3.1222 \(\int \frac{(a+b x+c x^2)^{5/2}}{b d+2 c d x} \, dx\)

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

[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)

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Rubi [A]  time = 0.12067, antiderivative size = 149, 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 = {685, 688, 205} \[ \frac{\left (b^2-4 a c\right )^2 \sqrt{a+b x+c x^2}}{32 c^3 d}-\frac{\left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}{24 c^2 d}-\frac{\left (b^2-4 a c\right )^{5/2} \tan ^{-1}\left (\frac{2 \sqrt{c} \sqrt{a+b x+c x^2}}{\sqrt{b^2-4 a c}}\right )}{64 c^{7/2} d}+\frac{\left (a+b x+c x^2\right )^{5/2}}{10 c d} \]

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 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]

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} \, 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*}

Mathematica [A]  time = 0.174842, size = 150, normalized size = 1.01 \[ \frac{\frac{\sqrt{a+x (b+c x)} \left (16 c^2 \left (23 a^2+11 a c x^2+3 c^2 x^4\right )+28 b^2 c \left (c x^2-5 a\right )+16 b c^2 x \left (11 a+6 c x^2\right )-20 b^3 c x+15 b^4\right )}{480 c^3}-\frac{\left (b^2-4 a c\right )^{5/2} \tan ^{-1}\left (\frac{2 \sqrt{c} \sqrt{a+x (b+c x)}}{\sqrt{b^2-4 a c}}\right )}{64 c^{7/2}}}{d} \]

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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Maple [B]  time = 0.19, size = 660, normalized size = 4.4 \begin{align*}{\frac{1}{10\,cd} \left ( \left ( x+{\frac{b}{2\,c}} \right ) ^{2}c+{\frac{4\,ac-{b}^{2}}{4\,c}} \right ) ^{{\frac{5}{2}}}}+{\frac{a}{6\,cd} \left ( \left ( x+{\frac{b}{2\,c}} \right ) ^{2}c+{\frac{4\,ac-{b}^{2}}{4\,c}} \right ) ^{{\frac{3}{2}}}}-{\frac{{b}^{2}}{24\,{c}^{2}d} \left ( \left ( x+{\frac{b}{2\,c}} \right ) ^{2}c+{\frac{4\,ac-{b}^{2}}{4\,c}} \right ) ^{{\frac{3}{2}}}}+{\frac{{a}^{2}}{4\,cd}\sqrt{4\, \left ( x+1/2\,{\frac{b}{c}} \right ) ^{2}c+{\frac{4\,ac-{b}^{2}}{c}}}}-{\frac{{b}^{2}a}{8\,{c}^{2}d}\sqrt{4\, \left ( x+1/2\,{\frac{b}{c}} \right ) ^{2}c+{\frac{4\,ac-{b}^{2}}{c}}}}+{\frac{{b}^{4}}{64\,d{c}^{3}}\sqrt{4\, \left ( x+1/2\,{\frac{b}{c}} \right ) ^{2}c+{\frac{4\,ac-{b}^{2}}{c}}}}-{\frac{{a}^{3}}{cd}\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{3\,{b}^{2}{a}^{2}}{4\,{c}^{2}d}\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{3\,a{b}^{4}}{16\,d{c}^{3}}\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}^{6}}{64\,d{c}^{4}}\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)^(5/2)/(2*c*d*x+b*d),x)

[Out]

1/10/d/c*((x+1/2*b/c)^2*c+1/4*(4*a*c-b^2)/c)^(5/2)+1/6/d/c*((x+1/2*b/c)^2*c+1/4*(4*a*c-b^2)/c)^(3/2)*a-1/24/d/
c^2*((x+1/2*b/c)^2*c+1/4*(4*a*c-b^2)/c)^(3/2)*b^2+1/4/d/c*(4*(x+1/2*b/c)^2*c+(4*a*c-b^2)/c)^(1/2)*a^2-1/8/d/c^
2*(4*(x+1/2*b/c)^2*c+(4*a*c-b^2)/c)^(1/2)*a*b^2+1/64/d/c^3*(4*(x+1/2*b/c)^2*c+(4*a*c-b^2)/c)^(1/2)*b^4-1/d/c/(
(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/d/c^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^2*b^2-3/16/d/c^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+1/64/d/c^4/((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),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 4.46849, size = 852, normalized size = 5.72 \begin{align*} \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{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),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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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{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),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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Giac [A]  time = 1.18425, size = 320, normalized size = 2.15 \begin{align*} \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{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),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)