### 3.1299 $$\int \frac{(b d+2 c d x)^{7/2}}{(a+b x+c x^2)^2} \, dx$$

Optimal. Leaf size=150 $-10 c d^{7/2} \sqrt [4]{b^2-4 a c} \tan ^{-1}\left (\frac{\sqrt{d (b+2 c x)}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right )-10 c d^{7/2} \sqrt [4]{b^2-4 a c} \tanh ^{-1}\left (\frac{\sqrt{d (b+2 c x)}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right )-\frac{d (b d+2 c d x)^{5/2}}{a+b x+c x^2}+20 c d^3 \sqrt{b d+2 c d x}$

[Out]

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

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Rubi [A]  time = 0.119651, antiderivative size = 150, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 26, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.269, Rules used = {686, 692, 694, 329, 212, 206, 203} $-10 c d^{7/2} \sqrt [4]{b^2-4 a c} \tan ^{-1}\left (\frac{\sqrt{d (b+2 c x)}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right )-10 c d^{7/2} \sqrt [4]{b^2-4 a c} \tanh ^{-1}\left (\frac{\sqrt{d (b+2 c x)}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right )-\frac{d (b d+2 c d x)^{5/2}}{a+b x+c x^2}+20 c d^3 \sqrt{b d+2 c d x}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

Rule 686

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(d*(d + e*x)^(m - 1)*
(a + b*x + c*x^2)^(p + 1))/(b*(p + 1)), x] - Dist[(d*e*(m - 1))/(b*(p + 1)), Int[(d + e*x)^(m - 2)*(a + b*x +
c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e}, 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] && IntegerQ[2*p]

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]

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 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 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 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])

Rubi steps

\begin{align*} \int \frac{(b d+2 c d x)^{7/2}}{\left (a+b x+c x^2\right )^2} \, dx &=-\frac{d (b d+2 c d x)^{5/2}}{a+b x+c x^2}+\left (5 c d^2\right ) \int \frac{(b d+2 c d x)^{3/2}}{a+b x+c x^2} \, dx\\ &=20 c d^3 \sqrt{b d+2 c d x}-\frac{d (b d+2 c d x)^{5/2}}{a+b x+c x^2}+\left (5 c \left (b^2-4 a c\right ) d^4\right ) \int \frac{1}{\sqrt{b d+2 c d x} \left (a+b x+c x^2\right )} \, dx\\ &=20 c d^3 \sqrt{b d+2 c d x}-\frac{d (b d+2 c d x)^{5/2}}{a+b x+c x^2}+\frac{1}{2} \left (5 \left (b^2-4 a c\right ) 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 )\\ &=20 c d^3 \sqrt{b d+2 c d x}-\frac{d (b d+2 c d x)^{5/2}}{a+b x+c x^2}+\left (5 \left (b^2-4 a c\right ) 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 )\\ &=20 c d^3 \sqrt{b d+2 c d x}-\frac{d (b d+2 c d x)^{5/2}}{a+b x+c x^2}-\left (10 c \sqrt{b^2-4 a c} 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 (10 c \sqrt{b^2-4 a c} 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 )\\ &=20 c d^3 \sqrt{b d+2 c d x}-\frac{d (b d+2 c d x)^{5/2}}{a+b x+c x^2}-10 c \sqrt [4]{b^2-4 a c} d^{7/2} \tan ^{-1}\left (\frac{\sqrt{d (b+2 c x)}}{\sqrt [4]{b^2-4 a c} \sqrt{d}}\right )-10 c \sqrt [4]{b^2-4 a c} 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*}

Mathematica [A]  time = 0.193973, size = 168, normalized size = 1.12 $-\frac{d^3 \sqrt{d (b+2 c x)} \left (\sqrt{b+2 c x} \left (-4 c \left (5 a+4 c x^2\right )+b^2-16 b c x\right )+10 c \sqrt [4]{b^2-4 a c} (a+x (b+c x)) \tan ^{-1}\left (\frac{\sqrt{b+2 c x}}{\sqrt [4]{b^2-4 a c}}\right )+10 c \sqrt [4]{b^2-4 a c} (a+x (b+c x)) \tanh ^{-1}\left (\frac{\sqrt{b+2 c x}}{\sqrt [4]{b^2-4 a c}}\right )\right )}{\sqrt{b+2 c x} (a+x (b+c x))}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

-((d^3*Sqrt[d*(b + 2*c*x)]*(Sqrt[b + 2*c*x]*(b^2 - 16*b*c*x - 4*c*(5*a + 4*c*x^2)) + 10*c*(b^2 - 4*a*c)^(1/4)*
(a + x*(b + c*x))*ArcTan[Sqrt[b + 2*c*x]/(b^2 - 4*a*c)^(1/4)] + 10*c*(b^2 - 4*a*c)^(1/4)*(a + x*(b + c*x))*Arc
Tanh[Sqrt[b + 2*c*x]/(b^2 - 4*a*c)^(1/4)]))/(Sqrt[b + 2*c*x]*(a + x*(b + c*x))))

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Maple [B]  time = 0.198, size = 693, normalized size = 4.6 \begin{align*} 16\,c{d}^{3}\sqrt{2\,cdx+bd}+16\,{\frac{{c}^{2}{d}^{5}a\sqrt{2\,cdx+bd}}{4\,{c}^{2}{d}^{2}{x}^{2}+4\,bc{d}^{2}x+4\,ac{d}^{2}}}-4\,{\frac{c{d}^{5}{b}^{2}\sqrt{2\,cdx+bd}}{4\,{c}^{2}{d}^{2}{x}^{2}+4\,bc{d}^{2}x+4\,ac{d}^{2}}}-20\,{\frac{{c}^{2}{d}^{5}\sqrt{2}a}{ \left ( 4\,ac{d}^{2}-{b}^{2}{d}^{2} \right ) ^{3/4}}\arctan \left ({\frac{\sqrt{2}\sqrt{2\,cdx+bd}}{\sqrt [4]{4\,ac{d}^{2}-{b}^{2}{d}^{2}}}}+1 \right ) }+5\,{\frac{c{d}^{5}\sqrt{2}{b}^{2}}{ \left ( 4\,ac{d}^{2}-{b}^{2}{d}^{2} \right ) ^{3/4}}\arctan \left ({\frac{\sqrt{2}\sqrt{2\,cdx+bd}}{\sqrt [4]{4\,ac{d}^{2}-{b}^{2}{d}^{2}}}}+1 \right ) }+20\,{\frac{{c}^{2}{d}^{5}\sqrt{2}a}{ \left ( 4\,ac{d}^{2}-{b}^{2}{d}^{2} \right ) ^{3/4}}\arctan \left ( -{\frac{\sqrt{2}\sqrt{2\,cdx+bd}}{\sqrt [4]{4\,ac{d}^{2}-{b}^{2}{d}^{2}}}}+1 \right ) }-5\,{\frac{c{d}^{5}\sqrt{2}{b}^{2}}{ \left ( 4\,ac{d}^{2}-{b}^{2}{d}^{2} \right ) ^{3/4}}\arctan \left ( -{\frac{\sqrt{2}\sqrt{2\,cdx+bd}}{\sqrt [4]{4\,ac{d}^{2}-{b}^{2}{d}^{2}}}}+1 \right ) }-10\,{\frac{{c}^{2}{d}^{5}\sqrt{2}a}{ \left ( 4\,ac{d}^{2}-{b}^{2}{d}^{2} \right ) ^{3/4}}\ln \left ({\frac{2\,cdx+bd+\sqrt [4]{4\,ac{d}^{2}-{b}^{2}{d}^{2}}\sqrt{2\,cdx+bd}\sqrt{2}+\sqrt{4\,ac{d}^{2}-{b}^{2}{d}^{2}}}{2\,cdx+bd-\sqrt [4]{4\,ac{d}^{2}-{b}^{2}{d}^{2}}\sqrt{2\,cdx+bd}\sqrt{2}+\sqrt{4\,ac{d}^{2}-{b}^{2}{d}^{2}}}} \right ) }+{\frac{5\,c{d}^{5}\sqrt{2}{b}^{2}}{2}\ln \left ({ \left ( 2\,cdx+bd+\sqrt [4]{4\,ac{d}^{2}-{b}^{2}{d}^{2}}\sqrt{2\,cdx+bd}\sqrt{2}+\sqrt{4\,ac{d}^{2}-{b}^{2}{d}^{2}} \right ) \left ( 2\,cdx+bd-\sqrt [4]{4\,ac{d}^{2}-{b}^{2}{d}^{2}}\sqrt{2\,cdx+bd}\sqrt{2}+\sqrt{4\,ac{d}^{2}-{b}^{2}{d}^{2}} \right ) ^{-1}} \right ) \left ( 4\,ac{d}^{2}-{b}^{2}{d}^{2} \right ) ^{-{\frac{3}{4}}}} \end{align*}

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)^2,x)

[Out]

16*c*d^3*(2*c*d*x+b*d)^(1/2)+16*c^2*d^5*(2*c*d*x+b*d)^(1/2)/(4*c^2*d^2*x^2+4*b*c*d^2*x+4*a*c*d^2)*a-4*c*d^5*(2
*c*d*x+b*d)^(1/2)/(4*c^2*d^2*x^2+4*b*c*d^2*x+4*a*c*d^2)*b^2-20*c^2*d^5/(4*a*c*d^2-b^2*d^2)^(3/4)*2^(1/2)*arcta
n(2^(1/2)/(4*a*c*d^2-b^2*d^2)^(1/4)*(2*c*d*x+b*d)^(1/2)+1)*a+5*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^2+20*c^2*d^5/(4*a*c*d^2-b^2*d^2)^(3/4)*2^(1/2)*arct
an(-2^(1/2)/(4*a*c*d^2-b^2*d^2)^(1/4)*(2*c*d*x+b*d)^(1/2)+1)*a-5*c*d^5/(4*a*c*d^2-b^2*d^2)^(3/4)*2^(1/2)*arcta
n(-2^(1/2)/(4*a*c*d^2-b^2*d^2)^(1/4)*(2*c*d*x+b*d)^(1/2)+1)*b^2-10*c^2*d^5/(4*a*c*d^2-b^2*d^2)^(3/4)*2^(1/2)*l
n((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+5/2*c*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^2

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.86165, size = 803, normalized size = 5.35 \begin{align*} -\frac{20 \, \left ({\left (b^{2} c^{4} - 4 \, a c^{5}\right )} d^{14}\right )^{\frac{1}{4}}{\left (c x^{2} + b x + a\right )} \arctan \left (\frac{\left ({\left (b^{2} c^{4} - 4 \, a c^{5}\right )} d^{14}\right )^{\frac{3}{4}} \sqrt{2 \, c d x + b d} c d^{3} - \left ({\left (b^{2} c^{4} - 4 \, a c^{5}\right )} d^{14}\right )^{\frac{3}{4}} \sqrt{2 \, c^{3} d^{7} x + b c^{2} d^{7} + \sqrt{{\left (b^{2} c^{4} - 4 \, a c^{5}\right )} d^{14}}}}{{\left (b^{2} c^{4} - 4 \, a c^{5}\right )} d^{14}}\right ) + 5 \, \left ({\left (b^{2} c^{4} - 4 \, a c^{5}\right )} d^{14}\right )^{\frac{1}{4}}{\left (c x^{2} + b x + a\right )} \log \left (5 \, \sqrt{2 \, c d x + b d} c d^{3} + 5 \, \left ({\left (b^{2} c^{4} - 4 \, a c^{5}\right )} d^{14}\right )^{\frac{1}{4}}\right ) - 5 \, \left ({\left (b^{2} c^{4} - 4 \, a c^{5}\right )} d^{14}\right )^{\frac{1}{4}}{\left (c x^{2} + b x + a\right )} \log \left (5 \, \sqrt{2 \, c d x + b d} c d^{3} - 5 \, \left ({\left (b^{2} c^{4} - 4 \, a c^{5}\right )} d^{14}\right )^{\frac{1}{4}}\right ) -{\left (16 \, c^{2} d^{3} x^{2} + 16 \, b c d^{3} x -{\left (b^{2} - 20 \, a c\right )} d^{3}\right )} \sqrt{2 \, c d x + b d}}{c x^{2} + b x + a} \end{align*}

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)^2,x, algorithm="fricas")

[Out]

-(20*((b^2*c^4 - 4*a*c^5)*d^14)^(1/4)*(c*x^2 + b*x + a)*arctan((((b^2*c^4 - 4*a*c^5)*d^14)^(3/4)*sqrt(2*c*d*x
+ b*d)*c*d^3 - ((b^2*c^4 - 4*a*c^5)*d^14)^(3/4)*sqrt(2*c^3*d^7*x + b*c^2*d^7 + sqrt((b^2*c^4 - 4*a*c^5)*d^14))
)/((b^2*c^4 - 4*a*c^5)*d^14)) + 5*((b^2*c^4 - 4*a*c^5)*d^14)^(1/4)*(c*x^2 + b*x + a)*log(5*sqrt(2*c*d*x + b*d)
*c*d^3 + 5*((b^2*c^4 - 4*a*c^5)*d^14)^(1/4)) - 5*((b^2*c^4 - 4*a*c^5)*d^14)^(1/4)*(c*x^2 + b*x + a)*log(5*sqrt
(2*c*d*x + b*d)*c*d^3 - 5*((b^2*c^4 - 4*a*c^5)*d^14)^(1/4)) - (16*c^2*d^3*x^2 + 16*b*c*d^3*x - (b^2 - 20*a*c)*
d^3)*sqrt(2*c*d*x + b*d))/(c*x^2 + b*x + a)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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)**2,x)

[Out]

Timed out

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Giac [B]  time = 1.19009, size = 595, normalized size = 3.97 \begin{align*} -5 \, \sqrt{2}{\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac{1}{4}} c d^{3} \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 ) - 5 \, \sqrt{2}{\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac{1}{4}} c d^{3} \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{5}{2} \, \sqrt{2}{\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac{1}{4}} c d^{3} \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{5}{2} \, \sqrt{2}{\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac{1}{4}} c d^{3} \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 ) + 16 \, \sqrt{2 \, c d x + b d} c d^{3} + \frac{4 \,{\left (\sqrt{2 \, c d x + b d} b^{2} c d^{5} - 4 \, \sqrt{2 \, c d x + b d} a c^{2} d^{5}\right )}}{b^{2} d^{2} - 4 \, a c d^{2} -{\left (2 \, c d x + b d\right )}^{2}} \end{align*}

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)^2,x, algorithm="giac")

[Out]

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