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

Optimal. Leaf size=129 \[ \frac {2 \tan ^{-1}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )}{d^{3/2} \left (b^2-4 a c\right )^{5/4}}-\frac {2 \tanh ^{-1}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )}{d^{3/2} \left (b^2-4 a c\right )^{5/4}}+\frac {4}{d \left (b^2-4 a c\right ) \sqrt {b d+2 c d x}} \]

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Rubi [A]  time = 0.11, antiderivative size = 129, normalized size of antiderivative = 1.00, number of steps used = 6, 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} \begin {gather*} \frac {2 \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 )^{5/4}}-\frac {2 \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 )^{5/4}}+\frac {4}{d \left (b^2-4 a c\right ) \sqrt {b d+2 c d x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [C]  time = 0.05, size = 54, normalized size = 0.42 \begin {gather*} \frac {4 \, _2F_1\left (-\frac {1}{4},1;\frac {3}{4};\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{d \left (b^2-4 a c\right ) \sqrt {d (b+2 c x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [C]  time = 0.38, size = 230, normalized size = 1.78 \begin {gather*} -\frac {(1+i) \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^{3/2} \left (b^2-4 a c\right )^{5/4}}-\frac {(1+i) \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^{3/2} \left (b^2-4 a c\right )^{5/4}}+\frac {4 \sqrt {b d+2 c d x}}{d^2 \left (b^2-4 a c\right ) (b+2 c x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(4*Sqrt[b*d + 2*c*d*x])/((b^2 - 4*a*c)*d^2*(b + 2*c*x)) - ((1 + I)*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)^(5/4)*d^(3/2)) - ((1 + I)*ArcTanh[((1 + I)*(b^2 - 4*a*c)^(1/4)*Sqrt[b*d + 2*c*d*x])/(S
qrt[d]*(I*b + Sqrt[b^2 - 4*a*c] + (2*I)*c*x))])/((b^2 - 4*a*c)^(5/4)*d^(3/2))

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fricas [B]  time = 0.45, size = 826, normalized size = 6.40 \begin {gather*} \frac {4 \, {\left (2 \, {\left (b^{2} c - 4 \, a c^{2}\right )} d^{2} x + {\left (b^{3} - 4 \, a b c\right )} d^{2}\right )} \left (\frac {1}{{\left (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}\right )} d^{6}}\right )^{\frac {1}{4}} \arctan \left (-\sqrt {{\left (b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}\right )} d^{4} \sqrt {\frac {1}{{\left (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}\right )} d^{6}}} + 2 \, c d x + b d} {\left (b^{2} - 4 \, a c\right )} d \left (\frac {1}{{\left (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}\right )} d^{6}}\right )^{\frac {1}{4}} + \sqrt {2 \, c d x + b d} {\left (b^{2} - 4 \, a c\right )} d \left (\frac {1}{{\left (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}\right )} d^{6}}\right )^{\frac {1}{4}}\right ) - {\left (2 \, {\left (b^{2} c - 4 \, a c^{2}\right )} d^{2} x + {\left (b^{3} - 4 \, a b c\right )} d^{2}\right )} \left (\frac {1}{{\left (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}\right )} d^{6}}\right )^{\frac {1}{4}} \log \left ({\left (b^{8} - 16 \, a b^{6} c + 96 \, a^{2} b^{4} c^{2} - 256 \, a^{3} b^{2} c^{3} + 256 \, a^{4} c^{4}\right )} d^{5} \left (\frac {1}{{\left (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}\right )} d^{6}}\right )^{\frac {3}{4}} + \sqrt {2 \, c d x + b d}\right ) + {\left (2 \, {\left (b^{2} c - 4 \, a c^{2}\right )} d^{2} x + {\left (b^{3} - 4 \, a b c\right )} d^{2}\right )} \left (\frac {1}{{\left (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}\right )} d^{6}}\right )^{\frac {1}{4}} \log \left (-{\left (b^{8} - 16 \, a b^{6} c + 96 \, a^{2} b^{4} c^{2} - 256 \, a^{3} b^{2} c^{3} + 256 \, a^{4} c^{4}\right )} d^{5} \left (\frac {1}{{\left (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}\right )} d^{6}}\right )^{\frac {3}{4}} + \sqrt {2 \, c d x + b d}\right ) + 4 \, \sqrt {2 \, c d x + b d}}{2 \, {\left (b^{2} c - 4 \, a c^{2}\right )} d^{2} x + {\left (b^{3} - 4 \, a b c\right )} d^{2}} \end {gather*}

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),x, algorithm="fricas")

[Out]

(4*(2*(b^2*c - 4*a*c^2)*d^2*x + (b^3 - 4*a*b*c)*d^2)*(1/((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^6))^(1/4)*arctan(-sqrt((b^6 - 12*a*b^4*c + 48*a^2*b^2*c^2 - 64*a^3*c^3)
*d^4*sqrt(1/((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^6)) +
2*c*d*x + b*d)*(b^2 - 4*a*c)*d*(1/((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^6))^(1/4) + sqrt(2*c*d*x + b*d)*(b^2 - 4*a*c)*d*(1/((b^10 - 20*a*b^8*c + 160*a^2*b^6*c^2 - 6
40*a^3*b^4*c^3 + 1280*a^4*b^2*c^4 - 1024*a^5*c^5)*d^6))^(1/4)) - (2*(b^2*c - 4*a*c^2)*d^2*x + (b^3 - 4*a*b*c)*
d^2)*(1/((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^6))^(1/4)
*log((b^8 - 16*a*b^6*c + 96*a^2*b^4*c^2 - 256*a^3*b^2*c^3 + 256*a^4*c^4)*d^5*(1/((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^6))^(3/4) + sqrt(2*c*d*x + b*d)) + (2*(b^2*c -
4*a*c^2)*d^2*x + (b^3 - 4*a*b*c)*d^2)*(1/((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^6))^(1/4)*log(-(b^8 - 16*a*b^6*c + 96*a^2*b^4*c^2 - 256*a^3*b^2*c^3 + 256*a^4*c^4)*d^5
*(1/((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^6))^(3/4) + s
qrt(2*c*d*x + b*d)) + 4*sqrt(2*c*d*x + b*d))/(2*(b^2*c - 4*a*c^2)*d^2*x + (b^3 - 4*a*b*c)*d^2)

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giac [B]  time = 0.23, size = 497, normalized size = 3.85 \begin {gather*} -\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^{4} d^{3} - 8 \, a b^{2} c d^{3} + 16 \, a^{2} c^{2} d^{3}} - \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^{4} d^{3} - 8 \, a b^{2} c d^{3} + 16 \, a^{2} c^{2} d^{3}} + \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^{4} d^{3} - 8 \, \sqrt {2} a b^{2} c d^{3} + 16 \, \sqrt {2} a^{2} c^{2} d^{3}} - \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^{4} d^{3} - 8 \, \sqrt {2} a b^{2} c d^{3} + 16 \, \sqrt {2} a^{2} c^{2} d^{3}} + \frac {4}{{\left (b^{2} d - 4 \, a c d\right )} \sqrt {2 \, c d x + b d}} \end {gather*}

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),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^4*d^3 - 8*a*b^2*c*d^3 + 16*a^2*c^2*d^3) - 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^4*d^3 - 8*a*b^2*c*d^3 + 16*a^2*c^2*d^3) + (-b^2*d^2 + 4*a*c*d^2)^(3/4)*log(2*c*d*x + b*d + s
qrt(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^4*d^3 - 8*sqr
t(2)*a*b^2*c*d^3 + 16*sqrt(2)*a^2*c^2*d^3) - (-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^4*d^3 - 8*sqrt(2)*a*b^2*c*d^
3 + 16*sqrt(2)*a^2*c^2*d^3) + 4/((b^2*d - 4*a*c*d)*sqrt(2*c*d*x + b*d))

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maple [B]  time = 0.06, size = 341, normalized size = 2.64 \begin {gather*} \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 ) \left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}} 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 ) \left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}} d}-\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 ) \left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}} d}-\frac {4}{\left (4 a c -b^{2}\right ) \sqrt {2 c d x +b d}\, d} \end {gather*}

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),x)

[Out]

-1/2/d/(4*a*c-b^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/(4*a*c-b^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/(4*a*c-b^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/d/(4*a*c-b^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 \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)^(3/2)/(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 = 0.61, size = 153, normalized size = 1.19 \begin {gather*} \frac {4}{\sqrt {b\,d+2\,c\,d\,x}\,\left (b^2\,d-4\,a\,c\,d\right )}+\frac {2\,\mathrm {atan}\left (\frac {b^2\,\sqrt {b\,d+2\,c\,d\,x}-4\,a\,c\,\sqrt {b\,d+2\,c\,d\,x}}{\sqrt {d}\,{\left (b^2-4\,a\,c\right )}^{5/4}}\right )}{d^{3/2}\,{\left (b^2-4\,a\,c\right )}^{5/4}}+\frac {\mathrm {atan}\left (\frac {b^2\,\sqrt {b\,d+2\,c\,d\,x}\,1{}\mathrm {i}-a\,c\,\sqrt {b\,d+2\,c\,d\,x}\,4{}\mathrm {i}}{\sqrt {d}\,{\left (b^2-4\,a\,c\right )}^{5/4}}\right )\,2{}\mathrm {i}}{d^{3/2}\,{\left (b^2-4\,a\,c\right )}^{5/4}} \end {gather*}

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)),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (d \left (b + 2 c x\right )\right )^{\frac {3}{2}} \left (a + b x + c x^{2}\right )}\, dx \end {gather*}

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),x)

[Out]

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

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