3.1312 \(\int \frac {(b d+2 c d x)^{7/2}}{(a+b x+c x^2)^3} \, dx\)

Optimal. Leaf size=170 \[ -\frac {5 c^2 d^{7/2} \tan ^{-1}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{3/4}}-\frac {5 c^2 d^{7/2} \tanh ^{-1}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{3/4}}-\frac {5 c d^3 \sqrt {b d+2 c d x}}{2 \left (a+b x+c x^2\right )}-\frac {d (b d+2 c d x)^{5/2}}{2 \left (a+b x+c x^2\right )^2} \]

[Out]

-1/2*d*(2*c*d*x+b*d)^(5/2)/(c*x^2+b*x+a)^2-5*c^2*d^(7/2)*arctan((d*(2*c*x+b))^(1/2)/(-4*a*c+b^2)^(1/4)/d^(1/2)
)/(-4*a*c+b^2)^(3/4)-5*c^2*d^(7/2)*arctanh((d*(2*c*x+b))^(1/2)/(-4*a*c+b^2)^(1/4)/d^(1/2))/(-4*a*c+b^2)^(3/4)-
5/2*c*d^3*(2*c*d*x+b*d)^(1/2)/(c*x^2+b*x+a)

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Rubi [A]  time = 0.12, antiderivative size = 170, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {686, 694, 329, 212, 206, 203} \[ -\frac {5 c^2 d^{7/2} \tan ^{-1}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{3/4}}-\frac {5 c^2 d^{7/2} \tanh ^{-1}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{3/4}}-\frac {5 c d^3 \sqrt {b d+2 c d x}}{2 \left (a+b x+c x^2\right )}-\frac {d (b d+2 c d x)^{5/2}}{2 \left (a+b x+c x^2\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 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 )^3} \, dx &=-\frac {d (b d+2 c d x)^{5/2}}{2 \left (a+b x+c x^2\right )^2}+\frac {1}{2} \left (5 c d^2\right ) \int \frac {(b d+2 c d x)^{3/2}}{\left (a+b x+c x^2\right )^2} \, dx\\ &=-\frac {d (b d+2 c d x)^{5/2}}{2 \left (a+b x+c x^2\right )^2}-\frac {5 c d^3 \sqrt {b d+2 c d x}}{2 \left (a+b x+c x^2\right )}+\frac {1}{2} \left (5 c^2 d^4\right ) \int \frac {1}{\sqrt {b d+2 c d x} \left (a+b x+c x^2\right )} \, dx\\ &=-\frac {d (b d+2 c d x)^{5/2}}{2 \left (a+b x+c x^2\right )^2}-\frac {5 c d^3 \sqrt {b d+2 c d x}}{2 \left (a+b x+c x^2\right )}+\frac {1}{4} \left (5 c 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 )\\ &=-\frac {d (b d+2 c d x)^{5/2}}{2 \left (a+b x+c x^2\right )^2}-\frac {5 c d^3 \sqrt {b d+2 c d x}}{2 \left (a+b x+c x^2\right )}+\frac {1}{2} \left (5 c 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 )\\ &=-\frac {d (b d+2 c d x)^{5/2}}{2 \left (a+b x+c x^2\right )^2}-\frac {5 c d^3 \sqrt {b d+2 c d x}}{2 \left (a+b x+c x^2\right )}-\frac {\left (5 c^2 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 )}{\sqrt {b^2-4 a c}}-\frac {\left (5 c^2 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 )}{\sqrt {b^2-4 a c}}\\ &=-\frac {d (b d+2 c d x)^{5/2}}{2 \left (a+b x+c x^2\right )^2}-\frac {5 c d^3 \sqrt {b d+2 c d x}}{2 \left (a+b x+c x^2\right )}-\frac {5 c^2 d^{7/2} \tan ^{-1}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )}{\left (b^2-4 a c\right )^{3/4}}-\frac {5 c^2 d^{7/2} \tanh ^{-1}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )}{\left (b^2-4 a c\right )^{3/4}}\\ \end {align*}

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Mathematica [A]  time = 0.28, size = 196, normalized size = 1.15 \[ \frac {(d (b+2 c x))^{7/2} \left (-64 \left (b^2-4 a c\right )^{3/4} (b+2 c x)^{5/2}+40 \left (b^2-4 a c\right )^{7/4} \sqrt {b+2 c x}+20 c (a+x (b+c x)) \left (2 \left (b^2-4 a c\right )^{3/4} \sqrt {b+2 c x}-12 c (a+x (b+c x)) \left (\tan ^{-1}\left (\frac {\sqrt {b+2 c x}}{\sqrt [4]{b^2-4 a c}}\right )+\tanh ^{-1}\left (\frac {\sqrt {b+2 c x}}{\sqrt [4]{b^2-4 a c}}\right )\right )\right )\right )}{48 \left (b^2-4 a c\right )^{3/4} (b+2 c x)^{7/2} (a+x (b+c x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((d*(b + 2*c*x))^(7/2)*(40*(b^2 - 4*a*c)^(7/4)*Sqrt[b + 2*c*x] - 64*(b^2 - 4*a*c)^(3/4)*(b + 2*c*x)^(5/2) + 20
*c*(a + x*(b + c*x))*(2*(b^2 - 4*a*c)^(3/4)*Sqrt[b + 2*c*x] - 12*c*(a + x*(b + c*x))*(ArcTan[Sqrt[b + 2*c*x]/(
b^2 - 4*a*c)^(1/4)] + ArcTanh[Sqrt[b + 2*c*x]/(b^2 - 4*a*c)^(1/4)]))))/(48*(b^2 - 4*a*c)^(3/4)*(b + 2*c*x)^(7/
2)*(a + x*(b + c*x))^2)

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fricas [B]  time = 0.75, size = 697, normalized size = 4.10 \[ \frac {20 \, \left (\frac {c^{8} d^{14}}{b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}}\right )^{\frac {1}{4}} {\left (c^{2} x^{4} + 2 \, b c x^{3} + 2 \, a b x + {\left (b^{2} + 2 \, a c\right )} x^{2} + a^{2}\right )} \arctan \left (-\frac {\left (\frac {c^{8} d^{14}}{b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}}\right )^{\frac {3}{4}} {\left (b^{4} c^{2} - 8 \, a b^{2} c^{3} + 16 \, a^{2} c^{4}\right )} \sqrt {2 \, c d x + b d} d^{3} - \left (\frac {c^{8} d^{14}}{b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}}\right )^{\frac {3}{4}} \sqrt {2 \, c^{5} d^{7} x + b c^{4} d^{7} + \sqrt {\frac {c^{8} d^{14}}{b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}}} {\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )}} {\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )}}{c^{8} d^{14}}\right ) - 5 \, \left (\frac {c^{8} d^{14}}{b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}}\right )^{\frac {1}{4}} {\left (c^{2} x^{4} + 2 \, b c x^{3} + 2 \, a b x + {\left (b^{2} + 2 \, a c\right )} x^{2} + a^{2}\right )} \log \left (5 \, \sqrt {2 \, c d x + b d} c^{2} d^{3} + 5 \, \left (\frac {c^{8} d^{14}}{b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}}\right )^{\frac {1}{4}} {\left (b^{2} - 4 \, a c\right )}\right ) + 5 \, \left (\frac {c^{8} d^{14}}{b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}}\right )^{\frac {1}{4}} {\left (c^{2} x^{4} + 2 \, b c x^{3} + 2 \, a b x + {\left (b^{2} + 2 \, a c\right )} x^{2} + a^{2}\right )} \log \left (5 \, \sqrt {2 \, c d x + b d} c^{2} d^{3} - 5 \, \left (\frac {c^{8} d^{14}}{b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}}\right )^{\frac {1}{4}} {\left (b^{2} - 4 \, a c\right )}\right ) - {\left (9 \, c^{2} d^{3} x^{2} + 9 \, b c d^{3} x + {\left (b^{2} + 5 \, a c\right )} d^{3}\right )} \sqrt {2 \, c d x + b d}}{2 \, {\left (c^{2} x^{4} + 2 \, b c x^{3} + 2 \, a b x + {\left (b^{2} + 2 \, a c\right )} x^{2} + a^{2}\right )}} \]

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)^3,x, algorithm="fricas")

[Out]

1/2*(20*(c^8*d^14/(b^6 - 12*a*b^4*c + 48*a^2*b^2*c^2 - 64*a^3*c^3))^(1/4)*(c^2*x^4 + 2*b*c*x^3 + 2*a*b*x + (b^
2 + 2*a*c)*x^2 + a^2)*arctan(-((c^8*d^14/(b^6 - 12*a*b^4*c + 48*a^2*b^2*c^2 - 64*a^3*c^3))^(3/4)*(b^4*c^2 - 8*
a*b^2*c^3 + 16*a^2*c^4)*sqrt(2*c*d*x + b*d)*d^3 - (c^8*d^14/(b^6 - 12*a*b^4*c + 48*a^2*b^2*c^2 - 64*a^3*c^3))^
(3/4)*sqrt(2*c^5*d^7*x + b*c^4*d^7 + sqrt(c^8*d^14/(b^6 - 12*a*b^4*c + 48*a^2*b^2*c^2 - 64*a^3*c^3))*(b^4 - 8*
a*b^2*c + 16*a^2*c^2))*(b^4 - 8*a*b^2*c + 16*a^2*c^2))/(c^8*d^14)) - 5*(c^8*d^14/(b^6 - 12*a*b^4*c + 48*a^2*b^
2*c^2 - 64*a^3*c^3))^(1/4)*(c^2*x^4 + 2*b*c*x^3 + 2*a*b*x + (b^2 + 2*a*c)*x^2 + a^2)*log(5*sqrt(2*c*d*x + b*d)
*c^2*d^3 + 5*(c^8*d^14/(b^6 - 12*a*b^4*c + 48*a^2*b^2*c^2 - 64*a^3*c^3))^(1/4)*(b^2 - 4*a*c)) + 5*(c^8*d^14/(b
^6 - 12*a*b^4*c + 48*a^2*b^2*c^2 - 64*a^3*c^3))^(1/4)*(c^2*x^4 + 2*b*c*x^3 + 2*a*b*x + (b^2 + 2*a*c)*x^2 + a^2
)*log(5*sqrt(2*c*d*x + b*d)*c^2*d^3 - 5*(c^8*d^14/(b^6 - 12*a*b^4*c + 48*a^2*b^2*c^2 - 64*a^3*c^3))^(1/4)*(b^2
- 4*a*c)) - (9*c^2*d^3*x^2 + 9*b*c*d^3*x + (b^2 + 5*a*c)*d^3)*sqrt(2*c*d*x + b*d))/(c^2*x^4 + 2*b*c*x^3 + 2*a
*b*x + (b^2 + 2*a*c)*x^2 + a^2)

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giac [B]  time = 0.29, size = 510, normalized size = 3.00 \[ -\frac {5 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} c^{2} d^{3} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} + 2 \, \sqrt {2 \, c d x + b d}\right )}}{2 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}}}\right )}{\sqrt {2} b^{2} - 4 \, \sqrt {2} a c} - \frac {5 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} c^{2} d^{3} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} - 2 \, \sqrt {2 \, c d x + b d}\right )}}{2 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}}}\right )}{\sqrt {2} b^{2} - 4 \, \sqrt {2} a c} - \frac {5 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} c^{2} d^{3} \log \left (2 \, c d x + b d + \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} \sqrt {2 \, c d x + b d} + \sqrt {-b^{2} d^{2} + 4 \, a c d^{2}}\right )}{2 \, {\left (\sqrt {2} b^{2} - 4 \, \sqrt {2} a c\right )}} + \frac {5 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} c^{2} d^{3} \log \left (2 \, c d x + b d - \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} \sqrt {2 \, c d x + b d} + \sqrt {-b^{2} d^{2} + 4 \, a c d^{2}}\right )}{2 \, {\left (\sqrt {2} b^{2} - 4 \, \sqrt {2} a c\right )}} + \frac {2 \, {\left (5 \, \sqrt {2 \, c d x + b d} b^{2} c^{2} d^{7} - 20 \, \sqrt {2 \, c d x + b d} a c^{3} d^{7} - 9 \, {\left (2 \, c d x + b d\right )}^{\frac {5}{2}} c^{2} d^{5}\right )}}{{\left (b^{2} d^{2} - 4 \, a c d^{2} - {\left (2 \, c d x + b d\right )}^{2}\right )}^{2}} \]

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)^3,x, algorithm="giac")

[Out]

-5*(-b^2*d^2 + 4*a*c*d^2)^(1/4)*c^2*d^3*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))/(sqrt(2)*b^2 - 4*sqrt(2)*a*c) - 5*(-b^2*d^2 + 4*a*c*d^2)^(1/4)*c^2*d
^3*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))/(sqrt(2)*b^2 - 4*sqrt(2)*a*c) - 5/2*(-b^2*d^2 + 4*a*c*d^2)^(1/4)*c^2*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))/(sqrt(2)*b^2 - 4*sqrt(2)*a*c) + 5/
2*(-b^2*d^2 + 4*a*c*d^2)^(1/4)*c^2*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))/(sqrt(2)*b^2 - 4*sqrt(2)*a*c) + 2*(5*sqrt(2*c*d*x + b*d)*b^2*c^2*d^7 - 20*
sqrt(2*c*d*x + b*d)*a*c^3*d^7 - 9*(2*c*d*x + b*d)^(5/2)*c^2*d^5)/(b^2*d^2 - 4*a*c*d^2 - (2*c*d*x + b*d)^2)^2

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maple [B]  time = 0.06, size = 435, normalized size = 2.56 \[ -\frac {40 \sqrt {2 c d x +b d}\, a \,c^{3} d^{7}}{\left (4 c^{2} d^{2} x^{2}+4 b c \,d^{2} x +4 a c \,d^{2}\right )^{2}}+\frac {10 \sqrt {2 c d x +b d}\, b^{2} c^{2} d^{7}}{\left (4 c^{2} d^{2} x^{2}+4 b c \,d^{2} x +4 a c \,d^{2}\right )^{2}}-\frac {5 \sqrt {2}\, c^{2} d^{5} \arctan \left (-\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )}{2 \left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {3}{4}}}+\frac {5 \sqrt {2}\, c^{2} d^{5} \arctan \left (\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )}{2 \left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {3}{4}}}+\frac {5 \sqrt {2}\, c^{2} d^{5} \ln \left (\frac {2 c d x +b d +\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a c \,d^{2}-b^{2} d^{2}}}{2 c d x +b d -\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a c \,d^{2}-b^{2} d^{2}}}\right )}{4 \left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {3}{4}}}-\frac {18 \left (2 c d x +b d \right )^{\frac {5}{2}} c^{2} d^{5}}{\left (4 c^{2} d^{2} x^{2}+4 b c \,d^{2} x +4 a c \,d^{2}\right )^{2}} \]

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)^3,x)

[Out]

-18*c^2*d^5/(4*c^2*d^2*x^2+4*b*c*d^2*x+4*a*c*d^2)^2*(2*c*d*x+b*d)^(5/2)-40*c^3*d^7/(4*c^2*d^2*x^2+4*b*c*d^2*x+
4*a*c*d^2)^2*(2*c*d*x+b*d)^(1/2)*a+10*c^2*d^7/(4*c^2*d^2*x^2+4*b*c*d^2*x+4*a*c*d^2)^2*(2*c*d*x+b*d)^(1/2)*b^2+
5/4*c^2*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)))+5/2*c^2*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)-5/2*c^2*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)

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

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)^3,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 = 0.59, size = 309, normalized size = 1.82 \[ -\frac {\sqrt {b\,d+2\,c\,d\,x}\,\left (40\,a\,c^3\,d^7-10\,b^2\,c^2\,d^7\right )+18\,c^2\,d^5\,{\left (b\,d+2\,c\,d\,x\right )}^{5/2}}{{\left (b\,d+2\,c\,d\,x\right )}^4-{\left (b\,d+2\,c\,d\,x\right )}^2\,\left (2\,b^2\,d^2-8\,a\,c\,d^2\right )+b^4\,d^4+16\,a^2\,c^2\,d^4-8\,a\,b^2\,c\,d^4}-\frac {5\,c^2\,d^{7/2}\,\mathrm {atan}\left (\frac {2000\,c^6\,d^{27/2}\,\sqrt {b\,d+2\,c\,d\,x}}{\left (\frac {2000\,b^2\,c^6\,d^{14}}{{\left (b^2-4\,a\,c\right )}^{3/2}}-\frac {8000\,a\,c^7\,d^{14}}{{\left (b^2-4\,a\,c\right )}^{3/2}}\right )\,{\left (b^2-4\,a\,c\right )}^{3/4}}\right )}{{\left (b^2-4\,a\,c\right )}^{3/4}}-\frac {5\,c^2\,d^{7/2}\,\mathrm {atanh}\left (\frac {2000\,c^6\,d^{27/2}\,\sqrt {b\,d+2\,c\,d\,x}}{\left (\frac {2000\,b^2\,c^6\,d^{14}}{{\left (b^2-4\,a\,c\right )}^{3/2}}-\frac {8000\,a\,c^7\,d^{14}}{{\left (b^2-4\,a\,c\right )}^{3/2}}\right )\,{\left (b^2-4\,a\,c\right )}^{3/4}}\right )}{{\left (b^2-4\,a\,c\right )}^{3/4}} \]

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)^3,x)

[Out]

- ((b*d + 2*c*d*x)^(1/2)*(40*a*c^3*d^7 - 10*b^2*c^2*d^7) + 18*c^2*d^5*(b*d + 2*c*d*x)^(5/2))/((b*d + 2*c*d*x)^
4 - (b*d + 2*c*d*x)^2*(2*b^2*d^2 - 8*a*c*d^2) + b^4*d^4 + 16*a^2*c^2*d^4 - 8*a*b^2*c*d^4) - (5*c^2*d^(7/2)*ata
n((2000*c^6*d^(27/2)*(b*d + 2*c*d*x)^(1/2))/(((2000*b^2*c^6*d^14)/(b^2 - 4*a*c)^(3/2) - (8000*a*c^7*d^14)/(b^2
- 4*a*c)^(3/2))*(b^2 - 4*a*c)^(3/4))))/(b^2 - 4*a*c)^(3/4) - (5*c^2*d^(7/2)*atanh((2000*c^6*d^(27/2)*(b*d + 2
*c*d*x)^(1/2))/(((2000*b^2*c^6*d^14)/(b^2 - 4*a*c)^(3/2) - (8000*a*c^7*d^14)/(b^2 - 4*a*c)^(3/2))*(b^2 - 4*a*c
)^(3/4))))/(b^2 - 4*a*c)^(3/4)

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

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)**3,x)

[Out]

Timed out

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