3.12.17 \(\int \frac {1}{(b d+2 c d x)^{5/2} (a+b x+c x^2)^2} \, dx\)

Optimal. Leaf size=174 \[ \frac {14 c \tan ^{-1}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )}{d^{5/2} \left (b^2-4 a c\right )^{11/4}}+\frac {14 c \tanh ^{-1}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )}{d^{5/2} \left (b^2-4 a c\right )^{11/4}}-\frac {1}{d \left (b^2-4 a c\right ) \left (a+b x+c x^2\right ) (b d+2 c d x)^{3/2}}-\frac {28 c}{3 d \left (b^2-4 a c\right )^2 (b d+2 c d x)^{3/2}} \]

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Rubi [A]  time = 0.14, antiderivative size = 174, 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, 212, 206, 203} \begin {gather*} \frac {14 c \tan ^{-1}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )}{d^{5/2} \left (b^2-4 a c\right )^{11/4}}+\frac {14 c \tanh ^{-1}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )}{d^{5/2} \left (b^2-4 a c\right )^{11/4}}-\frac {1}{d \left (b^2-4 a c\right ) \left (a+b x+c x^2\right ) (b d+2 c d x)^{3/2}}-\frac {28 c}{3 d \left (b^2-4 a c\right )^2 (b d+2 c d x)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((b*d + 2*c*d*x)^(5/2)*(a + b*x + c*x^2)^2),x]

[Out]

(-28*c)/(3*(b^2 - 4*a*c)^2*d*(b*d + 2*c*d*x)^(3/2)) - 1/((b^2 - 4*a*c)*d*(b*d + 2*c*d*x)^(3/2)*(a + b*x + c*x^
2)) + (14*c*ArcTan[Sqrt[d*(b + 2*c*x)]/((b^2 - 4*a*c)^(1/4)*Sqrt[d])])/((b^2 - 4*a*c)^(11/4)*d^(5/2)) + (14*c*
ArcTanh[Sqrt[d*(b + 2*c*x)]/((b^2 - 4*a*c)^(1/4)*Sqrt[d])])/((b^2 - 4*a*c)^(11/4)*d^(5/2))

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 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 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)^{5/2} \left (a+b x+c x^2\right )^2} \, dx &=-\frac {1}{\left (b^2-4 a c\right ) d (b d+2 c d x)^{3/2} \left (a+b x+c x^2\right )}-\frac {(7 c) \int \frac {1}{(b d+2 c d x)^{5/2} \left (a+b x+c x^2\right )} \, dx}{b^2-4 a c}\\ &=-\frac {28 c}{3 \left (b^2-4 a c\right )^2 d (b d+2 c d x)^{3/2}}-\frac {1}{\left (b^2-4 a c\right ) d (b d+2 c d x)^{3/2} \left (a+b x+c x^2\right )}-\frac {(7 c) \int \frac {1}{\sqrt {b d+2 c d x} \left (a+b x+c x^2\right )} \, dx}{\left (b^2-4 a c\right )^2 d^2}\\ &=-\frac {28 c}{3 \left (b^2-4 a c\right )^2 d (b d+2 c d x)^{3/2}}-\frac {1}{\left (b^2-4 a c\right ) d (b d+2 c d x)^{3/2} \left (a+b x+c x^2\right )}-\frac {7 \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 )}{2 \left (b^2-4 a c\right )^2 d^3}\\ &=-\frac {28 c}{3 \left (b^2-4 a c\right )^2 d (b d+2 c d x)^{3/2}}-\frac {1}{\left (b^2-4 a c\right ) d (b d+2 c d x)^{3/2} \left (a+b x+c x^2\right )}-\frac {7 \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 )}{\left (b^2-4 a c\right )^2 d^3}\\ &=-\frac {28 c}{3 \left (b^2-4 a c\right )^2 d (b d+2 c d x)^{3/2}}-\frac {1}{\left (b^2-4 a c\right ) d (b d+2 c d x)^{3/2} \left (a+b x+c x^2\right )}+\frac {(14 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 )^{5/2} d^2}+\frac {(14 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 )^{5/2} d^2}\\ &=-\frac {28 c}{3 \left (b^2-4 a c\right )^2 d (b d+2 c d x)^{3/2}}-\frac {1}{\left (b^2-4 a c\right ) d (b d+2 c d x)^{3/2} \left (a+b x+c x^2\right )}+\frac {14 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 )^{11/4} d^{5/2}}+\frac {14 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 )^{11/4} d^{5/2}}\\ \end {align*}

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Mathematica [C]  time = 0.06, size = 57, normalized size = 0.33 \begin {gather*} -\frac {16 c \, _2F_1\left (-\frac {3}{4},2;\frac {1}{4};\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{3 d \left (b^2-4 a c\right )^2 (d (b+2 c x))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/((b*d + 2*c*d*x)^(5/2)*(a + b*x + c*x^2)^2),x]

[Out]

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

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IntegrateAlgebraic [C]  time = 0.80, size = 269, normalized size = 1.55 \begin {gather*} \frac {\left (-16 a c-3 b^2-28 b c x-28 c^2 x^2\right ) \sqrt {b d+2 c d x}}{3 d^3 \left (b^2-4 a c\right )^2 (b+2 c x)^2 \left (a+b x+c x^2\right )}-\frac {(7-7 i) 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 )}{d^{5/2} \left (b^2-4 a c\right )^{11/4}}+\frac {(7-7 i) 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 )}{d^{5/2} \left (b^2-4 a c\right )^{11/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[1/((b*d + 2*c*d*x)^(5/2)*(a + b*x + c*x^2)^2),x]

[Out]

(Sqrt[b*d + 2*c*d*x]*(-3*b^2 - 16*a*c - 28*b*c*x - 28*c^2*x^2))/(3*(b^2 - 4*a*c)^2*d^3*(b + 2*c*x)^2*(a + b*x
+ c*x^2)) - ((7 - 7*I)*c*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)^(11/4)*d^(5/2)) +
((7 - 7*I)*c*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))])/((b^2 - 4*a*c)^(11/4)*d^(5/2))

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fricas [B]  time = 0.50, size = 2167, normalized size = 12.45

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(2*c*d*x+b*d)^(5/2)/(c*x^2+b*x+a)^2,x, algorithm="fricas")

[Out]

-1/3*(84*(4*(b^4*c^3 - 8*a*b^2*c^4 + 16*a^2*c^5)*d^3*x^4 + 8*(b^5*c^2 - 8*a*b^3*c^3 + 16*a^2*b*c^4)*d^3*x^3 +
(5*b^6*c - 36*a*b^4*c^2 + 48*a^2*b^2*c^3 + 64*a^3*c^4)*d^3*x^2 + (b^7 - 4*a*b^5*c - 16*a^2*b^3*c^2 + 64*a^3*b*
c^3)*d^3*x + (a*b^6 - 8*a^2*b^4*c + 16*a^3*b^2*c^2)*d^3)*(c^4/((b^22 - 44*a*b^20*c + 880*a^2*b^18*c^2 - 10560*
a^3*b^16*c^3 + 84480*a^4*b^14*c^4 - 473088*a^5*b^12*c^5 + 1892352*a^6*b^10*c^6 - 5406720*a^7*b^8*c^7 + 1081344
0*a^8*b^6*c^8 - 14417920*a^9*b^4*c^9 + 11534336*a^10*b^2*c^10 - 4194304*a^11*c^11)*d^10))^(1/4)*arctan(((b^16
- 32*a*b^14*c + 448*a^2*b^12*c^2 - 3584*a^3*b^10*c^3 + 17920*a^4*b^8*c^4 - 57344*a^5*b^6*c^5 + 114688*a^6*b^4*
c^6 - 131072*a^7*b^2*c^7 + 65536*a^8*c^8)*sqrt((b^12 - 24*a*b^10*c + 240*a^2*b^8*c^2 - 1280*a^3*b^6*c^3 + 3840
*a^4*b^4*c^4 - 6144*a^5*b^2*c^5 + 4096*a^6*c^6)*d^6*sqrt(c^4/((b^22 - 44*a*b^20*c + 880*a^2*b^18*c^2 - 10560*a
^3*b^16*c^3 + 84480*a^4*b^14*c^4 - 473088*a^5*b^12*c^5 + 1892352*a^6*b^10*c^6 - 5406720*a^7*b^8*c^7 + 10813440
*a^8*b^6*c^8 - 14417920*a^9*b^4*c^9 + 11534336*a^10*b^2*c^10 - 4194304*a^11*c^11)*d^10)) + 2*c^3*d*x + b*c^2*d
)*d^7*(c^4/((b^22 - 44*a*b^20*c + 880*a^2*b^18*c^2 - 10560*a^3*b^16*c^3 + 84480*a^4*b^14*c^4 - 473088*a^5*b^12
*c^5 + 1892352*a^6*b^10*c^6 - 5406720*a^7*b^8*c^7 + 10813440*a^8*b^6*c^8 - 14417920*a^9*b^4*c^9 + 11534336*a^1
0*b^2*c^10 - 4194304*a^11*c^11)*d^10))^(3/4) - (b^16*c - 32*a*b^14*c^2 + 448*a^2*b^12*c^3 - 3584*a^3*b^10*c^4
+ 17920*a^4*b^8*c^5 - 57344*a^5*b^6*c^6 + 114688*a^6*b^4*c^7 - 131072*a^7*b^2*c^8 + 65536*a^8*c^9)*sqrt(2*c*d*
x + b*d)*d^7*(c^4/((b^22 - 44*a*b^20*c + 880*a^2*b^18*c^2 - 10560*a^3*b^16*c^3 + 84480*a^4*b^14*c^4 - 473088*a
^5*b^12*c^5 + 1892352*a^6*b^10*c^6 - 5406720*a^7*b^8*c^7 + 10813440*a^8*b^6*c^8 - 14417920*a^9*b^4*c^9 + 11534
336*a^10*b^2*c^10 - 4194304*a^11*c^11)*d^10))^(3/4))/c^4) - 21*(4*(b^4*c^3 - 8*a*b^2*c^4 + 16*a^2*c^5)*d^3*x^4
+ 8*(b^5*c^2 - 8*a*b^3*c^3 + 16*a^2*b*c^4)*d^3*x^3 + (5*b^6*c - 36*a*b^4*c^2 + 48*a^2*b^2*c^3 + 64*a^3*c^4)*d
^3*x^2 + (b^7 - 4*a*b^5*c - 16*a^2*b^3*c^2 + 64*a^3*b*c^3)*d^3*x + (a*b^6 - 8*a^2*b^4*c + 16*a^3*b^2*c^2)*d^3)
*(c^4/((b^22 - 44*a*b^20*c + 880*a^2*b^18*c^2 - 10560*a^3*b^16*c^3 + 84480*a^4*b^14*c^4 - 473088*a^5*b^12*c^5
+ 1892352*a^6*b^10*c^6 - 5406720*a^7*b^8*c^7 + 10813440*a^8*b^6*c^8 - 14417920*a^9*b^4*c^9 + 11534336*a^10*b^2
*c^10 - 4194304*a^11*c^11)*d^10))^(1/4)*log(7*(b^6 - 12*a*b^4*c + 48*a^2*b^2*c^2 - 64*a^3*c^3)*d^3*(c^4/((b^22
- 44*a*b^20*c + 880*a^2*b^18*c^2 - 10560*a^3*b^16*c^3 + 84480*a^4*b^14*c^4 - 473088*a^5*b^12*c^5 + 1892352*a^
6*b^10*c^6 - 5406720*a^7*b^8*c^7 + 10813440*a^8*b^6*c^8 - 14417920*a^9*b^4*c^9 + 11534336*a^10*b^2*c^10 - 4194
304*a^11*c^11)*d^10))^(1/4) + 7*sqrt(2*c*d*x + b*d)*c) + 21*(4*(b^4*c^3 - 8*a*b^2*c^4 + 16*a^2*c^5)*d^3*x^4 +
8*(b^5*c^2 - 8*a*b^3*c^3 + 16*a^2*b*c^4)*d^3*x^3 + (5*b^6*c - 36*a*b^4*c^2 + 48*a^2*b^2*c^3 + 64*a^3*c^4)*d^3*
x^2 + (b^7 - 4*a*b^5*c - 16*a^2*b^3*c^2 + 64*a^3*b*c^3)*d^3*x + (a*b^6 - 8*a^2*b^4*c + 16*a^3*b^2*c^2)*d^3)*(c
^4/((b^22 - 44*a*b^20*c + 880*a^2*b^18*c^2 - 10560*a^3*b^16*c^3 + 84480*a^4*b^14*c^4 - 473088*a^5*b^12*c^5 + 1
892352*a^6*b^10*c^6 - 5406720*a^7*b^8*c^7 + 10813440*a^8*b^6*c^8 - 14417920*a^9*b^4*c^9 + 11534336*a^10*b^2*c^
10 - 4194304*a^11*c^11)*d^10))^(1/4)*log(-7*(b^6 - 12*a*b^4*c + 48*a^2*b^2*c^2 - 64*a^3*c^3)*d^3*(c^4/((b^22 -
44*a*b^20*c + 880*a^2*b^18*c^2 - 10560*a^3*b^16*c^3 + 84480*a^4*b^14*c^4 - 473088*a^5*b^12*c^5 + 1892352*a^6*
b^10*c^6 - 5406720*a^7*b^8*c^7 + 10813440*a^8*b^6*c^8 - 14417920*a^9*b^4*c^9 + 11534336*a^10*b^2*c^10 - 419430
4*a^11*c^11)*d^10))^(1/4) + 7*sqrt(2*c*d*x + b*d)*c) + (28*c^2*x^2 + 28*b*c*x + 3*b^2 + 16*a*c)*sqrt(2*c*d*x +
b*d))/(4*(b^4*c^3 - 8*a*b^2*c^4 + 16*a^2*c^5)*d^3*x^4 + 8*(b^5*c^2 - 8*a*b^3*c^3 + 16*a^2*b*c^4)*d^3*x^3 + (5
*b^6*c - 36*a*b^4*c^2 + 48*a^2*b^2*c^3 + 64*a^3*c^4)*d^3*x^2 + (b^7 - 4*a*b^5*c - 16*a^2*b^3*c^2 + 64*a^3*b*c^
3)*d^3*x + (a*b^6 - 8*a^2*b^4*c + 16*a^3*b^2*c^2)*d^3)

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giac [B]  time = 0.29, size = 646, normalized size = 3.71 \begin {gather*} \frac {7 \, \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{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 {7 \, \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{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 {7 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{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 {7 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{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 \, \sqrt {2 \, c d x + b d} c}{{\left (b^{4} d - 8 \, a b^{2} c d + 16 \, a^{2} c^{2} d\right )} {\left (b^{2} d^{2} - 4 \, a c d^{2} - {\left (2 \, c d x + b d\right )}^{2}\right )}} - \frac {16 \, c}{3 \, {\left (b^{4} d - 8 \, a b^{2} c d + 16 \, a^{2} c^{2} d\right )} {\left (2 \, c d x + b d\right )}^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(2*c*d*x+b*d)^(5/2)/(c*x^2+b*x+a)^2,x, algorithm="giac")

[Out]

7*sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(1/4)*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) +
7*sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(1/4)*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) +
7*(-b^2*d^2 + 4*a*c*d^2)^(1/4)*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) - 7*(-b^2*d^2 + 4*a*c*d^2)^(1/4)*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*sqrt(2*c*d*x + b*d)*c/((b^4*d - 8*a*b^2*c*d + 16*a^2*c^2*d)*(b^2*d^2
- 4*a*c*d^2 - (2*c*d*x + b*d)^2)) - 16/3*c/((b^4*d - 8*a*b^2*c*d + 16*a^2*c^2*d)*(2*c*d*x + b*d)^(3/2))

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maple [B]  time = 0.06, size = 404, normalized size = 2.32 \begin {gather*} \frac {7 \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 {3}{4}} d}-\frac {7 \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 {3}{4}} d}-\frac {7 \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 {3}{4}} d}-\frac {4 \sqrt {2 c d x +b d}\, 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}{3 \left (4 a c -b^{2}\right )^{2} \left (2 c d x +b d \right )^{\frac {3}{2}} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(2*c*d*x+b*d)^(5/2)/(c*x^2+b*x+a)^2,x)

[Out]

-4*c/d/(4*a*c-b^2)^2*(2*c*d*x+b*d)^(1/2)/(4*c^2*d^2*x^2+4*b*c*d^2*x+4*a*c*d^2)-7/2*c/d/(4*a*c-b^2)^2/(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)))-7*c
/d/(4*a*c-b^2)^2/(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)+7*c/d/(4*a*c-b^2)^2/(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)-16/3*c/d/(4*a*c-b^2)^2/(2*c*d*x+b*d)^(3/2)

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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(1/(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 = 0.90, size = 2280, normalized size = 13.10

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((b*d + 2*c*d*x)^(5/2)*(a + b*x + c*x^2)^2),x)

[Out]

((16*c*d)/(3*(4*a*c - b^2)) + (28*c*(b*d + 2*c*d*x)^2)/(3*d*(4*a*c - b^2)^2))/((b*d + 2*c*d*x)^(3/2)*(b^2*d^2
- 4*a*c*d^2) - (b*d + 2*c*d*x)^(7/2)) + (c*atan(((c*((b*d + 2*c*d*x)^(1/2)*(12845056*a^6*c^8*d^3 + 3136*b^12*c
^2*d^3 - 75264*a*b^10*c^3*d^3 + 752640*a^2*b^8*c^4*d^3 - 4014080*a^3*b^6*c^5*d^3 + 12042240*a^4*b^4*c^6*d^3 -
19267584*a^5*b^2*c^7*d^3) - (7*c*(448*b^18*c*d^6 - 117440512*a^9*c^10*d^6 - 16128*a*b^16*c^2*d^6 + 258048*a^2*
b^14*c^3*d^6 - 2408448*a^3*b^12*c^4*d^6 + 14450688*a^4*b^10*c^5*d^6 - 57802752*a^5*b^8*c^6*d^6 + 154140672*a^6
*b^6*c^7*d^6 - 264241152*a^7*b^4*c^8*d^6 + 264241152*a^8*b^2*c^9*d^6))/(d^(5/2)*(b^2 - 4*a*c)^(11/4)))*7i)/(d^
(5/2)*(b^2 - 4*a*c)^(11/4)) + (c*((b*d + 2*c*d*x)^(1/2)*(12845056*a^6*c^8*d^3 + 3136*b^12*c^2*d^3 - 75264*a*b^
10*c^3*d^3 + 752640*a^2*b^8*c^4*d^3 - 4014080*a^3*b^6*c^5*d^3 + 12042240*a^4*b^4*c^6*d^3 - 19267584*a^5*b^2*c^
7*d^3) + (7*c*(448*b^18*c*d^6 - 117440512*a^9*c^10*d^6 - 16128*a*b^16*c^2*d^6 + 258048*a^2*b^14*c^3*d^6 - 2408
448*a^3*b^12*c^4*d^6 + 14450688*a^4*b^10*c^5*d^6 - 57802752*a^5*b^8*c^6*d^6 + 154140672*a^6*b^6*c^7*d^6 - 2642
41152*a^7*b^4*c^8*d^6 + 264241152*a^8*b^2*c^9*d^6))/(d^(5/2)*(b^2 - 4*a*c)^(11/4)))*7i)/(d^(5/2)*(b^2 - 4*a*c)
^(11/4)))/((7*c*((b*d + 2*c*d*x)^(1/2)*(12845056*a^6*c^8*d^3 + 3136*b^12*c^2*d^3 - 75264*a*b^10*c^3*d^3 + 7526
40*a^2*b^8*c^4*d^3 - 4014080*a^3*b^6*c^5*d^3 + 12042240*a^4*b^4*c^6*d^3 - 19267584*a^5*b^2*c^7*d^3) - (7*c*(44
8*b^18*c*d^6 - 117440512*a^9*c^10*d^6 - 16128*a*b^16*c^2*d^6 + 258048*a^2*b^14*c^3*d^6 - 2408448*a^3*b^12*c^4*
d^6 + 14450688*a^4*b^10*c^5*d^6 - 57802752*a^5*b^8*c^6*d^6 + 154140672*a^6*b^6*c^7*d^6 - 264241152*a^7*b^4*c^8
*d^6 + 264241152*a^8*b^2*c^9*d^6))/(d^(5/2)*(b^2 - 4*a*c)^(11/4))))/(d^(5/2)*(b^2 - 4*a*c)^(11/4)) - (7*c*((b*
d + 2*c*d*x)^(1/2)*(12845056*a^6*c^8*d^3 + 3136*b^12*c^2*d^3 - 75264*a*b^10*c^3*d^3 + 752640*a^2*b^8*c^4*d^3 -
4014080*a^3*b^6*c^5*d^3 + 12042240*a^4*b^4*c^6*d^3 - 19267584*a^5*b^2*c^7*d^3) + (7*c*(448*b^18*c*d^6 - 11744
0512*a^9*c^10*d^6 - 16128*a*b^16*c^2*d^6 + 258048*a^2*b^14*c^3*d^6 - 2408448*a^3*b^12*c^4*d^6 + 14450688*a^4*b
^10*c^5*d^6 - 57802752*a^5*b^8*c^6*d^6 + 154140672*a^6*b^6*c^7*d^6 - 264241152*a^7*b^4*c^8*d^6 + 264241152*a^8
*b^2*c^9*d^6))/(d^(5/2)*(b^2 - 4*a*c)^(11/4))))/(d^(5/2)*(b^2 - 4*a*c)^(11/4))))*14i)/(d^(5/2)*(b^2 - 4*a*c)^(
11/4)) + (14*c*atan(((7*c*((b*d + 2*c*d*x)^(1/2)*(12845056*a^6*c^8*d^3 + 3136*b^12*c^2*d^3 - 75264*a*b^10*c^3*
d^3 + 752640*a^2*b^8*c^4*d^3 - 4014080*a^3*b^6*c^5*d^3 + 12042240*a^4*b^4*c^6*d^3 - 19267584*a^5*b^2*c^7*d^3)
- (c*(448*b^18*c*d^6 - 117440512*a^9*c^10*d^6 - 16128*a*b^16*c^2*d^6 + 258048*a^2*b^14*c^3*d^6 - 2408448*a^3*b
^12*c^4*d^6 + 14450688*a^4*b^10*c^5*d^6 - 57802752*a^5*b^8*c^6*d^6 + 154140672*a^6*b^6*c^7*d^6 - 264241152*a^7
*b^4*c^8*d^6 + 264241152*a^8*b^2*c^9*d^6)*7i)/(d^(5/2)*(b^2 - 4*a*c)^(11/4))))/(d^(5/2)*(b^2 - 4*a*c)^(11/4))
+ (7*c*((b*d + 2*c*d*x)^(1/2)*(12845056*a^6*c^8*d^3 + 3136*b^12*c^2*d^3 - 75264*a*b^10*c^3*d^3 + 752640*a^2*b^
8*c^4*d^3 - 4014080*a^3*b^6*c^5*d^3 + 12042240*a^4*b^4*c^6*d^3 - 19267584*a^5*b^2*c^7*d^3) + (c*(448*b^18*c*d^
6 - 117440512*a^9*c^10*d^6 - 16128*a*b^16*c^2*d^6 + 258048*a^2*b^14*c^3*d^6 - 2408448*a^3*b^12*c^4*d^6 + 14450
688*a^4*b^10*c^5*d^6 - 57802752*a^5*b^8*c^6*d^6 + 154140672*a^6*b^6*c^7*d^6 - 264241152*a^7*b^4*c^8*d^6 + 2642
41152*a^8*b^2*c^9*d^6)*7i)/(d^(5/2)*(b^2 - 4*a*c)^(11/4))))/(d^(5/2)*(b^2 - 4*a*c)^(11/4)))/((c*((b*d + 2*c*d*
x)^(1/2)*(12845056*a^6*c^8*d^3 + 3136*b^12*c^2*d^3 - 75264*a*b^10*c^3*d^3 + 752640*a^2*b^8*c^4*d^3 - 4014080*a
^3*b^6*c^5*d^3 + 12042240*a^4*b^4*c^6*d^3 - 19267584*a^5*b^2*c^7*d^3) - (c*(448*b^18*c*d^6 - 117440512*a^9*c^1
0*d^6 - 16128*a*b^16*c^2*d^6 + 258048*a^2*b^14*c^3*d^6 - 2408448*a^3*b^12*c^4*d^6 + 14450688*a^4*b^10*c^5*d^6
- 57802752*a^5*b^8*c^6*d^6 + 154140672*a^6*b^6*c^7*d^6 - 264241152*a^7*b^4*c^8*d^6 + 264241152*a^8*b^2*c^9*d^6
)*7i)/(d^(5/2)*(b^2 - 4*a*c)^(11/4)))*7i)/(d^(5/2)*(b^2 - 4*a*c)^(11/4)) - (c*((b*d + 2*c*d*x)^(1/2)*(12845056
*a^6*c^8*d^3 + 3136*b^12*c^2*d^3 - 75264*a*b^10*c^3*d^3 + 752640*a^2*b^8*c^4*d^3 - 4014080*a^3*b^6*c^5*d^3 + 1
2042240*a^4*b^4*c^6*d^3 - 19267584*a^5*b^2*c^7*d^3) + (c*(448*b^18*c*d^6 - 117440512*a^9*c^10*d^6 - 16128*a*b^
16*c^2*d^6 + 258048*a^2*b^14*c^3*d^6 - 2408448*a^3*b^12*c^4*d^6 + 14450688*a^4*b^10*c^5*d^6 - 57802752*a^5*b^8
*c^6*d^6 + 154140672*a^6*b^6*c^7*d^6 - 264241152*a^7*b^4*c^8*d^6 + 264241152*a^8*b^2*c^9*d^6)*7i)/(d^(5/2)*(b^
2 - 4*a*c)^(11/4)))*7i)/(d^(5/2)*(b^2 - 4*a*c)^(11/4)))))/(d^(5/2)*(b^2 - 4*a*c)^(11/4))

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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(1/(2*c*d*x+b*d)**(5/2)/(c*x**2+b*x+a)**2,x)

[Out]

Timed out

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