3.11.54 \(\int \frac {(b d+2 c d x)^4}{(a+b x+c x^2)^{3/2}} \, dx\)

Optimal. Leaf size=102 \[ 6 \sqrt {c} d^4 \left (b^2-4 a c\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )+12 c d^4 (b+2 c x) \sqrt {a+b x+c x^2}-\frac {2 d^4 (b+2 c x)^3}{\sqrt {a+b x+c x^2}} \]

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Rubi [A]  time = 0.05, antiderivative size = 102, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {686, 692, 621, 206} \begin {gather*} 6 \sqrt {c} d^4 \left (b^2-4 a c\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )+12 c d^4 (b+2 c x) \sqrt {a+b x+c x^2}-\frac {2 d^4 (b+2 c x)^3}{\sqrt {a+b x+c x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*d^4*(b + 2*c*x)^3)/Sqrt[a + b*x + c*x^2] + 12*c*d^4*(b + 2*c*x)*Sqrt[a + b*x + c*x^2] + 6*Sqrt[c]*(b^2 - 4
*a*c)*d^4*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^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 686

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(d*(d + e*x)^(m - 1)*
(a + b*x + c*x^2)^(p + 1))/(b*(p + 1)), x] - Dist[(d*e*(m - 1))/(b*(p + 1)), Int[(d + e*x)^(m - 2)*(a + b*x +
c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[2*c*d - b*e, 0] && NeQ[m + 2
*p + 3, 0] && LtQ[p, -1] && GtQ[m, 1] && 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 \frac {(b d+2 c d x)^4}{\left (a+b x+c x^2\right )^{3/2}} \, dx &=-\frac {2 d^4 (b+2 c x)^3}{\sqrt {a+b x+c x^2}}+\left (12 c d^2\right ) \int \frac {(b d+2 c d x)^2}{\sqrt {a+b x+c x^2}} \, dx\\ &=-\frac {2 d^4 (b+2 c x)^3}{\sqrt {a+b x+c x^2}}+12 c d^4 (b+2 c x) \sqrt {a+b x+c x^2}+\left (6 c \left (b^2-4 a c\right ) d^4\right ) \int \frac {1}{\sqrt {a+b x+c x^2}} \, dx\\ &=-\frac {2 d^4 (b+2 c x)^3}{\sqrt {a+b x+c x^2}}+12 c d^4 (b+2 c x) \sqrt {a+b x+c x^2}+\left (12 c \left (b^2-4 a c\right ) 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 )\\ &=-\frac {2 d^4 (b+2 c x)^3}{\sqrt {a+b x+c x^2}}+12 c d^4 (b+2 c x) \sqrt {a+b x+c x^2}+6 \sqrt {c} \left (b^2-4 a c\right ) d^4 \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )\\ \end {align*}

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Mathematica [A]  time = 0.40, size = 139, normalized size = 1.36 \begin {gather*} d^4 \left (-\frac {6 c^{3/2} (a+x (b+c x))^{3/2} \sinh ^{-1}\left (\frac {b+2 c x}{\sqrt {c} \sqrt {4 a-\frac {b^2}{c}}}\right )}{\sqrt {4 a-\frac {b^2}{c}} \left (\frac {c (a+x (b+c x))}{4 a c-b^2}\right )^{3/2}}-\frac {2 (b+2 c x) \left (-2 c \left (3 a+c x^2\right )+b^2-2 b c x\right )}{\sqrt {a+x (b+c x)}}\right ) \end {gather*}

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*((-2*(b + 2*c*x)*(b^2 - 2*b*c*x - 2*c*(3*a + c*x^2)))/Sqrt[a + x*(b + c*x)] - (6*c^(3/2)*(a + x*(b + c*x))
^(3/2)*ArcSinh[(b + 2*c*x)/(Sqrt[4*a - b^2/c]*Sqrt[c])])/(Sqrt[4*a - b^2/c]*((c*(a + x*(b + c*x)))/(-b^2 + 4*a
*c))^(3/2)))

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IntegrateAlgebraic [A]  time = 0.79, size = 120, normalized size = 1.18 \begin {gather*} -\frac {2 \left (-6 a b c d^4-12 a c^2 d^4 x+b^3 d^4-6 b c^2 d^4 x^2-4 c^3 d^4 x^3\right )}{\sqrt {a+b x+c x^2}}-6 \left (b^2 \sqrt {c} d^4-4 a c^{3/2} d^4\right ) \log \left (-2 \sqrt {c} \sqrt {a+b x+c x^2}+b+2 c x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(b^3*d^4 - 6*a*b*c*d^4 - 12*a*c^2*d^4*x - 6*b*c^2*d^4*x^2 - 4*c^3*d^4*x^3))/Sqrt[a + b*x + c*x^2] - 6*(b^2
*Sqrt[c]*d^4 - 4*a*c^(3/2)*d^4)*Log[b + 2*c*x - 2*Sqrt[c]*Sqrt[a + b*x + c*x^2]]

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fricas [A]  time = 0.56, size = 357, normalized size = 3.50 \begin {gather*} \left [-\frac {3 \, {\left ({\left (b^{2} c - 4 \, a c^{2}\right )} d^{4} x^{2} + {\left (b^{3} - 4 \, a b c\right )} d^{4} x + {\left (a b^{2} - 4 \, a^{2} c\right )} d^{4}\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 ) - 2 \, {\left (4 \, c^{3} d^{4} x^{3} + 6 \, b c^{2} d^{4} x^{2} + 12 \, a c^{2} d^{4} x - {\left (b^{3} - 6 \, a b c\right )} d^{4}\right )} \sqrt {c x^{2} + b x + a}}{c x^{2} + b x + a}, -\frac {2 \, {\left (3 \, {\left ({\left (b^{2} c - 4 \, a c^{2}\right )} d^{4} x^{2} + {\left (b^{3} - 4 \, a b c\right )} d^{4} x + {\left (a b^{2} - 4 \, a^{2} c\right )} d^{4}\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 ) - {\left (4 \, c^{3} d^{4} x^{3} + 6 \, b c^{2} d^{4} x^{2} + 12 \, a c^{2} d^{4} x - {\left (b^{3} - 6 \, a b c\right )} d^{4}\right )} \sqrt {c x^{2} + b x + a}\right )}}{c x^{2} + b x + a}\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)^(3/2),x, algorithm="fricas")

[Out]

[-(3*((b^2*c - 4*a*c^2)*d^4*x^2 + (b^3 - 4*a*b*c)*d^4*x + (a*b^2 - 4*a^2*c)*d^4)*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) - 2*(4*c^3*d^4*x^3 + 6*b*c^2*d^4*x^2 + 12*a*c
^2*d^4*x - (b^3 - 6*a*b*c)*d^4)*sqrt(c*x^2 + b*x + a))/(c*x^2 + b*x + a), -2*(3*((b^2*c - 4*a*c^2)*d^4*x^2 + (
b^3 - 4*a*b*c)*d^4*x + (a*b^2 - 4*a^2*c)*d^4)*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)) - (4*c^3*d^4*x^3 + 6*b*c^2*d^4*x^2 + 12*a*c^2*d^4*x - (b^3 - 6*a*b*c)*d^4)*sqrt(c*x^2
+ b*x + a))/(c*x^2 + b*x + a)]

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giac [B]  time = 0.27, size = 249, normalized size = 2.44 \begin {gather*} -\frac {6 \, {\left (b^{2} c d^{4} - 4 \, a c^{2} d^{4}\right )} \log \left ({\left | -2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} \sqrt {c} - b \right |}\right )}{\sqrt {c}} + \frac {2 \, {\left (2 \, {\left ({\left (\frac {2 \, {\left (b^{2} c^{5} d^{4} - 4 \, a c^{6} d^{4}\right )} x}{b^{2} c^{2} - 4 \, a c^{3}} + \frac {3 \, {\left (b^{3} c^{4} d^{4} - 4 \, a b c^{5} d^{4}\right )}}{b^{2} c^{2} - 4 \, a c^{3}}\right )} x + \frac {6 \, {\left (a b^{2} c^{4} d^{4} - 4 \, a^{2} c^{5} d^{4}\right )}}{b^{2} c^{2} - 4 \, a c^{3}}\right )} x - \frac {b^{5} c^{2} d^{4} - 10 \, a b^{3} c^{3} d^{4} + 24 \, a^{2} b c^{4} d^{4}}{b^{2} c^{2} - 4 \, a c^{3}}\right )}}{\sqrt {c x^{2} + b x + a}} \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)^(3/2),x, algorithm="giac")

[Out]

-6*(b^2*c*d^4 - 4*a*c^2*d^4)*log(abs(-2*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*sqrt(c) - b))/sqrt(c) + 2*(2*((2*(
b^2*c^5*d^4 - 4*a*c^6*d^4)*x/(b^2*c^2 - 4*a*c^3) + 3*(b^3*c^4*d^4 - 4*a*b*c^5*d^4)/(b^2*c^2 - 4*a*c^3))*x + 6*
(a*b^2*c^4*d^4 - 4*a^2*c^5*d^4)/(b^2*c^2 - 4*a*c^3))*x - (b^5*c^2*d^4 - 10*a*b^3*c^3*d^4 + 24*a^2*b*c^4*d^4)/(
b^2*c^2 - 4*a*c^3))/sqrt(c*x^2 + b*x + a)

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maple [B]  time = 0.06, size = 340, normalized size = 3.33 \begin {gather*} \frac {24 a \,b^{2} c^{2} d^{4} x}{\left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}}-\frac {6 b^{4} c \,d^{4} x}{\left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}}+\frac {8 c^{3} d^{4} x^{3}}{\sqrt {c \,x^{2}+b x +a}}+\frac {12 a \,b^{3} c \,d^{4}}{\left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}}-\frac {3 b^{5} d^{4}}{\left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}}+\frac {12 b \,c^{2} d^{4} x^{2}}{\sqrt {c \,x^{2}+b x +a}}+\frac {24 a \,c^{2} d^{4} x}{\sqrt {c \,x^{2}+b x +a}}-\frac {6 b^{2} c \,d^{4} x}{\sqrt {c \,x^{2}+b x +a}}-24 a \,c^{\frac {3}{2}} d^{4} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )+6 b^{2} \sqrt {c}\, d^{4} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )+\frac {12 a b c \,d^{4}}{\sqrt {c \,x^{2}+b x +a}}-\frac {5 b^{3} d^{4}}{\sqrt {c \,x^{2}+b x +a}} \end {gather*}

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*b^5/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)+12*d^4*c^2*b*x^2/(c*x^2+b*x+a)^(1/2)-6*d^4*c*b^2*x/(c*x^2+b*x+a)^(1
/2)+6*d^4*c^(1/2)*b^2*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))+12*d^4*c*b*a/(c*x^2+b*x+a)^(1/2)+24*d^4*c^2*
a*x/(c*x^2+b*x+a)^(1/2)-24*d^4*c^(3/2)*a*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))+24*d^4*c^2*b^2*a/(4*a*c-b
^2)/(c*x^2+b*x+a)^(1/2)*x-6*d^4*c*b^4/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)*x-5*d^4*b^3/(c*x^2+b*x+a)^(1/2)+8*d^4*c^
3*x^3/(c*x^2+b*x+a)^(1/2)+12*d^4*c*b^3*a/(4*a*c-b^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)^(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 zero or nonzero?

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (b\,d+2\,c\,d\,x\right )}^4}{{\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*d + 2*c*d*x)^4/(a + b*x + c*x^2)^(3/2),x)

[Out]

int((b*d + 2*c*d*x)^4/(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*} d^{4} \left (\int \frac {b^{4}}{a \sqrt {a + b x + c x^{2}} + b x \sqrt {a + b x + c x^{2}} + c x^{2} \sqrt {a + b x + c x^{2}}}\, dx + \int \frac {16 c^{4} x^{4}}{a \sqrt {a + b x + c x^{2}} + b x \sqrt {a + b x + c x^{2}} + c x^{2} \sqrt {a + b x + c x^{2}}}\, dx + \int \frac {32 b c^{3} x^{3}}{a \sqrt {a + b x + c x^{2}} + b x \sqrt {a + b x + c x^{2}} + c x^{2} \sqrt {a + b x + c x^{2}}}\, dx + \int \frac {24 b^{2} c^{2} x^{2}}{a \sqrt {a + b x + c x^{2}} + b x \sqrt {a + b x + c x^{2}} + c x^{2} \sqrt {a + b x + c x^{2}}}\, dx + \int \frac {8 b^{3} c x}{a \sqrt {a + b x + c x^{2}} + b x \sqrt {a + b x + c x^{2}} + c x^{2} \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)**(3/2),x)

[Out]

d**4*(Integral(b**4/(a*sqrt(a + b*x + c*x**2) + b*x*sqrt(a + b*x + c*x**2) + c*x**2*sqrt(a + b*x + c*x**2)), x
) + Integral(16*c**4*x**4/(a*sqrt(a + b*x + c*x**2) + b*x*sqrt(a + b*x + c*x**2) + c*x**2*sqrt(a + b*x + c*x**
2)), x) + Integral(32*b*c**3*x**3/(a*sqrt(a + b*x + c*x**2) + b*x*sqrt(a + b*x + c*x**2) + c*x**2*sqrt(a + b*x
+ c*x**2)), x) + Integral(24*b**2*c**2*x**2/(a*sqrt(a + b*x + c*x**2) + b*x*sqrt(a + b*x + c*x**2) + c*x**2*s
qrt(a + b*x + c*x**2)), x) + Integral(8*b**3*c*x/(a*sqrt(a + b*x + c*x**2) + b*x*sqrt(a + b*x + c*x**2) + c*x*
*2*sqrt(a + b*x + c*x**2)), x))

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