3.1287 \(\int \frac{(b d+2 c d x)^{7/2}}{a+b x+c x^2} \, dx\)

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

[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])]

________________________________________________________________________________________

Rubi [A]  time = 0.12059, antiderivative size = 145, 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 = {692, 694, 329, 212, 206, 203} \[ 4 d^3 \left (b^2-4 a c\right ) \sqrt{b d+2 c d x}-2 d^{7/2} \left (b^2-4 a c\right )^{5/4} \tan ^{-1}\left (\frac{\sqrt{d (b+2 c x)}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right )-2 d^{7/2} \left (b^2-4 a c\right )^{5/4} \tanh ^{-1}\left (\frac{\sqrt{d (b+2 c x)}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right )+\frac{4}{5} d (b d+2 c d x)^{5/2} \]

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 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}}{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*}

Mathematica [A]  time = 0.134548, size = 141, normalized size = 0.97 \[ \frac{2 d^3 \sqrt{d (b+2 c x)} \left (4 \sqrt{b+2 c x} \left (2 c \left (c x^2-5 a\right )+3 b^2+2 b c x\right )-5 \left (b^2-4 a c\right )^{5/4} \tan ^{-1}\left (\frac{\sqrt{b+2 c x}}{\sqrt [4]{b^2-4 a c}}\right )-5 \left (b^2-4 a c\right )^{5/4} \tanh ^{-1}\left (\frac{\sqrt{b+2 c x}}{\sqrt [4]{b^2-4 a c}}\right )\right )}{5 \sqrt{b+2 c x}} \]

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])

________________________________________________________________________________________

Maple [B]  time = 0.196, size = 922, normalized size = 6.4 \begin{align*} \text{result too large to display} \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),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^4-16*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

________________________________________________________________________________________

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),x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [B]  time = 2.10053, size = 1671, normalized size = 11.52 \begin{align*} \text{result too large to display} \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),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))

________________________________________________________________________________________

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),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [B]  time = 1.19183, size = 612, normalized size = 4.22 \begin{align*} 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 - \frac{1}{2} \, \sqrt{2}{\left (b^{2} d^{3} - 4 \, a c d^{3}\right )}{\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac{1}{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 ) + \frac{1}{2} \, \sqrt{2}{\left (b^{2} d^{3} - 4 \, a c d^{3}\right )}{\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac{1}{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 ) -{\left (\sqrt{2} b^{2} d^{3} - 4 \, \sqrt{2} a c d^{3}\right )}{\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac{1}{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 ) -{\left (\sqrt{2} b^{2} d^{3} - 4 \, \sqrt{2} a c d^{3}\right )}{\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac{1}{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 ) \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),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 - 1/2*sqrt(2)*(b^
2*d^3 - 4*a*c*d^3)*(-b^2*d^2 + 4*a*c*d^2)^(1/4)*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)) + 1/2*sqrt(2)*(b^2*d^3 - 4*a*c*d^3)*(-b^2*d^2 + 4*a*c*d^2)^(1/4)*
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^2*d^3 - 4*sqrt(2)*a*c*d^3)*(-b^2*d^2 + 4*a*c*d^2)^(1/4)*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)) - (sqrt(2)*b^2*d^3 - 4*sqrt(2)*a*c*d^3)*(-
b^2*d^2 + 4*a*c*d^2)^(1/4)*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))