a-\frac {b^2}{c}} (a+x (b+c x))^3 \sinh ^{-1}\left (\frac {b+2 c x}{\sqrt {c} \sqrt {4 a-\frac {b^2}{c}}}\right )}{2048 (b+2 c x) \left (\frac {c (a+x (b+c x))}{4 a c-b^2}\right )^{7/2}}+(a+x (b+c x))^3\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(b*d + 2*c*d*x)^2*(a + b*x + c*x^2)^(5/2),x]

[Out]

(d^2*(b + 2*c*x)*Sqrt[a + x*(b + c*x)]*((a + x*(b + c*x))^3 + ((b^2 - 4*a*c)*(15*b^4 - 40*b^3*c*x + 32*b*c^2*x
*(13*a + 8*c*x^2) + 8*b^2*c*(-20*a + 11*c*x^2) + 16*c^2*(33*a^2 + 26*a*c*x^2 + 8*c^2*x^4)))/(3072*c^3) - (5*Sq
rt[4*a - b^2/c]*Sqrt[c]*(a + x*(b + c*x))^3*ArcSinh[(b + 2*c*x)/(Sqrt[4*a - b^2/c]*Sqrt[c])])/(2048*(b + 2*c*x
)*((c*(a + x*(b + c*x)))/(-b^2 + 4*a*c))^(7/2))))/4

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IntegrateAlgebraic [A]  time = 1.11, size = 374, normalized size = 1.81 \begin {gather*} \frac {\sqrt {a+b x+c x^2} \left (960 a^3 b c^3 d^2+1920 a^3 c^4 d^2 x+1168 a^2 b^3 c^2 d^2+9888 a^2 b^2 c^3 d^2 x+22656 a^2 b c^4 d^2 x^2+15104 a^2 c^5 d^2 x^3-220 a b^5 c d^2+136 a b^4 c^2 d^2 x+10432 a b^3 c^3 d^2 x^2+35968 a b^2 c^4 d^2 x^3+43520 a b c^5 d^2 x^4+17408 a c^6 d^2 x^5+15 b^7 d^2-10 b^6 c d^2 x+8 b^5 c^2 d^2 x^2+3504 b^4 c^3 d^2 x^3+16000 b^3 c^4 d^2 x^4+27904 b^2 c^5 d^2 x^5+21504 b c^6 d^2 x^6+6144 c^7 d^2 x^7\right )}{12288 c^3}+\frac {5 \left (256 a^4 c^4 d^2-256 a^3 b^2 c^3 d^2+96 a^2 b^4 c^2 d^2-16 a b^6 c d^2+b^8 d^2\right ) \log \left (-2 \sqrt {c} \sqrt {a+b x+c x^2}+b+2 c x\right )}{8192 c^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(b*d + 2*c*d*x)^2*(a + b*x + c*x^2)^(5/2),x]

[Out]

(Sqrt[a + b*x + c*x^2]*(15*b^7*d^2 - 220*a*b^5*c*d^2 + 1168*a^2*b^3*c^2*d^2 + 960*a^3*b*c^3*d^2 - 10*b^6*c*d^2
*x + 136*a*b^4*c^2*d^2*x + 9888*a^2*b^2*c^3*d^2*x + 1920*a^3*c^4*d^2*x + 8*b^5*c^2*d^2*x^2 + 10432*a*b^3*c^3*d
^2*x^2 + 22656*a^2*b*c^4*d^2*x^2 + 3504*b^4*c^3*d^2*x^3 + 35968*a*b^2*c^4*d^2*x^3 + 15104*a^2*c^5*d^2*x^3 + 16
000*b^3*c^4*d^2*x^4 + 43520*a*b*c^5*d^2*x^4 + 27904*b^2*c^5*d^2*x^5 + 17408*a*c^6*d^2*x^5 + 21504*b*c^6*d^2*x^
6 + 6144*c^7*d^2*x^7))/(12288*c^3) + (5*(b^8*d^2 - 16*a*b^6*c*d^2 + 96*a^2*b^4*c^2*d^2 - 256*a^3*b^2*c^3*d^2 +
256*a^4*c^4*d^2)*Log[b + 2*c*x - 2*Sqrt[c]*Sqrt[a + b*x + c*x^2]])/(8192*c^(7/2))

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fricas [A]  time = 0.48, size = 675, normalized size = 3.26 \begin {gather*} \left [\frac {15 \, {\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^{2} \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 (6144 \, c^{8} d^{2} x^{7} + 21504 \, b c^{7} d^{2} x^{6} + 256 \, {\left (109 \, b^{2} c^{6} + 68 \, a c^{7}\right )} d^{2} x^{5} + 640 \, {\left (25 \, b^{3} c^{5} + 68 \, a b c^{6}\right )} d^{2} x^{4} + 16 \, {\left (219 \, b^{4} c^{4} + 2248 \, a b^{2} c^{5} + 944 \, a^{2} c^{6}\right )} d^{2} x^{3} + 8 \, {\left (b^{5} c^{3} + 1304 \, a b^{3} c^{4} + 2832 \, a^{2} b c^{5}\right )} d^{2} x^{2} - 2 \, {\left (5 \, b^{6} c^{2} - 68 \, a b^{4} c^{3} - 4944 \, a^{2} b^{2} c^{4} - 960 \, a^{3} c^{5}\right )} d^{2} x + {\left (15 \, b^{7} c - 220 \, a b^{5} c^{2} + 1168 \, a^{2} b^{3} c^{3} + 960 \, a^{3} b c^{4}\right )} d^{2}\right )} \sqrt {c x^{2} + b x + a}}{49152 \, c^{4}}, \frac {15 \, {\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^{2} \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 (6144 \, c^{8} d^{2} x^{7} + 21504 \, b c^{7} d^{2} x^{6} + 256 \, {\left (109 \, b^{2} c^{6} + 68 \, a c^{7}\right )} d^{2} x^{5} + 640 \, {\left (25 \, b^{3} c^{5} + 68 \, a b c^{6}\right )} d^{2} x^{4} + 16 \, {\left (219 \, b^{4} c^{4} + 2248 \, a b^{2} c^{5} + 944 \, a^{2} c^{6}\right )} d^{2} x^{3} + 8 \, {\left (b^{5} c^{3} + 1304 \, a b^{3} c^{4} + 2832 \, a^{2} b c^{5}\right )} d^{2} x^{2} - 2 \, {\left (5 \, b^{6} c^{2} - 68 \, a b^{4} c^{3} - 4944 \, a^{2} b^{2} c^{4} - 960 \, a^{3} c^{5}\right )} d^{2} x + {\left (15 \, b^{7} c - 220 \, a b^{5} c^{2} + 1168 \, a^{2} b^{3} c^{3} + 960 \, a^{3} b c^{4}\right )} d^{2}\right )} \sqrt {c x^{2} + b x + a}}{24576 \, c^{4}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*d*x+b*d)^2*(c*x^2+b*x+a)^(5/2),x, algorithm="fricas")

[Out]

[1/49152*(15*(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^2*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*(6144*c^8*d^2*x^7 + 21504*b*c^7*d^2*x
^6 + 256*(109*b^2*c^6 + 68*a*c^7)*d^2*x^5 + 640*(25*b^3*c^5 + 68*a*b*c^6)*d^2*x^4 + 16*(219*b^4*c^4 + 2248*a*b
^2*c^5 + 944*a^2*c^6)*d^2*x^3 + 8*(b^5*c^3 + 1304*a*b^3*c^4 + 2832*a^2*b*c^5)*d^2*x^2 - 2*(5*b^6*c^2 - 68*a*b^
4*c^3 - 4944*a^2*b^2*c^4 - 960*a^3*c^5)*d^2*x + (15*b^7*c - 220*a*b^5*c^2 + 1168*a^2*b^3*c^3 + 960*a^3*b*c^4)*
d^2)*sqrt(c*x^2 + b*x + a))/c^4, 1/24576*(15*(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^2*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*(6144*c^8*d
^2*x^7 + 21504*b*c^7*d^2*x^6 + 256*(109*b^2*c^6 + 68*a*c^7)*d^2*x^5 + 640*(25*b^3*c^5 + 68*a*b*c^6)*d^2*x^4 +
16*(219*b^4*c^4 + 2248*a*b^2*c^5 + 944*a^2*c^6)*d^2*x^3 + 8*(b^5*c^3 + 1304*a*b^3*c^4 + 2832*a^2*b*c^5)*d^2*x^
2 - 2*(5*b^6*c^2 - 68*a*b^4*c^3 - 4944*a^2*b^2*c^4 - 960*a^3*c^5)*d^2*x + (15*b^7*c - 220*a*b^5*c^2 + 1168*a^2
*b^3*c^3 + 960*a^3*b*c^4)*d^2)*sqrt(c*x^2 + b*x + a))/c^4]

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giac [B]  time = 0.26, size = 389, normalized size = 1.88 \begin {gather*} \frac {1}{12288} \, \sqrt {c x^{2} + b x + a} {\left (2 \, {\left (4 \, {\left (2 \, {\left (8 \, {\left (2 \, {\left (12 \, {\left (2 \, c^{4} d^{2} x + 7 \, b c^{3} d^{2}\right )} x + \frac {109 \, b^{2} c^{9} d^{2} + 68 \, a c^{10} d^{2}}{c^{7}}\right )} x + \frac {5 \, {\left (25 \, b^{3} c^{8} d^{2} + 68 \, a b c^{9} d^{2}\right )}}{c^{7}}\right )} x + \frac {219 \, b^{4} c^{7} d^{2} + 2248 \, a b^{2} c^{8} d^{2} + 944 \, a^{2} c^{9} d^{2}}{c^{7}}\right )} x + \frac {b^{5} c^{6} d^{2} + 1304 \, a b^{3} c^{7} d^{2} + 2832 \, a^{2} b c^{8} d^{2}}{c^{7}}\right )} x - \frac {5 \, b^{6} c^{5} d^{2} - 68 \, a b^{4} c^{6} d^{2} - 4944 \, a^{2} b^{2} c^{7} d^{2} - 960 \, a^{3} c^{8} d^{2}}{c^{7}}\right )} x + \frac {15 \, b^{7} c^{4} d^{2} - 220 \, a b^{5} c^{5} d^{2} + 1168 \, a^{2} b^{3} c^{6} d^{2} + 960 \, a^{3} b c^{7} d^{2}}{c^{7}}\right )} + \frac {5 \, {\left (b^{8} d^{2} - 16 \, a b^{6} c d^{2} + 96 \, a^{2} b^{4} c^{2} d^{2} - 256 \, a^{3} b^{2} c^{3} d^{2} + 256 \, a^{4} c^{4} d^{2}\right )} \log \left ({\left | -2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} \sqrt {c} - b \right |}\right )}{8192 \, c^{\frac {7}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*d*x+b*d)^2*(c*x^2+b*x+a)^(5/2),x, algorithm="giac")

[Out]

1/12288*sqrt(c*x^2 + b*x + a)*(2*(4*(2*(8*(2*(12*(2*c^4*d^2*x + 7*b*c^3*d^2)*x + (109*b^2*c^9*d^2 + 68*a*c^10*
d^2)/c^7)*x + 5*(25*b^3*c^8*d^2 + 68*a*b*c^9*d^2)/c^7)*x + (219*b^4*c^7*d^2 + 2248*a*b^2*c^8*d^2 + 944*a^2*c^9
*d^2)/c^7)*x + (b^5*c^6*d^2 + 1304*a*b^3*c^7*d^2 + 2832*a^2*b*c^8*d^2)/c^7)*x - (5*b^6*c^5*d^2 - 68*a*b^4*c^6*
d^2 - 4944*a^2*b^2*c^7*d^2 - 960*a^3*c^8*d^2)/c^7)*x + (15*b^7*c^4*d^2 - 220*a*b^5*c^5*d^2 + 1168*a^2*b^3*c^6*
d^2 + 960*a^3*b*c^7*d^2)/c^7) + 5/8192*(b^8*d^2 - 16*a*b^6*c*d^2 + 96*a^2*b^4*c^2*d^2 - 256*a^3*b^2*c^3*d^2 +
256*a^4*c^4*d^2)*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 = 634, normalized size = 3.06 \begin {gather*} -\frac {5 a^{4} \sqrt {c}\, d^{2} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{32}+\frac {5 a^{3} b^{2} d^{2} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{32 \sqrt {c}}-\frac {15 a^{2} b^{4} d^{2} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{256 c^{\frac {3}{2}}}+\frac {5 a \,b^{6} d^{2} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{512 c^{\frac {5}{2}}}-\frac {5 b^{8} d^{2} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{8192 c^{\frac {7}{2}}}-\frac {5 \sqrt {c \,x^{2}+b x +a}\, a^{3} c \,d^{2} x}{32}+\frac {15 \sqrt {c \,x^{2}+b x +a}\, a^{2} b^{2} d^{2} x}{128}-\frac {15 \sqrt {c \,x^{2}+b x +a}\, a \,b^{4} d^{2} x}{512 c}+\frac {5 \sqrt {c \,x^{2}+b x +a}\, b^{6} d^{2} x}{2048 c^{2}}-\frac {5 \sqrt {c \,x^{2}+b x +a}\, a^{3} b \,d^{2}}{64}+\frac {15 \sqrt {c \,x^{2}+b x +a}\, a^{2} b^{3} d^{2}}{256 c}-\frac {5 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} a^{2} c \,d^{2} x}{48}-\frac {15 \sqrt {c \,x^{2}+b x +a}\, a \,b^{5} d^{2}}{1024 c^{2}}+\frac {5 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} a \,b^{2} d^{2} x}{96}+\frac {5 \sqrt {c \,x^{2}+b x +a}\, b^{7} d^{2}}{4096 c^{3}}-\frac {5 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} b^{4} d^{2} x}{768 c}-\frac {5 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} a^{2} b \,d^{2}}{96}+\frac {5 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} a \,b^{3} d^{2}}{192 c}-\frac {\left (c \,x^{2}+b x +a \right )^{\frac {5}{2}} a c \,d^{2} x}{12}-\frac {5 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} b^{5} d^{2}}{1536 c^{2}}+\frac {\left (c \,x^{2}+b x +a \right )^{\frac {5}{2}} b^{2} d^{2} x}{48}-\frac {\left (c \,x^{2}+b x +a \right )^{\frac {5}{2}} a b \,d^{2}}{24}+\frac {\left (c \,x^{2}+b x +a \right )^{\frac {5}{2}} b^{3} d^{2}}{96 c}+\frac {\left (c \,x^{2}+b x +a \right )^{\frac {7}{2}} c \,d^{2} x}{2}+\frac {\left (c \,x^{2}+b x +a \right )^{\frac {7}{2}} b \,d^{2}}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-15/512*d^2/c*b^4*(c*x^2+b*x+a)^(1/2)*x*a-5/1536*d^2/c^2*b^5*(c*x^2+b*x+a)^(3/2)+5/4096*d^2/c^3*b^7*(c*x^2+b*x
+a)^(1/2)+1/96*d^2/c*b^3*(c*x^2+b*x+a)^(5/2)-5/96*d^2*a^2*(c*x^2+b*x+a)^(3/2)*b-5/64*d^2*a^3*(c*x^2+b*x+a)^(1/
2)*b-1/24*d^2*a*(c*x^2+b*x+a)^(5/2)*b+1/48*d^2*b^2*x*(c*x^2+b*x+a)^(5/2)-5/32*d^2*c^(1/2)*a^4*ln((c*x+1/2*b)/c
^(1/2)+(c*x^2+b*x+a)^(1/2))-5/8192*d^2*b^8/c^(7/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))+1/2*d^2*c*x*(c*
x^2+b*x+a)^(7/2)+5/192*d^2/c*b^3*(c*x^2+b*x+a)^(3/2)*a+5/96*d^2*b^2*(c*x^2+b*x+a)^(3/2)*x*a+15/128*d^2*b^2*(c*
x^2+b*x+a)^(1/2)*x*a^2-5/32*d^2*c*a^3*(c*x^2+b*x+a)^(1/2)*x-1/12*d^2*c*a*x*(c*x^2+b*x+a)^(5/2)-5/768*d^2/c*b^4
*(c*x^2+b*x+a)^(3/2)*x-5/48*d^2*c*a^2*(c*x^2+b*x+a)^(3/2)*x-15/256*d^2*b^4/c^(3/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x
^2+b*x+a)^(1/2))*a^2+5/512*d^2*b^6/c^(5/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*a-15/1024*d^2/c^2*b^5*(
c*x^2+b*x+a)^(1/2)*a+5/2048*d^2/c^2*b^6*(c*x^2+b*x+a)^(1/2)*x+15/256*d^2/c*b^3*(c*x^2+b*x+a)^(1/2)*a^2+5/32*d^
2*b^2/c^(1/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*a^3+1/4*d^2*b*(c*x^2+b*x+a)^(7/2)

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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*d*x+b*d)^2*(c*x^2+b*x+a)^(5/2),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 = -0.00 \begin {gather*} \int {\left (b\,d+2\,c\,d\,x\right )}^2\,{\left (c\,x^2+b\,x+a\right )}^{5/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} d^{2} \left (\int a^{2} b^{2} \sqrt {a + b x + c x^{2}}\, dx + \int b^{4} x^{2} \sqrt {a + b x + c x^{2}}\, dx + \int 4 c^{4} x^{6} \sqrt {a + b x + c x^{2}}\, dx + \int 2 a b^{3} x \sqrt {a + b x + c x^{2}}\, dx + \int 8 a c^{3} x^{4} \sqrt {a + b x + c x^{2}}\, dx + \int 4 a^{2} c^{2} x^{2} \sqrt {a + b x + c x^{2}}\, dx + \int 12 b c^{3} x^{5} \sqrt {a + b x + c x^{2}}\, dx + \int 13 b^{2} c^{2} x^{4} \sqrt {a + b x + c x^{2}}\, dx + \int 6 b^{3} c x^{3} \sqrt {a + b x + c x^{2}}\, dx + \int 16 a b c^{2} x^{3} \sqrt {a + b x + c x^{2}}\, dx + \int 10 a b^{2} c x^{2} \sqrt {a + b x + c x^{2}}\, dx + \int 4 a^{2} b c x \sqrt {a + b x + c x^{2}}\, dx\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*d*x+b*d)**2*(c*x**2+b*x+a)**(5/2),x)

[Out]

d**2*(Integral(a**2*b**2*sqrt(a + b*x + c*x**2), x) + Integral(b**4*x**2*sqrt(a + b*x + c*x**2), x) + Integral
(4*c**4*x**6*sqrt(a + b*x + c*x**2), x) + Integral(2*a*b**3*x*sqrt(a + b*x + c*x**2), x) + Integral(8*a*c**3*x
**4*sqrt(a + b*x + c*x**2), x) + Integral(4*a**2*c**2*x**2*sqrt(a + b*x + c*x**2), x) + Integral(12*b*c**3*x**
5*sqrt(a + b*x + c*x**2), x) + Integral(13*b**2*c**2*x**4*sqrt(a + b*x + c*x**2), x) + Integral(6*b**3*c*x**3*
sqrt(a + b*x + c*x**2), x) + Integral(16*a*b*c**2*x**3*sqrt(a + b*x + c*x**2), x) + Integral(10*a*b**2*c*x**2*
sqrt(a + b*x + c*x**2), x) + Integral(4*a**2*b*c*x*sqrt(a + b*x + c*x**2), x))

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