3.1294 \(\int \frac {1}{(b d+2 c d x)^{7/2} (a+b x+c x^2)} \, dx\)

Optimal. Leaf size=159 \[ \frac {2 \tan ^{-1}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )}{d^{7/2} \left (b^2-4 a c\right )^{9/4}}-\frac {2 \tanh ^{-1}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )}{d^{7/2} \left (b^2-4 a c\right )^{9/4}}+\frac {4}{d^3 \left (b^2-4 a c\right )^2 \sqrt {b d+2 c d x}}+\frac {4}{5 d \left (b^2-4 a c\right ) (b d+2 c d x)^{5/2}} \]

[Out]

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

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Rubi [A]  time = 0.13, antiderivative size = 159, 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 = {693, 694, 329, 298, 203, 206} \[ \frac {4}{d^3 \left (b^2-4 a c\right )^2 \sqrt {b d+2 c d x}}+\frac {2 \tan ^{-1}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )}{d^{7/2} \left (b^2-4 a c\right )^{9/4}}-\frac {2 \tanh ^{-1}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )}{d^{7/2} \left (b^2-4 a c\right )^{9/4}}+\frac {4}{5 d \left (b^2-4 a c\right ) (b d+2 c d x)^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

4/(5*(b^2 - 4*a*c)*d*(b*d + 2*c*d*x)^(5/2)) + 4/((b^2 - 4*a*c)^2*d^3*Sqrt[b*d + 2*c*d*x]) + (2*ArcTan[Sqrt[d*(
b + 2*c*x)]/((b^2 - 4*a*c)^(1/4)*Sqrt[d])])/((b^2 - 4*a*c)^(9/4)*d^(7/2)) - (2*ArcTanh[Sqrt[d*(b + 2*c*x)]/((b
^2 - 4*a*c)^(1/4)*Sqrt[d])])/((b^2 - 4*a*c)^(9/4)*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 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 )} \, dx &=\frac {4}{5 \left (b^2-4 a c\right ) d (b d+2 c d x)^{5/2}}+\frac {\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 ) d^2}\\ &=\frac {4}{5 \left (b^2-4 a c\right ) d (b d+2 c d x)^{5/2}}+\frac {4}{\left (b^2-4 a c\right )^2 d^3 \sqrt {b d+2 c d x}}+\frac {\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^4}\\ &=\frac {4}{5 \left (b^2-4 a c\right ) d (b d+2 c d x)^{5/2}}+\frac {4}{\left (b^2-4 a c\right )^2 d^3 \sqrt {b d+2 c d x}}+\frac {\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 c \left (b^2-4 a c\right )^2 d^5}\\ &=\frac {4}{5 \left (b^2-4 a c\right ) d (b d+2 c d x)^{5/2}}+\frac {4}{\left (b^2-4 a c\right )^2 d^3 \sqrt {b d+2 c d x}}+\frac {\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 )}{c \left (b^2-4 a c\right )^2 d^5}\\ &=\frac {4}{5 \left (b^2-4 a c\right ) d (b d+2 c d x)^{5/2}}+\frac {4}{\left (b^2-4 a c\right )^2 d^3 \sqrt {b d+2 c d x}}-\frac {2 \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^3}+\frac {2 \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^3}\\ &=\frac {4}{5 \left (b^2-4 a c\right ) d (b d+2 c d x)^{5/2}}+\frac {4}{\left (b^2-4 a c\right )^2 d^3 \sqrt {b d+2 c d x}}+\frac {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 )^{9/4} d^{7/2}}-\frac {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 )^{9/4} d^{7/2}}\\ \end {align*}

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Mathematica [C]  time = 0.06, size = 56, normalized size = 0.35 \[ \frac {4 \, _2F_1\left (-\frac {5}{4},1;-\frac {1}{4};\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{5 d \left (b^2-4 a c\right ) (d (b+2 c x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.79, size = 1693, normalized size = 10.65 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/5*(20*(8*(b^4*c^3 - 8*a*b^2*c^4 + 16*a^2*c^5)*d^4*x^3 + 12*(b^5*c^2 - 8*a*b^3*c^3 + 16*a^2*b*c^4)*d^4*x^2 +
6*(b^6*c - 8*a*b^4*c^2 + 16*a^2*b^2*c^3)*d^4*x + (b^7 - 8*a*b^5*c + 16*a^2*b^3*c^2)*d^4)*(1/((b^18 - 36*a*b^1
6*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 - 58
9824*a^7*b^4*c^7 + 589824*a^8*b^2*c^8 - 262144*a^9*c^9)*d^14))^(1/4)*arctan(sqrt((b^10 - 20*a*b^8*c + 160*a^2*
b^6*c^2 - 640*a^3*b^4*c^3 + 1280*a^4*b^2*c^4 - 1024*a^5*c^5)*d^8*sqrt(1/((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 + 58
9824*a^8*b^2*c^8 - 262144*a^9*c^9)*d^14)) + 2*c*d*x + b*d)*(b^4 - 8*a*b^2*c + 16*a^2*c^2)*d^3*(1/((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^14))^(1/4) - (b^4 - 8*a*b^2*c + 16*a^2*c^2)*sqrt
(2*c*d*x + b*d)*d^3*(1/((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 - 1290
24*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^14))^(1/4))
+ 5*(8*(b^4*c^3 - 8*a*b^2*c^4 + 16*a^2*c^5)*d^4*x^3 + 12*(b^5*c^2 - 8*a*b^3*c^3 + 16*a^2*b*c^4)*d^4*x^2 + 6*(b
^6*c - 8*a*b^4*c^2 + 16*a^2*b^2*c^3)*d^4*x + (b^7 - 8*a*b^5*c + 16*a^2*b^3*c^2)*d^4)*(1/((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^14))^(1/4)*log((b^14 - 28*a*b^12*c + 336*a^2*b^10*c^2 - 2
240*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^11*(1/((b^18 - 3
6*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^14))^(3/4) + sqrt(2*c*d*x + b*d)) - 5*(8*(b^4
*c^3 - 8*a*b^2*c^4 + 16*a^2*c^5)*d^4*x^3 + 12*(b^5*c^2 - 8*a*b^3*c^3 + 16*a^2*b*c^4)*d^4*x^2 + 6*(b^6*c - 8*a*
b^4*c^2 + 16*a^2*b^2*c^3)*d^4*x + (b^7 - 8*a*b^5*c + 16*a^2*b^3*c^2)*d^4)*(1/((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^14))^(1/4)*log(-(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^11*(1/((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 - 58982
4*a^7*b^4*c^7 + 589824*a^8*b^2*c^8 - 262144*a^9*c^9)*d^14))^(3/4) + sqrt(2*c*d*x + b*d)) - 8*(10*c^2*x^2 + 10*
b*c*x + 3*b^2 - 2*a*c)*sqrt(2*c*d*x + b*d))/(8*(b^4*c^3 - 8*a*b^2*c^4 + 16*a^2*c^5)*d^4*x^3 + 12*(b^5*c^2 - 8*
a*b^3*c^3 + 16*a^2*b*c^4)*d^4*x^2 + 6*(b^6*c - 8*a*b^4*c^2 + 16*a^2*b^2*c^3)*d^4*x + (b^7 - 8*a*b^5*c + 16*a^2
*b^3*c^2)*d^4)

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giac [B]  time = 0.27, size = 605, normalized size = 3.81 \[ -\frac {\sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {3}{4}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} + 2 \, \sqrt {2 \, c d x + b d}\right )}}{2 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}}}\right )}{b^{6} d^{5} - 12 \, a b^{4} c d^{5} + 48 \, a^{2} b^{2} c^{2} d^{5} - 64 \, a^{3} c^{3} d^{5}} - \frac {\sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {3}{4}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} - 2 \, \sqrt {2 \, c d x + b d}\right )}}{2 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}}}\right )}{b^{6} d^{5} - 12 \, a b^{4} c d^{5} + 48 \, a^{2} b^{2} c^{2} d^{5} - 64 \, a^{3} c^{3} d^{5}} + \frac {{\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {3}{4}} \log \left (2 \, c d x + b d + \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} \sqrt {2 \, c d x + b d} + \sqrt {-b^{2} d^{2} + 4 \, a c d^{2}}\right )}{\sqrt {2} b^{6} d^{5} - 12 \, \sqrt {2} a b^{4} c d^{5} + 48 \, \sqrt {2} a^{2} b^{2} c^{2} d^{5} - 64 \, \sqrt {2} a^{3} c^{3} d^{5}} - \frac {{\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {3}{4}} \log \left (2 \, c d x + b d - \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} \sqrt {2 \, c d x + b d} + \sqrt {-b^{2} d^{2} + 4 \, a c d^{2}}\right )}{\sqrt {2} b^{6} d^{5} - 12 \, \sqrt {2} a b^{4} c d^{5} + 48 \, \sqrt {2} a^{2} b^{2} c^{2} d^{5} - 64 \, \sqrt {2} a^{3} c^{3} d^{5}} + \frac {4 \, {\left (b^{2} d^{2} - 4 \, a c d^{2} + 5 \, {\left (2 \, c d x + b d\right )}^{2}\right )}}{5 \, {\left (b^{4} d^{3} - 8 \, a b^{2} c d^{3} + 16 \, a^{2} c^{2} d^{3}\right )} {\left (2 \, c d x + b d\right )}^{\frac {5}{2}}} \]

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),x, algorithm="giac")

[Out]

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

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maple [B]  time = 0.06, size = 369, normalized size = 2.32 \[ -\frac {4}{5 \left (4 a c -b^{2}\right ) \left (2 c d x +b d \right )^{\frac {5}{2}} d}-\frac {\sqrt {2}\, \arctan \left (-\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )}{\left (4 a c -b^{2}\right )^{2} \left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}} d^{3}}+\frac {\sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )}{\left (4 a c -b^{2}\right )^{2} \left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}} d^{3}}+\frac {\sqrt {2}\, \ln \left (\frac {2 c d x +b d -\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a c \,d^{2}-b^{2} d^{2}}}{2 c d x +b d +\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a c \,d^{2}-b^{2} d^{2}}}\right )}{2 \left (4 a c -b^{2}\right )^{2} \left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}} d^{3}}+\frac {4}{\left (4 a c -b^{2}\right )^{2} \sqrt {2 c d x +b d}\, d^{3}} \]

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),x)

[Out]

1/2/d^3/(4*a*c-b^2)^2/(4*a*c*d^2-b^2*d^2)^(1/4)*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)))+1/d^3/(4*a*c-b^2)^2/(4*a*c*d^2-b^2*d^2)^(1/4)*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)-1/d^3/(4*a*c-b^2)^2/(4*a*c*d^2-b^2*d^2)^(1/4)*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)-4/5/d/(4*a*c-b^2)/(2*c*d*x+b*d)^(5/2)+4/d^3/(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)^(7/2)/(c*x^2+b*x+a),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 = 0.26, size = 230, normalized size = 1.45 \[ \frac {\frac {4}{5\,\left (b^2\,d-4\,a\,c\,d\right )}+\frac {4\,{\left (b\,d+2\,c\,d\,x\right )}^2}{d\,{\left (b^2\,d-4\,a\,c\,d\right )}^2}}{{\left (b\,d+2\,c\,d\,x\right )}^{5/2}}+\frac {2\,\mathrm {atan}\left (\frac {b^4\,\sqrt {b\,d+2\,c\,d\,x}+16\,a^2\,c^2\,\sqrt {b\,d+2\,c\,d\,x}-8\,a\,b^2\,c\,\sqrt {b\,d+2\,c\,d\,x}}{\sqrt {d}\,{\left (b^2-4\,a\,c\right )}^{9/4}}\right )}{d^{7/2}\,{\left (b^2-4\,a\,c\right )}^{9/4}}+\frac {\mathrm {atan}\left (\frac {b^4\,\sqrt {b\,d+2\,c\,d\,x}\,1{}\mathrm {i}+a^2\,c^2\,\sqrt {b\,d+2\,c\,d\,x}\,16{}\mathrm {i}-a\,b^2\,c\,\sqrt {b\,d+2\,c\,d\,x}\,8{}\mathrm {i}}{\sqrt {d}\,{\left (b^2-4\,a\,c\right )}^{9/4}}\right )\,2{}\mathrm {i}}{d^{7/2}\,{\left (b^2-4\,a\,c\right )}^{9/4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(4/(5*(b^2*d - 4*a*c*d)) + (4*(b*d + 2*c*d*x)^2)/(d*(b^2*d - 4*a*c*d)^2))/(b*d + 2*c*d*x)^(5/2) + (2*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^(7/2)*(b^2 - 4*a*c)^(9/4)) + (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)))*2i)/(d^(7/2)*(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)**(7/2)/(c*x**2+b*x+a),x)

[Out]

Timed out

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