3.12.91 \(\int (b d+2 c d x)^4 \sqrt {a+b x+c x^2} \, dx\) [1191]

Optimal. Leaf size=165 \[ -\frac {\left (b^2-4 a c\right )^2 d^4 (b+2 c x) \sqrt {a+b x+c x^2}}{32 c}-\frac {\left (b^2-4 a c\right ) d^4 (b+2 c x)^3 \sqrt {a+b x+c x^2}}{48 c}+\frac {d^4 (b+2 c x)^5 \sqrt {a+b x+c x^2}}{12 c}-\frac {\left (b^2-4 a c\right )^3 d^4 \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{64 c^{3/2}} \]

[Out]

-1/64*(-4*a*c+b^2)^3*d^4*arctanh(1/2*(2*c*x+b)/c^(1/2)/(c*x^2+b*x+a)^(1/2))/c^(3/2)-1/32*(-4*a*c+b^2)^2*d^4*(2
*c*x+b)*(c*x^2+b*x+a)^(1/2)/c-1/48*(-4*a*c+b^2)*d^4*(2*c*x+b)^3*(c*x^2+b*x+a)^(1/2)/c+1/12*d^4*(2*c*x+b)^5*(c*
x^2+b*x+a)^(1/2)/c

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Rubi [A]
time = 0.07, antiderivative size = 165, 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 = {699, 706, 635, 212} \begin {gather*} -\frac {d^4 \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 )}{64 c^{3/2}}-\frac {d^4 \left (b^2-4 a c\right ) (b+2 c x)^3 \sqrt {a+b x+c x^2}}{48 c}-\frac {d^4 \left (b^2-4 a c\right )^2 (b+2 c x) \sqrt {a+b x+c x^2}}{32 c}+\frac {d^4 (b+2 c x)^5 \sqrt {a+b x+c x^2}}{12 c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/32*((b^2 - 4*a*c)^2*d^4*(b + 2*c*x)*Sqrt[a + b*x + c*x^2])/c - ((b^2 - 4*a*c)*d^4*(b + 2*c*x)^3*Sqrt[a + b*
x + c*x^2])/(48*c) + (d^4*(b + 2*c*x)^5*Sqrt[a + b*x + c*x^2])/(12*c) - ((b^2 - 4*a*c)^3*d^4*ArcTanh[(b + 2*c*
x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(64*c^(3/2))

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 635

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 699

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 706

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)^4 \sqrt {a+b x+c x^2} \, dx &=\frac {d^4 (b+2 c x)^5 \sqrt {a+b x+c x^2}}{12 c}-\frac {\left (b^2-4 a c\right ) \int \frac {(b d+2 c d x)^4}{\sqrt {a+b x+c x^2}} \, dx}{24 c}\\ &=-\frac {\left (b^2-4 a c\right ) d^4 (b+2 c x)^3 \sqrt {a+b x+c x^2}}{48 c}+\frac {d^4 (b+2 c x)^5 \sqrt {a+b x+c x^2}}{12 c}-\frac {\left (\left (b^2-4 a c\right )^2 d^2\right ) \int \frac {(b d+2 c d x)^2}{\sqrt {a+b x+c x^2}} \, dx}{32 c}\\ &=-\frac {\left (b^2-4 a c\right )^2 d^4 (b+2 c x) \sqrt {a+b x+c x^2}}{32 c}-\frac {\left (b^2-4 a c\right ) d^4 (b+2 c x)^3 \sqrt {a+b x+c x^2}}{48 c}+\frac {d^4 (b+2 c x)^5 \sqrt {a+b x+c x^2}}{12 c}-\frac {\left (\left (b^2-4 a c\right )^3 d^4\right ) \int \frac {1}{\sqrt {a+b x+c x^2}} \, dx}{64 c}\\ &=-\frac {\left (b^2-4 a c\right )^2 d^4 (b+2 c x) \sqrt {a+b x+c x^2}}{32 c}-\frac {\left (b^2-4 a c\right ) d^4 (b+2 c x)^3 \sqrt {a+b x+c x^2}}{48 c}+\frac {d^4 (b+2 c x)^5 \sqrt {a+b x+c x^2}}{12 c}-\frac {\left (\left (b^2-4 a c\right )^3 d^4\right ) \text {Subst}\left (\int \frac {1}{4 c-x^2} \, dx,x,\frac {b+2 c x}{\sqrt {a+b x+c x^2}}\right )}{32 c}\\ &=-\frac {\left (b^2-4 a c\right )^2 d^4 (b+2 c x) \sqrt {a+b x+c x^2}}{32 c}-\frac {\left (b^2-4 a c\right ) d^4 (b+2 c x)^3 \sqrt {a+b x+c x^2}}{48 c}+\frac {d^4 (b+2 c x)^5 \sqrt {a+b x+c x^2}}{12 c}-\frac {\left (b^2-4 a c\right )^3 d^4 \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{64 c^{3/2}}\\ \end {align*}

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Mathematica [A]
time = 0.55, size = 147, normalized size = 0.89 \begin {gather*} d^4 \left (\frac {(b+2 c x) \sqrt {a+x (b+c x)} \left (3 b^4+56 b^3 c x+32 b c^2 x \left (a+8 c x^2\right )+8 b^2 c \left (4 a+23 c x^2\right )+16 c^2 \left (-3 a^2+2 a c x^2+8 c^2 x^4\right )\right )}{96 c}+\frac {\left (b^2-4 a c\right )^3 \log \left (b+2 c x-2 \sqrt {c} \sqrt {a+x (b+c x)}\right )}{64 c^{3/2}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

d^4*(((b + 2*c*x)*Sqrt[a + x*(b + c*x)]*(3*b^4 + 56*b^3*c*x + 32*b*c^2*x*(a + 8*c*x^2) + 8*b^2*c*(4*a + 23*c*x
^2) + 16*c^2*(-3*a^2 + 2*a*c*x^2 + 8*c^2*x^4)))/(96*c) + ((b^2 - 4*a*c)^3*Log[b + 2*c*x - 2*Sqrt[c]*Sqrt[a + x
*(b + c*x)]])/(64*c^(3/2)))

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(1193\) vs. \(2(143)=286\).
time = 0.70, size = 1194, normalized size = 7.24

method	result	size
risch	\(-\frac {\left (-256 c^{5} x^{5}-640 b \,c^{4} x^{4}-64 a \,c^{4} x^{3}-624 b^{2} c^{3} x^{3}-96 a b \,c^{3} x^{2}-296 b^{3} c^{2} x^{2}+96 a^{2} c^{3} x -96 b^{2} c^{2} a x -62 c \,b^{4} x +48 a^{2} b \,c^{2}-32 a \,b^{3} c -3 b^{5}\right ) \sqrt {c \,x^{2}+b x +a}\, d^{4}}{96 c}+\left (c^{\frac {3}{2}} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right ) a^{3}-\frac {3 \sqrt {c}\, \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right ) a^{2} b^{2}}{4}+\frac {3 \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right ) a \,b^{4}}{16 \sqrt {c}}-\frac {\ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right ) b^{6}}{64 c^{\frac {3}{2}}}\right ) d^{4}\)	\(268\)
default	\(\text {Expression too large to display}\)	\(1194\)

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)^(1/2),x,method=_RETURNVERBOSE)

[Out]

d^4*(16*c^4*(1/6*x^3*(c*x^2+b*x+a)^(3/2)/c-3/4*b/c*(1/5*x^2*(c*x^2+b*x+a)^(3/2)/c-7/10*b/c*(1/4*x*(c*x^2+b*x+a
)^(3/2)/c-5/8*b/c*(1/3*(c*x^2+b*x+a)^(3/2)/c-1/2*b/c*(1/4*(2*c*x+b)*(c*x^2+b*x+a)^(1/2)/c+1/8*(4*a*c-b^2)/c^(3
/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))))-1/4*a/c*(1/4*(2*c*x+b)*(c*x^2+b*x+a)^(1/2)/c+1/8*(4*a*c-b^2)
/c^(3/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))))-2/5*a/c*(1/3*(c*x^2+b*x+a)^(3/2)/c-1/2*b/c*(1/4*(2*c*x+
b)*(c*x^2+b*x+a)^(1/2)/c+1/8*(4*a*c-b^2)/c^(3/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2)))))-1/2*a/c*(1/4*x
*(c*x^2+b*x+a)^(3/2)/c-5/8*b/c*(1/3*(c*x^2+b*x+a)^(3/2)/c-1/2*b/c*(1/4*(2*c*x+b)*(c*x^2+b*x+a)^(1/2)/c+1/8*(4*
a*c-b^2)/c^(3/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))))-1/4*a/c*(1/4*(2*c*x+b)*(c*x^2+b*x+a)^(1/2)/c+1/
8*(4*a*c-b^2)/c^(3/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2)))))+32*b*c^3*(1/5*x^2*(c*x^2+b*x+a)^(3/2)/c-7
/10*b/c*(1/4*x*(c*x^2+b*x+a)^(3/2)/c-5/8*b/c*(1/3*(c*x^2+b*x+a)^(3/2)/c-1/2*b/c*(1/4*(2*c*x+b)*(c*x^2+b*x+a)^(
1/2)/c+1/8*(4*a*c-b^2)/c^(3/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))))-1/4*a/c*(1/4*(2*c*x+b)*(c*x^2+b*x
+a)^(1/2)/c+1/8*(4*a*c-b^2)/c^(3/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))))-2/5*a/c*(1/3*(c*x^2+b*x+a)^(
3/2)/c-1/2*b/c*(1/4*(2*c*x+b)*(c*x^2+b*x+a)^(1/2)/c+1/8*(4*a*c-b^2)/c^(3/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+
a)^(1/2)))))+24*b^2*c^2*(1/4*x*(c*x^2+b*x+a)^(3/2)/c-5/8*b/c*(1/3*(c*x^2+b*x+a)^(3/2)/c-1/2*b/c*(1/4*(2*c*x+b)
*(c*x^2+b*x+a)^(1/2)/c+1/8*(4*a*c-b^2)/c^(3/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))))-1/4*a/c*(1/4*(2*c
*x+b)*(c*x^2+b*x+a)^(1/2)/c+1/8*(4*a*c-b^2)/c^(3/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))))+8*b^3*c*(1/3
*(c*x^2+b*x+a)^(3/2)/c-1/2*b/c*(1/4*(2*c*x+b)*(c*x^2+b*x+a)^(1/2)/c+1/8*(4*a*c-b^2)/c^(3/2)*ln((1/2*b+c*x)/c^(
1/2)+(c*x^2+b*x+a)^(1/2))))+b^4*(1/4*(2*c*x+b)*(c*x^2+b*x+a)^(1/2)/c+1/8*(4*a*c-b^2)/c^(3/2)*ln((1/2*b+c*x)/c^
(1/2)+(c*x^2+b*x+a)^(1/2))))

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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)^4*(c*x^2+b*x+a)^(1/2),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 deta

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Fricas [A]
time = 1.77, size = 473, normalized size = 2.87 \begin {gather*} \left [-\frac {3 \, {\left (b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}\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 (256 \, c^{6} d^{4} x^{5} + 640 \, b c^{5} d^{4} x^{4} + 16 \, {\left (39 \, b^{2} c^{4} + 4 \, a c^{5}\right )} d^{4} x^{3} + 8 \, {\left (37 \, b^{3} c^{3} + 12 \, a b c^{4}\right )} d^{4} x^{2} + 2 \, {\left (31 \, b^{4} c^{2} + 48 \, a b^{2} c^{3} - 48 \, a^{2} c^{4}\right )} d^{4} x + {\left (3 \, b^{5} c + 32 \, a b^{3} c^{2} - 48 \, a^{2} b c^{3}\right )} d^{4}\right )} \sqrt {c x^{2} + b x + a}}{384 \, c^{2}}, \frac {3 \, {\left (b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}\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 (256 \, c^{6} d^{4} x^{5} + 640 \, b c^{5} d^{4} x^{4} + 16 \, {\left (39 \, b^{2} c^{4} + 4 \, a c^{5}\right )} d^{4} x^{3} + 8 \, {\left (37 \, b^{3} c^{3} + 12 \, a b c^{4}\right )} d^{4} x^{2} + 2 \, {\left (31 \, b^{4} c^{2} + 48 \, a b^{2} c^{3} - 48 \, a^{2} c^{4}\right )} d^{4} x + {\left (3 \, b^{5} c + 32 \, a b^{3} c^{2} - 48 \, a^{2} b c^{3}\right )} d^{4}\right )} \sqrt {c x^{2} + b x + a}}{192 \, c^{2}}\right ] \end {gather*}

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)^(1/2),x, algorithm="fricas")

[Out]

[-1/384*(3*(b^6 - 12*a*b^4*c + 48*a^2*b^2*c^2 - 64*a^3*c^3)*sqrt(c)*d^4*log(-8*c^2*x^2 - 8*b*c*x - b^2 - 4*sqr
t(c*x^2 + b*x + a)*(2*c*x + b)*sqrt(c) - 4*a*c) - 4*(256*c^6*d^4*x^5 + 640*b*c^5*d^4*x^4 + 16*(39*b^2*c^4 + 4*
a*c^5)*d^4*x^3 + 8*(37*b^3*c^3 + 12*a*b*c^4)*d^4*x^2 + 2*(31*b^4*c^2 + 48*a*b^2*c^3 - 48*a^2*c^4)*d^4*x + (3*b
^5*c + 32*a*b^3*c^2 - 48*a^2*b*c^3)*d^4)*sqrt(c*x^2 + b*x + a))/c^2, 1/192*(3*(b^6 - 12*a*b^4*c + 48*a^2*b^2*c
^2 - 64*a^3*c^3)*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*(256*c^6*d^4*x^5 + 640*b*c^5*d^4*x^4 + 16*(39*b^2*c^4 + 4*a*c^5)*d^4*x^3 + 8*(37*b^3*c^3 + 12*a*b*c^4)*d^4*
x^2 + 2*(31*b^4*c^2 + 48*a*b^2*c^3 - 48*a^2*c^4)*d^4*x + (3*b^5*c + 32*a*b^3*c^2 - 48*a^2*b*c^3)*d^4)*sqrt(c*x
^2 + b*x + a))/c^2]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} d^{4} \left (\int b^{4} \sqrt {a + b x + c x^{2}}\, dx + \int 16 c^{4} x^{4} \sqrt {a + b x + c x^{2}}\, dx + \int 32 b c^{3} x^{3} \sqrt {a + b x + c x^{2}}\, dx + \int 24 b^{2} c^{2} x^{2} \sqrt {a + b x + c x^{2}}\, dx + \int 8 b^{3} 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)**4*(c*x**2+b*x+a)**(1/2),x)

[Out]

d**4*(Integral(b**4*sqrt(a + b*x + c*x**2), x) + Integral(16*c**4*x**4*sqrt(a + b*x + c*x**2), x) + Integral(3
2*b*c**3*x**3*sqrt(a + b*x + c*x**2), x) + Integral(24*b**2*c**2*x**2*sqrt(a + b*x + c*x**2), x) + Integral(8*
b**3*c*x*sqrt(a + b*x + c*x**2), x))

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Giac [A]
time = 1.40, size = 259, normalized size = 1.57 \begin {gather*} \frac {1}{96} \, \sqrt {c x^{2} + b x + a} {\left (2 \, {\left (4 \, {\left (2 \, {\left (8 \, {\left (2 \, c^{4} d^{4} x + 5 \, b c^{3} d^{4}\right )} x + \frac {39 \, b^{2} c^{7} d^{4} + 4 \, a c^{8} d^{4}}{c^{5}}\right )} x + \frac {37 \, b^{3} c^{6} d^{4} + 12 \, a b c^{7} d^{4}}{c^{5}}\right )} x + \frac {31 \, b^{4} c^{5} d^{4} + 48 \, a b^{2} c^{6} d^{4} - 48 \, a^{2} c^{7} d^{4}}{c^{5}}\right )} x + \frac {3 \, b^{5} c^{4} d^{4} + 32 \, a b^{3} c^{5} d^{4} - 48 \, a^{2} b c^{6} d^{4}}{c^{5}}\right )} + \frac {{\left (b^{6} d^{4} - 12 \, a b^{4} c d^{4} + 48 \, a^{2} b^{2} c^{2} d^{4} - 64 \, a^{3} c^{3} d^{4}\right )} \log \left ({\left | -2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} \sqrt {c} - b \right |}\right )}{64 \, c^{\frac {3}{2}}} \end {gather*}

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)^(1/2),x, algorithm="giac")

[Out]

1/96*sqrt(c*x^2 + b*x + a)*(2*(4*(2*(8*(2*c^4*d^4*x + 5*b*c^3*d^4)*x + (39*b^2*c^7*d^4 + 4*a*c^8*d^4)/c^5)*x +
(37*b^3*c^6*d^4 + 12*a*b*c^7*d^4)/c^5)*x + (31*b^4*c^5*d^4 + 48*a*b^2*c^6*d^4 - 48*a^2*c^7*d^4)/c^5)*x + (3*b
^5*c^4*d^4 + 32*a*b^3*c^5*d^4 - 48*a^2*b*c^6*d^4)/c^5) + 1/64*(b^6*d^4 - 12*a*b^4*c*d^4 + 48*a^2*b^2*c^2*d^4 -
64*a^3*c^3*d^4)*log(abs(-2*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*sqrt(c) - b))/c^(3/2)

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Mupad [B]
time = 2.19, size = 1144, normalized size = 6.93 \begin {gather*} 8\,a\,c^3\,d^4\,\left (\frac {5\,b\,\left (\frac {\ln \left (\frac {b+2\,c\,x}{\sqrt {c}}+2\,\sqrt {c\,x^2+b\,x+a}\right )\,\left (b^3-4\,a\,b\,c\right )}{16\,c^{5/2}}+\frac {\left (-3\,b^2+2\,c\,x\,b+8\,c\,\left (c\,x^2+a\right )\right )\,\sqrt {c\,x^2+b\,x+a}}{24\,c^2}\right )}{8\,c}-\frac {x\,{\left (c\,x^2+b\,x+a\right )}^{3/2}}{4\,c}+\frac {a\,\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 )}{4\,c}\right )-12\,b\,c^3\,d^4\,\left (\frac {7\,b\,\left (\frac {5\,b\,\left (\frac {\ln \left (\frac {b+2\,c\,x}{\sqrt {c}}+2\,\sqrt {c\,x^2+b\,x+a}\right )\,\left (b^3-4\,a\,b\,c\right )}{16\,c^{5/2}}+\frac {\left (-3\,b^2+2\,c\,x\,b+8\,c\,\left (c\,x^2+a\right )\right )\,\sqrt {c\,x^2+b\,x+a}}{24\,c^2}\right )}{8\,c}-\frac {x\,{\left (c\,x^2+b\,x+a\right )}^{3/2}}{4\,c}+\frac {a\,\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 )}{4\,c}\right )}{10\,c}-\frac {2\,a\,\left (\frac {\ln \left (\frac {b+2\,c\,x}{\sqrt {c}}+2\,\sqrt {c\,x^2+b\,x+a}\right )\,\left (b^3-4\,a\,b\,c\right )}{16\,c^{5/2}}+\frac {\left (-3\,b^2+2\,c\,x\,b+8\,c\,\left (c\,x^2+a\right )\right )\,\sqrt {c\,x^2+b\,x+a}}{24\,c^2}\right )}{5\,c}+\frac {x^2\,{\left (c\,x^2+b\,x+a\right )}^{3/2}}{5\,c}\right )+b^4\,d^4\,\left (\frac {x}{2}+\frac {b}{4\,c}\right )\,\sqrt {c\,x^2+b\,x+a}+\frac {8\,c^3\,d^4\,x^3\,{\left (c\,x^2+b\,x+a\right )}^{3/2}}{3}+\frac {112\,b^2\,c^2\,d^4\,\left (\frac {5\,b\,\left (\frac {\ln \left (\frac {b+2\,c\,x}{\sqrt {c}}+2\,\sqrt {c\,x^2+b\,x+a}\right )\,\left (b^3-4\,a\,b\,c\right )}{16\,c^{5/2}}+\frac {\left (-3\,b^2+2\,c\,x\,b+8\,c\,\left (c\,x^2+a\right )\right )\,\sqrt {c\,x^2+b\,x+a}}{24\,c^2}\right )}{8\,c}-\frac {x\,{\left (c\,x^2+b\,x+a\right )}^{3/2}}{4\,c}+\frac {a\,\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 )}{4\,c}\right )}{5}-15\,b^3\,c\,d^4\,\left (\frac {\ln \left (\frac {b+2\,c\,x}{\sqrt {c}}+2\,\sqrt {c\,x^2+b\,x+a}\right )\,\left (b^3-4\,a\,b\,c\right )}{16\,c^{5/2}}+\frac {\left (-3\,b^2+2\,c\,x\,b+8\,c\,\left (c\,x^2+a\right )\right )\,\sqrt {c\,x^2+b\,x+a}}{24\,c^2}\right )+\frac {32\,b\,c^2\,d^4\,x^2\,{\left (c\,x^2+b\,x+a\right )}^{3/2}}{5}-6\,a\,b^2\,c\,d^4\,\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 )-\frac {64\,a\,b\,c^2\,d^4\,\left (\frac {\ln \left (\frac {b+2\,c\,x}{\sqrt {c}}+2\,\sqrt {c\,x^2+b\,x+a}\right )\,\left (b^3-4\,a\,b\,c\right )}{16\,c^{5/2}}+\frac {\left (-3\,b^2+2\,c\,x\,b+8\,c\,\left (c\,x^2+a\right )\right )\,\sqrt {c\,x^2+b\,x+a}}{24\,c^2}\right )}{5}+\frac {b^4\,d^4\,\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}}+\frac {b^3\,d^4\,\ln \left (\frac {b+2\,c\,x}{\sqrt {c}}+2\,\sqrt {c\,x^2+b\,x+a}\right )\,\left (b^3-4\,a\,b\,c\right )}{2\,c^{3/2}}+6\,b^2\,c\,d^4\,x\,{\left (c\,x^2+b\,x+a\right )}^{3/2}+\frac {b^3\,d^4\,\left (-3\,b^2+2\,c\,x\,b+8\,c\,\left (c\,x^2+a\right )\right )\,\sqrt {c\,x^2+b\,x+a}}{3\,c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*d + 2*c*d*x)^4*(a + b*x + c*x^2)^(1/2),x)

[Out]

8*a*c^3*d^4*((5*b*((log((b + 2*c*x)/c^(1/2) + 2*(a + b*x + c*x^2)^(1/2))*(b^3 - 4*a*b*c))/(16*c^(5/2)) + ((8*c
*(a + c*x^2) - 3*b^2 + 2*b*c*x)*(a + b*x + c*x^2)^(1/2))/(24*c^2)))/(8*c) - (x*(a + b*x + c*x^2)^(3/2))/(4*c)
+ (a*((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))))/(4*c)) - 12*b*c^3*d^4*((7*b*((5*b*((log((b + 2*c*x)/c^(1/2) + 2*(a + b*x + c*x^2)^(1/2))*(b
^3 - 4*a*b*c))/(16*c^(5/2)) + ((8*c*(a + c*x^2) - 3*b^2 + 2*b*c*x)*(a + b*x + c*x^2)^(1/2))/(24*c^2)))/(8*c) -
(x*(a + b*x + c*x^2)^(3/2))/(4*c) + (a*((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))))/(4*c)))/(10*c) - (2*a*((log((b + 2*c*x)/c^(1/2) + 2*(a +
b*x + c*x^2)^(1/2))*(b^3 - 4*a*b*c))/(16*c^(5/2)) + ((8*c*(a + c*x^2) - 3*b^2 + 2*b*c*x)*(a + b*x + c*x^2)^(1
/2))/(24*c^2)))/(5*c) + (x^2*(a + b*x + c*x^2)^(3/2))/(5*c)) + b^4*d^4*(x/2 + b/(4*c))*(a + b*x + c*x^2)^(1/2)
+ (8*c^3*d^4*x^3*(a + b*x + c*x^2)^(3/2))/3 + (112*b^2*c^2*d^4*((5*b*((log((b + 2*c*x)/c^(1/2) + 2*(a + b*x +
c*x^2)^(1/2))*(b^3 - 4*a*b*c))/(16*c^(5/2)) + ((8*c*(a + c*x^2) - 3*b^2 + 2*b*c*x)*(a + b*x + c*x^2)^(1/2))/(
24*c^2)))/(8*c) - (x*(a + b*x + c*x^2)^(3/2))/(4*c) + (a*((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))))/(4*c)))/5 - 15*b^3*c*d^4*((log((b + 2*c
*x)/c^(1/2) + 2*(a + b*x + c*x^2)^(1/2))*(b^3 - 4*a*b*c))/(16*c^(5/2)) + ((8*c*(a + c*x^2) - 3*b^2 + 2*b*c*x)*
(a + b*x + c*x^2)^(1/2))/(24*c^2)) + (32*b*c^2*d^4*x^2*(a + b*x + c*x^2)^(3/2))/5 - 6*a*b^2*c*d^4*((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)))
- (64*a*b*c^2*d^4*((log((b + 2*c*x)/c^(1/2) + 2*(a + b*x + c*x^2)^(1/2))*(b^3 - 4*a*b*c))/(16*c^(5/2)) + ((8*
c*(a + c*x^2) - 3*b^2 + 2*b*c*x)*(a + b*x + c*x^2)^(1/2))/(24*c^2)))/5 + (b^4*d^4*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)) + (b^3*d^4*log((b + 2*c*x)/c^(1/2) + 2*(a + b*x + c*x^2)^(1/
2))*(b^3 - 4*a*b*c))/(2*c^(3/2)) + 6*b^2*c*d^4*x*(a + b*x + c*x^2)^(3/2) + (b^3*d^4*(8*c*(a + c*x^2) - 3*b^2 +
2*b*c*x)*(a + b*x + c*x^2)^(1/2))/(3*c)

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