3.1310 \(\int \frac {(b d+2 c d x)^{11/2}}{(a+b x+c x^2)^3} \, dx\)

Optimal. Leaf size=191 \[ -45 c^2 d^{11/2} \sqrt [4]{b^2-4 a c} \tan ^{-1}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )-45 c^2 d^{11/2} \sqrt [4]{b^2-4 a c} \tanh ^{-1}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )-\frac {9 c d^3 (b d+2 c d x)^{5/2}}{2 \left (a+b x+c x^2\right )}-\frac {d (b d+2 c d x)^{9/2}}{2 \left (a+b x+c x^2\right )^2}+90 c^2 d^5 \sqrt {b d+2 c d x} \]

[Out]

-1/2*d*(2*c*d*x+b*d)^(9/2)/(c*x^2+b*x+a)^2-9/2*c*d^3*(2*c*d*x+b*d)^(5/2)/(c*x^2+b*x+a)-45*c^2*(-4*a*c+b^2)^(1/
4)*d^(11/2)*arctan((d*(2*c*x+b))^(1/2)/(-4*a*c+b^2)^(1/4)/d^(1/2))-45*c^2*(-4*a*c+b^2)^(1/4)*d^(11/2)*arctanh(
(d*(2*c*x+b))^(1/2)/(-4*a*c+b^2)^(1/4)/d^(1/2))+90*c^2*d^5*(2*c*d*x+b*d)^(1/2)

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Rubi [A]  time = 0.15, antiderivative size = 191, normalized size of antiderivative = 1.00, number of steps used = 8, 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} \[ -45 c^2 d^{11/2} \sqrt [4]{b^2-4 a c} \tan ^{-1}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )-45 c^2 d^{11/2} \sqrt [4]{b^2-4 a c} \tanh ^{-1}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )-\frac {9 c d^3 (b d+2 c d x)^{5/2}}{2 \left (a+b x+c x^2\right )}-\frac {d (b d+2 c d x)^{9/2}}{2 \left (a+b x+c x^2\right )^2}+90 c^2 d^5 \sqrt {b d+2 c d x} \]

Antiderivative was successfully verified.

[In]

Int[(b*d + 2*c*d*x)^(11/2)/(a + b*x + c*x^2)^3,x]

[Out]

90*c^2*d^5*Sqrt[b*d + 2*c*d*x] - (d*(b*d + 2*c*d*x)^(9/2))/(2*(a + b*x + c*x^2)^2) - (9*c*d^3*(b*d + 2*c*d*x)^
(5/2))/(2*(a + b*x + c*x^2)) - 45*c^2*(b^2 - 4*a*c)^(1/4)*d^(11/2)*ArcTan[Sqrt[d*(b + 2*c*x)]/((b^2 - 4*a*c)^(
1/4)*Sqrt[d])] - 45*c^2*(b^2 - 4*a*c)^(1/4)*d^(11/2)*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)^{11/2}}{\left (a+b x+c x^2\right )^3} \, dx &=-\frac {d (b d+2 c d x)^{9/2}}{2 \left (a+b x+c x^2\right )^2}+\frac {1}{2} \left (9 c d^2\right ) \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)^{9/2}}{2 \left (a+b x+c x^2\right )^2}-\frac {9 c d^3 (b d+2 c d x)^{5/2}}{2 \left (a+b x+c x^2\right )}+\frac {1}{2} \left (45 c^2 d^4\right ) \int \frac {(b d+2 c d x)^{3/2}}{a+b x+c x^2} \, dx\\ &=90 c^2 d^5 \sqrt {b d+2 c d x}-\frac {d (b d+2 c d x)^{9/2}}{2 \left (a+b x+c x^2\right )^2}-\frac {9 c d^3 (b d+2 c d x)^{5/2}}{2 \left (a+b x+c x^2\right )}+\frac {1}{2} \left (45 c^2 \left (b^2-4 a c\right ) d^6\right ) \int \frac {1}{\sqrt {b d+2 c d x} \left (a+b x+c x^2\right )} \, dx\\ &=90 c^2 d^5 \sqrt {b d+2 c d x}-\frac {d (b d+2 c d x)^{9/2}}{2 \left (a+b x+c x^2\right )^2}-\frac {9 c d^3 (b d+2 c d x)^{5/2}}{2 \left (a+b x+c x^2\right )}+\frac {1}{4} \left (45 c \left (b^2-4 a c\right ) d^5\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 )\\ &=90 c^2 d^5 \sqrt {b d+2 c d x}-\frac {d (b d+2 c d x)^{9/2}}{2 \left (a+b x+c x^2\right )^2}-\frac {9 c d^3 (b d+2 c d x)^{5/2}}{2 \left (a+b x+c x^2\right )}+\frac {1}{2} \left (45 c \left (b^2-4 a c\right ) d^5\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 )\\ &=90 c^2 d^5 \sqrt {b d+2 c d x}-\frac {d (b d+2 c d x)^{9/2}}{2 \left (a+b x+c x^2\right )^2}-\frac {9 c d^3 (b d+2 c d x)^{5/2}}{2 \left (a+b x+c x^2\right )}-\left (45 c^2 \sqrt {b^2-4 a c} d^6\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 (45 c^2 \sqrt {b^2-4 a c} d^6\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 )\\ &=90 c^2 d^5 \sqrt {b d+2 c d x}-\frac {d (b d+2 c d x)^{9/2}}{2 \left (a+b x+c x^2\right )^2}-\frac {9 c d^3 (b d+2 c d x)^{5/2}}{2 \left (a+b x+c x^2\right )}-45 c^2 \sqrt [4]{b^2-4 a c} d^{11/2} \tan ^{-1}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )-45 c^2 \sqrt [4]{b^2-4 a c} d^{11/2} \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 = 0.33, size = 226, normalized size = 1.18 \[ -\frac {d^5 \sqrt {d (b+2 c x)} \left (\sqrt {b+2 c x} \left (-4 c^2 \left (45 a^2+81 a c x^2+32 c^2 x^4\right )+3 b^2 c \left (3 a-37 c x^2\right )-4 b c^2 x \left (81 a+64 c x^2\right )+b^4+17 b^3 c x\right )+90 c^2 \sqrt [4]{b^2-4 a c} (a+x (b+c x))^2 \tan ^{-1}\left (\frac {\sqrt {b+2 c x}}{\sqrt [4]{b^2-4 a c}}\right )+90 c^2 \sqrt [4]{b^2-4 a c} (a+x (b+c x))^2 \tanh ^{-1}\left (\frac {\sqrt {b+2 c x}}{\sqrt [4]{b^2-4 a c}}\right )\right )}{2 \sqrt {b+2 c x} (a+x (b+c x))^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*d + 2*c*d*x)^(11/2)/(a + b*x + c*x^2)^3,x]

[Out]

-1/2*(d^5*Sqrt[d*(b + 2*c*x)]*(Sqrt[b + 2*c*x]*(b^4 + 17*b^3*c*x + 3*b^2*c*(3*a - 37*c*x^2) - 4*b*c^2*x*(81*a
+ 64*c*x^2) - 4*c^2*(45*a^2 + 81*a*c*x^2 + 32*c^2*x^4)) + 90*c^2*(b^2 - 4*a*c)^(1/4)*(a + x*(b + c*x))^2*ArcTa
n[Sqrt[b + 2*c*x]/(b^2 - 4*a*c)^(1/4)] + 90*c^2*(b^2 - 4*a*c)^(1/4)*(a + x*(b + c*x))^2*ArcTanh[Sqrt[b + 2*c*x
]/(b^2 - 4*a*c)^(1/4)]))/(Sqrt[b + 2*c*x]*(a + x*(b + c*x))^2)

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fricas [B]  time = 0.80, size = 523, normalized size = 2.74 \[ -\frac {180 \, \left ({\left (b^{2} c^{8} - 4 \, a c^{9}\right )} d^{22}\right )^{\frac {1}{4}} {\left (c^{2} x^{4} + 2 \, b c x^{3} + 2 \, a b x + {\left (b^{2} + 2 \, a c\right )} x^{2} + a^{2}\right )} \arctan \left (\frac {\left ({\left (b^{2} c^{8} - 4 \, a c^{9}\right )} d^{22}\right )^{\frac {3}{4}} \sqrt {2 \, c d x + b d} c^{2} d^{5} - \left ({\left (b^{2} c^{8} - 4 \, a c^{9}\right )} d^{22}\right )^{\frac {3}{4}} \sqrt {2 \, c^{5} d^{11} x + b c^{4} d^{11} + \sqrt {{\left (b^{2} c^{8} - 4 \, a c^{9}\right )} d^{22}}}}{{\left (b^{2} c^{8} - 4 \, a c^{9}\right )} d^{22}}\right ) + 45 \, \left ({\left (b^{2} c^{8} - 4 \, a c^{9}\right )} d^{22}\right )^{\frac {1}{4}} {\left (c^{2} x^{4} + 2 \, b c x^{3} + 2 \, a b x + {\left (b^{2} + 2 \, a c\right )} x^{2} + a^{2}\right )} \log \left (45 \, \sqrt {2 \, c d x + b d} c^{2} d^{5} + 45 \, \left ({\left (b^{2} c^{8} - 4 \, a c^{9}\right )} d^{22}\right )^{\frac {1}{4}}\right ) - 45 \, \left ({\left (b^{2} c^{8} - 4 \, a c^{9}\right )} d^{22}\right )^{\frac {1}{4}} {\left (c^{2} x^{4} + 2 \, b c x^{3} + 2 \, a b x + {\left (b^{2} + 2 \, a c\right )} x^{2} + a^{2}\right )} \log \left (45 \, \sqrt {2 \, c d x + b d} c^{2} d^{5} - 45 \, \left ({\left (b^{2} c^{8} - 4 \, a c^{9}\right )} d^{22}\right )^{\frac {1}{4}}\right ) - {\left (128 \, c^{4} d^{5} x^{4} + 256 \, b c^{3} d^{5} x^{3} + 3 \, {\left (37 \, b^{2} c^{2} + 108 \, a c^{3}\right )} d^{5} x^{2} - {\left (17 \, b^{3} c - 324 \, a b c^{2}\right )} d^{5} x - {\left (b^{4} + 9 \, a b^{2} c - 180 \, a^{2} c^{2}\right )} d^{5}\right )} \sqrt {2 \, c d x + b d}}{2 \, {\left (c^{2} x^{4} + 2 \, b c x^{3} + 2 \, a b x + {\left (b^{2} + 2 \, a c\right )} x^{2} + a^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*d*x+b*d)^(11/2)/(c*x^2+b*x+a)^3,x, algorithm="fricas")

[Out]

-1/2*(180*((b^2*c^8 - 4*a*c^9)*d^22)^(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(((
(b^2*c^8 - 4*a*c^9)*d^22)^(3/4)*sqrt(2*c*d*x + b*d)*c^2*d^5 - ((b^2*c^8 - 4*a*c^9)*d^22)^(3/4)*sqrt(2*c^5*d^11
*x + b*c^4*d^11 + sqrt((b^2*c^8 - 4*a*c^9)*d^22)))/((b^2*c^8 - 4*a*c^9)*d^22)) + 45*((b^2*c^8 - 4*a*c^9)*d^22)
^(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(45*sqrt(2*c*d*x + b*d)*c^2*d^5 + 45*((b^2
*c^8 - 4*a*c^9)*d^22)^(1/4)) - 45*((b^2*c^8 - 4*a*c^9)*d^22)^(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(45*sqrt(2*c*d*x + b*d)*c^2*d^5 - 45*((b^2*c^8 - 4*a*c^9)*d^22)^(1/4)) - (128*c^4*d^5*x^4 +
256*b*c^3*d^5*x^3 + 3*(37*b^2*c^2 + 108*a*c^3)*d^5*x^2 - (17*b^3*c - 324*a*b*c^2)*d^5*x - (b^4 + 9*a*b^2*c - 1
80*a^2*c^2)*d^5)*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 = 0.35, size = 521, normalized size = 2.73 \[ -\frac {45}{2} \, \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} c^{2} d^{5} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} + 2 \, \sqrt {2 \, c d x + b d}\right )}}{2 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}}}\right ) - \frac {45}{2} \, \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} c^{2} d^{5} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} - 2 \, \sqrt {2 \, c d x + b d}\right )}}{2 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}}}\right ) - \frac {45}{4} \, \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} c^{2} d^{5} \log \left (2 \, c d x + b d + \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} \sqrt {2 \, c d x + b d} + \sqrt {-b^{2} d^{2} + 4 \, a c d^{2}}\right ) + \frac {45}{4} \, \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} c^{2} d^{5} \log \left (2 \, c d x + b d - \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} \sqrt {2 \, c d x + b d} + \sqrt {-b^{2} d^{2} + 4 \, a c d^{2}}\right ) + 64 \, \sqrt {2 \, c d x + b d} c^{2} d^{5} + \frac {2 \, {\left (13 \, \sqrt {2 \, c d x + b d} b^{4} c^{2} d^{9} - 104 \, \sqrt {2 \, c d x + b d} a b^{2} c^{3} d^{9} + 208 \, \sqrt {2 \, c d x + b d} a^{2} c^{4} d^{9} - 17 \, {\left (2 \, c d x + b d\right )}^{\frac {5}{2}} b^{2} c^{2} d^{7} + 68 \, {\left (2 \, c d x + b d\right )}^{\frac {5}{2}} a c^{3} d^{7}\right )}}{{\left (b^{2} d^{2} - 4 \, a c d^{2} - {\left (2 \, c d x + b d\right )}^{2}\right )}^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*d*x+b*d)^(11/2)/(c*x^2+b*x+a)^3,x, algorithm="giac")

[Out]

-45/2*sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(1/4)*c^2*d^5*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)) - 45/2*sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(1/4)*c^2*d^5*arcta
n(-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)) -
45/4*sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(1/4)*c^2*d^5*log(2*c*d*x + b*d + sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(1/4)*sqr
t(2*c*d*x + b*d) + sqrt(-b^2*d^2 + 4*a*c*d^2)) + 45/4*sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(1/4)*c^2*d^5*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)) + 64*sqrt(2*c*
d*x + b*d)*c^2*d^5 + 2*(13*sqrt(2*c*d*x + b*d)*b^4*c^2*d^9 - 104*sqrt(2*c*d*x + b*d)*a*b^2*c^3*d^9 + 208*sqrt(
2*c*d*x + b*d)*a^2*c^4*d^9 - 17*(2*c*d*x + b*d)^(5/2)*b^2*c^2*d^7 + 68*(2*c*d*x + b*d)^(5/2)*a*c^3*d^7)/(b^2*d
^2 - 4*a*c*d^2 - (2*c*d*x + b*d)^2)^2

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maple [B]  time = 0.07, size = 857, normalized size = 4.49 \[ \frac {416 \sqrt {2 c d x +b d}\, a^{2} c^{4} d^{9}}{\left (4 c^{2} d^{2} x^{2}+4 b c \,d^{2} x +4 a c \,d^{2}\right )^{2}}-\frac {208 \sqrt {2 c d x +b d}\, a \,b^{2} c^{3} d^{9}}{\left (4 c^{2} d^{2} x^{2}+4 b c \,d^{2} x +4 a c \,d^{2}\right )^{2}}+\frac {26 \sqrt {2 c d x +b d}\, b^{4} c^{2} d^{9}}{\left (4 c^{2} d^{2} x^{2}+4 b c \,d^{2} x +4 a c \,d^{2}\right )^{2}}+\frac {90 \sqrt {2}\, a \,c^{3} d^{7} \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 {90 \sqrt {2}\, a \,c^{3} d^{7} \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 {45 \sqrt {2}\, a \,c^{3} d^{7} \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 {45 \sqrt {2}\, b^{2} c^{2} d^{7} \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 {45 \sqrt {2}\, b^{2} c^{2} d^{7} \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 {45 \sqrt {2}\, b^{2} c^{2} d^{7} \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 {136 \left (2 c d x +b d \right )^{\frac {5}{2}} 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 {34 \left (2 c d x +b d \right )^{\frac {5}{2}} 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}}+64 \sqrt {2 c d x +b d}\, c^{2} d^{5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*c*d*x+b*d)^(11/2)/(c*x^2+b*x+a)^3,x)

[Out]

64*c^2*d^5*(2*c*d*x+b*d)^(1/2)+136*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)^(5/2)*a-34*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)^(5/2)*b^2+416*c^4*d^9/(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^2-208*c^3*d^9/(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*
b^2+26*c^2*d^9/(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^4-90*c^3*d^7/(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)*a+45/2*c^2*d^7/(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+90*c^3*d^7/(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)*a-45/2*c^2*d^7/(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^2-45*c^3*d^7/(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)))*
a+45/4*c^2*d^7/(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 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*d*x+b*d)^(11/2)/(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 = 0.21, size = 720, normalized size = 3.77 \[ \frac {{\left (b\,d+2\,c\,d\,x\right )}^{5/2}\,\left (136\,a\,c^3\,d^7-34\,b^2\,c^2\,d^7\right )+\sqrt {b\,d+2\,c\,d\,x}\,\left (416\,a^2\,c^4\,d^9-208\,a\,b^2\,c^3\,d^9+26\,b^4\,c^2\,d^9\right )}{{\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}+64\,c^2\,d^5\,\sqrt {b\,d+2\,c\,d\,x}-45\,c^2\,d^{11/2}\,\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^2\,d^{11/2}\,\mathrm {atan}\left (\frac {\frac {c^2\,d^{11/2}\,\left (\sqrt {b\,d+2\,c\,d\,x}\,\left (518400\,a^2\,c^6\,d^{14}-259200\,a\,b^2\,c^5\,d^{14}+32400\,b^4\,c^4\,d^{14}\right )-\frac {45\,c^2\,d^{11/2}\,{\left (b^2-4\,a\,c\right )}^{1/4}\,\left (23040\,a^2\,c^4\,d^9-11520\,a\,b^2\,c^3\,d^9+1440\,b^4\,c^2\,d^9\right )}{2}\right )\,{\left (b^2-4\,a\,c\right )}^{1/4}\,45{}\mathrm {i}}{2}+\frac {c^2\,d^{11/2}\,\left (\sqrt {b\,d+2\,c\,d\,x}\,\left (518400\,a^2\,c^6\,d^{14}-259200\,a\,b^2\,c^5\,d^{14}+32400\,b^4\,c^4\,d^{14}\right )+\frac {45\,c^2\,d^{11/2}\,{\left (b^2-4\,a\,c\right )}^{1/4}\,\left (23040\,a^2\,c^4\,d^9-11520\,a\,b^2\,c^3\,d^9+1440\,b^4\,c^2\,d^9\right )}{2}\right )\,{\left (b^2-4\,a\,c\right )}^{1/4}\,45{}\mathrm {i}}{2}}{\frac {45\,c^2\,d^{11/2}\,\left (\sqrt {b\,d+2\,c\,d\,x}\,\left (518400\,a^2\,c^6\,d^{14}-259200\,a\,b^2\,c^5\,d^{14}+32400\,b^4\,c^4\,d^{14}\right )-\frac {45\,c^2\,d^{11/2}\,{\left (b^2-4\,a\,c\right )}^{1/4}\,\left (23040\,a^2\,c^4\,d^9-11520\,a\,b^2\,c^3\,d^9+1440\,b^4\,c^2\,d^9\right )}{2}\right )\,{\left (b^2-4\,a\,c\right )}^{1/4}}{2}-\frac {45\,c^2\,d^{11/2}\,\left (\sqrt {b\,d+2\,c\,d\,x}\,\left (518400\,a^2\,c^6\,d^{14}-259200\,a\,b^2\,c^5\,d^{14}+32400\,b^4\,c^4\,d^{14}\right )+\frac {45\,c^2\,d^{11/2}\,{\left (b^2-4\,a\,c\right )}^{1/4}\,\left (23040\,a^2\,c^4\,d^9-11520\,a\,b^2\,c^3\,d^9+1440\,b^4\,c^2\,d^9\right )}{2}\right )\,{\left (b^2-4\,a\,c\right )}^{1/4}}{2}}\right )\,{\left (b^2-4\,a\,c\right )}^{1/4}\,45{}\mathrm {i} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*d + 2*c*d*x)^(11/2)/(a + b*x + c*x^2)^3,x)

[Out]

((b*d + 2*c*d*x)^(5/2)*(136*a*c^3*d^7 - 34*b^2*c^2*d^7) + (b*d + 2*c*d*x)^(1/2)*(416*a^2*c^4*d^9 + 26*b^4*c^2*
d^9 - 208*a*b^2*c^3*d^9))/((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) + 64*c^2*d^5*(b*d + 2*c*d*x)^(1/2) - 45*c^2*d^(11/2)*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) - c^2*d^(11/2)*atan(((c^2*d^(11/2)*((b*d + 2*c*d*x)^(1/2)*(518400*
a^2*c^6*d^14 + 32400*b^4*c^4*d^14 - 259200*a*b^2*c^5*d^14) - (45*c^2*d^(11/2)*(b^2 - 4*a*c)^(1/4)*(23040*a^2*c
^4*d^9 + 1440*b^4*c^2*d^9 - 11520*a*b^2*c^3*d^9))/2)*(b^2 - 4*a*c)^(1/4)*45i)/2 + (c^2*d^(11/2)*((b*d + 2*c*d*
x)^(1/2)*(518400*a^2*c^6*d^14 + 32400*b^4*c^4*d^14 - 259200*a*b^2*c^5*d^14) + (45*c^2*d^(11/2)*(b^2 - 4*a*c)^(
1/4)*(23040*a^2*c^4*d^9 + 1440*b^4*c^2*d^9 - 11520*a*b^2*c^3*d^9))/2)*(b^2 - 4*a*c)^(1/4)*45i)/2)/((45*c^2*d^(
11/2)*((b*d + 2*c*d*x)^(1/2)*(518400*a^2*c^6*d^14 + 32400*b^4*c^4*d^14 - 259200*a*b^2*c^5*d^14) - (45*c^2*d^(1
1/2)*(b^2 - 4*a*c)^(1/4)*(23040*a^2*c^4*d^9 + 1440*b^4*c^2*d^9 - 11520*a*b^2*c^3*d^9))/2)*(b^2 - 4*a*c)^(1/4))
/2 - (45*c^2*d^(11/2)*((b*d + 2*c*d*x)^(1/2)*(518400*a^2*c^6*d^14 + 32400*b^4*c^4*d^14 - 259200*a*b^2*c^5*d^14
) + (45*c^2*d^(11/2)*(b^2 - 4*a*c)^(1/4)*(23040*a^2*c^4*d^9 + 1440*b^4*c^2*d^9 - 11520*a*b^2*c^3*d^9))/2)*(b^2
- 4*a*c)^(1/4))/2))*(b^2 - 4*a*c)^(1/4)*45i

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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)**(11/2)/(c*x**2+b*x+a)**3,x)

[Out]

Timed out

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