### 3.1294 $$\int \frac{1}{(b d+2 c d x)^{7/2} (a+b x+c x^2)} \, dx$$

Optimal. Leaf size=159 $\frac{4}{d^3 \left (b^2-4 a c\right )^2 \sqrt{b d+2 c d x}}+\frac{2 \tan ^{-1}\left (\frac{\sqrt{d (b+2 c x)}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right )}{d^{7/2} \left (b^2-4 a c\right )^{9/4}}-\frac{2 \tanh ^{-1}\left (\frac{\sqrt{d (b+2 c x)}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right )}{d^{7/2} \left (b^2-4 a c\right )^{9/4}}+\frac{4}{5 d \left (b^2-4 a c\right ) (b d+2 c d x)^{5/2}}$

[Out]

4/(5*(b^2 - 4*a*c)*d*(b*d + 2*c*d*x)^(5/2)) + 4/((b^2 - 4*a*c)^2*d^3*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)^(9/4)*d^(7/2)) - (2*ArcTanh[Sqrt[d*(b + 2*c*x)]/((b
^2 - 4*a*c)^(1/4)*Sqrt[d])])/((b^2 - 4*a*c)^(9/4)*d^(7/2))

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Rubi [A]  time = 0.131502, antiderivative size = 159, normalized size of antiderivative = 1., 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 = {693, 694, 329, 298, 203, 206} $\frac{4}{d^3 \left (b^2-4 a c\right )^2 \sqrt{b d+2 c d x}}+\frac{2 \tan ^{-1}\left (\frac{\sqrt{d (b+2 c x)}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right )}{d^{7/2} \left (b^2-4 a c\right )^{9/4}}-\frac{2 \tanh ^{-1}\left (\frac{\sqrt{d (b+2 c x)}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right )}{d^{7/2} \left (b^2-4 a c\right )^{9/4}}+\frac{4}{5 d \left (b^2-4 a c\right ) (b d+2 c d x)^{5/2}}$

Antiderivative was successfully veriﬁed.

[In]

Int[1/((b*d + 2*c*d*x)^(7/2)*(a + b*x + c*x^2)),x]

[Out]

4/(5*(b^2 - 4*a*c)*d*(b*d + 2*c*d*x)^(5/2)) + 4/((b^2 - 4*a*c)^2*d^3*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)^(9/4)*d^(7/2)) - (2*ArcTanh[Sqrt[d*(b + 2*c*x)]/((b
^2 - 4*a*c)^(1/4)*Sqrt[d])])/((b^2 - 4*a*c)^(9/4)*d^(7/2))

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]

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 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 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])

Rubi steps

\begin{align*} \int \frac{1}{(b d+2 c d x)^{7/2} \left (a+b x+c x^2\right )} \, dx &=\frac{4}{5 \left (b^2-4 a c\right ) d (b d+2 c d x)^{5/2}}+\frac{\int \frac{1}{(b d+2 c d x)^{3/2} \left (a+b x+c x^2\right )} \, dx}{\left (b^2-4 a c\right ) d^2}\\ &=\frac{4}{5 \left (b^2-4 a c\right ) d (b d+2 c d x)^{5/2}}+\frac{4}{\left (b^2-4 a c\right )^2 d^3 \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 )^2 d^4}\\ &=\frac{4}{5 \left (b^2-4 a c\right ) d (b d+2 c d x)^{5/2}}+\frac{4}{\left (b^2-4 a c\right )^2 d^3 \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 )^2 d^5}\\ &=\frac{4}{5 \left (b^2-4 a c\right ) d (b d+2 c d x)^{5/2}}+\frac{4}{\left (b^2-4 a c\right )^2 d^3 \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 )^2 d^5}\\ &=\frac{4}{5 \left (b^2-4 a c\right ) d (b d+2 c d x)^{5/2}}+\frac{4}{\left (b^2-4 a c\right )^2 d^3 \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 )^2 d^3}+\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 )^2 d^3}\\ &=\frac{4}{5 \left (b^2-4 a c\right ) d (b d+2 c d x)^{5/2}}+\frac{4}{\left (b^2-4 a c\right )^2 d^3 \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 )^{9/4} d^{7/2}}-\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 )^{9/4} d^{7/2}}\\ \end{align*}

Mathematica [C]  time = 0.0660311, size = 56, normalized size = 0.35 $\frac{4 \, _2F_1\left (-\frac{5}{4},1;-\frac{1}{4};\frac{(b+2 c x)^2}{b^2-4 a c}\right )}{5 d \left (b^2-4 a c\right ) (d (b+2 c x))^{5/2}}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[1/((b*d + 2*c*d*x)^(7/2)*(a + b*x + c*x^2)),x]

[Out]

(4*Hypergeometric2F1[-5/4, 1, -1/4, (b + 2*c*x)^2/(b^2 - 4*a*c)])/(5*(b^2 - 4*a*c)*d*(d*(b + 2*c*x))^(5/2))

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Maple [B]  time = 0.197, size = 369, normalized size = 2.3 \begin{align*} -{\frac{4}{5\,d \left ( 4\,ac-{b}^{2} \right ) } \left ( 2\,cdx+bd \right ) ^{-{\frac{5}{2}}}}+4\,{\frac{1}{{d}^{3} \left ( 4\,ac-{b}^{2} \right ) ^{2}\sqrt{2\,cdx+bd}}}+{\frac{\sqrt{2}}{2\,{d}^{3} \left ( 4\,ac-{b}^{2} \right ) ^{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 ){\frac{1}{\sqrt [4]{4\,ac{d}^{2}-{b}^{2}{d}^{2}}}}}+{\frac{\sqrt{2}}{{d}^{3} \left ( 4\,ac-{b}^{2} \right ) ^{2}}\arctan \left ({\sqrt{2}\sqrt{2\,cdx+bd}{\frac{1}{\sqrt [4]{4\,ac{d}^{2}-{b}^{2}{d}^{2}}}}}+1 \right ){\frac{1}{\sqrt [4]{4\,ac{d}^{2}-{b}^{2}{d}^{2}}}}}-{\frac{\sqrt{2}}{{d}^{3} \left ( 4\,ac-{b}^{2} \right ) ^{2}}\arctan \left ( -{\sqrt{2}\sqrt{2\,cdx+bd}{\frac{1}{\sqrt [4]{4\,ac{d}^{2}-{b}^{2}{d}^{2}}}}}+1 \right ){\frac{1}{\sqrt [4]{4\,ac{d}^{2}-{b}^{2}{d}^{2}}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(1/(2*c*d*x+b*d)^(7/2)/(c*x^2+b*x+a),x)

[Out]

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

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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(1/(2*c*d*x+b*d)^(7/2)/(c*x^2+b*x+a),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.31129, size = 4031, normalized size = 25.35 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(2*c*d*x+b*d)^(7/2)/(c*x^2+b*x+a),x, algorithm="fricas")

[Out]

-1/5*(20*(8*(b^4*c^3 - 8*a*b^2*c^4 + 16*a^2*c^5)*d^4*x^3 + 12*(b^5*c^2 - 8*a*b^3*c^3 + 16*a^2*b*c^4)*d^4*x^2 +
6*(b^6*c - 8*a*b^4*c^2 + 16*a^2*b^2*c^3)*d^4*x + (b^7 - 8*a*b^5*c + 16*a^2*b^3*c^2)*d^4)*(1/((b^18 - 36*a*b^1
6*c + 576*a^2*b^14*c^2 - 5376*a^3*b^12*c^3 + 32256*a^4*b^10*c^4 - 129024*a^5*b^8*c^5 + 344064*a^6*b^6*c^6 - 58
9824*a^7*b^4*c^7 + 589824*a^8*b^2*c^8 - 262144*a^9*c^9)*d^14))^(1/4)*arctan(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^8*sqrt(1/((b^18 - 36*a*b^16*c + 576*a^2*b^14*c^
2 - 5376*a^3*b^12*c^3 + 32256*a^4*b^10*c^4 - 129024*a^5*b^8*c^5 + 344064*a^6*b^6*c^6 - 589824*a^7*b^4*c^7 + 58
9824*a^8*b^2*c^8 - 262144*a^9*c^9)*d^14)) + 2*c*d*x + b*d)*(b^4 - 8*a*b^2*c + 16*a^2*c^2)*d^3*(1/((b^18 - 36*a
*b^16*c + 576*a^2*b^14*c^2 - 5376*a^3*b^12*c^3 + 32256*a^4*b^10*c^4 - 129024*a^5*b^8*c^5 + 344064*a^6*b^6*c^6
- 589824*a^7*b^4*c^7 + 589824*a^8*b^2*c^8 - 262144*a^9*c^9)*d^14))^(1/4) - (b^4 - 8*a*b^2*c + 16*a^2*c^2)*sqrt
(2*c*d*x + b*d)*d^3*(1/((b^18 - 36*a*b^16*c + 576*a^2*b^14*c^2 - 5376*a^3*b^12*c^3 + 32256*a^4*b^10*c^4 - 1290
24*a^5*b^8*c^5 + 344064*a^6*b^6*c^6 - 589824*a^7*b^4*c^7 + 589824*a^8*b^2*c^8 - 262144*a^9*c^9)*d^14))^(1/4))
+ 5*(8*(b^4*c^3 - 8*a*b^2*c^4 + 16*a^2*c^5)*d^4*x^3 + 12*(b^5*c^2 - 8*a*b^3*c^3 + 16*a^2*b*c^4)*d^4*x^2 + 6*(b
^6*c - 8*a*b^4*c^2 + 16*a^2*b^2*c^3)*d^4*x + (b^7 - 8*a*b^5*c + 16*a^2*b^3*c^2)*d^4)*(1/((b^18 - 36*a*b^16*c +
576*a^2*b^14*c^2 - 5376*a^3*b^12*c^3 + 32256*a^4*b^10*c^4 - 129024*a^5*b^8*c^5 + 344064*a^6*b^6*c^6 - 589824*
a^7*b^4*c^7 + 589824*a^8*b^2*c^8 - 262144*a^9*c^9)*d^14))^(1/4)*log((b^14 - 28*a*b^12*c + 336*a^2*b^10*c^2 - 2
240*a^3*b^8*c^3 + 8960*a^4*b^6*c^4 - 21504*a^5*b^4*c^5 + 28672*a^6*b^2*c^6 - 16384*a^7*c^7)*d^11*(1/((b^18 - 3
6*a*b^16*c + 576*a^2*b^14*c^2 - 5376*a^3*b^12*c^3 + 32256*a^4*b^10*c^4 - 129024*a^5*b^8*c^5 + 344064*a^6*b^6*c
^6 - 589824*a^7*b^4*c^7 + 589824*a^8*b^2*c^8 - 262144*a^9*c^9)*d^14))^(3/4) + sqrt(2*c*d*x + b*d)) - 5*(8*(b^4
*c^3 - 8*a*b^2*c^4 + 16*a^2*c^5)*d^4*x^3 + 12*(b^5*c^2 - 8*a*b^3*c^3 + 16*a^2*b*c^4)*d^4*x^2 + 6*(b^6*c - 8*a*
b^4*c^2 + 16*a^2*b^2*c^3)*d^4*x + (b^7 - 8*a*b^5*c + 16*a^2*b^3*c^2)*d^4)*(1/((b^18 - 36*a*b^16*c + 576*a^2*b^
14*c^2 - 5376*a^3*b^12*c^3 + 32256*a^4*b^10*c^4 - 129024*a^5*b^8*c^5 + 344064*a^6*b^6*c^6 - 589824*a^7*b^4*c^7
+ 589824*a^8*b^2*c^8 - 262144*a^9*c^9)*d^14))^(1/4)*log(-(b^14 - 28*a*b^12*c + 336*a^2*b^10*c^2 - 2240*a^3*b^
8*c^3 + 8960*a^4*b^6*c^4 - 21504*a^5*b^4*c^5 + 28672*a^6*b^2*c^6 - 16384*a^7*c^7)*d^11*(1/((b^18 - 36*a*b^16*c
+ 576*a^2*b^14*c^2 - 5376*a^3*b^12*c^3 + 32256*a^4*b^10*c^4 - 129024*a^5*b^8*c^5 + 344064*a^6*b^6*c^6 - 58982
4*a^7*b^4*c^7 + 589824*a^8*b^2*c^8 - 262144*a^9*c^9)*d^14))^(3/4) + sqrt(2*c*d*x + b*d)) - 8*(10*c^2*x^2 + 10*
b*c*x + 3*b^2 - 2*a*c)*sqrt(2*c*d*x + b*d))/(8*(b^4*c^3 - 8*a*b^2*c^4 + 16*a^2*c^5)*d^4*x^3 + 12*(b^5*c^2 - 8*
a*b^3*c^3 + 16*a^2*b*c^4)*d^4*x^2 + 6*(b^6*c - 8*a*b^4*c^2 + 16*a^2*b^2*c^3)*d^4*x + (b^7 - 8*a*b^5*c + 16*a^2
*b^3*c^2)*d^4)

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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(1/(2*c*d*x+b*d)**(7/2)/(c*x**2+b*x+a),x)

[Out]

Timed out

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Giac [B]  time = 1.20625, size = 817, normalized size = 5.14 \begin{align*} -\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^{6} d^{5} - 12 \, a b^{4} c d^{5} + 48 \, a^{2} b^{2} c^{2} d^{5} - 64 \, a^{3} c^{3} d^{5}} - \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^{6} d^{5} - 12 \, a b^{4} c d^{5} + 48 \, a^{2} b^{2} c^{2} d^{5} - 64 \, a^{3} c^{3} d^{5}} + \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^{6} d^{5} - 12 \, \sqrt{2} a b^{4} c d^{5} + 48 \, \sqrt{2} a^{2} b^{2} c^{2} d^{5} - 64 \, \sqrt{2} a^{3} c^{3} d^{5}} - \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^{6} d^{5} - 12 \, \sqrt{2} a b^{4} c d^{5} + 48 \, \sqrt{2} a^{2} b^{2} c^{2} d^{5} - 64 \, \sqrt{2} a^{3} c^{3} d^{5}} + \frac{4 \,{\left (b^{2} d^{2} - 4 \, a c d^{2} + 5 \,{\left (2 \, c d x + b d\right )}^{2}\right )}}{5 \,{\left (b^{4} d^{3} - 8 \, a b^{2} c d^{3} + 16 \, a^{2} c^{2} d^{3}\right )}{\left (2 \, c d x + b d\right )}^{\frac{5}{2}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(2*c*d*x+b*d)^(7/2)/(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^6*d^5 - 12*a*b^4*c*d^5 + 48*a^2*b^2*c^2*d^5 - 64*a^3*c^3*d^5) - sqr
t(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^6*d^5 - 12*a*b^4*c*d^5 + 48*a^2*b^2*c^2*d^5 - 64*a^3*c^3*d^5) + (-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^6*d^5 - 12*sqrt(2)*a*b^4*c*d^5 + 48*sqrt(2)*a^2*b^2*c^2*d^5 - 64*sqrt(2)*a^3*c^
3*d^5) - (-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^6*d^5 - 12*sqrt(2)*a*b^4*c*d^5 + 48*sqrt(2)*a^2*b^2*c^2*d^5 - 64
*sqrt(2)*a^3*c^3*d^5) + 4/5*(b^2*d^2 - 4*a*c*d^2 + 5*(2*c*d*x + b*d)^2)/((b^4*d^3 - 8*a*b^2*c*d^3 + 16*a^2*c^2
*d^3)*(2*c*d*x + b*d)^(5/2))