3.12.11 \(\int \frac {(b d+2 c d x)^{7/2}}{(a+b x+c x^2)^2} \, dx\)

Optimal. Leaf size=150 \[ -10 c d^{7/2} \sqrt [4]{b^2-4 a c} \tan ^{-1}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )-10 c d^{7/2} \sqrt [4]{b^2-4 a c} \tanh ^{-1}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )-\frac {d (b d+2 c d x)^{5/2}}{a+b x+c x^2}+20 c d^3 \sqrt {b d+2 c d x} \]

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Rubi [A]  time = 0.12, antiderivative size = 150, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.269, Rules used = {686, 692, 694, 329, 212, 206, 203} \begin {gather*} -10 c d^{7/2} \sqrt [4]{b^2-4 a c} \tan ^{-1}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )-10 c d^{7/2} \sqrt [4]{b^2-4 a c} \tanh ^{-1}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )-\frac {d (b d+2 c d x)^{5/2}}{a+b x+c x^2}+20 c d^3 \sqrt {b d+2 c d x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(b*d + 2*c*d*x)^(7/2)/(a + b*x + c*x^2)^2,x]

[Out]

20*c*d^3*Sqrt[b*d + 2*c*d*x] - (d*(b*d + 2*c*d*x)^(5/2))/(a + b*x + c*x^2) - 10*c*(b^2 - 4*a*c)^(1/4)*d^(7/2)*
ArcTan[Sqrt[d*(b + 2*c*x)]/((b^2 - 4*a*c)^(1/4)*Sqrt[d])] - 10*c*(b^2 - 4*a*c)^(1/4)*d^(7/2)*ArcTanh[Sqrt[d*(b
+ 2*c*x)]/((b^2 - 4*a*c)^(1/4)*Sqrt[d])]

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 212

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]], s = Denominator[Rt[-(a/b), 2]
]}, Dist[r/(2*a), Int[1/(r - s*x^2), x], x] + Dist[r/(2*a), 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 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 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)^{7/2}}{\left (a+b x+c x^2\right )^2} \, dx &=-\frac {d (b d+2 c d x)^{5/2}}{a+b x+c x^2}+\left (5 c d^2\right ) \int \frac {(b d+2 c d x)^{3/2}}{a+b x+c x^2} \, dx\\ &=20 c d^3 \sqrt {b d+2 c d x}-\frac {d (b d+2 c d x)^{5/2}}{a+b x+c x^2}+\left (5 c \left (b^2-4 a c\right ) d^4\right ) \int \frac {1}{\sqrt {b d+2 c d x} \left (a+b x+c x^2\right )} \, dx\\ &=20 c d^3 \sqrt {b d+2 c d x}-\frac {d (b d+2 c d x)^{5/2}}{a+b x+c x^2}+\frac {1}{2} \left (5 \left (b^2-4 a c\right ) d^3\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {x} \left (a-\frac {b^2}{4 c}+\frac {x^2}{4 c d^2}\right )} \, dx,x,b d+2 c d x\right )\\ &=20 c d^3 \sqrt {b d+2 c d x}-\frac {d (b d+2 c d x)^{5/2}}{a+b x+c x^2}+\left (5 \left (b^2-4 a c\right ) d^3\right ) \operatorname {Subst}\left (\int \frac {1}{a-\frac {b^2}{4 c}+\frac {x^4}{4 c d^2}} \, dx,x,\sqrt {d (b+2 c x)}\right )\\ &=20 c d^3 \sqrt {b d+2 c d x}-\frac {d (b d+2 c d x)^{5/2}}{a+b x+c x^2}-\left (10 c \sqrt {b^2-4 a c} d^4\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 (10 c \sqrt {b^2-4 a c} d^4\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 )\\ &=20 c d^3 \sqrt {b d+2 c d x}-\frac {d (b d+2 c d x)^{5/2}}{a+b x+c x^2}-10 c \sqrt [4]{b^2-4 a c} d^{7/2} \tan ^{-1}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )-10 c \sqrt [4]{b^2-4 a c} d^{7/2} \tanh ^{-1}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )\\ \end {align*}

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Mathematica [A]  time = 0.18, size = 168, normalized size = 1.12 \begin {gather*} -\frac {d^3 \sqrt {d (b+2 c x)} \left (\sqrt {b+2 c x} \left (-4 c \left (5 a+4 c x^2\right )+b^2-16 b c x\right )+10 c \sqrt [4]{b^2-4 a c} (a+x (b+c x)) \tan ^{-1}\left (\frac {\sqrt {b+2 c x}}{\sqrt [4]{b^2-4 a c}}\right )+10 c \sqrt [4]{b^2-4 a c} (a+x (b+c x)) \tanh ^{-1}\left (\frac {\sqrt {b+2 c x}}{\sqrt [4]{b^2-4 a c}}\right )\right )}{\sqrt {b+2 c x} (a+x (b+c x))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(b*d + 2*c*d*x)^(7/2)/(a + b*x + c*x^2)^2,x]

[Out]

-((d^3*Sqrt[d*(b + 2*c*x)]*(Sqrt[b + 2*c*x]*(b^2 - 16*b*c*x - 4*c*(5*a + 4*c*x^2)) + 10*c*(b^2 - 4*a*c)^(1/4)*
(a + x*(b + c*x))*ArcTan[Sqrt[b + 2*c*x]/(b^2 - 4*a*c)^(1/4)] + 10*c*(b^2 - 4*a*c)^(1/4)*(a + x*(b + c*x))*Arc
Tanh[Sqrt[b + 2*c*x]/(b^2 - 4*a*c)^(1/4)]))/(Sqrt[b + 2*c*x]*(a + x*(b + c*x))))

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IntegrateAlgebraic [C]  time = 0.68, size = 257, normalized size = 1.71 \begin {gather*} \frac {\sqrt {b d+2 c d x} \left (20 a c d^3-b^2 d^3+16 b c d^3 x+16 c^2 d^3 x^2\right )}{a+b x+c x^2}+(5-5 i) c d^{7/2} \sqrt [4]{b^2-4 a c} \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 )-(5-5 i) c d^{7/2} \sqrt [4]{b^2-4 a c} \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 ) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(b*d + 2*c*d*x)^(7/2)/(a + b*x + c*x^2)^2,x]

[Out]

(Sqrt[b*d + 2*c*d*x]*(-(b^2*d^3) + 20*a*c*d^3 + 16*b*c*d^3*x + 16*c^2*d^3*x^2))/(a + b*x + c*x^2) + (5 - 5*I)*
c*(b^2 - 4*a*c)^(1/4)*d^(7/2)*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]] - (5 - 5*I)*c*(b^2 - 4*a*c)^(
1/4)*d^(7/2)*ArcTanh[((1 + I)*(b^2 - 4*a*c)^(1/4)*Sqrt[b*d + 2*c*d*x])/(Sqrt[d]*(I*b + Sqrt[b^2 - 4*a*c] + (2*
I)*c*x))]

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fricas [B]  time = 0.44, size = 359, normalized size = 2.39 \begin {gather*} -\frac {20 \, \left ({\left (b^{2} c^{4} - 4 \, a c^{5}\right )} d^{14}\right )^{\frac {1}{4}} {\left (c x^{2} + b x + a\right )} \arctan \left (\frac {\left ({\left (b^{2} c^{4} - 4 \, a c^{5}\right )} d^{14}\right )^{\frac {3}{4}} \sqrt {2 \, c d x + b d} c d^{3} - \left ({\left (b^{2} c^{4} - 4 \, a c^{5}\right )} d^{14}\right )^{\frac {3}{4}} \sqrt {2 \, c^{3} d^{7} x + b c^{2} d^{7} + \sqrt {{\left (b^{2} c^{4} - 4 \, a c^{5}\right )} d^{14}}}}{{\left (b^{2} c^{4} - 4 \, a c^{5}\right )} d^{14}}\right ) + 5 \, \left ({\left (b^{2} c^{4} - 4 \, a c^{5}\right )} d^{14}\right )^{\frac {1}{4}} {\left (c x^{2} + b x + a\right )} \log \left (5 \, \sqrt {2 \, c d x + b d} c d^{3} + 5 \, \left ({\left (b^{2} c^{4} - 4 \, a c^{5}\right )} d^{14}\right )^{\frac {1}{4}}\right ) - 5 \, \left ({\left (b^{2} c^{4} - 4 \, a c^{5}\right )} d^{14}\right )^{\frac {1}{4}} {\left (c x^{2} + b x + a\right )} \log \left (5 \, \sqrt {2 \, c d x + b d} c d^{3} - 5 \, \left ({\left (b^{2} c^{4} - 4 \, a c^{5}\right )} d^{14}\right )^{\frac {1}{4}}\right ) - {\left (16 \, c^{2} d^{3} x^{2} + 16 \, b c d^{3} x - {\left (b^{2} - 20 \, a c\right )} d^{3}\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)^(7/2)/(c*x^2+b*x+a)^2,x, algorithm="fricas")

[Out]

-(20*((b^2*c^4 - 4*a*c^5)*d^14)^(1/4)*(c*x^2 + b*x + a)*arctan((((b^2*c^4 - 4*a*c^5)*d^14)^(3/4)*sqrt(2*c*d*x
+ b*d)*c*d^3 - ((b^2*c^4 - 4*a*c^5)*d^14)^(3/4)*sqrt(2*c^3*d^7*x + b*c^2*d^7 + sqrt((b^2*c^4 - 4*a*c^5)*d^14))
)/((b^2*c^4 - 4*a*c^5)*d^14)) + 5*((b^2*c^4 - 4*a*c^5)*d^14)^(1/4)*(c*x^2 + b*x + a)*log(5*sqrt(2*c*d*x + b*d)
*c*d^3 + 5*((b^2*c^4 - 4*a*c^5)*d^14)^(1/4)) - 5*((b^2*c^4 - 4*a*c^5)*d^14)^(1/4)*(c*x^2 + b*x + a)*log(5*sqrt
(2*c*d*x + b*d)*c*d^3 - 5*((b^2*c^4 - 4*a*c^5)*d^14)^(1/4)) - (16*c^2*d^3*x^2 + 16*b*c*d^3*x - (b^2 - 20*a*c)*
d^3)*sqrt(2*c*d*x + b*d))/(c*x^2 + b*x + a)

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giac [B]  time = 0.25, size = 441, normalized size = 2.94 \begin {gather*} -5 \, \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} c d^{3} \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 ) - 5 \, \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} c d^{3} \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 ) - \frac {5}{2} \, \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} c d^{3} \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 ) + \frac {5}{2} \, \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} c d^{3} \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 ) + 16 \, \sqrt {2 \, c d x + b d} c d^{3} + \frac {4 \, {\left (\sqrt {2 \, c d x + b d} b^{2} c d^{5} - 4 \, \sqrt {2 \, c d x + b d} a c^{2} d^{5}\right )}}{b^{2} d^{2} - 4 \, a c d^{2} - {\left (2 \, c d x + b d\right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*d*x+b*d)^(7/2)/(c*x^2+b*x+a)^2,x, algorithm="giac")

[Out]

-5*sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(1/4)*c*d^3*arctan(1/2*sqrt(2)*(sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(1/4) + 2*sqr
t(2*c*d*x + b*d))/(-b^2*d^2 + 4*a*c*d^2)^(1/4)) - 5*sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(1/4)*c*d^3*arctan(-1/2*sqr
t(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)) - 5/2*sqrt(2
)*(-b^2*d^2 + 4*a*c*d^2)^(1/4)*c*d^3*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)) + 5/2*sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(1/4)*c*d^3*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)) + 16*sqrt(2*c*d*x + b*d)*c*d^3
+ 4*(sqrt(2*c*d*x + b*d)*b^2*c*d^5 - 4*sqrt(2*c*d*x + b*d)*a*c^2*d^5)/(b^2*d^2 - 4*a*c*d^2 - (2*c*d*x + b*d)^
2)

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maple [B]  time = 0.06, size = 693, normalized size = 4.62 \begin {gather*} \frac {20 \sqrt {2}\, a \,c^{2} d^{5} \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 {3}{4}}}-\frac {20 \sqrt {2}\, a \,c^{2} d^{5} \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 {3}{4}}}-\frac {10 \sqrt {2}\, a \,c^{2} d^{5} \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 )}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {3}{4}}}-\frac {5 \sqrt {2}\, b^{2} c \,d^{5} \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 {3}{4}}}+\frac {5 \sqrt {2}\, b^{2} c \,d^{5} \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 {3}{4}}}+\frac {5 \sqrt {2}\, b^{2} c \,d^{5} \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 {3}{4}}}+\frac {16 \sqrt {2 c d x +b d}\, a \,c^{2} d^{5}}{4 c^{2} d^{2} x^{2}+4 b c \,d^{2} x +4 a c \,d^{2}}-\frac {4 \sqrt {2 c d x +b d}\, b^{2} c \,d^{5}}{4 c^{2} d^{2} x^{2}+4 b c \,d^{2} x +4 a c \,d^{2}}+16 \sqrt {2 c d x +b d}\, c \,d^{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*c*d*x+b*d)^(7/2)/(c*x^2+b*x+a)^2,x)

[Out]

16*c*d^3*(2*c*d*x+b*d)^(1/2)+16*c^2*d^5*(2*c*d*x+b*d)^(1/2)/(4*c^2*d^2*x^2+4*b*c*d^2*x+4*a*c*d^2)*a-4*c*d^5*(2
*c*d*x+b*d)^(1/2)/(4*c^2*d^2*x^2+4*b*c*d^2*x+4*a*c*d^2)*b^2-20*c^2*d^5/(4*a*c*d^2-b^2*d^2)^(3/4)*2^(1/2)*arcta
n(2^(1/2)/(4*a*c*d^2-b^2*d^2)^(1/4)*(2*c*d*x+b*d)^(1/2)+1)*a+5*c*d^5/(4*a*c*d^2-b^2*d^2)^(3/4)*2^(1/2)*arctan(
2^(1/2)/(4*a*c*d^2-b^2*d^2)^(1/4)*(2*c*d*x+b*d)^(1/2)+1)*b^2+20*c^2*d^5/(4*a*c*d^2-b^2*d^2)^(3/4)*2^(1/2)*arct
an(-2^(1/2)/(4*a*c*d^2-b^2*d^2)^(1/4)*(2*c*d*x+b*d)^(1/2)+1)*a-5*c*d^5/(4*a*c*d^2-b^2*d^2)^(3/4)*2^(1/2)*arcta
n(-2^(1/2)/(4*a*c*d^2-b^2*d^2)^(1/4)*(2*c*d*x+b*d)^(1/2)+1)*b^2-10*c^2*d^5/(4*a*c*d^2-b^2*d^2)^(3/4)*2^(1/2)*l
n((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)))*a+5/2*c*d^5/(4*a*c*d^2-b^2*d^
2)^(3/4)*2^(1/2)*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)))*b^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)^(7/2)/(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 = 0.62, size = 597, normalized size = 3.98 \begin {gather*} 16\,c\,d^3\,\sqrt {b\,d+2\,c\,d\,x}+\frac {\sqrt {b\,d+2\,c\,d\,x}\,\left (16\,a\,c^2\,d^5-4\,b^2\,c\,d^5\right )}{{\left (b\,d+2\,c\,d\,x\right )}^2-b^2\,d^2+4\,a\,c\,d^2}-10\,c\,d^{7/2}\,\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}-c\,d^{7/2}\,\mathrm {atan}\left (\frac {c\,d^{7/2}\,\left (\sqrt {b\,d+2\,c\,d\,x}\,\left (25600\,a^2\,c^4\,d^{10}-12800\,a\,b^2\,c^3\,d^{10}+1600\,b^4\,c^2\,d^{10}\right )-5\,c\,d^{7/2}\,{\left (b^2-4\,a\,c\right )}^{1/4}\,\left (5120\,a^2\,c^3\,d^7-2560\,a\,b^2\,c^2\,d^7+320\,b^4\,c\,d^7\right )\right )\,{\left (b^2-4\,a\,c\right )}^{1/4}\,5{}\mathrm {i}+c\,d^{7/2}\,\left (\sqrt {b\,d+2\,c\,d\,x}\,\left (25600\,a^2\,c^4\,d^{10}-12800\,a\,b^2\,c^3\,d^{10}+1600\,b^4\,c^2\,d^{10}\right )+5\,c\,d^{7/2}\,{\left (b^2-4\,a\,c\right )}^{1/4}\,\left (5120\,a^2\,c^3\,d^7-2560\,a\,b^2\,c^2\,d^7+320\,b^4\,c\,d^7\right )\right )\,{\left (b^2-4\,a\,c\right )}^{1/4}\,5{}\mathrm {i}}{5\,c\,d^{7/2}\,\left (\sqrt {b\,d+2\,c\,d\,x}\,\left (25600\,a^2\,c^4\,d^{10}-12800\,a\,b^2\,c^3\,d^{10}+1600\,b^4\,c^2\,d^{10}\right )-5\,c\,d^{7/2}\,{\left (b^2-4\,a\,c\right )}^{1/4}\,\left (5120\,a^2\,c^3\,d^7-2560\,a\,b^2\,c^2\,d^7+320\,b^4\,c\,d^7\right )\right )\,{\left (b^2-4\,a\,c\right )}^{1/4}-5\,c\,d^{7/2}\,\left (\sqrt {b\,d+2\,c\,d\,x}\,\left (25600\,a^2\,c^4\,d^{10}-12800\,a\,b^2\,c^3\,d^{10}+1600\,b^4\,c^2\,d^{10}\right )+5\,c\,d^{7/2}\,{\left (b^2-4\,a\,c\right )}^{1/4}\,\left (5120\,a^2\,c^3\,d^7-2560\,a\,b^2\,c^2\,d^7+320\,b^4\,c\,d^7\right )\right )\,{\left (b^2-4\,a\,c\right )}^{1/4}}\right )\,{\left (b^2-4\,a\,c\right )}^{1/4}\,10{}\mathrm {i} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*d + 2*c*d*x)^(7/2)/(a + b*x + c*x^2)^2,x)

[Out]

16*c*d^3*(b*d + 2*c*d*x)^(1/2) + ((b*d + 2*c*d*x)^(1/2)*(16*a*c^2*d^5 - 4*b^2*c*d^5))/((b*d + 2*c*d*x)^2 - b^2
*d^2 + 4*a*c*d^2) - c*d^(7/2)*atan((c*d^(7/2)*((b*d + 2*c*d*x)^(1/2)*(25600*a^2*c^4*d^10 + 1600*b^4*c^2*d^10 -
12800*a*b^2*c^3*d^10) - 5*c*d^(7/2)*(b^2 - 4*a*c)^(1/4)*(320*b^4*c*d^7 + 5120*a^2*c^3*d^7 - 2560*a*b^2*c^2*d^
7))*(b^2 - 4*a*c)^(1/4)*5i + c*d^(7/2)*((b*d + 2*c*d*x)^(1/2)*(25600*a^2*c^4*d^10 + 1600*b^4*c^2*d^10 - 12800*
a*b^2*c^3*d^10) + 5*c*d^(7/2)*(b^2 - 4*a*c)^(1/4)*(320*b^4*c*d^7 + 5120*a^2*c^3*d^7 - 2560*a*b^2*c^2*d^7))*(b^
2 - 4*a*c)^(1/4)*5i)/(5*c*d^(7/2)*((b*d + 2*c*d*x)^(1/2)*(25600*a^2*c^4*d^10 + 1600*b^4*c^2*d^10 - 12800*a*b^2
*c^3*d^10) - 5*c*d^(7/2)*(b^2 - 4*a*c)^(1/4)*(320*b^4*c*d^7 + 5120*a^2*c^3*d^7 - 2560*a*b^2*c^2*d^7))*(b^2 - 4
*a*c)^(1/4) - 5*c*d^(7/2)*((b*d + 2*c*d*x)^(1/2)*(25600*a^2*c^4*d^10 + 1600*b^4*c^2*d^10 - 12800*a*b^2*c^3*d^1
0) + 5*c*d^(7/2)*(b^2 - 4*a*c)^(1/4)*(320*b^4*c*d^7 + 5120*a^2*c^3*d^7 - 2560*a*b^2*c^2*d^7))*(b^2 - 4*a*c)^(1
/4)))*(b^2 - 4*a*c)^(1/4)*10i - 10*c*d^(7/2)*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)

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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)**(7/2)/(c*x**2+b*x+a)**2,x)

[Out]

Timed out

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