3.1304 \(\int \frac {1}{(b d+2 c d x)^{3/2} (a+b x+c x^2)^2} \, dx\)

Optimal. Leaf size=172 \[ -\frac {10 c \tan ^{-1}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )}{d^{3/2} \left (b^2-4 a c\right )^{9/4}}+\frac {10 c \tanh ^{-1}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )}{d^{3/2} \left (b^2-4 a c\right )^{9/4}}-\frac {1}{d \left (b^2-4 a c\right ) \left (a+b x+c x^2\right ) \sqrt {b d+2 c d x}}-\frac {20 c}{d \left (b^2-4 a c\right )^2 \sqrt {b d+2 c d x}} \]

[Out]

-10*c*arctan((d*(2*c*x+b))^(1/2)/(-4*a*c+b^2)^(1/4)/d^(1/2))/(-4*a*c+b^2)^(9/4)/d^(3/2)+10*c*arctanh((d*(2*c*x
+b))^(1/2)/(-4*a*c+b^2)^(1/4)/d^(1/2))/(-4*a*c+b^2)^(9/4)/d^(3/2)-20*c/(-4*a*c+b^2)^2/d/(2*c*d*x+b*d)^(1/2)-1/
(-4*a*c+b^2)/d/(c*x^2+b*x+a)/(2*c*d*x+b*d)^(1/2)

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Rubi [A]  time = 0.12, antiderivative size = 172, 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 = {687, 693, 694, 329, 298, 203, 206} \[ -\frac {10 c \tan ^{-1}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )}{d^{3/2} \left (b^2-4 a c\right )^{9/4}}+\frac {10 c \tanh ^{-1}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )}{d^{3/2} \left (b^2-4 a c\right )^{9/4}}-\frac {1}{d \left (b^2-4 a c\right ) \left (a+b x+c x^2\right ) \sqrt {b d+2 c d x}}-\frac {20 c}{d \left (b^2-4 a c\right )^2 \sqrt {b d+2 c d x}} \]

Antiderivative was successfully verified.

[In]

Int[1/((b*d + 2*c*d*x)^(3/2)*(a + b*x + c*x^2)^2),x]

[Out]

(-20*c)/((b^2 - 4*a*c)^2*d*Sqrt[b*d + 2*c*d*x]) - 1/((b^2 - 4*a*c)*d*Sqrt[b*d + 2*c*d*x]*(a + b*x + c*x^2)) -
(10*c*ArcTan[Sqrt[d*(b + 2*c*x)]/((b^2 - 4*a*c)^(1/4)*Sqrt[d])])/((b^2 - 4*a*c)^(9/4)*d^(3/2)) + (10*c*ArcTanh
[Sqrt[d*(b + 2*c*x)]/((b^2 - 4*a*c)^(1/4)*Sqrt[d])])/((b^2 - 4*a*c)^(9/4)*d^(3/2))

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 687

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(2*c*(d + e*x)^(m +
1)*(a + b*x + c*x^2)^(p + 1))/(e*(p + 1)*(b^2 - 4*a*c)), x] - Dist[(2*c*e*(m + 2*p + 3))/(e*(p + 1)*(b^2 - 4*a
*c)), Int[(d + e*x)^m*(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, m}, 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] && RationalQ[m] && IntegerQ[2*p]

Rule 693

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(-2*b*d*(d + e*x)^(m
+ 1)*(a + b*x + c*x^2)^(p + 1))/(d^2*(m + 1)*(b^2 - 4*a*c)), x] + Dist[(b^2*(m + 2*p + 3))/(d^2*(m + 1)*(b^2
- 4*a*c)), Int[(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] && LtQ[m, -1] && (IntegerQ[2*p] || (IntegerQ[m] && Rationa
lQ[p]) || IntegerQ[(m + 2*p + 3)/2])

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 {1}{(b d+2 c d x)^{3/2} \left (a+b x+c x^2\right )^2} \, dx &=-\frac {1}{\left (b^2-4 a c\right ) d \sqrt {b d+2 c d x} \left (a+b x+c x^2\right )}-\frac {(5 c) \int \frac {1}{(b d+2 c d x)^{3/2} \left (a+b x+c x^2\right )} \, dx}{b^2-4 a c}\\ &=-\frac {20 c}{\left (b^2-4 a c\right )^2 d \sqrt {b d+2 c d x}}-\frac {1}{\left (b^2-4 a c\right ) d \sqrt {b d+2 c d x} \left (a+b x+c x^2\right )}-\frac {(5 c) \int \frac {\sqrt {b d+2 c d x}}{a+b x+c x^2} \, dx}{\left (b^2-4 a c\right )^2 d^2}\\ &=-\frac {20 c}{\left (b^2-4 a c\right )^2 d \sqrt {b d+2 c d x}}-\frac {1}{\left (b^2-4 a c\right ) d \sqrt {b d+2 c d x} \left (a+b x+c x^2\right )}-\frac {5 \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 )}{2 \left (b^2-4 a c\right )^2 d^3}\\ &=-\frac {20 c}{\left (b^2-4 a c\right )^2 d \sqrt {b d+2 c d x}}-\frac {1}{\left (b^2-4 a c\right ) d \sqrt {b d+2 c d x} \left (a+b x+c x^2\right )}-\frac {5 \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 )}{\left (b^2-4 a c\right )^2 d^3}\\ &=-\frac {20 c}{\left (b^2-4 a c\right )^2 d \sqrt {b d+2 c d x}}-\frac {1}{\left (b^2-4 a c\right ) d \sqrt {b d+2 c d x} \left (a+b x+c x^2\right )}+\frac {(10 c) \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 (b^2-4 a c\right )^2 d}-\frac {(10 c) \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 (b^2-4 a c\right )^2 d}\\ &=-\frac {20 c}{\left (b^2-4 a c\right )^2 d \sqrt {b d+2 c d x}}-\frac {1}{\left (b^2-4 a c\right ) d \sqrt {b d+2 c d x} \left (a+b x+c x^2\right )}-\frac {10 c \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 )^{9/4} d^{3/2}}+\frac {10 c \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 )^{9/4} d^{3/2}}\\ \end {align*}

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Mathematica [C]  time = 0.05, size = 55, normalized size = 0.32 \[ -\frac {16 c \, _2F_1\left (-\frac {1}{4},2;\frac {3}{4};\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{d \left (b^2-4 a c\right )^2 \sqrt {d (b+2 c x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((b*d + 2*c*d*x)^(3/2)*(a + b*x + c*x^2)^2),x]

[Out]

(-16*c*Hypergeometric2F1[-1/4, 2, 3/4, (b + 2*c*x)^2/(b^2 - 4*a*c)])/((b^2 - 4*a*c)^2*d*Sqrt[d*(b + 2*c*x)])

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fricas [B]  time = 0.75, size = 1710, normalized size = 9.94 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(2*c*d*x+b*d)^(3/2)/(c*x^2+b*x+a)^2,x, algorithm="fricas")

[Out]

(20*(2*(b^4*c^2 - 8*a*b^2*c^3 + 16*a^2*c^4)*d^2*x^3 + 3*(b^5*c - 8*a*b^3*c^2 + 16*a^2*b*c^3)*d^2*x^2 + (b^6 -
6*a*b^4*c + 32*a^3*c^3)*d^2*x + (a*b^5 - 8*a^2*b^3*c + 16*a^3*b*c^2)*d^2)*(c^4/((b^18 - 36*a*b^16*c + 576*a^2*
b^14*c^2 - 5376*a^3*b^12*c^3 + 32256*a^4*b^10*c^4 - 129024*a^5*b^8*c^5 + 344064*a^6*b^6*c^6 - 589824*a^7*b^4*c
^7 + 589824*a^8*b^2*c^8 - 262144*a^9*c^9)*d^6))^(1/4)*arctan((sqrt(2*c^7*d*x + b*c^6*d + (b^10*c^4 - 20*a*b^8*
c^5 + 160*a^2*b^6*c^6 - 640*a^3*b^4*c^7 + 1280*a^4*b^2*c^8 - 1024*a^5*c^9)*d^4*sqrt(c^4/((b^18 - 36*a*b^16*c +
576*a^2*b^14*c^2 - 5376*a^3*b^12*c^3 + 32256*a^4*b^10*c^4 - 129024*a^5*b^8*c^5 + 344064*a^6*b^6*c^6 - 589824*
a^7*b^4*c^7 + 589824*a^8*b^2*c^8 - 262144*a^9*c^9)*d^6)))*(b^4 - 8*a*b^2*c + 16*a^2*c^2)*d*(c^4/((b^18 - 36*a*
b^16*c + 576*a^2*b^14*c^2 - 5376*a^3*b^12*c^3 + 32256*a^4*b^10*c^4 - 129024*a^5*b^8*c^5 + 344064*a^6*b^6*c^6 -
589824*a^7*b^4*c^7 + 589824*a^8*b^2*c^8 - 262144*a^9*c^9)*d^6))^(1/4) - (b^4*c^3 - 8*a*b^2*c^4 + 16*a^2*c^5)*
sqrt(2*c*d*x + b*d)*d*(c^4/((b^18 - 36*a*b^16*c + 576*a^2*b^14*c^2 - 5376*a^3*b^12*c^3 + 32256*a^4*b^10*c^4 -
129024*a^5*b^8*c^5 + 344064*a^6*b^6*c^6 - 589824*a^7*b^4*c^7 + 589824*a^8*b^2*c^8 - 262144*a^9*c^9)*d^6))^(1/4
))/c^4) + 5*(2*(b^4*c^2 - 8*a*b^2*c^3 + 16*a^2*c^4)*d^2*x^3 + 3*(b^5*c - 8*a*b^3*c^2 + 16*a^2*b*c^3)*d^2*x^2 +
(b^6 - 6*a*b^4*c + 32*a^3*c^3)*d^2*x + (a*b^5 - 8*a^2*b^3*c + 16*a^3*b*c^2)*d^2)*(c^4/((b^18 - 36*a*b^16*c +
576*a^2*b^14*c^2 - 5376*a^3*b^12*c^3 + 32256*a^4*b^10*c^4 - 129024*a^5*b^8*c^5 + 344064*a^6*b^6*c^6 - 589824*a
^7*b^4*c^7 + 589824*a^8*b^2*c^8 - 262144*a^9*c^9)*d^6))^(1/4)*log(125*(b^14 - 28*a*b^12*c + 336*a^2*b^10*c^2 -
2240*a^3*b^8*c^3 + 8960*a^4*b^6*c^4 - 21504*a^5*b^4*c^5 + 28672*a^6*b^2*c^6 - 16384*a^7*c^7)*d^5*(c^4/((b^18
- 36*a*b^16*c + 576*a^2*b^14*c^2 - 5376*a^3*b^12*c^3 + 32256*a^4*b^10*c^4 - 129024*a^5*b^8*c^5 + 344064*a^6*b^
6*c^6 - 589824*a^7*b^4*c^7 + 589824*a^8*b^2*c^8 - 262144*a^9*c^9)*d^6))^(3/4) + 125*sqrt(2*c*d*x + b*d)*c^3) -
5*(2*(b^4*c^2 - 8*a*b^2*c^3 + 16*a^2*c^4)*d^2*x^3 + 3*(b^5*c - 8*a*b^3*c^2 + 16*a^2*b*c^3)*d^2*x^2 + (b^6 - 6
*a*b^4*c + 32*a^3*c^3)*d^2*x + (a*b^5 - 8*a^2*b^3*c + 16*a^3*b*c^2)*d^2)*(c^4/((b^18 - 36*a*b^16*c + 576*a^2*b
^14*c^2 - 5376*a^3*b^12*c^3 + 32256*a^4*b^10*c^4 - 129024*a^5*b^8*c^5 + 344064*a^6*b^6*c^6 - 589824*a^7*b^4*c^
7 + 589824*a^8*b^2*c^8 - 262144*a^9*c^9)*d^6))^(1/4)*log(-125*(b^14 - 28*a*b^12*c + 336*a^2*b^10*c^2 - 2240*a^
3*b^8*c^3 + 8960*a^4*b^6*c^4 - 21504*a^5*b^4*c^5 + 28672*a^6*b^2*c^6 - 16384*a^7*c^7)*d^5*(c^4/((b^18 - 36*a*b
^16*c + 576*a^2*b^14*c^2 - 5376*a^3*b^12*c^3 + 32256*a^4*b^10*c^4 - 129024*a^5*b^8*c^5 + 344064*a^6*b^6*c^6 -
589824*a^7*b^4*c^7 + 589824*a^8*b^2*c^8 - 262144*a^9*c^9)*d^6))^(3/4) + 125*sqrt(2*c*d*x + b*d)*c^3) - (20*c^2
*x^2 + 20*b*c*x + b^2 + 16*a*c)*sqrt(2*c*d*x + b*d))/(2*(b^4*c^2 - 8*a*b^2*c^3 + 16*a^2*c^4)*d^2*x^3 + 3*(b^5*
c - 8*a*b^3*c^2 + 16*a^2*b*c^3)*d^2*x^2 + (b^6 - 6*a*b^4*c + 32*a^3*c^3)*d^2*x + (a*b^5 - 8*a^2*b^3*c + 16*a^3
*b*c^2)*d^2)

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giac [B]  time = 0.29, size = 650, normalized size = 3.78 \[ \frac {5 \, \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {3}{4}} c \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} + 2 \, \sqrt {2 \, c d x + b d}\right )}}{2 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}}}\right )}{b^{6} d^{3} - 12 \, a b^{4} c d^{3} + 48 \, a^{2} b^{2} c^{2} d^{3} - 64 \, a^{3} c^{3} d^{3}} + \frac {5 \, \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {3}{4}} c \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} - 2 \, \sqrt {2 \, c d x + b d}\right )}}{2 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}}}\right )}{b^{6} d^{3} - 12 \, a b^{4} c d^{3} + 48 \, a^{2} b^{2} c^{2} d^{3} - 64 \, a^{3} c^{3} d^{3}} - \frac {5 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {3}{4}} c \log \left (2 \, c d x + b d + \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} \sqrt {2 \, c d x + b d} + \sqrt {-b^{2} d^{2} + 4 \, a c d^{2}}\right )}{\sqrt {2} b^{6} d^{3} - 12 \, \sqrt {2} a b^{4} c d^{3} + 48 \, \sqrt {2} a^{2} b^{2} c^{2} d^{3} - 64 \, \sqrt {2} a^{3} c^{3} d^{3}} + \frac {5 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {3}{4}} c \log \left (2 \, c d x + b d - \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} \sqrt {2 \, c d x + b d} + \sqrt {-b^{2} d^{2} + 4 \, a c d^{2}}\right )}{\sqrt {2} b^{6} d^{3} - 12 \, \sqrt {2} a b^{4} c d^{3} + 48 \, \sqrt {2} a^{2} b^{2} c^{2} d^{3} - 64 \, \sqrt {2} a^{3} c^{3} d^{3}} - \frac {4 \, {\left (4 \, b^{2} c d^{2} - 16 \, a c^{2} d^{2} - 5 \, {\left (2 \, c d x + b d\right )}^{2} c\right )}}{{\left (b^{4} d - 8 \, a b^{2} c d + 16 \, a^{2} c^{2} d\right )} {\left (\sqrt {2 \, c d x + b d} b^{2} d^{2} - 4 \, \sqrt {2 \, c d x + b d} a c d^{2} - {\left (2 \, c d x + b d\right )}^{\frac {5}{2}}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(2*c*d*x+b*d)^(3/2)/(c*x^2+b*x+a)^2,x, algorithm="giac")

[Out]

5*sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(3/4)*c*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^6*d^3 - 12*a*b^4*c*d^3 + 48*a^2*b^2*c^2*d^3 - 64*a^3*c^3*d^3) +
5*sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(3/4)*c*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^6*d^3 - 12*a*b^4*c*d^3 + 48*a^2*b^2*c^2*d^3 - 64*a^3*c^3*d^3) -
5*(-b^2*d^2 + 4*a*c*d^2)^(3/4)*c*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^6*d^3 - 12*sqrt(2)*a*b^4*c*d^3 + 48*sqrt(2)*a^2*b^2*c^2*d^3 - 64*sqr
t(2)*a^3*c^3*d^3) + 5*(-b^2*d^2 + 4*a*c*d^2)^(3/4)*c*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^6*d^3 - 12*sqrt(2)*a*b^4*c*d^3 + 48*sqrt(2)*a^2*b
^2*c^2*d^3 - 64*sqrt(2)*a^3*c^3*d^3) - 4*(4*b^2*c*d^2 - 16*a*c^2*d^2 - 5*(2*c*d*x + b*d)^2*c)/((b^4*d - 8*a*b^
2*c*d + 16*a^2*c^2*d)*(sqrt(2*c*d*x + b*d)*b^2*d^2 - 4*sqrt(2*c*d*x + b*d)*a*c*d^2 - (2*c*d*x + b*d)^(5/2)))

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maple [B]  time = 0.07, size = 404, normalized size = 2.35 \[ \frac {5 \sqrt {2}\, c \arctan \left (-\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )}{\left (4 a c -b^{2}\right )^{2} \left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}} d}-\frac {5 \sqrt {2}\, c \arctan \left (\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )}{\left (4 a c -b^{2}\right )^{2} \left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}} d}-\frac {5 \sqrt {2}\, c \ln \left (\frac {2 c d x +b d -\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a c \,d^{2}-b^{2} d^{2}}}{2 c d x +b d +\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a c \,d^{2}-b^{2} d^{2}}}\right )}{2 \left (4 a c -b^{2}\right )^{2} \left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}} d}-\frac {4 \left (2 c d x +b d \right )^{\frac {3}{2}} c}{\left (4 a c -b^{2}\right )^{2} \left (4 c^{2} d^{2} x^{2}+4 b c \,d^{2} x +4 a c \,d^{2}\right ) d}-\frac {16 c}{\left (4 a c -b^{2}\right )^{2} \sqrt {2 c d x +b d}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(2*c*d*x+b*d)^(3/2)/(c*x^2+b*x+a)^2,x)

[Out]

-4*c/d/(4*a*c-b^2)^2*(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)-5/2*c/d/(4*a*c-b^2)^2*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)))-5*c
/d/(4*a*c-b^2)^2*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)+5*c/d/(4*a*c-b^2)^2*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)-16*c/d/(4*a*c-b^2)^2/(2*c*d*x+b*d)^(1/2)

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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(1/(2*c*d*x+b*d)^(3/2)/(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 = 0.74, size = 264, normalized size = 1.53 \[ \frac {\frac {16\,c\,d}{4\,a\,c-b^2}+\frac {20\,c\,{\left (b\,d+2\,c\,d\,x\right )}^2}{d\,{\left (4\,a\,c-b^2\right )}^2}}{\sqrt {b\,d+2\,c\,d\,x}\,\left (b^2\,d^2-4\,a\,c\,d^2\right )-{\left (b\,d+2\,c\,d\,x\right )}^{5/2}}-\frac {10\,c\,\mathrm {atan}\left (\frac {b^4\,\sqrt {b\,d+2\,c\,d\,x}+16\,a^2\,c^2\,\sqrt {b\,d+2\,c\,d\,x}-8\,a\,b^2\,c\,\sqrt {b\,d+2\,c\,d\,x}}{\sqrt {d}\,{\left (b^2-4\,a\,c\right )}^{9/4}}\right )}{d^{3/2}\,{\left (b^2-4\,a\,c\right )}^{9/4}}-\frac {c\,\mathrm {atan}\left (\frac {b^4\,\sqrt {b\,d+2\,c\,d\,x}\,1{}\mathrm {i}+a^2\,c^2\,\sqrt {b\,d+2\,c\,d\,x}\,16{}\mathrm {i}-a\,b^2\,c\,\sqrt {b\,d+2\,c\,d\,x}\,8{}\mathrm {i}}{\sqrt {d}\,{\left (b^2-4\,a\,c\right )}^{9/4}}\right )\,10{}\mathrm {i}}{d^{3/2}\,{\left (b^2-4\,a\,c\right )}^{9/4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((b*d + 2*c*d*x)^(3/2)*(a + b*x + c*x^2)^2),x)

[Out]

((16*c*d)/(4*a*c - b^2) + (20*c*(b*d + 2*c*d*x)^2)/(d*(4*a*c - b^2)^2))/((b*d + 2*c*d*x)^(1/2)*(b^2*d^2 - 4*a*
c*d^2) - (b*d + 2*c*d*x)^(5/2)) - (10*c*atan((b^4*(b*d + 2*c*d*x)^(1/2) + 16*a^2*c^2*(b*d + 2*c*d*x)^(1/2) - 8
*a*b^2*c*(b*d + 2*c*d*x)^(1/2))/(d^(1/2)*(b^2 - 4*a*c)^(9/4))))/(d^(3/2)*(b^2 - 4*a*c)^(9/4)) - (c*atan((b^4*(
b*d + 2*c*d*x)^(1/2)*1i + a^2*c^2*(b*d + 2*c*d*x)^(1/2)*16i - a*b^2*c*(b*d + 2*c*d*x)^(1/2)*8i)/(d^(1/2)*(b^2
- 4*a*c)^(9/4)))*10i)/(d^(3/2)*(b^2 - 4*a*c)^(9/4))

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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(1/(2*c*d*x+b*d)**(3/2)/(c*x**2+b*x+a)**2,x)

[Out]

Timed out

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