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

Optimal. Leaf size=149 \[ -\frac {\left (b^2-4 a c\right )^{5/2} \tan ^{-1}\left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right )}{64 c^{7/2} d}+\frac {\left (b^2-4 a c\right )^2 \sqrt {a+b x+c x^2}}{32 c^3 d}-\frac {\left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}{24 c^2 d}+\frac {\left (a+b x+c x^2\right )^{5/2}}{10 c d} \]

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Rubi [A]  time = 0.12, antiderivative size = 149, normalized size of antiderivative = 1.00, 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 = {685, 688, 205} \begin {gather*} \frac {\left (b^2-4 a c\right )^2 \sqrt {a+b x+c x^2}}{32 c^3 d}-\frac {\left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}{24 c^2 d}-\frac {\left (b^2-4 a c\right )^{5/2} \tan ^{-1}\left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right )}{64 c^{7/2} d}+\frac {\left (a+b x+c x^2\right )^{5/2}}{10 c d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((b^2 - 4*a*c)^2*Sqrt[a + b*x + c*x^2])/(32*c^3*d) - ((b^2 - 4*a*c)*(a + b*x + c*x^2)^(3/2))/(24*c^2*d) + (a +
b*x + c*x^2)^(5/2)/(10*c*d) - ((b^2 - 4*a*c)^(5/2)*ArcTan[(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])/Sqrt[b^2 - 4*a*c]
])/(64*c^(7/2)*d)

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 685

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 + 2*p + 1)), x] - Dist[(d*p*(b^2 - 4*a*c))/(b*e*(m + 2*p + 1)), Int[(d + e*x)^m*(a +
b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, d, e, m}, 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] &&  !(IGtQ[(m - 1)/2, 0] && ( !IntegerQ[p] || LtQ[m, 2*p]))
&& RationalQ[m] && 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]

Rubi steps

\begin {align*} \int \frac {\left (a+b x+c x^2\right )^{5/2}}{b d+2 c d x} \, dx &=\frac {\left (a+b x+c x^2\right )^{5/2}}{10 c d}-\frac {\left (b^2-4 a c\right ) \int \frac {\left (a+b x+c x^2\right )^{3/2}}{b d+2 c d x} \, dx}{4 c}\\ &=-\frac {\left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}{24 c^2 d}+\frac {\left (a+b x+c x^2\right )^{5/2}}{10 c d}+\frac {\left (b^2-4 a c\right )^2 \int \frac {\sqrt {a+b x+c x^2}}{b d+2 c d x} \, dx}{16 c^2}\\ &=\frac {\left (b^2-4 a c\right )^2 \sqrt {a+b x+c x^2}}{32 c^3 d}-\frac {\left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}{24 c^2 d}+\frac {\left (a+b x+c x^2\right )^{5/2}}{10 c d}-\frac {\left (b^2-4 a c\right )^3 \int \frac {1}{(b d+2 c d x) \sqrt {a+b x+c x^2}} \, dx}{64 c^3}\\ &=\frac {\left (b^2-4 a c\right )^2 \sqrt {a+b x+c x^2}}{32 c^3 d}-\frac {\left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}{24 c^2 d}+\frac {\left (a+b x+c x^2\right )^{5/2}}{10 c d}-\frac {\left (b^2-4 a c\right )^3 \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 )}{16 c^2}\\ &=\frac {\left (b^2-4 a c\right )^2 \sqrt {a+b x+c x^2}}{32 c^3 d}-\frac {\left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}{24 c^2 d}+\frac {\left (a+b x+c x^2\right )^{5/2}}{10 c d}-\frac {\left (b^2-4 a c\right )^{5/2} \tan ^{-1}\left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right )}{64 c^{7/2} d}\\ \end {align*}

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Mathematica [A]  time = 0.17, size = 150, normalized size = 1.01 \begin {gather*} \frac {\frac {\sqrt {a+x (b+c x)} \left (16 c^2 \left (23 a^2+11 a c x^2+3 c^2 x^4\right )+28 b^2 c \left (c x^2-5 a\right )+16 b c^2 x \left (11 a+6 c x^2\right )+15 b^4-20 b^3 c x\right )}{480 c^3}-\frac {\left (b^2-4 a c\right )^{5/2} \tan ^{-1}\left (\frac {2 \sqrt {c} \sqrt {a+x (b+c x)}}{\sqrt {b^2-4 a c}}\right )}{64 c^{7/2}}}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((Sqrt[a + x*(b + c*x)]*(15*b^4 - 20*b^3*c*x + 28*b^2*c*(-5*a + c*x^2) + 16*b*c^2*x*(11*a + 6*c*x^2) + 16*c^2*
(23*a^2 + 11*a*c*x^2 + 3*c^2*x^4)))/(480*c^3) - ((b^2 - 4*a*c)^(5/2)*ArcTan[(2*Sqrt[c]*Sqrt[a + x*(b + c*x)])/
Sqrt[b^2 - 4*a*c]])/(64*c^(7/2)))/d

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IntegrateAlgebraic [A]  time = 0.89, size = 182, normalized size = 1.22 \begin {gather*} \frac {\sqrt {b^2-4 a c} \left (16 a^2 c^2-8 a b^2 c+b^4\right ) \tan ^{-1}\left (\frac {-2 \sqrt {c} \sqrt {a+b x+c x^2}+b+2 c x}{\sqrt {b^2-4 a c}}\right )}{32 c^{7/2} d}+\frac {\sqrt {a+b x+c x^2} \left (368 a^2 c^2-140 a b^2 c+176 a b c^2 x+176 a c^3 x^2+15 b^4-20 b^3 c x+28 b^2 c^2 x^2+96 b c^3 x^3+48 c^4 x^4\right )}{480 c^3 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + b*x + c*x^2]*(15*b^4 - 140*a*b^2*c + 368*a^2*c^2 - 20*b^3*c*x + 176*a*b*c^2*x + 28*b^2*c^2*x^2 + 176
*a*c^3*x^2 + 96*b*c^3*x^3 + 48*c^4*x^4))/(480*c^3*d) + (Sqrt[b^2 - 4*a*c]*(b^4 - 8*a*b^2*c + 16*a^2*c^2)*ArcTa
n[(b + 2*c*x - 2*Sqrt[c]*Sqrt[a + b*x + c*x^2])/Sqrt[b^2 - 4*a*c]])/(32*c^(7/2)*d)

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fricas [A]  time = 0.55, size = 372, normalized size = 2.50 \begin {gather*} \left [\frac {15 \, {\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} \sqrt {-\frac {b^{2} - 4 \, a c}{c}} \log \left (-\frac {4 \, c^{2} x^{2} + 4 \, b c x - b^{2} + 8 \, a c - 4 \, \sqrt {c x^{2} + b x + a} c \sqrt {-\frac {b^{2} - 4 \, a c}{c}}}{4 \, c^{2} x^{2} + 4 \, b c x + b^{2}}\right ) + 4 \, {\left (48 \, c^{4} x^{4} + 96 \, b c^{3} x^{3} + 15 \, b^{4} - 140 \, a b^{2} c + 368 \, a^{2} c^{2} + 4 \, {\left (7 \, b^{2} c^{2} + 44 \, a c^{3}\right )} x^{2} - 4 \, {\left (5 \, b^{3} c - 44 \, a b c^{2}\right )} x\right )} \sqrt {c x^{2} + b x + a}}{1920 \, c^{3} d}, \frac {15 \, {\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} \sqrt {\frac {b^{2} - 4 \, a c}{c}} \arctan \left (\frac {\sqrt {\frac {b^{2} - 4 \, a c}{c}}}{2 \, \sqrt {c x^{2} + b x + a}}\right ) + 2 \, {\left (48 \, c^{4} x^{4} + 96 \, b c^{3} x^{3} + 15 \, b^{4} - 140 \, a b^{2} c + 368 \, a^{2} c^{2} + 4 \, {\left (7 \, b^{2} c^{2} + 44 \, a c^{3}\right )} x^{2} - 4 \, {\left (5 \, b^{3} c - 44 \, a b c^{2}\right )} x\right )} \sqrt {c x^{2} + b x + a}}{960 \, c^{3} d}\right ] \end {gather*}

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),x, algorithm="fricas")

[Out]

[1/1920*(15*(b^4 - 8*a*b^2*c + 16*a^2*c^2)*sqrt(-(b^2 - 4*a*c)/c)*log(-(4*c^2*x^2 + 4*b*c*x - b^2 + 8*a*c - 4*
sqrt(c*x^2 + b*x + a)*c*sqrt(-(b^2 - 4*a*c)/c))/(4*c^2*x^2 + 4*b*c*x + b^2)) + 4*(48*c^4*x^4 + 96*b*c^3*x^3 +
15*b^4 - 140*a*b^2*c + 368*a^2*c^2 + 4*(7*b^2*c^2 + 44*a*c^3)*x^2 - 4*(5*b^3*c - 44*a*b*c^2)*x)*sqrt(c*x^2 + b
*x + a))/(c^3*d), 1/960*(15*(b^4 - 8*a*b^2*c + 16*a^2*c^2)*sqrt((b^2 - 4*a*c)/c)*arctan(1/2*sqrt((b^2 - 4*a*c)
/c)/sqrt(c*x^2 + b*x + a)) + 2*(48*c^4*x^4 + 96*b*c^3*x^3 + 15*b^4 - 140*a*b^2*c + 368*a^2*c^2 + 4*(7*b^2*c^2
+ 44*a*c^3)*x^2 - 4*(5*b^3*c - 44*a*b*c^2)*x)*sqrt(c*x^2 + b*x + a))/(c^3*d)]

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giac [A]  time = 0.24, size = 237, normalized size = 1.59 \begin {gather*} \frac {1}{480} \, \sqrt {c x^{2} + b x + a} {\left (4 \, {\left ({\left (12 \, {\left (\frac {c x}{d} + \frac {2 \, b}{d}\right )} x + \frac {7 \, b^{2} c^{9} d^{5} + 44 \, a c^{10} d^{5}}{c^{10} d^{6}}\right )} x - \frac {5 \, b^{3} c^{8} d^{5} - 44 \, a b c^{9} d^{5}}{c^{10} d^{6}}\right )} x + \frac {15 \, b^{4} c^{7} d^{5} - 140 \, a b^{2} c^{8} d^{5} + 368 \, a^{2} c^{9} d^{5}}{c^{10} d^{6}}\right )} - \frac {{\left (b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}\right )} \arctan \left (-\frac {2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} c + b \sqrt {c}}{\sqrt {b^{2} c - 4 \, a c^{2}}}\right )}{32 \, \sqrt {b^{2} c - 4 \, a c^{2}} c^{3} d} \end {gather*}

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),x, algorithm="giac")

[Out]

1/480*sqrt(c*x^2 + b*x + a)*(4*((12*(c*x/d + 2*b/d)*x + (7*b^2*c^9*d^5 + 44*a*c^10*d^5)/(c^10*d^6))*x - (5*b^3
*c^8*d^5 - 44*a*b*c^9*d^5)/(c^10*d^6))*x + (15*b^4*c^7*d^5 - 140*a*b^2*c^8*d^5 + 368*a^2*c^9*d^5)/(c^10*d^6))
- 1/32*(b^6 - 12*a*b^4*c + 48*a^2*b^2*c^2 - 64*a^3*c^3)*arctan(-(2*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*c + b*s
qrt(c))/sqrt(b^2*c - 4*a*c^2))/(sqrt(b^2*c - 4*a*c^2)*c^3*d)

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maple [B]  time = 0.05, size = 660, normalized size = 4.43 \begin {gather*} -\frac {a^{3} \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 )}{\sqrt {\frac {4 a c -b^{2}}{c}}\, c d}+\frac {3 a^{2} 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 )}{4 \sqrt {\frac {4 a c -b^{2}}{c}}\, c^{2} d}-\frac {3 a \,b^{4} \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 )}{16 \sqrt {\frac {4 a c -b^{2}}{c}}\, c^{3} d}+\frac {b^{6} \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 )}{64 \sqrt {\frac {4 a c -b^{2}}{c}}\, c^{4} d}+\frac {\sqrt {4 \left (x +\frac {b}{2 c}\right )^{2} c +\frac {4 a c -b^{2}}{c}}\, a^{2}}{4 c d}-\frac {\sqrt {4 \left (x +\frac {b}{2 c}\right )^{2} c +\frac {4 a c -b^{2}}{c}}\, a \,b^{2}}{8 c^{2} d}+\frac {\sqrt {4 \left (x +\frac {b}{2 c}\right )^{2} c +\frac {4 a c -b^{2}}{c}}\, b^{4}}{64 c^{3} d}+\frac {\left (\left (x +\frac {b}{2 c}\right )^{2} c +\frac {4 a c -b^{2}}{4 c}\right )^{\frac {3}{2}} a}{6 c d}-\frac {\left (\left (x +\frac {b}{2 c}\right )^{2} c +\frac {4 a c -b^{2}}{4 c}\right )^{\frac {3}{2}} b^{2}}{24 c^{2} d}+\frac {\left (\left (x +\frac {b}{2 c}\right )^{2} c +\frac {4 a c -b^{2}}{4 c}\right )^{\frac {5}{2}}}{10 c d} \end {gather*}

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),x)

[Out]

1/10/c/d*((x+1/2*b/c)^2*c+1/4*(4*a*c-b^2)/c)^(5/2)+1/6/c/d*((x+1/2*b/c)^2*c+1/4*(4*a*c-b^2)/c)^(3/2)*a-1/24/c^
2/d*((x+1/2*b/c)^2*c+1/4*(4*a*c-b^2)/c)^(3/2)*b^2+1/4/c/d*(4*(x+1/2*b/c)^2*c+(4*a*c-b^2)/c)^(1/2)*a^2-1/8/c^2/
d*(4*(x+1/2*b/c)^2*c+(4*a*c-b^2)/c)^(1/2)*a*b^2+1/64/c^3/d*(4*(x+1/2*b/c)^2*c+(4*a*c-b^2)/c)^(1/2)*b^4-1/c/d/(
(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+3/4/c^2/d/((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^2*b^2-3/16/c^3/d/((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+1/64/c^4/d/((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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

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),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 \begin {gather*} \int \frac {{\left (c\,x^2+b\,x+a\right )}^{5/2}}{b\,d+2\,c\,d\,x} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\int \frac {a^{2} \sqrt {a + b x + c x^{2}}}{b + 2 c x}\, dx + \int \frac {b^{2} x^{2} \sqrt {a + b x + c x^{2}}}{b + 2 c x}\, dx + \int \frac {c^{2} x^{4} \sqrt {a + b x + c x^{2}}}{b + 2 c x}\, dx + \int \frac {2 a b x \sqrt {a + b x + c x^{2}}}{b + 2 c x}\, dx + \int \frac {2 a c x^{2} \sqrt {a + b x + c x^{2}}}{b + 2 c x}\, dx + \int \frac {2 b c x^{3} \sqrt {a + b x + c x^{2}}}{b + 2 c x}\, dx}{d} \end {gather*}

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),x)

[Out]

(Integral(a**2*sqrt(a + b*x + c*x**2)/(b + 2*c*x), x) + Integral(b**2*x**2*sqrt(a + b*x + c*x**2)/(b + 2*c*x),
x) + Integral(c**2*x**4*sqrt(a + b*x + c*x**2)/(b + 2*c*x), x) + Integral(2*a*b*x*sqrt(a + b*x + c*x**2)/(b +
2*c*x), x) + Integral(2*a*c*x**2*sqrt(a + b*x + c*x**2)/(b + 2*c*x), x) + Integral(2*b*c*x**3*sqrt(a + b*x +
c*x**2)/(b + 2*c*x), x))/d

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