3.1199 \(\int \frac{\sqrt{a+b x+c x^2}}{(b d+2 c d x)^5} \, dx\)

Optimal. Leaf size=133 \[ \frac{\tan ^{-1}\left (\frac{2 \sqrt{c} \sqrt{a+b x+c x^2}}{\sqrt{b^2-4 a c}}\right )}{32 c^{3/2} d^5 \left (b^2-4 a c\right )^{3/2}}+\frac{\sqrt{a+b x+c x^2}}{16 c d^5 \left (b^2-4 a c\right ) (b+2 c x)^2}-\frac{\sqrt{a+b x+c x^2}}{8 c d^5 (b+2 c x)^4} \]

[Out]

-Sqrt[a + b*x + c*x^2]/(8*c*d^5*(b + 2*c*x)^4) + Sqrt[a + b*x + c*x^2]/(16*c*(b^2 - 4*a*c)*d^5*(b + 2*c*x)^2)
+ ArcTan[(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])/Sqrt[b^2 - 4*a*c]]/(32*c^(3/2)*(b^2 - 4*a*c)^(3/2)*d^5)

________________________________________________________________________________________

Rubi [A]  time = 0.0892683, antiderivative size = 133, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {684, 693, 688, 205} \[ \frac{\tan ^{-1}\left (\frac{2 \sqrt{c} \sqrt{a+b x+c x^2}}{\sqrt{b^2-4 a c}}\right )}{32 c^{3/2} d^5 \left (b^2-4 a c\right )^{3/2}}+\frac{\sqrt{a+b x+c x^2}}{16 c d^5 \left (b^2-4 a c\right ) (b+2 c x)^2}-\frac{\sqrt{a+b x+c x^2}}{8 c d^5 (b+2 c x)^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-Sqrt[a + b*x + c*x^2]/(8*c*d^5*(b + 2*c*x)^4) + Sqrt[a + b*x + c*x^2]/(16*c*(b^2 - 4*a*c)*d^5*(b + 2*c*x)^2)
+ ArcTan[(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])/Sqrt[b^2 - 4*a*c]]/(32*c^(3/2)*(b^2 - 4*a*c)^(3/2)*d^5)

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 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 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{\sqrt{a+b x+c x^2}}{(b d+2 c d x)^5} \, dx &=-\frac{\sqrt{a+b x+c x^2}}{8 c d^5 (b+2 c x)^4}+\frac{\int \frac{1}{(b d+2 c d x)^3 \sqrt{a+b x+c x^2}} \, dx}{16 c d^2}\\ &=-\frac{\sqrt{a+b x+c x^2}}{8 c d^5 (b+2 c x)^4}+\frac{\sqrt{a+b x+c x^2}}{16 c \left (b^2-4 a c\right ) d^5 (b+2 c x)^2}+\frac{\int \frac{1}{(b d+2 c d x) \sqrt{a+b x+c x^2}} \, dx}{32 c \left (b^2-4 a c\right ) d^4}\\ &=-\frac{\sqrt{a+b x+c x^2}}{8 c d^5 (b+2 c x)^4}+\frac{\sqrt{a+b x+c x^2}}{16 c \left (b^2-4 a c\right ) d^5 (b+2 c x)^2}+\frac{\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 )}{8 \left (b^2-4 a c\right ) d^4}\\ &=-\frac{\sqrt{a+b x+c x^2}}{8 c d^5 (b+2 c x)^4}+\frac{\sqrt{a+b x+c x^2}}{16 c \left (b^2-4 a c\right ) d^5 (b+2 c x)^2}+\frac{\tan ^{-1}\left (\frac{2 \sqrt{c} \sqrt{a+b x+c x^2}}{\sqrt{b^2-4 a c}}\right )}{32 c^{3/2} \left (b^2-4 a c\right )^{3/2} d^5}\\ \end{align*}

Mathematica [C]  time = 0.0289667, size = 62, normalized size = 0.47 \[ \frac{2 (a+x (b+c x))^{3/2} \, _2F_1\left (\frac{3}{2},3;\frac{5}{2};\frac{4 c (a+x (b+c x))}{4 a c-b^2}\right )}{3 d^5 \left (b^2-4 a c\right )^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(a + x*(b + c*x))^(3/2)*Hypergeometric2F1[3/2, 3, 5/2, (4*c*(a + x*(b + c*x)))/(-b^2 + 4*a*c)])/(3*(b^2 - 4
*a*c)^3*d^5)

________________________________________________________________________________________

Maple [B]  time = 0.197, size = 400, normalized size = 3. \begin{align*} -{\frac{1}{32\,{c}^{4}{d}^{5} \left ( 4\,ac-{b}^{2} \right ) } \left ( \left ( x+{\frac{b}{2\,c}} \right ) ^{2}c+{\frac{4\,ac-{b}^{2}}{4\,c}} \right ) ^{{\frac{3}{2}}} \left ( x+{\frac{b}{2\,c}} \right ) ^{-4}}+{\frac{1}{16\,{c}^{2}{d}^{5} \left ( 4\,ac-{b}^{2} \right ) ^{2}} \left ( \left ( x+{\frac{b}{2\,c}} \right ) ^{2}c+{\frac{4\,ac-{b}^{2}}{4\,c}} \right ) ^{{\frac{3}{2}}} \left ( x+{\frac{b}{2\,c}} \right ) ^{-2}}-{\frac{1}{32\,{d}^{5}c \left ( 4\,ac-{b}^{2} \right ) ^{2}}\sqrt{4\, \left ( x+1/2\,{\frac{b}{c}} \right ) ^{2}c+{\frac{4\,ac-{b}^{2}}{c}}}}+{\frac{a}{8\,{d}^{5}c \left ( 4\,ac-{b}^{2} \right ) ^{2}}\ln \left ({ \left ({\frac{4\,ac-{b}^{2}}{2\,c}}+{\frac{1}{2}\sqrt{{\frac{4\,ac-{b}^{2}}{c}}}\sqrt{4\, \left ( x+1/2\,{\frac{b}{c}} \right ) ^{2}c+{\frac{4\,ac-{b}^{2}}{c}}}} \right ) \left ( x+{\frac{b}{2\,c}} \right ) ^{-1}} \right ){\frac{1}{\sqrt{{\frac{4\,ac-{b}^{2}}{c}}}}}}-{\frac{{b}^{2}}{32\,{c}^{2}{d}^{5} \left ( 4\,ac-{b}^{2} \right ) ^{2}}\ln \left ({ \left ({\frac{4\,ac-{b}^{2}}{2\,c}}+{\frac{1}{2}\sqrt{{\frac{4\,ac-{b}^{2}}{c}}}\sqrt{4\, \left ( x+1/2\,{\frac{b}{c}} \right ) ^{2}c+{\frac{4\,ac-{b}^{2}}{c}}}} \right ) \left ( x+{\frac{b}{2\,c}} \right ) ^{-1}} \right ){\frac{1}{\sqrt{{\frac{4\,ac-{b}^{2}}{c}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^2+b*x+a)^(1/2)/(2*c*d*x+b*d)^5,x)

[Out]

-1/32/d^5/c^4/(4*a*c-b^2)/(x+1/2*b/c)^4*((x+1/2*b/c)^2*c+1/4*(4*a*c-b^2)/c)^(3/2)+1/16/d^5/c^2/(4*a*c-b^2)^2/(
x+1/2*b/c)^2*((x+1/2*b/c)^2*c+1/4*(4*a*c-b^2)/c)^(3/2)-1/32/d^5/c/(4*a*c-b^2)^2*(4*(x+1/2*b/c)^2*c+(4*a*c-b^2)
/c)^(1/2)+1/8/d^5/c/(4*a*c-b^2)^2/((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-1/32/d^5/c^2/(4*a*c-b^2)^2/((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^2

________________________________________________________________________________________

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)^(1/2)/(2*c*d*x+b*d)^5,x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [B]  time = 10.4778, size = 1532, normalized size = 11.52 \begin{align*} \left [\frac{{\left (16 \, c^{4} x^{4} + 32 \, b c^{3} x^{3} + 24 \, b^{2} c^{2} x^{2} + 8 \, b^{3} c x + b^{4}\right )} \sqrt{-b^{2} c + 4 \, a c^{2}} \log \left (-\frac{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}}\right ) - 4 \,{\left (b^{4} c - 12 \, a b^{2} c^{2} + 32 \, a^{2} c^{3} - 4 \,{\left (b^{2} c^{3} - 4 \, a c^{4}\right )} x^{2} - 4 \,{\left (b^{3} c^{2} - 4 \, a b c^{3}\right )} x\right )} \sqrt{c x^{2} + b x + a}}{64 \,{\left (16 \,{\left (b^{4} c^{6} - 8 \, a b^{2} c^{7} + 16 \, a^{2} c^{8}\right )} d^{5} x^{4} + 32 \,{\left (b^{5} c^{5} - 8 \, a b^{3} c^{6} + 16 \, a^{2} b c^{7}\right )} d^{5} x^{3} + 24 \,{\left (b^{6} c^{4} - 8 \, a b^{4} c^{5} + 16 \, a^{2} b^{2} c^{6}\right )} d^{5} x^{2} + 8 \,{\left (b^{7} c^{3} - 8 \, a b^{5} c^{4} + 16 \, a^{2} b^{3} c^{5}\right )} d^{5} x +{\left (b^{8} c^{2} - 8 \, a b^{6} c^{3} + 16 \, a^{2} b^{4} c^{4}\right )} d^{5}\right )}}, -\frac{{\left (16 \, c^{4} x^{4} + 32 \, b c^{3} x^{3} + 24 \, b^{2} c^{2} x^{2} + 8 \, b^{3} c x + b^{4}\right )} \sqrt{b^{2} c - 4 \, a c^{2}} \arctan \left (\frac{\sqrt{b^{2} c - 4 \, a c^{2}} \sqrt{c x^{2} + b x + a}}{2 \,{\left (c^{2} x^{2} + b c x + a c\right )}}\right ) + 2 \,{\left (b^{4} c - 12 \, a b^{2} c^{2} + 32 \, a^{2} c^{3} - 4 \,{\left (b^{2} c^{3} - 4 \, a c^{4}\right )} x^{2} - 4 \,{\left (b^{3} c^{2} - 4 \, a b c^{3}\right )} x\right )} \sqrt{c x^{2} + b x + a}}{32 \,{\left (16 \,{\left (b^{4} c^{6} - 8 \, a b^{2} c^{7} + 16 \, a^{2} c^{8}\right )} d^{5} x^{4} + 32 \,{\left (b^{5} c^{5} - 8 \, a b^{3} c^{6} + 16 \, a^{2} b c^{7}\right )} d^{5} x^{3} + 24 \,{\left (b^{6} c^{4} - 8 \, a b^{4} c^{5} + 16 \, a^{2} b^{2} c^{6}\right )} d^{5} x^{2} + 8 \,{\left (b^{7} c^{3} - 8 \, a b^{5} c^{4} + 16 \, a^{2} b^{3} c^{5}\right )} d^{5} x +{\left (b^{8} c^{2} - 8 \, a b^{6} c^{3} + 16 \, a^{2} b^{4} c^{4}\right )} d^{5}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x+a)^(1/2)/(2*c*d*x+b*d)^5,x, algorithm="fricas")

[Out]

[1/64*((16*c^4*x^4 + 32*b*c^3*x^3 + 24*b^2*c^2*x^2 + 8*b^3*c*x + b^4)*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*(b^4
*c - 12*a*b^2*c^2 + 32*a^2*c^3 - 4*(b^2*c^3 - 4*a*c^4)*x^2 - 4*(b^3*c^2 - 4*a*b*c^3)*x)*sqrt(c*x^2 + b*x + a))
/(16*(b^4*c^6 - 8*a*b^2*c^7 + 16*a^2*c^8)*d^5*x^4 + 32*(b^5*c^5 - 8*a*b^3*c^6 + 16*a^2*b*c^7)*d^5*x^3 + 24*(b^
6*c^4 - 8*a*b^4*c^5 + 16*a^2*b^2*c^6)*d^5*x^2 + 8*(b^7*c^3 - 8*a*b^5*c^4 + 16*a^2*b^3*c^5)*d^5*x + (b^8*c^2 -
8*a*b^6*c^3 + 16*a^2*b^4*c^4)*d^5), -1/32*((16*c^4*x^4 + 32*b*c^3*x^3 + 24*b^2*c^2*x^2 + 8*b^3*c*x + b^4)*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*(b^4*c -
12*a*b^2*c^2 + 32*a^2*c^3 - 4*(b^2*c^3 - 4*a*c^4)*x^2 - 4*(b^3*c^2 - 4*a*b*c^3)*x)*sqrt(c*x^2 + b*x + a))/(16
*(b^4*c^6 - 8*a*b^2*c^7 + 16*a^2*c^8)*d^5*x^4 + 32*(b^5*c^5 - 8*a*b^3*c^6 + 16*a^2*b*c^7)*d^5*x^3 + 24*(b^6*c^
4 - 8*a*b^4*c^5 + 16*a^2*b^2*c^6)*d^5*x^2 + 8*(b^7*c^3 - 8*a*b^5*c^4 + 16*a^2*b^3*c^5)*d^5*x + (b^8*c^2 - 8*a*
b^6*c^3 + 16*a^2*b^4*c^4)*d^5)]

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{\sqrt{a + b x + c x^{2}}}{b^{5} + 10 b^{4} c x + 40 b^{3} c^{2} x^{2} + 80 b^{2} c^{3} x^{3} + 80 b c^{4} x^{4} + 32 c^{5} x^{5}}\, dx}{d^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**2+b*x+a)**(1/2)/(2*c*d*x+b*d)**5,x)

[Out]

Integral(sqrt(a + b*x + c*x**2)/(b**5 + 10*b**4*c*x + 40*b**3*c**2*x**2 + 80*b**2*c**3*x**3 + 80*b*c**4*x**4 +
32*c**5*x**5), x)/d**5

________________________________________________________________________________________

Giac [B]  time = 1.69006, size = 859, normalized size = 6.46 \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)^(1/2)/(2*c*d*x+b*d)^5,x, algorithm="giac")

[Out]

-1/12288*(sqrt(-b^2*c + 4*a*c^2)*log(abs(c))*sgn(1/(2*c*d*x + b*d))*sgn(c)*sgn(d)/(b^8*c^9*d^13 - 16*a*b^6*c^1
0*d^13 + 96*a^2*b^4*c^11*d^13 - 256*a^3*b^2*c^12*d^13 + 256*a^4*c^13*d^13) - 2*sqrt(-b^2*c*d^2/(2*c*d*x + b*d)
^2 + 4*a*c^2*d^2/(2*c*d*x + b*d)^2 + c)*((b^2*c^3*d^5*abs(c)*sgn(1/(2*c*d*x + b*d))*sgn(c)*sgn(d) - 4*a*c^4*d^
5*abs(c)*sgn(1/(2*c*d*x + b*d))*sgn(c)*sgn(d))/(b^8*c^12*d^16 - 16*a*b^6*c^13*d^16 + 96*a^2*b^4*c^14*d^16 - 25
6*a^3*b^2*c^15*d^16 + 256*a^4*c^16*d^16) - 2*(b^4*c^5*d^9*abs(c)*sgn(1/(2*c*d*x + b*d))*sgn(c)*sgn(d) - 8*a*b^
2*c^6*d^9*abs(c)*sgn(1/(2*c*d*x + b*d))*sgn(c)*sgn(d) + 16*a^2*c^7*d^9*abs(c)*sgn(1/(2*c*d*x + b*d))*sgn(c)*sg
n(d))/((b^8*c^12*d^16 - 16*a*b^6*c^13*d^16 + 96*a^2*b^4*c^14*d^16 - 256*a^3*b^2*c^15*d^16 + 256*a^4*c^16*d^16)
*(2*c*d*x + b*d)^2*c^2*d^2))/((2*c*d*x + b*d)*c*d) - 2*sqrt(-b^2*c + 4*a*c^2)*log(abs(sqrt(-b^2*c*d^2/(2*c*d*x
+ b*d)^2 + 4*a*c^2*d^2/(2*c*d*x + b*d)^2 + c) + sqrt(-b^2*c^3*d^4 + 4*a*c^4*d^4)/((2*c*d*x + b*d)*c*d)))*sgn(
1/(2*c*d*x + b*d))*sgn(c)*sgn(d)/((b^8*c^9 - 16*a*b^6*c^10 + 96*a^2*b^4*c^11 - 256*a^3*b^2*c^12 + 256*a^4*c^13
)*d^13))*d^2*abs(c)