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

Optimal. Leaf size=145 \[ -2 d^{7/2} \left (b^2-4 a c\right )^{5/4} \tan ^{-1}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )-2 d^{7/2} \left (b^2-4 a c\right )^{5/4} \tanh ^{-1}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )+4 d^3 \left (b^2-4 a c\right ) \sqrt {b d+2 c d x}+\frac {4}{5} d (b d+2 c d x)^{5/2} \]

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Rubi [A]  time = 0.12, antiderivative size = 145, 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 = {692, 694, 329, 212, 206, 203} \begin {gather*} 4 d^3 \left (b^2-4 a c\right ) \sqrt {b d+2 c d x}-2 d^{7/2} \left (b^2-4 a c\right )^{5/4} \tan ^{-1}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )-2 d^{7/2} \left (b^2-4 a c\right )^{5/4} \tanh ^{-1}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )+\frac {4}{5} d (b d+2 c d x)^{5/2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 692

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(2*d*(d + e*x)^(m -
1)*(a + b*x + c*x^2)^(p + 1))/(b*(m + 2*p + 1)), x] + Dist[(d^2*(m - 1)*(b^2 - 4*a*c))/(b^2*(m + 2*p + 1)), In
t[(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] && GtQ[m, 1] && NeQ[m + 2*p + 1, 0] && (IntegerQ[2*p] || (IntegerQ[m] &
& RationalQ[p]) || OddQ[m])

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 {(b d+2 c d x)^{7/2}}{a+b x+c x^2} \, dx &=\frac {4}{5} d (b d+2 c d x)^{5/2}+\left (\left (b^2-4 a c\right ) d^2\right ) \int \frac {(b d+2 c d x)^{3/2}}{a+b x+c x^2} \, dx\\ &=4 \left (b^2-4 a c\right ) d^3 \sqrt {b d+2 c d x}+\frac {4}{5} d (b d+2 c d x)^{5/2}+\left (\left (b^2-4 a c\right )^2 d^4\right ) \int \frac {1}{\sqrt {b d+2 c d x} \left (a+b x+c x^2\right )} \, dx\\ &=4 \left (b^2-4 a c\right ) d^3 \sqrt {b d+2 c d x}+\frac {4}{5} d (b d+2 c d x)^{5/2}+\frac {\left (\left (b^2-4 a c\right )^2 d^3\right ) \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 c}\\ &=4 \left (b^2-4 a c\right ) d^3 \sqrt {b d+2 c d x}+\frac {4}{5} d (b d+2 c d x)^{5/2}+\frac {\left (\left (b^2-4 a c\right )^2 d^3\right ) \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 )}{c}\\ &=4 \left (b^2-4 a c\right ) d^3 \sqrt {b d+2 c d x}+\frac {4}{5} d (b d+2 c d x)^{5/2}-\left (2 \left (b^2-4 a c\right )^{3/2} d^4\right ) \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 (2 \left (b^2-4 a c\right )^{3/2} d^4\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {b^2-4 a c} d+x^2} \, dx,x,\sqrt {d (b+2 c x)}\right )\\ &=4 \left (b^2-4 a c\right ) d^3 \sqrt {b d+2 c d x}+\frac {4}{5} d (b d+2 c d x)^{5/2}-2 \left (b^2-4 a c\right )^{5/4} d^{7/2} \tan ^{-1}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )-2 \left (b^2-4 a c\right )^{5/4} d^{7/2} \tanh ^{-1}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )\\ \end {align*}

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Mathematica [A]  time = 0.13, size = 141, normalized size = 0.97 \begin {gather*} \frac {2 d^3 \sqrt {d (b+2 c x)} \left (4 \sqrt {b+2 c x} \left (2 c \left (c x^2-5 a\right )+3 b^2+2 b c x\right )-5 \left (b^2-4 a c\right )^{5/4} \tan ^{-1}\left (\frac {\sqrt {b+2 c x}}{\sqrt [4]{b^2-4 a c}}\right )-5 \left (b^2-4 a c\right )^{5/4} \tanh ^{-1}\left (\frac {\sqrt {b+2 c x}}{\sqrt [4]{b^2-4 a c}}\right )\right )}{5 \sqrt {b+2 c x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [C]  time = 0.33, size = 246, normalized size = 1.70 \begin {gather*} \frac {8}{5} \sqrt {b d+2 c d x} \left (-10 a c d^3+3 b^2 d^3+2 b c d^3 x+2 c^2 d^3 x^2\right )+(1-i) d^{7/2} \left (b^2-4 a c\right )^{5/4} \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 )-(1-i) d^{7/2} \left (b^2-4 a c\right )^{5/4} \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 ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(8*Sqrt[b*d + 2*c*d*x]*(3*b^2*d^3 - 10*a*c*d^3 + 2*b*c*d^3*x + 2*c^2*d^3*x^2))/5 + (1 - I)*(b^2 - 4*a*c)^(5/4)
*d^(7/2)*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]] - (1 - I)*(b^2 - 4*a*c)^(5/4)*d^(7/2)*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))]

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fricas [B]  time = 0.44, size = 712, normalized size = 4.91 \begin {gather*} \frac {8}{5} \, {\left (2 \, c^{2} d^{3} x^{2} + 2 \, b c d^{3} x + {\left (3 \, b^{2} - 10 \, a c\right )} d^{3}\right )} \sqrt {2 \, c d x + b d} + 4 \, \left ({\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^{14}\right )^{\frac {1}{4}} \arctan \left (-\frac {\left ({\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^{14}\right )^{\frac {3}{4}} \sqrt {2 \, c d x + b d} {\left (b^{2} - 4 \, a c\right )} d^{3} + \left ({\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^{14}\right )^{\frac {3}{4}} \sqrt {2 \, {\left (b^{4} c - 8 \, a b^{2} c^{2} + 16 \, a^{2} c^{3}\right )} d^{7} x + {\left (b^{5} - 8 \, a b^{3} c + 16 \, a^{2} b c^{2}\right )} d^{7} + \sqrt {{\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^{14}}}}{{\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^{14}}\right ) + \left ({\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^{14}\right )^{\frac {1}{4}} \log \left (-\sqrt {2 \, c d x + b d} {\left (b^{2} - 4 \, a c\right )} d^{3} + \left ({\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^{14}\right )^{\frac {1}{4}}\right ) - \left ({\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^{14}\right )^{\frac {1}{4}} \log \left (-\sqrt {2 \, c d x + b d} {\left (b^{2} - 4 \, a c\right )} d^{3} - \left ({\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^{14}\right )^{\frac {1}{4}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

8/5*(2*c^2*d^3*x^2 + 2*b*c*d^3*x + (3*b^2 - 10*a*c)*d^3)*sqrt(2*c*d*x + b*d) + 4*((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^14)^(1/4)*arctan(-(((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^14)^(3/4)*sqrt(2*c*d*x + b*d)*(b^2 - 4*a*c)*
d^3 + ((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^14)^(3/4)*s
qrt(2*(b^4*c - 8*a*b^2*c^2 + 16*a^2*c^3)*d^7*x + (b^5 - 8*a*b^3*c + 16*a^2*b*c^2)*d^7 + 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^14)))/((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^14)) + ((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^14)^(1/4)*log(-sqrt(2*c*d*x + b*d)*(b^2 - 4*a*c)*d^3 + (
(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^14)^(1/4)) - ((b^1
0 - 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^14)^(1/4)*log(-sqrt(2*
c*d*x + b*d)*(b^2 - 4*a*c)*d^3 - ((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^14)^(1/4))

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giac [B]  time = 0.24, size = 531, normalized size = 3.66 \begin {gather*} 4 \, \sqrt {2 \, c d x + b d} b^{2} d^{3} - 16 \, \sqrt {2 \, c d x + b d} a c d^{3} + \frac {4}{5} \, {\left (2 \, c d x + b d\right )}^{\frac {5}{2}} d - {\left (\sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} b^{2} d^{3} - 4 \, \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} a c d^{3}\right )} \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 ) - {\left (\sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} b^{2} d^{3} - 4 \, \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} a c d^{3}\right )} \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 ) - \frac {1}{2} \, {\left (\sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} b^{2} d^{3} - 4 \, \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} a c d^{3}\right )} \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 ) + \frac {1}{2} \, {\left (\sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} b^{2} d^{3} - 4 \, \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} a c d^{3}\right )} \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 ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.06, size = 922, normalized size = 6.36 \begin {gather*} -\frac {16 \sqrt {2}\, a^{2} c^{2} d^{5} \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 {3}{4}}}+\frac {16 \sqrt {2}\, a^{2} c^{2} d^{5} \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 {3}{4}}}+\frac {8 \sqrt {2}\, a^{2} c^{2} d^{5} \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 )}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {3}{4}}}+\frac {8 \sqrt {2}\, a \,b^{2} c \,d^{5} \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 {3}{4}}}-\frac {8 \sqrt {2}\, a \,b^{2} c \,d^{5} \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 {3}{4}}}-\frac {4 \sqrt {2}\, a \,b^{2} c \,d^{5} \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 )}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {3}{4}}}-\frac {\sqrt {2}\, b^{4} d^{5} \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 {3}{4}}}+\frac {\sqrt {2}\, b^{4} d^{5} \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 {3}{4}}}+\frac {\sqrt {2}\, b^{4} d^{5} \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 {3}{4}}}-16 \sqrt {2 c d x +b d}\, a c \,d^{3}+4 \sqrt {2 c d x +b d}\, b^{2} d^{3}+\frac {4 \left (2 c d x +b d \right )^{\frac {5}{2}} d}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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)^(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 = 0.64, size = 167, normalized size = 1.15 \begin {gather*} \frac {4\,d\,{\left (b\,d+2\,c\,d\,x\right )}^{5/2}}{5}-4\,d^3\,\sqrt {b\,d+2\,c\,d\,x}\,\left (4\,a\,c-b^2\right )-2\,d^{7/2}\,\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 )\,{\left (b^2-4\,a\,c\right )}^{5/4}+d^{7/2}\,\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 )\,{\left (b^2-4\,a\,c\right )}^{5/4}\,2{}\mathrm {i} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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