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]

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

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Rubi [A]  time = 0.09, antiderivative size = 133, normalized size of antiderivative = 1.00, 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 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]

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 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])

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*}

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Mathematica [C]  time = 0.03, 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)

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fricas [B]  time = 2.67, size = 728, normalized size = 5.47 \[ \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 ] \]

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)]

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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]

Timed out

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maple [B]  time = 0.06, size = 400, normalized size = 3.01 \[ \frac {a \ln \left (\frac {\frac {4 a c -b^{2}}{2 c}+\frac {\sqrt {\frac {4 a c -b^{2}}{c}}\, \sqrt {4 \left (x +\frac {b}{2 c}\right )^{2} c +\frac {4 a c -b^{2}}{c}}}{2}}{x +\frac {b}{2 c}}\right )}{8 \left (4 a c -b^{2}\right )^{2} \sqrt {\frac {4 a c -b^{2}}{c}}\, c \,d^{5}}-\frac {b^{2} \ln \left (\frac {\frac {4 a c -b^{2}}{2 c}+\frac {\sqrt {\frac {4 a c -b^{2}}{c}}\, \sqrt {4 \left (x +\frac {b}{2 c}\right )^{2} c +\frac {4 a c -b^{2}}{c}}}{2}}{x +\frac {b}{2 c}}\right )}{32 \left (4 a c -b^{2}\right )^{2} \sqrt {\frac {4 a c -b^{2}}{c}}\, c^{2} d^{5}}-\frac {\sqrt {4 \left (x +\frac {b}{2 c}\right )^{2} c +\frac {4 a c -b^{2}}{c}}}{32 \left (4 a c -b^{2}\right )^{2} c \,d^{5}}+\frac {\left (\left (x +\frac {b}{2 c}\right )^{2} c +\frac {4 a c -b^{2}}{4 c}\right )^{\frac {3}{2}}}{16 \left (4 a c -b^{2}\right )^{2} \left (x +\frac {b}{2 c}\right )^{2} c^{2} d^{5}}-\frac {\left (\left (x +\frac {b}{2 c}\right )^{2} c +\frac {4 a c -b^{2}}{4 c}\right )^{\frac {3}{2}}}{32 \left (4 a c -b^{2}\right ) \left (x +\frac {b}{2 c}\right )^{4} c^{4} d^{5}} \]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

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 >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a*c-b^2>0)', see `assume?` f
or more details)Is 4*a*c-b^2 zero or nonzero?

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\sqrt {c\,x^2+b\,x+a}}{{\left (b\,d+2\,c\,d\,x\right )}^5} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {\sqrt {a + b x + c x^{2}}}{b^{5} + 10 b^{4} c x + 40 b^{3} c^{2} x^{2} + 80 b^{2} c^{3} x^{3} + 80 b c^{4} x^{4} + 32 c^{5} x^{5}}\, dx}{d^{5}} \]

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

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