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

Optimal. Leaf size=131 \[ \frac {6 c d^{5/2} \tan ^{-1}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )}{\sqrt [4]{b^2-4 a c}}-\frac {6 c d^{5/2} \tanh ^{-1}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )}{\sqrt [4]{b^2-4 a c}}-\frac {d (b d+2 c d x)^{3/2}}{a+b x+c x^2} \]

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Rubi [A]  time = 0.10, antiderivative size = 131, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {686, 694, 329, 298, 203, 206} \begin {gather*} \frac {6 c d^{5/2} \tan ^{-1}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )}{\sqrt [4]{b^2-4 a c}}-\frac {6 c d^{5/2} \tanh ^{-1}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )}{\sqrt [4]{b^2-4 a c}}-\frac {d (b d+2 c d x)^{3/2}}{a+b x+c x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((d*(b*d + 2*c*d*x)^(3/2))/(a + b*x + c*x^2)) + (6*c*d^(5/2)*ArcTan[Sqrt[d*(b + 2*c*x)]/((b^2 - 4*a*c)^(1/4)*
Sqrt[d])])/(b^2 - 4*a*c)^(1/4) - (6*c*d^(5/2)*ArcTanh[Sqrt[d*(b + 2*c*x)]/((b^2 - 4*a*c)^(1/4)*Sqrt[d])])/(b^2
- 4*a*c)^(1/4)

Rule 203

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

Rule 298

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]], s = Denominator[Rt[-(a/b),
2]]}, Dist[s/(2*b), Int[1/(r + s*x^2), x], x] - Dist[s/(2*b), Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] &
&  !GtQ[a/b, 0]

Rule 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

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 694

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

Rubi steps

\begin {align*} \int \frac {(b d+2 c d x)^{5/2}}{\left (a+b x+c x^2\right )^2} \, dx &=-\frac {d (b d+2 c d x)^{3/2}}{a+b x+c x^2}+\left (3 c d^2\right ) \int \frac {\sqrt {b d+2 c d x}}{a+b x+c x^2} \, dx\\ &=-\frac {d (b d+2 c d x)^{3/2}}{a+b x+c x^2}+\frac {1}{2} (3 d) \operatorname {Subst}\left (\int \frac {\sqrt {x}}{a-\frac {b^2}{4 c}+\frac {x^2}{4 c d^2}} \, dx,x,b d+2 c d x\right )\\ &=-\frac {d (b d+2 c d x)^{3/2}}{a+b x+c x^2}+(3 d) \operatorname {Subst}\left (\int \frac {x^2}{a-\frac {b^2}{4 c}+\frac {x^4}{4 c d^2}} \, dx,x,\sqrt {d (b+2 c x)}\right )\\ &=-\frac {d (b d+2 c d x)^{3/2}}{a+b x+c x^2}-\left (6 c d^3\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {b^2-4 a c} d-x^2} \, dx,x,\sqrt {d (b+2 c x)}\right )+\left (6 c d^3\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {b^2-4 a c} d+x^2} \, dx,x,\sqrt {d (b+2 c x)}\right )\\ &=-\frac {d (b d+2 c d x)^{3/2}}{a+b x+c x^2}+\frac {6 c d^{5/2} \tan ^{-1}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )}{\sqrt [4]{b^2-4 a c}}-\frac {6 c d^{5/2} \tanh ^{-1}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )}{\sqrt [4]{b^2-4 a c}}\\ \end {align*}

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Mathematica [C]  time = 0.08, size = 83, normalized size = 0.63 \begin {gather*} -\frac {4 d (d (b+2 c x))^{3/2} \left (4 c (a+x (b+c x)) \, _2F_1\left (\frac {3}{4},2;\frac {7}{4};\frac {(b+2 c x)^2}{b^2-4 a c}\right )-4 a c+b^2\right )}{\left (b^2-4 a c\right ) (a+x (b+c x))} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*d*(d*(b + 2*c*x))^(3/2)*(b^2 - 4*a*c + 4*c*(a + x*(b + c*x))*Hypergeometric2F1[3/4, 2, 7/4, (b + 2*c*x)^2/
(b^2 - 4*a*c)]))/((b^2 - 4*a*c)*(a + x*(b + c*x)))

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IntegrateAlgebraic [C]  time = 0.80, size = 236, normalized size = 1.80 \begin {gather*} -\frac {(3+3 i) c d^{5/2} \tan ^{-1}\left (\frac {-\frac {(1+i) c \sqrt {d} x}{\sqrt [4]{b^2-4 a c}}-\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) b \sqrt {d}}{\sqrt [4]{b^2-4 a c}}+\left (\frac {1}{2}-\frac {i}{2}\right ) \sqrt {d} \sqrt [4]{b^2-4 a c}}{\sqrt {b d+2 c d x}}\right )}{\sqrt [4]{b^2-4 a c}}-\frac {(3+3 i) c d^{5/2} \tanh ^{-1}\left (\frac {(1+i) \sqrt [4]{b^2-4 a c} \sqrt {b d+2 c d x}}{\sqrt {d} \left (\sqrt {b^2-4 a c}+i b+2 i c x\right )}\right )}{\sqrt [4]{b^2-4 a c}}+\frac {\sqrt {b d+2 c d x} \left (-b d^2-2 c d^2 x\right )}{a+b x+c x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[b*d + 2*c*d*x]*(-(b*d^2) - 2*c*d^2*x))/(a + b*x + c*x^2) - ((3 + 3*I)*c*d^(5/2)*ArcTan[(((-1/2 - I/2)*b*
Sqrt[d])/(b^2 - 4*a*c)^(1/4) + (1/2 - I/2)*(b^2 - 4*a*c)^(1/4)*Sqrt[d] - ((1 + I)*c*Sqrt[d]*x)/(b^2 - 4*a*c)^(
1/4))/Sqrt[b*d + 2*c*d*x]])/(b^2 - 4*a*c)^(1/4) - ((3 + 3*I)*c*d^(5/2)*ArcTanh[((1 + I)*(b^2 - 4*a*c)^(1/4)*Sq
rt[b*d + 2*c*d*x])/(Sqrt[d]*(I*b + Sqrt[b^2 - 4*a*c] + (2*I)*c*x))])/(b^2 - 4*a*c)^(1/4)

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fricas [B]  time = 0.45, size = 358, normalized size = 2.73 \begin {gather*} -\frac {12 \, \left (\frac {c^{4} d^{10}}{b^{2} - 4 \, a c}\right )^{\frac {1}{4}} {\left (c x^{2} + b x + a\right )} \arctan \left (-\frac {\left (\frac {c^{4} d^{10}}{b^{2} - 4 \, a c}\right )^{\frac {1}{4}} \sqrt {2 \, c d x + b d} c^{3} d^{7} - \sqrt {2 \, c^{7} d^{15} x + b c^{6} d^{15} + \sqrt {\frac {c^{4} d^{10}}{b^{2} - 4 \, a c}} {\left (b^{2} c^{4} - 4 \, a c^{5}\right )} d^{10}} \left (\frac {c^{4} d^{10}}{b^{2} - 4 \, a c}\right )^{\frac {1}{4}}}{c^{4} d^{10}}\right ) + 3 \, \left (\frac {c^{4} d^{10}}{b^{2} - 4 \, a c}\right )^{\frac {1}{4}} {\left (c x^{2} + b x + a\right )} \log \left (27 \, \sqrt {2 \, c d x + b d} c^{3} d^{7} + 27 \, \left (\frac {c^{4} d^{10}}{b^{2} - 4 \, a c}\right )^{\frac {3}{4}} {\left (b^{2} - 4 \, a c\right )}\right ) - 3 \, \left (\frac {c^{4} d^{10}}{b^{2} - 4 \, a c}\right )^{\frac {1}{4}} {\left (c x^{2} + b x + a\right )} \log \left (27 \, \sqrt {2 \, c d x + b d} c^{3} d^{7} - 27 \, \left (\frac {c^{4} d^{10}}{b^{2} - 4 \, a c}\right )^{\frac {3}{4}} {\left (b^{2} - 4 \, a c\right )}\right ) + {\left (2 \, c d^{2} x + b d^{2}\right )} \sqrt {2 \, c d x + b d}}{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)^(5/2)/(c*x^2+b*x+a)^2,x, algorithm="fricas")

[Out]

-(12*(c^4*d^10/(b^2 - 4*a*c))^(1/4)*(c*x^2 + b*x + a)*arctan(-((c^4*d^10/(b^2 - 4*a*c))^(1/4)*sqrt(2*c*d*x + b
*d)*c^3*d^7 - sqrt(2*c^7*d^15*x + b*c^6*d^15 + sqrt(c^4*d^10/(b^2 - 4*a*c))*(b^2*c^4 - 4*a*c^5)*d^10)*(c^4*d^1
0/(b^2 - 4*a*c))^(1/4))/(c^4*d^10)) + 3*(c^4*d^10/(b^2 - 4*a*c))^(1/4)*(c*x^2 + b*x + a)*log(27*sqrt(2*c*d*x +
b*d)*c^3*d^7 + 27*(c^4*d^10/(b^2 - 4*a*c))^(3/4)*(b^2 - 4*a*c)) - 3*(c^4*d^10/(b^2 - 4*a*c))^(1/4)*(c*x^2 + b
*x + a)*log(27*sqrt(2*c*d*x + b*d)*c^3*d^7 - 27*(c^4*d^10/(b^2 - 4*a*c))^(3/4)*(b^2 - 4*a*c)) + (2*c*d^2*x + b
*d^2)*sqrt(2*c*d*x + b*d))/(c*x^2 + b*x + a)

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giac [B]  time = 0.26, size = 439, normalized size = 3.35 \begin {gather*} \frac {4 \, {\left (2 \, c d x + b d\right )}^{\frac {3}{2}} c d^{3}}{b^{2} d^{2} - 4 \, a c d^{2} - {\left (2 \, c d x + b d\right )}^{2}} - \frac {3 \, \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {3}{4}} c d \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} + 2 \, \sqrt {2 \, c d x + b d}\right )}}{2 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}}}\right )}{b^{2} - 4 \, a c} - \frac {3 \, \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {3}{4}} c d \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} - 2 \, \sqrt {2 \, c d x + b d}\right )}}{2 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}}}\right )}{b^{2} - 4 \, a c} + \frac {3 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {3}{4}} c d \log \left (2 \, c d x + b d + \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} \sqrt {2 \, c d x + b d} + \sqrt {-b^{2} d^{2} + 4 \, a c d^{2}}\right )}{\sqrt {2} b^{2} - 4 \, \sqrt {2} a c} - \frac {3 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {3}{4}} c d \log \left (2 \, c d x + b d - \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} \sqrt {2 \, c d x + b d} + \sqrt {-b^{2} d^{2} + 4 \, a c d^{2}}\right )}{\sqrt {2} b^{2} - 4 \, \sqrt {2} a c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4*(2*c*d*x + b*d)^(3/2)*c*d^3/(b^2*d^2 - 4*a*c*d^2 - (2*c*d*x + b*d)^2) - 3*sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(3/
4)*c*d*arctan(1/2*sqrt(2)*(sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(1/4) + 2*sqrt(2*c*d*x + b*d))/(-b^2*d^2 + 4*a*c*d^2
)^(1/4))/(b^2 - 4*a*c) - 3*sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(3/4)*c*d*arctan(-1/2*sqrt(2)*(sqrt(2)*(-b^2*d^2 + 4
*a*c*d^2)^(1/4) - 2*sqrt(2*c*d*x + b*d))/(-b^2*d^2 + 4*a*c*d^2)^(1/4))/(b^2 - 4*a*c) + 3*(-b^2*d^2 + 4*a*c*d^2
)^(3/4)*c*d*log(2*c*d*x + b*d + sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(1/4)*sqrt(2*c*d*x + b*d) + sqrt(-b^2*d^2 + 4*a
*c*d^2))/(sqrt(2)*b^2 - 4*sqrt(2)*a*c) - 3*(-b^2*d^2 + 4*a*c*d^2)^(3/4)*c*d*log(2*c*d*x + b*d - sqrt(2)*(-b^2*
d^2 + 4*a*c*d^2)^(1/4)*sqrt(2*c*d*x + b*d) + sqrt(-b^2*d^2 + 4*a*c*d^2))/(sqrt(2)*b^2 - 4*sqrt(2)*a*c)

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maple [B]  time = 0.06, size = 327, normalized size = 2.50 \begin {gather*} -\frac {3 \sqrt {2}\, c \,d^{3} \arctan \left (-\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}+\frac {3 \sqrt {2}\, c \,d^{3} \arctan \left (\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}+\frac {3 \sqrt {2}\, c \,d^{3} \ln \left (\frac {2 c d x +b d -\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a c \,d^{2}-b^{2} d^{2}}}{2 c d x +b d +\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a c \,d^{2}-b^{2} d^{2}}}\right )}{2 \left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}-\frac {4 \left (2 c d x +b d \right )^{\frac {3}{2}} c \,d^{3}}{4 c^{2} d^{2} x^{2}+4 b c \,d^{2} x +4 a c \,d^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4*c*d^3*(2*c*d*x+b*d)^(3/2)/(4*c^2*d^2*x^2+4*b*c*d^2*x+4*a*c*d^2)+3/2*c*d^3*2^(1/2)/(4*a*c*d^2-b^2*d^2)^(1/4)
*ln((2*c*d*x+b*d-(4*a*c*d^2-b^2*d^2)^(1/4)*(2*c*d*x+b*d)^(1/2)*2^(1/2)+(4*a*c*d^2-b^2*d^2)^(1/2))/(2*c*d*x+b*d
+(4*a*c*d^2-b^2*d^2)^(1/4)*(2*c*d*x+b*d)^(1/2)*2^(1/2)+(4*a*c*d^2-b^2*d^2)^(1/2)))+3*c*d^3*2^(1/2)/(4*a*c*d^2-
b^2*d^2)^(1/4)*arctan(2^(1/2)/(4*a*c*d^2-b^2*d^2)^(1/4)*(2*c*d*x+b*d)^(1/2)+1)-3*c*d^3*2^(1/2)/(4*a*c*d^2-b^2*
d^2)^(1/4)*arctan(-2^(1/2)/(4*a*c*d^2-b^2*d^2)^(1/4)*(2*c*d*x+b*d)^(1/2)+1)

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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)^(5/2)/(c*x^2+b*x+a)^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 [B]  time = 0.18, size = 131, normalized size = 1.00 \begin {gather*} \frac {6\,c\,d^{5/2}\,\mathrm {atan}\left (\frac {\sqrt {b\,d+2\,c\,d\,x}}{\sqrt {d}\,{\left (b^2-4\,a\,c\right )}^{1/4}}\right )}{{\left (b^2-4\,a\,c\right )}^{1/4}}-\frac {4\,c\,d^3\,{\left (b\,d+2\,c\,d\,x\right )}^{3/2}}{{\left (b\,d+2\,c\,d\,x\right )}^2-b^2\,d^2+4\,a\,c\,d^2}-\frac {6\,c\,d^{5/2}\,\mathrm {atanh}\left (\frac {\sqrt {b\,d+2\,c\,d\,x}}{\sqrt {d}\,{\left (b^2-4\,a\,c\right )}^{1/4}}\right )}{{\left (b^2-4\,a\,c\right )}^{1/4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(6*c*d^(5/2)*atan((b*d + 2*c*d*x)^(1/2)/(d^(1/2)*(b^2 - 4*a*c)^(1/4))))/(b^2 - 4*a*c)^(1/4) - (4*c*d^3*(b*d +
2*c*d*x)^(3/2))/((b*d + 2*c*d*x)^2 - b^2*d^2 + 4*a*c*d^2) - (6*c*d^(5/2)*atanh((b*d + 2*c*d*x)^(1/2)/(d^(1/2)*
(b^2 - 4*a*c)^(1/4))))/(b^2 - 4*a*c)^(1/4)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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