3.12.14 \(\int \frac {\sqrt {b d+2 c d x}}{(a+b x+c x^2)^2} \, dx\)

Optimal. Leaf size=143 \[ -\frac {(b d+2 c d x)^{3/2}}{d \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}-\frac {2 c \sqrt {d} \tan ^{-1}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{5/4}}+\frac {2 c \sqrt {d} \tanh ^{-1}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{5/4}} \]

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Rubi [A]  time = 0.10, antiderivative size = 143, 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 = {687, 694, 329, 298, 203, 206} \begin {gather*} -\frac {(b d+2 c d x)^{3/2}}{d \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}-\frac {2 c \sqrt {d} \tan ^{-1}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{5/4}}+\frac {2 c \sqrt {d} \tanh ^{-1}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{5/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [C]  time = 0.04, size = 57, normalized size = 0.40 \begin {gather*} \frac {16 c (d (b+2 c x))^{3/2} \, _2F_1\left (\frac {3}{4},2;\frac {7}{4};\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{3 d \left (b^2-4 a c\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(16*c*(d*(b + 2*c*x))^(3/2)*Hypergeometric2F1[3/4, 2, 7/4, (b + 2*c*x)^2/(b^2 - 4*a*c)])/(3*(b^2 - 4*a*c)^2*d)

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IntegrateAlgebraic [C]  time = 0.47, size = 240, normalized size = 1.68 \begin {gather*} \frac {(-b-2 c x) \sqrt {b d+2 c d x}}{\left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}+\frac {(1+i) c \sqrt {d} \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 )}{\left (b^2-4 a c\right )^{5/4}}+\frac {(1+i) c \sqrt {d} \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 )}{\left (b^2-4 a c\right )^{5/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.45, size = 915, normalized size = 6.40 \begin {gather*} -\frac {4 \, \left (\frac {c^{4} d^{2}}{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 )^{\frac {1}{4}} {\left (a b^{2} - 4 \, a^{2} c + {\left (b^{2} c - 4 \, a c^{2}\right )} x^{2} + {\left (b^{3} - 4 \, a b c\right )} x\right )} \arctan \left (\frac {{\left (b^{2} c^{3} - 4 \, a c^{4}\right )} \left (\frac {c^{4} d^{2}}{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 )^{\frac {1}{4}} \sqrt {2 \, c d x + b d} d - \sqrt {2 \, c^{7} d^{3} x + b c^{6} d^{3} + {\left (b^{6} c^{4} - 12 \, a b^{4} c^{5} + 48 \, a^{2} b^{2} c^{6} - 64 \, a^{3} c^{7}\right )} \sqrt {\frac {c^{4} d^{2}}{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^{2}} \left (\frac {c^{4} d^{2}}{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 )^{\frac {1}{4}} {\left (b^{2} - 4 \, a c\right )}}{c^{4} d^{2}}\right ) - \left (\frac {c^{4} d^{2}}{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 )^{\frac {1}{4}} {\left (a b^{2} - 4 \, a^{2} c + {\left (b^{2} c - 4 \, a c^{2}\right )} x^{2} + {\left (b^{3} - 4 \, a b c\right )} x\right )} \log \left (\sqrt {2 \, c d x + b d} c^{3} d + {\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 )} \left (\frac {c^{4} d^{2}}{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 )^{\frac {3}{4}}\right ) + \left (\frac {c^{4} d^{2}}{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 )^{\frac {1}{4}} {\left (a b^{2} - 4 \, a^{2} c + {\left (b^{2} c - 4 \, a c^{2}\right )} x^{2} + {\left (b^{3} - 4 \, a b c\right )} x\right )} \log \left (\sqrt {2 \, c d x + b d} c^{3} d - {\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 )} \left (\frac {c^{4} d^{2}}{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 )^{\frac {3}{4}}\right ) + \sqrt {2 \, c d x + b d} {\left (2 \, c x + b\right )}}{a b^{2} - 4 \, a^{2} c + {\left (b^{2} c - 4 \, a c^{2}\right )} x^{2} + {\left (b^{3} - 4 \, a b c\right )} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(4*(c^4*d^2/(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))^(1/4)*
(a*b^2 - 4*a^2*c + (b^2*c - 4*a*c^2)*x^2 + (b^3 - 4*a*b*c)*x)*arctan(((b^2*c^3 - 4*a*c^4)*(c^4*d^2/(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))^(1/4)*sqrt(2*c*d*x + b*d)*d -
sqrt(2*c^7*d^3*x + b*c^6*d^3 + (b^6*c^4 - 12*a*b^4*c^5 + 48*a^2*b^2*c^6 - 64*a^3*c^7)*sqrt(c^4*d^2/(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^2)*(c^4*d^2/(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))^(1/4)*(b^2 - 4*a*c))/(c^4*d^2)) - (
c^4*d^2/(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))^(1/4)*(a*b^
2 - 4*a^2*c + (b^2*c - 4*a*c^2)*x^2 + (b^3 - 4*a*b*c)*x)*log(sqrt(2*c*d*x + b*d)*c^3*d + (b^8 - 16*a*b^6*c + 9
6*a^2*b^4*c^2 - 256*a^3*b^2*c^3 + 256*a^4*c^4)*(c^4*d^2/(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))^(3/4)) + (c^4*d^2/(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))^(1/4)*(a*b^2 - 4*a^2*c + (b^2*c - 4*a*c^2)*x^2 + (b^3 - 4*a*b*c)*x)*log(sq
rt(2*c*d*x + b*d)*c^3*d - (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)*(c^4*d^2/(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))^(3/4)) + sqrt(2*c*d*x + b*
d)*(2*c*x + b))/(a*b^2 - 4*a^2*c + (b^2*c - 4*a*c^2)*x^2 + (b^3 - 4*a*b*c)*x)

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giac [B]  time = 0.23, size = 504, normalized size = 3.52 \begin {gather*} \frac {\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^{4} d - 8 \, a b^{2} c d + 16 \, a^{2} c^{2} d} + \frac {\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^{4} d - 8 \, a b^{2} c d + 16 \, a^{2} c^{2} d} - \frac {{\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^{4} d - 8 \, \sqrt {2} a b^{2} c d + 16 \, \sqrt {2} a^{2} c^{2} d} + \frac {{\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^{4} d - 8 \, \sqrt {2} a b^{2} c d + 16 \, \sqrt {2} a^{2} c^{2} d} + \frac {4 \, {\left (2 \, c d x + b d\right )}^{\frac {3}{2}} c d}{{\left (b^{2} d^{2} - 4 \, a c d^{2} - {\left (2 \, c d x + b d\right )}^{2}\right )} {\left (b^{2} - 4 \, a c\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.06, size = 344, normalized size = 2.41 \begin {gather*} -\frac {\sqrt {2}\, c \,d^{3} \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 \,d^{2}-b^{2} d^{2}\right )^{\frac {5}{4}}}+\frac {\sqrt {2}\, c \,d^{3} \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 \,d^{2}-b^{2} d^{2}\right )^{\frac {5}{4}}}+\frac {\sqrt {2}\, c \,d^{3} \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 \,d^{2}-b^{2} d^{2}\right )^{\frac {5}{4}}}+\frac {4 \left (2 c d x +b d \right )^{\frac {3}{2}} c \,d^{3}}{\left (4 a c \,d^{2}-b^{2} d^{2}\right ) \left (4 c^{2} d^{2} x^{2}+4 b c \,d^{2} x +4 a c \,d^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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((2*c*d*x+b*d)^(1/2)/(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 = 0.14, size = 179, normalized size = 1.25 \begin {gather*} \frac {2\,c\,\sqrt {d}\,\mathrm {atanh}\left (\frac {\sqrt {b\,d+2\,c\,d\,x}\,\left (16\,a^2\,c^2-8\,a\,b^2\,c+b^4\right )}{\sqrt {d}\,{\left (b^2-4\,a\,c\right )}^{9/4}}\right )}{{\left (b^2-4\,a\,c\right )}^{5/4}}-\frac {2\,c\,\sqrt {d}\,\mathrm {atan}\left (\frac {\sqrt {b\,d+2\,c\,d\,x}\,\left (16\,a^2\,c^2-8\,a\,b^2\,c+b^4\right )}{\sqrt {d}\,{\left (b^2-4\,a\,c\right )}^{9/4}}\right )}{{\left (b^2-4\,a\,c\right )}^{5/4}}+\frac {4\,c\,d\,{\left (b\,d+2\,c\,d\,x\right )}^{3/2}}{\left (4\,a\,c-b^2\right )\,\left ({\left (b\,d+2\,c\,d\,x\right )}^2-b^2\,d^2+4\,a\,c\,d^2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [B]  time = 53.14, size = 279, normalized size = 1.95 \begin {gather*} \frac {16 c d^{3} \left (b d + 2 c d x\right )^{\frac {3}{2}}}{64 a^{2} c^{2} d^{4} - 32 a b^{2} c d^{4} + 16 a c d^{2} \left (b d + 2 c d x\right )^{2} + 4 b^{4} d^{4} - 4 b^{2} d^{2} \left (b d + 2 c d x\right )^{2}} + 16 c d^{3} \operatorname {RootSum} {\left (t^{4} \left (67108864 a^{5} c^{5} d^{10} - 83886080 a^{4} b^{2} c^{4} d^{10} + 41943040 a^{3} b^{4} c^{3} d^{10} - 10485760 a^{2} b^{6} c^{2} d^{10} + 1310720 a b^{8} c d^{10} - 65536 b^{10} d^{10}\right ) + 1, \left (t \mapsto t \log {\left (1048576 t^{3} a^{4} c^{4} d^{8} - 1048576 t^{3} a^{3} b^{2} c^{3} d^{8} + 393216 t^{3} a^{2} b^{4} c^{2} d^{8} - 65536 t^{3} a b^{6} c d^{8} + 4096 t^{3} b^{8} d^{8} + \sqrt {b d + 2 c d x} \right )} \right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

16*c*d**3*(b*d + 2*c*d*x)**(3/2)/(64*a**2*c**2*d**4 - 32*a*b**2*c*d**4 + 16*a*c*d**2*(b*d + 2*c*d*x)**2 + 4*b*
*4*d**4 - 4*b**2*d**2*(b*d + 2*c*d*x)**2) + 16*c*d**3*RootSum(_t**4*(67108864*a**5*c**5*d**10 - 83886080*a**4*
b**2*c**4*d**10 + 41943040*a**3*b**4*c**3*d**10 - 10485760*a**2*b**6*c**2*d**10 + 1310720*a*b**8*c*d**10 - 655
36*b**10*d**10) + 1, Lambda(_t, _t*log(1048576*_t**3*a**4*c**4*d**8 - 1048576*_t**3*a**3*b**2*c**3*d**8 + 3932
16*_t**3*a**2*b**4*c**2*d**8 - 65536*_t**3*a*b**6*c*d**8 + 4096*_t**3*b**8*d**8 + sqrt(b*d + 2*c*d*x))))

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