3.13.87 \(\int \frac {(b d+2 c d x)^{7/2}}{a+b x+c x^2} \, dx\) [1287]

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

[Out]

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

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Rubi [A]
time = 0.09, 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 = {706, 708, 335, 218, 212, 209} \begin {gather*} -2 d^{7/2} \left (b^2-4 a c\right )^{5/4} \text {ArcTan}\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} \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 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 218

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] &&  !Gt
Q[a/b, 0]

Rule 335

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 706

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 708

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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 [C] Result contains complex when optimal does not.
time = 0.32, size = 215, normalized size = 1.48 \begin {gather*} \frac {\left (\frac {1}{5}-\frac {i}{5}\right ) (d (b+2 c x))^{7/2} \left ((2+2 i) \sqrt {b+2 c x} \left (5 b^2-20 a c+(b+2 c x)^2\right )+5 \left (b^2-4 a c\right )^{5/4} \tan ^{-1}\left (1-\frac {(1+i) \sqrt {b+2 c x}}{\sqrt [4]{b^2-4 a c}}\right )-5 \left (b^2-4 a c\right )^{5/4} \tan ^{-1}\left (1+\frac {(1+i) \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 {(1+i) \sqrt [4]{b^2-4 a c} \sqrt {b+2 c x}}{\sqrt {b^2-4 a c}+i (b+2 c x)}\right )\right )}{(b+2 c x)^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((1/5 - I/5)*(d*(b + 2*c*x))^(7/2)*((2 + 2*I)*Sqrt[b + 2*c*x]*(5*b^2 - 20*a*c + (b + 2*c*x)^2) + 5*(b^2 - 4*a*
c)^(5/4)*ArcTan[1 - ((1 + I)*Sqrt[b + 2*c*x])/(b^2 - 4*a*c)^(1/4)] - 5*(b^2 - 4*a*c)^(5/4)*ArcTan[1 + ((1 + I)
*Sqrt[b + 2*c*x])/(b^2 - 4*a*c)^(1/4)] - 5*(b^2 - 4*a*c)^(5/4)*ArcTanh[((1 + I)*(b^2 - 4*a*c)^(1/4)*Sqrt[b + 2
*c*x])/(Sqrt[b^2 - 4*a*c] + I*(b + 2*c*x))]))/(b + 2*c*x)^(7/2)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(301\) vs. \(2(119)=238\).
time = 0.84, size = 302, normalized size = 2.08

method	result	size
derivativedivides	\(4 d \left (-4 a c \,d^{2} \sqrt {2 c d x +b d}+b^{2} d^{2} \sqrt {2 c d x +b d}+\frac {\left (2 c d x +b d \right )^{\frac {5}{2}}}{5}+\frac {d^{4} \left (16 a^{2} c^{2}-8 a c \,b^{2}+b^{4}\right ) \sqrt {2}\, \left (\ln \left (\frac {2 c d x +b d +\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a c \,d^{2}-b^{2} d^{2}}}{2 c d x +b d -\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a c \,d^{2}-b^{2} d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 \left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {3}{4}}}\right )\)	\(302\)
default	\(4 d \left (-4 a c \,d^{2} \sqrt {2 c d x +b d}+b^{2} d^{2} \sqrt {2 c d x +b d}+\frac {\left (2 c d x +b d \right )^{\frac {5}{2}}}{5}+\frac {d^{4} \left (16 a^{2} c^{2}-8 a c \,b^{2}+b^{4}\right ) \sqrt {2}\, \left (\ln \left (\frac {2 c d x +b d +\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a c \,d^{2}-b^{2} d^{2}}}{2 c d x +b d -\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a c \,d^{2}-b^{2} d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 \left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {3}{4}}}\right )\)	\(302\)

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,method=_RETURNVERBOSE)

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 712 vs. \(2 (119) = 238\).
time = 3.78, 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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Sympy [F(-1)] Timed out
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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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 531 vs. \(2 (119) = 238\).
time = 1.55, 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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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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