3.1228 \(\int \frac{(a+b x+c x^2)^{5/2}}{(b d+2 c d x)^7} \, dx\)

Optimal. Leaf size=155 \[ \frac{5 \tan ^{-1}\left (\frac{2 \sqrt{c} \sqrt{a+b x+c x^2}}{\sqrt{b^2-4 a c}}\right )}{1024 c^{7/2} d^7 \sqrt{b^2-4 a c}}-\frac{5 \left (a+b x+c x^2\right )^{3/2}}{192 c^2 d^7 (b+2 c x)^4}-\frac{5 \sqrt{a+b x+c x^2}}{512 c^3 d^7 (b+2 c x)^2}-\frac{\left (a+b x+c x^2\right )^{5/2}}{12 c d^7 (b+2 c x)^6} \]

[Out]

(-5*Sqrt[a + b*x + c*x^2])/(512*c^3*d^7*(b + 2*c*x)^2) - (5*(a + b*x + c*x^2)^(3/2))/(192*c^2*d^7*(b + 2*c*x)^
4) - (a + b*x + c*x^2)^(5/2)/(12*c*d^7*(b + 2*c*x)^6) + (5*ArcTan[(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])/Sqrt[b^2 -
 4*a*c]])/(1024*c^(7/2)*Sqrt[b^2 - 4*a*c]*d^7)

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Rubi [A]  time = 0.0988183, antiderivative size = 155, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.115, Rules used = {684, 688, 205} \[ \frac{5 \tan ^{-1}\left (\frac{2 \sqrt{c} \sqrt{a+b x+c x^2}}{\sqrt{b^2-4 a c}}\right )}{1024 c^{7/2} d^7 \sqrt{b^2-4 a c}}-\frac{5 \left (a+b x+c x^2\right )^{3/2}}{192 c^2 d^7 (b+2 c x)^4}-\frac{5 \sqrt{a+b x+c x^2}}{512 c^3 d^7 (b+2 c x)^2}-\frac{\left (a+b x+c x^2\right )^{5/2}}{12 c d^7 (b+2 c x)^6} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x + c*x^2)^(5/2)/(b*d + 2*c*d*x)^7,x]

[Out]

(-5*Sqrt[a + b*x + c*x^2])/(512*c^3*d^7*(b + 2*c*x)^2) - (5*(a + b*x + c*x^2)^(3/2))/(192*c^2*d^7*(b + 2*c*x)^
4) - (a + b*x + c*x^2)^(5/2)/(12*c*d^7*(b + 2*c*x)^6) + (5*ArcTan[(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])/Sqrt[b^2 -
 4*a*c]])/(1024*c^(7/2)*Sqrt[b^2 - 4*a*c]*d^7)

Rule 684

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x)^(m + 1)*(
a + b*x + c*x^2)^p)/(e*(m + 1)), x] - Dist[(b*p)/(d*e*(m + 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] &&
 GtQ[p, 0] && LtQ[m, -1] &&  !(IntegerQ[m/2] && LtQ[m + 2*p + 3, 0]) && IntegerQ[2*p]

Rule 688

Int[1/(((d_) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[4*c, Subst[Int[1/(b^2*e
 - 4*a*c*e + 4*c*e*x^2), x], x, Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0]
 && EqQ[2*c*d - b*e, 0]

Rule 205

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

Rubi steps

\begin{align*} \int \frac{\left (a+b x+c x^2\right )^{5/2}}{(b d+2 c d x)^7} \, dx &=-\frac{\left (a+b x+c x^2\right )^{5/2}}{12 c d^7 (b+2 c x)^6}+\frac{5 \int \frac{\left (a+b x+c x^2\right )^{3/2}}{(b d+2 c d x)^5} \, dx}{24 c d^2}\\ &=-\frac{5 \left (a+b x+c x^2\right )^{3/2}}{192 c^2 d^7 (b+2 c x)^4}-\frac{\left (a+b x+c x^2\right )^{5/2}}{12 c d^7 (b+2 c x)^6}+\frac{5 \int \frac{\sqrt{a+b x+c x^2}}{(b d+2 c d x)^3} \, dx}{128 c^2 d^4}\\ &=-\frac{5 \sqrt{a+b x+c x^2}}{512 c^3 d^7 (b+2 c x)^2}-\frac{5 \left (a+b x+c x^2\right )^{3/2}}{192 c^2 d^7 (b+2 c x)^4}-\frac{\left (a+b x+c x^2\right )^{5/2}}{12 c d^7 (b+2 c x)^6}+\frac{5 \int \frac{1}{(b d+2 c d x) \sqrt{a+b x+c x^2}} \, dx}{1024 c^3 d^6}\\ &=-\frac{5 \sqrt{a+b x+c x^2}}{512 c^3 d^7 (b+2 c x)^2}-\frac{5 \left (a+b x+c x^2\right )^{3/2}}{192 c^2 d^7 (b+2 c x)^4}-\frac{\left (a+b x+c x^2\right )^{5/2}}{12 c d^7 (b+2 c x)^6}+\frac{5 \operatorname{Subst}\left (\int \frac{1}{2 b^2 c d-8 a c^2 d+8 c^2 d x^2} \, dx,x,\sqrt{a+b x+c x^2}\right )}{256 c^2 d^6}\\ &=-\frac{5 \sqrt{a+b x+c x^2}}{512 c^3 d^7 (b+2 c x)^2}-\frac{5 \left (a+b x+c x^2\right )^{3/2}}{192 c^2 d^7 (b+2 c x)^4}-\frac{\left (a+b x+c x^2\right )^{5/2}}{12 c d^7 (b+2 c x)^6}+\frac{5 \tan ^{-1}\left (\frac{2 \sqrt{c} \sqrt{a+b x+c x^2}}{\sqrt{b^2-4 a c}}\right )}{1024 c^{7/2} \sqrt{b^2-4 a c} d^7}\\ \end{align*}

Mathematica [A]  time = 0.385353, size = 233, normalized size = 1.5 \[ \frac{-2 c \left (8 a^2 c \left (5 b^2+68 b c x+68 c^2 x^2\right )+128 a^3 c^2+a \left (1144 b^2 c^2 x^2+200 b^3 c x+15 b^4+1888 b c^3 x^3+944 c^4 x^4\right )+x \left (848 b^3 c^2 x^2+1744 b^2 c^3 x^3+175 b^4 c x+15 b^5+1584 b c^4 x^4+528 c^5 x^5\right )\right )-15 (b+2 c x)^6 \sqrt{\frac{c (a+x (b+c x))}{4 a c-b^2}} \tanh ^{-1}\left (2 \sqrt{\frac{c (a+x (b+c x))}{4 a c-b^2}}\right )}{3072 c^4 d^7 (b+2 c x)^6 \sqrt{a+x (b+c x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x + c*x^2)^(5/2)/(b*d + 2*c*d*x)^7,x]

[Out]

(-2*c*(128*a^3*c^2 + 8*a^2*c*(5*b^2 + 68*b*c*x + 68*c^2*x^2) + a*(15*b^4 + 200*b^3*c*x + 1144*b^2*c^2*x^2 + 18
88*b*c^3*x^3 + 944*c^4*x^4) + x*(15*b^5 + 175*b^4*c*x + 848*b^3*c^2*x^2 + 1744*b^2*c^3*x^3 + 1584*b*c^4*x^4 +
528*c^5*x^5)) - 15*(b + 2*c*x)^6*Sqrt[(c*(a + x*(b + c*x)))/(-b^2 + 4*a*c)]*ArcTanh[2*Sqrt[(c*(a + x*(b + c*x)
))/(-b^2 + 4*a*c)]])/(3072*c^4*d^7*(b + 2*c*x)^6*Sqrt[a + x*(b + c*x)])

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Maple [B]  time = 0.2, size = 960, normalized size = 6.2 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^2+b*x+a)^(5/2)/(2*c*d*x+b*d)^7,x)

[Out]

-1/192/d^7/c^6/(4*a*c-b^2)/(x+1/2*b/c)^6*((x+1/2*b/c)^2*c+1/4*(4*a*c-b^2)/c)^(7/2)-1/192/d^7/c^4/(4*a*c-b^2)^2
/(x+1/2*b/c)^4*((x+1/2*b/c)^2*c+1/4*(4*a*c-b^2)/c)^(7/2)-1/32/d^7/c^2/(4*a*c-b^2)^3/(x+1/2*b/c)^2*((x+1/2*b/c)
^2*c+1/4*(4*a*c-b^2)/c)^(7/2)+1/32/d^7/c/(4*a*c-b^2)^3*((x+1/2*b/c)^2*c+1/4*(4*a*c-b^2)/c)^(5/2)+5/96/d^7/c/(4
*a*c-b^2)^3*((x+1/2*b/c)^2*c+1/4*(4*a*c-b^2)/c)^(3/2)*a-5/384/d^7/c^2/(4*a*c-b^2)^3*((x+1/2*b/c)^2*c+1/4*(4*a*
c-b^2)/c)^(3/2)*b^2+5/64/d^7/c/(4*a*c-b^2)^3*(4*(x+1/2*b/c)^2*c+(4*a*c-b^2)/c)^(1/2)*a^2-5/128/d^7/c^2/(4*a*c-
b^2)^3*(4*(x+1/2*b/c)^2*c+(4*a*c-b^2)/c)^(1/2)*a*b^2+5/1024/d^7/c^3/(4*a*c-b^2)^3*(4*(x+1/2*b/c)^2*c+(4*a*c-b^
2)/c)^(1/2)*b^4-5/16/d^7/c/(4*a*c-b^2)^3/((4*a*c-b^2)/c)^(1/2)*ln((1/2*(4*a*c-b^2)/c+1/2*((4*a*c-b^2)/c)^(1/2)
*(4*(x+1/2*b/c)^2*c+(4*a*c-b^2)/c)^(1/2))/(x+1/2*b/c))*a^3+15/64/d^7/c^2/(4*a*c-b^2)^3/((4*a*c-b^2)/c)^(1/2)*l
n((1/2*(4*a*c-b^2)/c+1/2*((4*a*c-b^2)/c)^(1/2)*(4*(x+1/2*b/c)^2*c+(4*a*c-b^2)/c)^(1/2))/(x+1/2*b/c))*a^2*b^2-1
5/256/d^7/c^3/(4*a*c-b^2)^3/((4*a*c-b^2)/c)^(1/2)*ln((1/2*(4*a*c-b^2)/c+1/2*((4*a*c-b^2)/c)^(1/2)*(4*(x+1/2*b/
c)^2*c+(4*a*c-b^2)/c)^(1/2))/(x+1/2*b/c))*a*b^4+5/1024/d^7/c^4/(4*a*c-b^2)^3/((4*a*c-b^2)/c)^(1/2)*ln((1/2*(4*
a*c-b^2)/c+1/2*((4*a*c-b^2)/c)^(1/2)*(4*(x+1/2*b/c)^2*c+(4*a*c-b^2)/c)^(1/2))/(x+1/2*b/c))*b^6

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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((c*x^2+b*x+a)^(5/2)/(2*c*d*x+b*d)^7,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 46.172, size = 1985, normalized size = 12.81 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x+a)^(5/2)/(2*c*d*x+b*d)^7,x, algorithm="fricas")

[Out]

[-1/6144*(15*(64*c^6*x^6 + 192*b*c^5*x^5 + 240*b^2*c^4*x^4 + 160*b^3*c^3*x^3 + 60*b^4*c^2*x^2 + 12*b^5*c*x + b
^6)*sqrt(-b^2*c + 4*a*c^2)*log(-(4*c^2*x^2 + 4*b*c*x - b^2 + 8*a*c - 4*sqrt(-b^2*c + 4*a*c^2)*sqrt(c*x^2 + b*x
 + a))/(4*c^2*x^2 + 4*b*c*x + b^2)) + 4*(15*b^6*c - 20*a*b^4*c^2 - 32*a^2*b^2*c^3 - 512*a^3*c^4 + 528*(b^2*c^5
 - 4*a*c^6)*x^4 + 1056*(b^3*c^4 - 4*a*b*c^5)*x^3 + 16*(43*b^4*c^3 - 146*a*b^2*c^4 - 104*a^2*c^5)*x^2 + 32*(5*b
^5*c^2 - 7*a*b^3*c^3 - 52*a^2*b*c^4)*x)*sqrt(c*x^2 + b*x + a))/(64*(b^2*c^10 - 4*a*c^11)*d^7*x^6 + 192*(b^3*c^
9 - 4*a*b*c^10)*d^7*x^5 + 240*(b^4*c^8 - 4*a*b^2*c^9)*d^7*x^4 + 160*(b^5*c^7 - 4*a*b^3*c^8)*d^7*x^3 + 60*(b^6*
c^6 - 4*a*b^4*c^7)*d^7*x^2 + 12*(b^7*c^5 - 4*a*b^5*c^6)*d^7*x + (b^8*c^4 - 4*a*b^6*c^5)*d^7), -1/3072*(15*(64*
c^6*x^6 + 192*b*c^5*x^5 + 240*b^2*c^4*x^4 + 160*b^3*c^3*x^3 + 60*b^4*c^2*x^2 + 12*b^5*c*x + b^6)*sqrt(b^2*c -
4*a*c^2)*arctan(1/2*sqrt(b^2*c - 4*a*c^2)*sqrt(c*x^2 + b*x + a)/(c^2*x^2 + b*c*x + a*c)) + 2*(15*b^6*c - 20*a*
b^4*c^2 - 32*a^2*b^2*c^3 - 512*a^3*c^4 + 528*(b^2*c^5 - 4*a*c^6)*x^4 + 1056*(b^3*c^4 - 4*a*b*c^5)*x^3 + 16*(43
*b^4*c^3 - 146*a*b^2*c^4 - 104*a^2*c^5)*x^2 + 32*(5*b^5*c^2 - 7*a*b^3*c^3 - 52*a^2*b*c^4)*x)*sqrt(c*x^2 + b*x
+ a))/(64*(b^2*c^10 - 4*a*c^11)*d^7*x^6 + 192*(b^3*c^9 - 4*a*b*c^10)*d^7*x^5 + 240*(b^4*c^8 - 4*a*b^2*c^9)*d^7
*x^4 + 160*(b^5*c^7 - 4*a*b^3*c^8)*d^7*x^3 + 60*(b^6*c^6 - 4*a*b^4*c^7)*d^7*x^2 + 12*(b^7*c^5 - 4*a*b^5*c^6)*d
^7*x + (b^8*c^4 - 4*a*b^6*c^5)*d^7)]

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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((c*x**2+b*x+a)**(5/2)/(2*c*d*x+b*d)**7,x)

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x+a)^(5/2)/(2*c*d*x+b*d)^7,x, algorithm="giac")

[Out]

Exception raised: TypeError