3.939 \(\int (A+B x) (a+b x+c x^2)^{5/2} \, dx\)

Optimal. Leaf size=203 \[ \frac{5 \left (b^2-4 a c\right ) (b+2 c x) \left (a+b x+c x^2\right )^{3/2} (b B-2 A c)}{384 c^3}-\frac{5 \left (b^2-4 a c\right )^2 (b+2 c x) \sqrt{a+b x+c x^2} (b B-2 A c)}{1024 c^4}+\frac{5 \left (b^2-4 a c\right )^3 (b B-2 A c) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{2048 c^{9/2}}-\frac{(b+2 c x) \left (a+b x+c x^2\right )^{5/2} (b B-2 A c)}{24 c^2}+\frac{B \left (a+b x+c x^2\right )^{7/2}}{7 c} \]

[Out]

(-5*(b^2 - 4*a*c)^2*(b*B - 2*A*c)*(b + 2*c*x)*Sqrt[a + b*x + c*x^2])/(1024*c^4) + (5*(b^2 - 4*a*c)*(b*B - 2*A*
c)*(b + 2*c*x)*(a + b*x + c*x^2)^(3/2))/(384*c^3) - ((b*B - 2*A*c)*(b + 2*c*x)*(a + b*x + c*x^2)^(5/2))/(24*c^
2) + (B*(a + b*x + c*x^2)^(7/2))/(7*c) + (5*(b^2 - 4*a*c)^3*(b*B - 2*A*c)*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[
a + b*x + c*x^2])])/(2048*c^(9/2))

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Rubi [A]  time = 0.095175, antiderivative size = 203, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {640, 612, 621, 206} \[ \frac{5 \left (b^2-4 a c\right ) (b+2 c x) \left (a+b x+c x^2\right )^{3/2} (b B-2 A c)}{384 c^3}-\frac{5 \left (b^2-4 a c\right )^2 (b+2 c x) \sqrt{a+b x+c x^2} (b B-2 A c)}{1024 c^4}+\frac{5 \left (b^2-4 a c\right )^3 (b B-2 A c) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{2048 c^{9/2}}-\frac{(b+2 c x) \left (a+b x+c x^2\right )^{5/2} (b B-2 A c)}{24 c^2}+\frac{B \left (a+b x+c x^2\right )^{7/2}}{7 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-5*(b^2 - 4*a*c)^2*(b*B - 2*A*c)*(b + 2*c*x)*Sqrt[a + b*x + c*x^2])/(1024*c^4) + (5*(b^2 - 4*a*c)*(b*B - 2*A*
c)*(b + 2*c*x)*(a + b*x + c*x^2)^(3/2))/(384*c^3) - ((b*B - 2*A*c)*(b + 2*c*x)*(a + b*x + c*x^2)^(5/2))/(24*c^
2) + (B*(a + b*x + c*x^2)^(7/2))/(7*c) + (5*(b^2 - 4*a*c)^3*(b*B - 2*A*c)*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[
a + b*x + c*x^2])])/(2048*c^(9/2))

Rule 640

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

Rule 612

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((b + 2*c*x)*(a + b*x + c*x^2)^p)/(2*c*(2*p +
1)), x] - Dist[(p*(b^2 - 4*a*c))/(2*c*(2*p + 1)), Int[(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c}, x]
&& NeQ[b^2 - 4*a*c, 0] && GtQ[p, 0] && IntegerQ[4*p]

Rule 621

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

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int (A+B x) \left (a+b x+c x^2\right )^{5/2} \, dx &=\frac{B \left (a+b x+c x^2\right )^{7/2}}{7 c}+\frac{(-b B+2 A c) \int \left (a+b x+c x^2\right )^{5/2} \, dx}{2 c}\\ &=-\frac{(b B-2 A c) (b+2 c x) \left (a+b x+c x^2\right )^{5/2}}{24 c^2}+\frac{B \left (a+b x+c x^2\right )^{7/2}}{7 c}+\frac{\left (5 \left (b^2-4 a c\right ) (b B-2 A c)\right ) \int \left (a+b x+c x^2\right )^{3/2} \, dx}{48 c^2}\\ &=\frac{5 \left (b^2-4 a c\right ) (b B-2 A c) (b+2 c x) \left (a+b x+c x^2\right )^{3/2}}{384 c^3}-\frac{(b B-2 A c) (b+2 c x) \left (a+b x+c x^2\right )^{5/2}}{24 c^2}+\frac{B \left (a+b x+c x^2\right )^{7/2}}{7 c}-\frac{\left (5 \left (b^2-4 a c\right )^2 (b B-2 A c)\right ) \int \sqrt{a+b x+c x^2} \, dx}{256 c^3}\\ &=-\frac{5 \left (b^2-4 a c\right )^2 (b B-2 A c) (b+2 c x) \sqrt{a+b x+c x^2}}{1024 c^4}+\frac{5 \left (b^2-4 a c\right ) (b B-2 A c) (b+2 c x) \left (a+b x+c x^2\right )^{3/2}}{384 c^3}-\frac{(b B-2 A c) (b+2 c x) \left (a+b x+c x^2\right )^{5/2}}{24 c^2}+\frac{B \left (a+b x+c x^2\right )^{7/2}}{7 c}+\frac{\left (5 \left (b^2-4 a c\right )^3 (b B-2 A c)\right ) \int \frac{1}{\sqrt{a+b x+c x^2}} \, dx}{2048 c^4}\\ &=-\frac{5 \left (b^2-4 a c\right )^2 (b B-2 A c) (b+2 c x) \sqrt{a+b x+c x^2}}{1024 c^4}+\frac{5 \left (b^2-4 a c\right ) (b B-2 A c) (b+2 c x) \left (a+b x+c x^2\right )^{3/2}}{384 c^3}-\frac{(b B-2 A c) (b+2 c x) \left (a+b x+c x^2\right )^{5/2}}{24 c^2}+\frac{B \left (a+b x+c x^2\right )^{7/2}}{7 c}+\frac{\left (5 \left (b^2-4 a c\right )^3 (b B-2 A c)\right ) \operatorname{Subst}\left (\int \frac{1}{4 c-x^2} \, dx,x,\frac{b+2 c x}{\sqrt{a+b x+c x^2}}\right )}{1024 c^4}\\ &=-\frac{5 \left (b^2-4 a c\right )^2 (b B-2 A c) (b+2 c x) \sqrt{a+b x+c x^2}}{1024 c^4}+\frac{5 \left (b^2-4 a c\right ) (b B-2 A c) (b+2 c x) \left (a+b x+c x^2\right )^{3/2}}{384 c^3}-\frac{(b B-2 A c) (b+2 c x) \left (a+b x+c x^2\right )^{5/2}}{24 c^2}+\frac{B \left (a+b x+c x^2\right )^{7/2}}{7 c}+\frac{5 \left (b^2-4 a c\right )^3 (b B-2 A c) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{2048 c^{9/2}}\\ \end{align*}

Mathematica [A]  time = 0.284057, size = 179, normalized size = 0.88 \[ \frac{B (a+x (b+c x))^{7/2}}{7 c}-\frac{(b B-2 A c) \left (256 c^{5/2} (b+2 c x) (a+x (b+c x))^{5/2}-5 \left (b^2-4 a c\right ) \left (16 c^{3/2} (b+2 c x) (a+x (b+c x))^{3/2}-3 \left (b^2-4 a c\right ) \left (2 \sqrt{c} (b+2 c x) \sqrt{a+x (b+c x)}-\left (b^2-4 a c\right ) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+x (b+c x)}}\right )\right )\right )\right )}{6144 c^{9/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(B*(a + x*(b + c*x))^(7/2))/(7*c) - ((b*B - 2*A*c)*(256*c^(5/2)*(b + 2*c*x)*(a + x*(b + c*x))^(5/2) - 5*(b^2 -
4*a*c)*(16*c^(3/2)*(b + 2*c*x)*(a + x*(b + c*x))^(3/2) - 3*(b^2 - 4*a*c)*(2*Sqrt[c]*(b + 2*c*x)*Sqrt[a + x*(b
+ c*x)] - (b^2 - 4*a*c)*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + x*(b + c*x)])]))))/(6144*c^(9/2))

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Maple [B]  time = 0.006, size = 807, normalized size = 4. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-5/48*B*b/c*(c*x^2+b*x+a)^(3/2)*x*a-5/32*B*b/c*(c*x^2+b*x+a)^(1/2)*x*a^2+1/7*B*(c*x^2+b*x+a)^(7/2)/c-5/512*B*b
^5/c^3*(c*x^2+b*x+a)^(1/2)*x-5/96*B*b^2/c^2*(c*x^2+b*x+a)^(3/2)*a-1/12*B*b/c*(c*x^2+b*x+a)^(5/2)*x+5/192*B*b^3
/c^2*(c*x^2+b*x+a)^(3/2)*x-5/64*B*b^2/c^2*(c*x^2+b*x+a)^(1/2)*a^2+5/128*B*b^4/c^3*(c*x^2+b*x+a)^(1/2)*a-5/32*B
*b/c^(3/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*a^3+15/128*B*b^3/c^(5/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+
b*x+a)^(1/2))*a^2+1/6*A*x*(c*x^2+b*x+a)^(5/2)-15/512*B*b^5/c^(7/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))
*a+5/48*A/c*(c*x^2+b*x+a)^(3/2)*b*a+5/256*A/c^2*(c*x^2+b*x+a)^(1/2)*x*b^4-5/96*A/c*(c*x^2+b*x+a)^(3/2)*x*b^2+5
/32*A/c*(c*x^2+b*x+a)^(1/2)*b*a^2-5/64*A/c^2*(c*x^2+b*x+a)^(1/2)*b^3*a+15/256*A/c^(5/2)*ln((1/2*b+c*x)/c^(1/2)
+(c*x^2+b*x+a)^(1/2))*b^4*a-15/64*A/c^(3/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*b^2*a^2-5/32*A/c*(c*x^
2+b*x+a)^(1/2)*x*a*b^2+5/64*B*b^3/c^2*(c*x^2+b*x+a)^(1/2)*x*a+1/12*A/c*(c*x^2+b*x+a)^(5/2)*b+5/24*A*(c*x^2+b*x
+a)^(3/2)*x*a-5/192*A/c^2*(c*x^2+b*x+a)^(3/2)*b^3+5/2048*B*b^7/c^(9/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1
/2))-1/24*B*b^2/c^2*(c*x^2+b*x+a)^(5/2)+5/384*B*b^4/c^3*(c*x^2+b*x+a)^(3/2)-5/1024*B*b^6/c^4*(c*x^2+b*x+a)^(1/
2)+5/512*A/c^3*(c*x^2+b*x+a)^(1/2)*b^5+5/16*A/c^(1/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*a^3-5/1024*A
/c^(7/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*b^6+5/16*A*(c*x^2+b*x+a)^(1/2)*x*a^2

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.8315, size = 1982, normalized size = 9.76 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/86016*(105*(B*b^7 + 128*A*a^3*c^4 - 32*(2*B*a^3*b + 3*A*a^2*b^2)*c^3 + 24*(2*B*a^2*b^3 + A*a*b^4)*c^2 - 2*
(6*B*a*b^5 + A*b^6)*c)*sqrt(c)*log(-8*c^2*x^2 - 8*b*c*x - b^2 + 4*sqrt(c*x^2 + b*x + a)*(2*c*x + b)*sqrt(c) -
4*a*c) - 4*(3072*B*c^7*x^6 - 105*B*b^6*c + 256*(29*B*b*c^6 + 14*A*c^7)*x^5 + 96*(32*B*a^3 + 77*A*a^2*b)*c^4 +
128*(37*B*b^2*c^5 + 2*(36*B*a + 35*A*b)*c^6)*x^4 - 112*(33*B*a^2*b^2 + 20*A*a*b^3)*c^3 + 16*(3*B*b^3*c^4 + 728
*A*a*c^6 + 2*(394*B*a*b + 189*A*b^2)*c^5)*x^3 + 70*(16*B*a*b^4 + 3*A*b^5)*c^2 - 8*(7*B*b^4*c^3 - 24*(48*B*a^2
+ 91*A*a*b)*c^5 - 2*(30*B*a*b^2 + 7*A*b^3)*c^4)*x^2 + 2*(35*B*b^5*c^2 + 7392*A*a^2*c^5 + 48*(19*B*a^2*b + 14*A
*a*b^2)*c^4 - 14*(24*B*a*b^3 + 5*A*b^4)*c^3)*x)*sqrt(c*x^2 + b*x + a))/c^5, -1/43008*(105*(B*b^7 + 128*A*a^3*c
^4 - 32*(2*B*a^3*b + 3*A*a^2*b^2)*c^3 + 24*(2*B*a^2*b^3 + A*a*b^4)*c^2 - 2*(6*B*a*b^5 + A*b^6)*c)*sqrt(-c)*arc
tan(1/2*sqrt(c*x^2 + b*x + a)*(2*c*x + b)*sqrt(-c)/(c^2*x^2 + b*c*x + a*c)) - 2*(3072*B*c^7*x^6 - 105*B*b^6*c
+ 256*(29*B*b*c^6 + 14*A*c^7)*x^5 + 96*(32*B*a^3 + 77*A*a^2*b)*c^4 + 128*(37*B*b^2*c^5 + 2*(36*B*a + 35*A*b)*c
^6)*x^4 - 112*(33*B*a^2*b^2 + 20*A*a*b^3)*c^3 + 16*(3*B*b^3*c^4 + 728*A*a*c^6 + 2*(394*B*a*b + 189*A*b^2)*c^5)
*x^3 + 70*(16*B*a*b^4 + 3*A*b^5)*c^2 - 8*(7*B*b^4*c^3 - 24*(48*B*a^2 + 91*A*a*b)*c^5 - 2*(30*B*a*b^2 + 7*A*b^3
)*c^4)*x^2 + 2*(35*B*b^5*c^2 + 7392*A*a^2*c^5 + 48*(19*B*a^2*b + 14*A*a*b^2)*c^4 - 14*(24*B*a*b^3 + 5*A*b^4)*c
^3)*x)*sqrt(c*x^2 + b*x + a))/c^5]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (A + B x\right ) \left (a + b x + c x^{2}\right )^{\frac{5}{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x**2+b*x+a)**(5/2),x)

[Out]

Integral((A + B*x)*(a + b*x + c*x**2)**(5/2), x)

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Giac [B]  time = 1.32425, size = 574, normalized size = 2.83 \begin{align*} \frac{1}{21504} \, \sqrt{c x^{2} + b x + a}{\left (2 \,{\left (4 \,{\left (2 \,{\left (8 \,{\left (2 \,{\left (12 \, B c^{2} x + \frac{29 \, B b c^{7} + 14 \, A c^{8}}{c^{6}}\right )} x + \frac{37 \, B b^{2} c^{6} + 72 \, B a c^{7} + 70 \, A b c^{7}}{c^{6}}\right )} x + \frac{3 \, B b^{3} c^{5} + 788 \, B a b c^{6} + 378 \, A b^{2} c^{6} + 728 \, A a c^{7}}{c^{6}}\right )} x - \frac{7 \, B b^{4} c^{4} - 60 \, B a b^{2} c^{5} - 14 \, A b^{3} c^{5} - 1152 \, B a^{2} c^{6} - 2184 \, A a b c^{6}}{c^{6}}\right )} x + \frac{35 \, B b^{5} c^{3} - 336 \, B a b^{3} c^{4} - 70 \, A b^{4} c^{4} + 912 \, B a^{2} b c^{5} + 672 \, A a b^{2} c^{5} + 7392 \, A a^{2} c^{6}}{c^{6}}\right )} x - \frac{105 \, B b^{6} c^{2} - 1120 \, B a b^{4} c^{3} - 210 \, A b^{5} c^{3} + 3696 \, B a^{2} b^{2} c^{4} + 2240 \, A a b^{3} c^{4} - 3072 \, B a^{3} c^{5} - 7392 \, A a^{2} b c^{5}}{c^{6}}\right )} - \frac{5 \,{\left (B b^{7} - 12 \, B a b^{5} c - 2 \, A b^{6} c + 48 \, B a^{2} b^{3} c^{2} + 24 \, A a b^{4} c^{2} - 64 \, B a^{3} b c^{3} - 96 \, A a^{2} b^{2} c^{3} + 128 \, A a^{3} c^{4}\right )} \log \left ({\left | -2 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )} \sqrt{c} - b \right |}\right )}{2048 \, c^{\frac{9}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/21504*sqrt(c*x^2 + b*x + a)*(2*(4*(2*(8*(2*(12*B*c^2*x + (29*B*b*c^7 + 14*A*c^8)/c^6)*x + (37*B*b^2*c^6 + 72
*B*a*c^7 + 70*A*b*c^7)/c^6)*x + (3*B*b^3*c^5 + 788*B*a*b*c^6 + 378*A*b^2*c^6 + 728*A*a*c^7)/c^6)*x - (7*B*b^4*
c^4 - 60*B*a*b^2*c^5 - 14*A*b^3*c^5 - 1152*B*a^2*c^6 - 2184*A*a*b*c^6)/c^6)*x + (35*B*b^5*c^3 - 336*B*a*b^3*c^
4 - 70*A*b^4*c^4 + 912*B*a^2*b*c^5 + 672*A*a*b^2*c^5 + 7392*A*a^2*c^6)/c^6)*x - (105*B*b^6*c^2 - 1120*B*a*b^4*
c^3 - 210*A*b^5*c^3 + 3696*B*a^2*b^2*c^4 + 2240*A*a*b^3*c^4 - 3072*B*a^3*c^5 - 7392*A*a^2*b*c^5)/c^6) - 5/2048
*(B*b^7 - 12*B*a*b^5*c - 2*A*b^6*c + 48*B*a^2*b^3*c^2 + 24*A*a*b^4*c^2 - 64*B*a^3*b*c^3 - 96*A*a^2*b^2*c^3 + 1
28*A*a^3*c^4)*log(abs(-2*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*sqrt(c) - b))/c^(9/2)