3.12.18 \(\int \frac {1}{(b d+2 c d x)^{7/2} (a+b x+c x^2)^2} \, dx\)

Optimal. Leaf size=203 \[ -\frac {18 c \tan ^{-1}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )}{d^{7/2} \left (b^2-4 a c\right )^{13/4}}+\frac {18 c \tanh ^{-1}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )}{d^{7/2} \left (b^2-4 a c\right )^{13/4}}-\frac {36 c}{d^3 \left (b^2-4 a c\right )^3 \sqrt {b d+2 c d x}}-\frac {1}{d \left (b^2-4 a c\right ) \left (a+b x+c x^2\right ) (b d+2 c d x)^{5/2}}-\frac {36 c}{5 d \left (b^2-4 a c\right )^2 (b d+2 c d x)^{5/2}} \]

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Rubi [A]  time = 0.17, antiderivative size = 203, 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 = {687, 693, 694, 329, 298, 203, 206} \begin {gather*} -\frac {36 c}{d^3 \left (b^2-4 a c\right )^3 \sqrt {b d+2 c d x}}-\frac {18 c \tan ^{-1}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )}{d^{7/2} \left (b^2-4 a c\right )^{13/4}}+\frac {18 c \tanh ^{-1}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )}{d^{7/2} \left (b^2-4 a c\right )^{13/4}}-\frac {1}{d \left (b^2-4 a c\right ) \left (a+b x+c x^2\right ) (b d+2 c d x)^{5/2}}-\frac {36 c}{5 d \left (b^2-4 a c\right )^2 (b d+2 c d x)^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((b*d + 2*c*d*x)^(7/2)*(a + b*x + c*x^2)^2),x]

[Out]

(-36*c)/(5*(b^2 - 4*a*c)^2*d*(b*d + 2*c*d*x)^(5/2)) - (36*c)/((b^2 - 4*a*c)^3*d^3*Sqrt[b*d + 2*c*d*x]) - 1/((b
^2 - 4*a*c)*d*(b*d + 2*c*d*x)^(5/2)*(a + b*x + c*x^2)) - (18*c*ArcTan[Sqrt[d*(b + 2*c*x)]/((b^2 - 4*a*c)^(1/4)
*Sqrt[d])])/((b^2 - 4*a*c)^(13/4)*d^(7/2)) + (18*c*ArcTanh[Sqrt[d*(b + 2*c*x)]/((b^2 - 4*a*c)^(1/4)*Sqrt[d])])
/((b^2 - 4*a*c)^(13/4)*d^(7/2))

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)^{7/2} \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)^{5/2} \left (a+b x+c x^2\right )}-\frac {(9 c) \int \frac {1}{(b d+2 c d x)^{7/2} \left (a+b x+c x^2\right )} \, dx}{b^2-4 a c}\\ &=-\frac {36 c}{5 \left (b^2-4 a c\right )^2 d (b d+2 c d x)^{5/2}}-\frac {1}{\left (b^2-4 a c\right ) d (b d+2 c d x)^{5/2} \left (a+b x+c x^2\right )}-\frac {(9 c) \int \frac {1}{(b d+2 c d x)^{3/2} \left (a+b x+c x^2\right )} \, dx}{\left (b^2-4 a c\right )^2 d^2}\\ &=-\frac {36 c}{5 \left (b^2-4 a c\right )^2 d (b d+2 c d x)^{5/2}}-\frac {36 c}{\left (b^2-4 a c\right )^3 d^3 \sqrt {b d+2 c d x}}-\frac {1}{\left (b^2-4 a c\right ) d (b d+2 c d x)^{5/2} \left (a+b x+c x^2\right )}-\frac {(9 c) \int \frac {\sqrt {b d+2 c d x}}{a+b x+c x^2} \, dx}{\left (b^2-4 a c\right )^3 d^4}\\ &=-\frac {36 c}{5 \left (b^2-4 a c\right )^2 d (b d+2 c d x)^{5/2}}-\frac {36 c}{\left (b^2-4 a c\right )^3 d^3 \sqrt {b d+2 c d x}}-\frac {1}{\left (b^2-4 a c\right ) d (b d+2 c d x)^{5/2} \left (a+b x+c x^2\right )}-\frac {9 \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 )^3 d^5}\\ &=-\frac {36 c}{5 \left (b^2-4 a c\right )^2 d (b d+2 c d x)^{5/2}}-\frac {36 c}{\left (b^2-4 a c\right )^3 d^3 \sqrt {b d+2 c d x}}-\frac {1}{\left (b^2-4 a c\right ) d (b d+2 c d x)^{5/2} \left (a+b x+c x^2\right )}-\frac {9 \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 )^3 d^5}\\ &=-\frac {36 c}{5 \left (b^2-4 a c\right )^2 d (b d+2 c d x)^{5/2}}-\frac {36 c}{\left (b^2-4 a c\right )^3 d^3 \sqrt {b d+2 c d x}}-\frac {1}{\left (b^2-4 a c\right ) d (b d+2 c d x)^{5/2} \left (a+b x+c x^2\right )}+\frac {(18 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 )^3 d^3}-\frac {(18 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 )^3 d^3}\\ &=-\frac {36 c}{5 \left (b^2-4 a c\right )^2 d (b d+2 c d x)^{5/2}}-\frac {36 c}{\left (b^2-4 a c\right )^3 d^3 \sqrt {b d+2 c d x}}-\frac {1}{\left (b^2-4 a c\right ) d (b d+2 c d x)^{5/2} \left (a+b x+c x^2\right )}-\frac {18 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 )^{13/4} d^{7/2}}+\frac {18 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 )^{13/4} d^{7/2}}\\ \end {align*}

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Mathematica [C]  time = 0.07, size = 57, normalized size = 0.28 \begin {gather*} -\frac {16 c \, _2F_1\left (-\frac {5}{4},2;-\frac {1}{4};\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{5 d \left (b^2-4 a c\right )^2 (d (b+2 c x))^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/((b*d + 2*c*d*x)^(7/2)*(a + b*x + c*x^2)^2),x]

[Out]

(-16*c*Hypergeometric2F1[-5/4, 2, -1/4, (b + 2*c*x)^2/(b^2 - 4*a*c)])/(5*(b^2 - 4*a*c)^2*d*(d*(b + 2*c*x))^(5/
2))

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IntegrateAlgebraic [C]  time = 0.93, size = 319, normalized size = 1.57 \begin {gather*} \frac {\left (64 a^2 c^2-176 a b^2 c-576 a b c^2 x-576 a c^3 x^2-5 b^4-216 b^3 c x-936 b^2 c^2 x^2-1440 b c^3 x^3-720 c^4 x^4\right ) \sqrt {b d+2 c d x}}{5 d^4 \left (b^2-4 a c\right )^3 (b+2 c x)^3 \left (a+b x+c x^2\right )}+\frac {(9+9 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^{7/2} \left (b^2-4 a c\right )^{13/4}}+\frac {(9+9 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^{7/2} \left (b^2-4 a c\right )^{13/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[1/((b*d + 2*c*d*x)^(7/2)*(a + b*x + c*x^2)^2),x]

[Out]

(Sqrt[b*d + 2*c*d*x]*(-5*b^4 - 176*a*b^2*c + 64*a^2*c^2 - 216*b^3*c*x - 576*a*b*c^2*x - 936*b^2*c^2*x^2 - 576*
a*c^3*x^2 - 1440*b*c^3*x^3 - 720*c^4*x^4))/(5*(b^2 - 4*a*c)^3*d^4*(b + 2*c*x)^3*(a + b*x + c*x^2)) + ((9 + 9*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)^(13/4)*d^(7/2)) + ((9 + 9*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)^(13/4)*d^(7/2))

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fricas [B]  time = 0.53, size = 2919, normalized size = 14.38

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(2*c*d*x+b*d)^(7/2)/(c*x^2+b*x+a)^2,x, algorithm="fricas")

[Out]

-1/5*(180*(8*(b^6*c^4 - 12*a*b^4*c^5 + 48*a^2*b^2*c^6 - 64*a^3*c^7)*d^4*x^5 + 20*(b^7*c^3 - 12*a*b^5*c^4 + 48*
a^2*b^3*c^5 - 64*a^3*b*c^6)*d^4*x^4 + 2*(9*b^8*c^2 - 104*a*b^6*c^3 + 384*a^2*b^4*c^4 - 384*a^3*b^2*c^5 - 256*a
^4*c^6)*d^4*x^3 + (7*b^9*c - 72*a*b^7*c^2 + 192*a^2*b^5*c^3 + 128*a^3*b^3*c^4 - 768*a^4*b*c^5)*d^4*x^2 + (b^10
- 6*a*b^8*c - 24*a^2*b^6*c^2 + 224*a^3*b^4*c^3 - 384*a^4*b^2*c^4)*d^4*x + (a*b^9 - 12*a^2*b^7*c + 48*a^3*b^5*
c^2 - 64*a^4*b^3*c^3)*d^4)*(c^4/((b^26 - 52*a*b^24*c + 1248*a^2*b^22*c^2 - 18304*a^3*b^20*c^3 + 183040*a^4*b^1
8*c^4 - 1317888*a^5*b^16*c^5 + 7028736*a^6*b^14*c^6 - 28114944*a^7*b^12*c^7 + 84344832*a^8*b^10*c^8 - 18743296
0*a^9*b^8*c^9 + 299892736*a^10*b^6*c^10 - 327155712*a^11*b^4*c^11 + 218103808*a^12*b^2*c^12 - 67108864*a^13*c^
13)*d^14))^(1/4)*arctan(-(sqrt((b^14*c^4 - 28*a*b^12*c^5 + 336*a^2*b^10*c^6 - 2240*a^3*b^8*c^7 + 8960*a^4*b^6*
c^8 - 21504*a^5*b^4*c^9 + 28672*a^6*b^2*c^10 - 16384*a^7*c^11)*d^8*sqrt(c^4/((b^26 - 52*a*b^24*c + 1248*a^2*b^
22*c^2 - 18304*a^3*b^20*c^3 + 183040*a^4*b^18*c^4 - 1317888*a^5*b^16*c^5 + 7028736*a^6*b^14*c^6 - 28114944*a^7
*b^12*c^7 + 84344832*a^8*b^10*c^8 - 187432960*a^9*b^8*c^9 + 299892736*a^10*b^6*c^10 - 327155712*a^11*b^4*c^11
+ 218103808*a^12*b^2*c^12 - 67108864*a^13*c^13)*d^14)) + 2*c^7*d*x + b*c^6*d)*(b^6 - 12*a*b^4*c + 48*a^2*b^2*c
^2 - 64*a^3*c^3)*d^3*(c^4/((b^26 - 52*a*b^24*c + 1248*a^2*b^22*c^2 - 18304*a^3*b^20*c^3 + 183040*a^4*b^18*c^4
- 1317888*a^5*b^16*c^5 + 7028736*a^6*b^14*c^6 - 28114944*a^7*b^12*c^7 + 84344832*a^8*b^10*c^8 - 187432960*a^9*
b^8*c^9 + 299892736*a^10*b^6*c^10 - 327155712*a^11*b^4*c^11 + 218103808*a^12*b^2*c^12 - 67108864*a^13*c^13)*d^
14))^(1/4) - (b^6*c^3 - 12*a*b^4*c^4 + 48*a^2*b^2*c^5 - 64*a^3*c^6)*sqrt(2*c*d*x + b*d)*d^3*(c^4/((b^26 - 52*a
*b^24*c + 1248*a^2*b^22*c^2 - 18304*a^3*b^20*c^3 + 183040*a^4*b^18*c^4 - 1317888*a^5*b^16*c^5 + 7028736*a^6*b^
14*c^6 - 28114944*a^7*b^12*c^7 + 84344832*a^8*b^10*c^8 - 187432960*a^9*b^8*c^9 + 299892736*a^10*b^6*c^10 - 327
155712*a^11*b^4*c^11 + 218103808*a^12*b^2*c^12 - 67108864*a^13*c^13)*d^14))^(1/4))/c^4) - 45*(8*(b^6*c^4 - 12*
a*b^4*c^5 + 48*a^2*b^2*c^6 - 64*a^3*c^7)*d^4*x^5 + 20*(b^7*c^3 - 12*a*b^5*c^4 + 48*a^2*b^3*c^5 - 64*a^3*b*c^6)
*d^4*x^4 + 2*(9*b^8*c^2 - 104*a*b^6*c^3 + 384*a^2*b^4*c^4 - 384*a^3*b^2*c^5 - 256*a^4*c^6)*d^4*x^3 + (7*b^9*c
- 72*a*b^7*c^2 + 192*a^2*b^5*c^3 + 128*a^3*b^3*c^4 - 768*a^4*b*c^5)*d^4*x^2 + (b^10 - 6*a*b^8*c - 24*a^2*b^6*c
^2 + 224*a^3*b^4*c^3 - 384*a^4*b^2*c^4)*d^4*x + (a*b^9 - 12*a^2*b^7*c + 48*a^3*b^5*c^2 - 64*a^4*b^3*c^3)*d^4)*
(c^4/((b^26 - 52*a*b^24*c + 1248*a^2*b^22*c^2 - 18304*a^3*b^20*c^3 + 183040*a^4*b^18*c^4 - 1317888*a^5*b^16*c^
5 + 7028736*a^6*b^14*c^6 - 28114944*a^7*b^12*c^7 + 84344832*a^8*b^10*c^8 - 187432960*a^9*b^8*c^9 + 299892736*a
^10*b^6*c^10 - 327155712*a^11*b^4*c^11 + 218103808*a^12*b^2*c^12 - 67108864*a^13*c^13)*d^14))^(1/4)*log(729*(b
^20 - 40*a*b^18*c + 720*a^2*b^16*c^2 - 7680*a^3*b^14*c^3 + 53760*a^4*b^12*c^4 - 258048*a^5*b^10*c^5 + 860160*a
^6*b^8*c^6 - 1966080*a^7*b^6*c^7 + 2949120*a^8*b^4*c^8 - 2621440*a^9*b^2*c^9 + 1048576*a^10*c^10)*d^11*(c^4/((
b^26 - 52*a*b^24*c + 1248*a^2*b^22*c^2 - 18304*a^3*b^20*c^3 + 183040*a^4*b^18*c^4 - 1317888*a^5*b^16*c^5 + 702
8736*a^6*b^14*c^6 - 28114944*a^7*b^12*c^7 + 84344832*a^8*b^10*c^8 - 187432960*a^9*b^8*c^9 + 299892736*a^10*b^6
*c^10 - 327155712*a^11*b^4*c^11 + 218103808*a^12*b^2*c^12 - 67108864*a^13*c^13)*d^14))^(3/4) + 729*sqrt(2*c*d*
x + b*d)*c^3) + 45*(8*(b^6*c^4 - 12*a*b^4*c^5 + 48*a^2*b^2*c^6 - 64*a^3*c^7)*d^4*x^5 + 20*(b^7*c^3 - 12*a*b^5*
c^4 + 48*a^2*b^3*c^5 - 64*a^3*b*c^6)*d^4*x^4 + 2*(9*b^8*c^2 - 104*a*b^6*c^3 + 384*a^2*b^4*c^4 - 384*a^3*b^2*c^
5 - 256*a^4*c^6)*d^4*x^3 + (7*b^9*c - 72*a*b^7*c^2 + 192*a^2*b^5*c^3 + 128*a^3*b^3*c^4 - 768*a^4*b*c^5)*d^4*x^
2 + (b^10 - 6*a*b^8*c - 24*a^2*b^6*c^2 + 224*a^3*b^4*c^3 - 384*a^4*b^2*c^4)*d^4*x + (a*b^9 - 12*a^2*b^7*c + 48
*a^3*b^5*c^2 - 64*a^4*b^3*c^3)*d^4)*(c^4/((b^26 - 52*a*b^24*c + 1248*a^2*b^22*c^2 - 18304*a^3*b^20*c^3 + 18304
0*a^4*b^18*c^4 - 1317888*a^5*b^16*c^5 + 7028736*a^6*b^14*c^6 - 28114944*a^7*b^12*c^7 + 84344832*a^8*b^10*c^8 -
187432960*a^9*b^8*c^9 + 299892736*a^10*b^6*c^10 - 327155712*a^11*b^4*c^11 + 218103808*a^12*b^2*c^12 - 6710886
4*a^13*c^13)*d^14))^(1/4)*log(-729*(b^20 - 40*a*b^18*c + 720*a^2*b^16*c^2 - 7680*a^3*b^14*c^3 + 53760*a^4*b^12
*c^4 - 258048*a^5*b^10*c^5 + 860160*a^6*b^8*c^6 - 1966080*a^7*b^6*c^7 + 2949120*a^8*b^4*c^8 - 2621440*a^9*b^2*
c^9 + 1048576*a^10*c^10)*d^11*(c^4/((b^26 - 52*a*b^24*c + 1248*a^2*b^22*c^2 - 18304*a^3*b^20*c^3 + 183040*a^4*
b^18*c^4 - 1317888*a^5*b^16*c^5 + 7028736*a^6*b^14*c^6 - 28114944*a^7*b^12*c^7 + 84344832*a^8*b^10*c^8 - 18743
2960*a^9*b^8*c^9 + 299892736*a^10*b^6*c^10 - 327155712*a^11*b^4*c^11 + 218103808*a^12*b^2*c^12 - 67108864*a^13
*c^13)*d^14))^(3/4) + 729*sqrt(2*c*d*x + b*d)*c^3) + (720*c^4*x^4 + 1440*b*c^3*x^3 + 5*b^4 + 176*a*b^2*c - 64*
a^2*c^2 + 72*(13*b^2*c^2 + 8*a*c^3)*x^2 + 72*(3*b^3*c + 8*a*b*c^2)*x)*sqrt(2*c*d*x + b*d))/(8*(b^6*c^4 - 12*a*
b^4*c^5 + 48*a^2*b^2*c^6 - 64*a^3*c^7)*d^4*x^5 + 20*(b^7*c^3 - 12*a*b^5*c^4 + 48*a^2*b^3*c^5 - 64*a^3*b*c^6)*d
^4*x^4 + 2*(9*b^8*c^2 - 104*a*b^6*c^3 + 384*a^2*b^4*c^4 - 384*a^3*b^2*c^5 - 256*a^4*c^6)*d^4*x^3 + (7*b^9*c -
72*a*b^7*c^2 + 192*a^2*b^5*c^3 + 128*a^3*b^3*c^4 - 768*a^4*b*c^5)*d^4*x^2 + (b^10 - 6*a*b^8*c - 24*a^2*b^6*c^2
+ 224*a^3*b^4*c^3 - 384*a^4*b^2*c^4)*d^4*x + (a*b^9 - 12*a^2*b^7*c + 48*a^3*b^5*c^2 - 64*a^4*b^3*c^3)*d^4)

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giac [B]  time = 0.33, size = 779, normalized size = 3.84 \begin {gather*} \frac {9 \, \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^{8} d^{5} - 16 \, a b^{6} c d^{5} + 96 \, a^{2} b^{4} c^{2} d^{5} - 256 \, a^{3} b^{2} c^{3} d^{5} + 256 \, a^{4} c^{4} d^{5}} + \frac {9 \, \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^{8} d^{5} - 16 \, a b^{6} c d^{5} + 96 \, a^{2} b^{4} c^{2} d^{5} - 256 \, a^{3} b^{2} c^{3} d^{5} + 256 \, a^{4} c^{4} d^{5}} - \frac {9 \, {\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^{8} d^{5} - 16 \, \sqrt {2} a b^{6} c d^{5} + 96 \, \sqrt {2} a^{2} b^{4} c^{2} d^{5} - 256 \, \sqrt {2} a^{3} b^{2} c^{3} d^{5} + 256 \, \sqrt {2} a^{4} c^{4} d^{5}} + \frac {9 \, {\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^{8} d^{5} - 16 \, \sqrt {2} a b^{6} c d^{5} + 96 \, \sqrt {2} a^{2} b^{4} c^{2} d^{5} - 256 \, \sqrt {2} a^{3} b^{2} c^{3} d^{5} + 256 \, \sqrt {2} a^{4} c^{4} d^{5}} + \frac {4 \, {\left (2 \, c d x + b d\right )}^{\frac {3}{2}} c}{{\left (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}\right )} {\left (b^{2} d^{2} - 4 \, a c d^{2} - {\left (2 \, c d x + b d\right )}^{2}\right )}} - \frac {16 \, {\left (b^{2} c d^{2} - 4 \, a c^{2} d^{2} + 10 \, {\left (2 \, c d x + b d\right )}^{2} c\right )}}{5 \, {\left (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}\right )} {\left (2 \, c d x + b d\right )}^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(2*c*d*x+b*d)^(7/2)/(c*x^2+b*x+a)^2,x, algorithm="giac")

[Out]

9*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^8*d^5 - 16*a*b^6*c*d^5 + 96*a^2*b^4*c^2*d^5 - 256*a^3*b^2*c^3*d^
5 + 256*a^4*c^4*d^5) + 9*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^8*d^5 - 16*a*b^6*c*d^5 + 96*a^2*b^4*c^2*
d^5 - 256*a^3*b^2*c^3*d^5 + 256*a^4*c^4*d^5) - 9*(-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^8*d^5 - 16*sqrt(2)*a*b
^6*c*d^5 + 96*sqrt(2)*a^2*b^4*c^2*d^5 - 256*sqrt(2)*a^3*b^2*c^3*d^5 + 256*sqrt(2)*a^4*c^4*d^5) + 9*(-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^8*d^5 - 16*sqrt(2)*a*b^6*c*d^5 + 96*sqrt(2)*a^2*b^4*c^2*d^5 - 256*sqrt(2)*a^3*b^2*
c^3*d^5 + 256*sqrt(2)*a^4*c^4*d^5) + 4*(2*c*d*x + b*d)^(3/2)*c/((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)*(b^2*d^2 - 4*a*c*d^2 - (2*c*d*x + b*d)^2)) - 16/5*(b^2*c*d^2 - 4*a*c^2*d^2 + 10*(2*c*d*x +
b*d)^2*c)/((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)*(2*c*d*x + b*d)^(5/2))

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maple [B]  time = 0.07, size = 433, normalized size = 2.13 \begin {gather*} -\frac {16 c}{5 \left (4 a c -b^{2}\right )^{2} \left (2 c d x +b d \right )^{\frac {5}{2}} d}-\frac {9 \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 )^{3} \left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}} d^{3}}+\frac {9 \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 )^{3} \left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}} d^{3}}+\frac {9 \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 )^{3} \left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}} d^{3}}+\frac {4 \left (2 c d x +b d \right )^{\frac {3}{2}} c}{\left (4 a c -b^{2}\right )^{3} \left (4 c^{2} d^{2} x^{2}+4 b c \,d^{2} x +4 a c \,d^{2}\right ) d^{3}}+\frac {32 c}{\left (4 a c -b^{2}\right )^{3} \sqrt {2 c d x +b d}\, d^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(2*c*d*x+b*d)^(7/2)/(c*x^2+b*x+a)^2,x)

[Out]

4*c/d^3/(4*a*c-b^2)^3*(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)+9/2*c/d^3/(4*a*c-b^2)^3*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)))+
9*c/d^3/(4*a*c-b^2)^3*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)-9*c/d^3/(4*a*c-b^2)^3*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)-16/5*c/d/(4*a*c-b^2)^2/(2*c*d*x+b*d)^(5/2)+32*c/d^3/(4*a*c-b^2)^3/(2*c*d*x+b*d)^(1/2)

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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)^(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 = 0.89, size = 336, normalized size = 1.66 \begin {gather*} \frac {\frac {36\,c\,{\left (b\,d+2\,c\,d\,x\right )}^4}{{\left (b^2\,d-4\,a\,c\,d\right )}^3}+\frac {16\,c\,d}{5\,\left (4\,a\,c-b^2\right )}-\frac {144\,c\,{\left (b\,d+2\,c\,d\,x\right )}^2}{5\,d\,{\left (4\,a\,c-b^2\right )}^2}}{{\left (b\,d+2\,c\,d\,x\right )}^{5/2}\,\left (b^2\,d^2-4\,a\,c\,d^2\right )-{\left (b\,d+2\,c\,d\,x\right )}^{9/2}}-\frac {18\,c\,\mathrm {atan}\left (\frac {b^6\,\sqrt {b\,d+2\,c\,d\,x}-64\,a^3\,c^3\,\sqrt {b\,d+2\,c\,d\,x}+48\,a^2\,b^2\,c^2\,\sqrt {b\,d+2\,c\,d\,x}-12\,a\,b^4\,c\,\sqrt {b\,d+2\,c\,d\,x}}{\sqrt {d}\,{\left (b^2-4\,a\,c\right )}^{13/4}}\right )}{d^{7/2}\,{\left (b^2-4\,a\,c\right )}^{13/4}}-\frac {c\,\mathrm {atan}\left (\frac {b^6\,\sqrt {b\,d+2\,c\,d\,x}\,1{}\mathrm {i}-a^3\,c^3\,\sqrt {b\,d+2\,c\,d\,x}\,64{}\mathrm {i}+a^2\,b^2\,c^2\,\sqrt {b\,d+2\,c\,d\,x}\,48{}\mathrm {i}-a\,b^4\,c\,\sqrt {b\,d+2\,c\,d\,x}\,12{}\mathrm {i}}{\sqrt {d}\,{\left (b^2-4\,a\,c\right )}^{13/4}}\right )\,18{}\mathrm {i}}{d^{7/2}\,{\left (b^2-4\,a\,c\right )}^{13/4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((b*d + 2*c*d*x)^(7/2)*(a + b*x + c*x^2)^2),x)

[Out]

((36*c*(b*d + 2*c*d*x)^4)/(b^2*d - 4*a*c*d)^3 + (16*c*d)/(5*(4*a*c - b^2)) - (144*c*(b*d + 2*c*d*x)^2)/(5*d*(4
*a*c - b^2)^2))/((b*d + 2*c*d*x)^(5/2)*(b^2*d^2 - 4*a*c*d^2) - (b*d + 2*c*d*x)^(9/2)) - (18*c*atan((b^6*(b*d +
2*c*d*x)^(1/2) - 64*a^3*c^3*(b*d + 2*c*d*x)^(1/2) + 48*a^2*b^2*c^2*(b*d + 2*c*d*x)^(1/2) - 12*a*b^4*c*(b*d +
2*c*d*x)^(1/2))/(d^(1/2)*(b^2 - 4*a*c)^(13/4))))/(d^(7/2)*(b^2 - 4*a*c)^(13/4)) - (c*atan((b^6*(b*d + 2*c*d*x)
^(1/2)*1i - a^3*c^3*(b*d + 2*c*d*x)^(1/2)*64i + a^2*b^2*c^2*(b*d + 2*c*d*x)^(1/2)*48i - a*b^4*c*(b*d + 2*c*d*x
)^(1/2)*12i)/(d^(1/2)*(b^2 - 4*a*c)^(13/4)))*18i)/(d^(7/2)*(b^2 - 4*a*c)^(13/4))

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

[Out]

Timed out

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