3.1185 \(\int (A+B x) (b x+c x^2)^{5/2} \, dx\)

Optimal. Leaf size=171 \[ -\frac{5 b^4 (b+2 c x) \sqrt{b x+c x^2} (b B-2 A c)}{1024 c^4}+\frac{5 b^2 (b+2 c x) \left (b x+c x^2\right )^{3/2} (b B-2 A c)}{384 c^3}+\frac{5 b^6 (b B-2 A c) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{1024 c^{9/2}}-\frac{(b+2 c x) \left (b x+c x^2\right )^{5/2} (b B-2 A c)}{24 c^2}+\frac{B \left (b x+c x^2\right )^{7/2}}{7 c} \]

[Out]

(-5*b^4*(b*B - 2*A*c)*(b + 2*c*x)*Sqrt[b*x + c*x^2])/(1024*c^4) + (5*b^2*(b*B - 2*A*c)*(b + 2*c*x)*(b*x + c*x^
2)^(3/2))/(384*c^3) - ((b*B - 2*A*c)*(b + 2*c*x)*(b*x + c*x^2)^(5/2))/(24*c^2) + (B*(b*x + c*x^2)^(7/2))/(7*c)
+ (5*b^6*(b*B - 2*A*c)*ArcTanh[(Sqrt[c]*x)/Sqrt[b*x + c*x^2]])/(1024*c^(9/2))

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Rubi [A]  time = 0.0729027, antiderivative size = 171, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, Rules used = {640, 612, 620, 206} \[ -\frac{5 b^4 (b+2 c x) \sqrt{b x+c x^2} (b B-2 A c)}{1024 c^4}+\frac{5 b^2 (b+2 c x) \left (b x+c x^2\right )^{3/2} (b B-2 A c)}{384 c^3}+\frac{5 b^6 (b B-2 A c) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{1024 c^{9/2}}-\frac{(b+2 c x) \left (b x+c x^2\right )^{5/2} (b B-2 A c)}{24 c^2}+\frac{B \left (b x+c x^2\right )^{7/2}}{7 c} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*x)*(b*x + c*x^2)^(5/2),x]

[Out]

(-5*b^4*(b*B - 2*A*c)*(b + 2*c*x)*Sqrt[b*x + c*x^2])/(1024*c^4) + (5*b^2*(b*B - 2*A*c)*(b + 2*c*x)*(b*x + c*x^
2)^(3/2))/(384*c^3) - ((b*B - 2*A*c)*(b + 2*c*x)*(b*x + c*x^2)^(5/2))/(24*c^2) + (B*(b*x + c*x^2)^(7/2))/(7*c)
+ (5*b^6*(b*B - 2*A*c)*ArcTanh[(Sqrt[c]*x)/Sqrt[b*x + c*x^2]])/(1024*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 620

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

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 (b x+c x^2\right )^{5/2} \, dx &=\frac{B \left (b x+c x^2\right )^{7/2}}{7 c}+\frac{(-b B+2 A c) \int \left (b x+c x^2\right )^{5/2} \, dx}{2 c}\\ &=-\frac{(b B-2 A c) (b+2 c x) \left (b x+c x^2\right )^{5/2}}{24 c^2}+\frac{B \left (b x+c x^2\right )^{7/2}}{7 c}+\frac{\left (5 b^2 (b B-2 A c)\right ) \int \left (b x+c x^2\right )^{3/2} \, dx}{48 c^2}\\ &=\frac{5 b^2 (b B-2 A c) (b+2 c x) \left (b x+c x^2\right )^{3/2}}{384 c^3}-\frac{(b B-2 A c) (b+2 c x) \left (b x+c x^2\right )^{5/2}}{24 c^2}+\frac{B \left (b x+c x^2\right )^{7/2}}{7 c}-\frac{\left (5 b^4 (b B-2 A c)\right ) \int \sqrt{b x+c x^2} \, dx}{256 c^3}\\ &=-\frac{5 b^4 (b B-2 A c) (b+2 c x) \sqrt{b x+c x^2}}{1024 c^4}+\frac{5 b^2 (b B-2 A c) (b+2 c x) \left (b x+c x^2\right )^{3/2}}{384 c^3}-\frac{(b B-2 A c) (b+2 c x) \left (b x+c x^2\right )^{5/2}}{24 c^2}+\frac{B \left (b x+c x^2\right )^{7/2}}{7 c}+\frac{\left (5 b^6 (b B-2 A c)\right ) \int \frac{1}{\sqrt{b x+c x^2}} \, dx}{2048 c^4}\\ &=-\frac{5 b^4 (b B-2 A c) (b+2 c x) \sqrt{b x+c x^2}}{1024 c^4}+\frac{5 b^2 (b B-2 A c) (b+2 c x) \left (b x+c x^2\right )^{3/2}}{384 c^3}-\frac{(b B-2 A c) (b+2 c x) \left (b x+c x^2\right )^{5/2}}{24 c^2}+\frac{B \left (b x+c x^2\right )^{7/2}}{7 c}+\frac{\left (5 b^6 (b B-2 A c)\right ) \operatorname{Subst}\left (\int \frac{1}{1-c x^2} \, dx,x,\frac{x}{\sqrt{b x+c x^2}}\right )}{1024 c^4}\\ &=-\frac{5 b^4 (b B-2 A c) (b+2 c x) \sqrt{b x+c x^2}}{1024 c^4}+\frac{5 b^2 (b B-2 A c) (b+2 c x) \left (b x+c x^2\right )^{3/2}}{384 c^3}-\frac{(b B-2 A c) (b+2 c x) \left (b x+c x^2\right )^{5/2}}{24 c^2}+\frac{B \left (b x+c x^2\right )^{7/2}}{7 c}+\frac{5 b^6 (b B-2 A c) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{1024 c^{9/2}}\\ \end{align*}

Mathematica [A]  time = 0.375875, size = 171, normalized size = 1. \[ \frac{(x (b+c x))^{7/2} \left (\frac{49 (b B-2 A c) \left (15 b^{11/2} \sinh ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{b}}\right )-\sqrt{c} \sqrt{x} \sqrt{\frac{c x}{b}+1} \left (8 b^3 c^2 x^2+432 b^2 c^3 x^3-10 b^4 c x+15 b^5+640 b c^4 x^4+256 c^5 x^5\right )\right )}{3072 c^{7/2} x^{7/2} \sqrt{\frac{c x}{b}+1}}+7 B (b+c x)^3\right )}{49 c (b+c x)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((x*(b + c*x))^(7/2)*(7*B*(b + c*x)^3 + (49*(b*B - 2*A*c)*(-(Sqrt[c]*Sqrt[x]*Sqrt[1 + (c*x)/b]*(15*b^5 - 10*b^
4*c*x + 8*b^3*c^2*x^2 + 432*b^2*c^3*x^3 + 640*b*c^4*x^4 + 256*c^5*x^5)) + 15*b^(11/2)*ArcSinh[(Sqrt[c]*Sqrt[x]
)/Sqrt[b]]))/(3072*c^(7/2)*x^(7/2)*Sqrt[1 + (c*x)/b])))/(49*c*(b + c*x)^3)

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Maple [B]  time = 0.006, size = 321, normalized size = 1.9 \begin{align*}{\frac{B}{7\,c} \left ( c{x}^{2}+bx \right ) ^{{\frac{7}{2}}}}-{\frac{bBx}{12\,c} \left ( c{x}^{2}+bx \right ) ^{{\frac{5}{2}}}}-{\frac{{b}^{2}B}{24\,{c}^{2}} \left ( c{x}^{2}+bx \right ) ^{{\frac{5}{2}}}}+{\frac{5\,{b}^{3}Bx}{192\,{c}^{2}} \left ( c{x}^{2}+bx \right ) ^{{\frac{3}{2}}}}+{\frac{5\,{b}^{4}B}{384\,{c}^{3}} \left ( c{x}^{2}+bx \right ) ^{{\frac{3}{2}}}}-{\frac{5\,B{b}^{5}x}{512\,{c}^{3}}\sqrt{c{x}^{2}+bx}}-{\frac{5\,B{b}^{6}}{1024\,{c}^{4}}\sqrt{c{x}^{2}+bx}}+{\frac{5\,B{b}^{7}}{2048}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx} \right ){c}^{-{\frac{9}{2}}}}+{\frac{Ax}{6} \left ( c{x}^{2}+bx \right ) ^{{\frac{5}{2}}}}+{\frac{Ab}{12\,c} \left ( c{x}^{2}+bx \right ) ^{{\frac{5}{2}}}}-{\frac{5\,A{b}^{2}x}{96\,c} \left ( c{x}^{2}+bx \right ) ^{{\frac{3}{2}}}}-{\frac{5\,A{b}^{3}}{192\,{c}^{2}} \left ( c{x}^{2}+bx \right ) ^{{\frac{3}{2}}}}+{\frac{5\,A{b}^{4}x}{256\,{c}^{2}}\sqrt{c{x}^{2}+bx}}+{\frac{5\,A{b}^{5}}{512\,{c}^{3}}\sqrt{c{x}^{2}+bx}}-{\frac{5\,A{b}^{6}}{1024}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx} \right ){c}^{-{\frac{7}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.60738, size = 914, normalized size = 5.35 \begin{align*} \left [-\frac{105 \,{\left (B b^{7} - 2 \, A b^{6} c\right )} \sqrt{c} \log \left (2 \, c x + b - 2 \, \sqrt{c x^{2} + b x} \sqrt{c}\right ) - 2 \,{\left (3072 \, B c^{7} x^{6} - 105 \, B b^{6} c + 210 \, A b^{5} c^{2} + 256 \,{\left (29 \, B b c^{6} + 14 \, A c^{7}\right )} x^{5} + 128 \,{\left (37 \, B b^{2} c^{5} + 70 \, A b c^{6}\right )} x^{4} + 48 \,{\left (B b^{3} c^{4} + 126 \, A b^{2} c^{5}\right )} x^{3} - 56 \,{\left (B b^{4} c^{3} - 2 \, A b^{3} c^{4}\right )} x^{2} + 70 \,{\left (B b^{5} c^{2} - 2 \, A b^{4} c^{3}\right )} x\right )} \sqrt{c x^{2} + b x}}{43008 \, c^{5}}, -\frac{105 \,{\left (B b^{7} - 2 \, A b^{6} c\right )} \sqrt{-c} \arctan \left (\frac{\sqrt{c x^{2} + b x} \sqrt{-c}}{c x}\right ) -{\left (3072 \, B c^{7} x^{6} - 105 \, B b^{6} c + 210 \, A b^{5} c^{2} + 256 \,{\left (29 \, B b c^{6} + 14 \, A c^{7}\right )} x^{5} + 128 \,{\left (37 \, B b^{2} c^{5} + 70 \, A b c^{6}\right )} x^{4} + 48 \,{\left (B b^{3} c^{4} + 126 \, A b^{2} c^{5}\right )} x^{3} - 56 \,{\left (B b^{4} c^{3} - 2 \, A b^{3} c^{4}\right )} x^{2} + 70 \,{\left (B b^{5} c^{2} - 2 \, A b^{4} c^{3}\right )} x\right )} \sqrt{c x^{2} + b x}}{21504 \, c^{5}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/43008*(105*(B*b^7 - 2*A*b^6*c)*sqrt(c)*log(2*c*x + b - 2*sqrt(c*x^2 + b*x)*sqrt(c)) - 2*(3072*B*c^7*x^6 -
105*B*b^6*c + 210*A*b^5*c^2 + 256*(29*B*b*c^6 + 14*A*c^7)*x^5 + 128*(37*B*b^2*c^5 + 70*A*b*c^6)*x^4 + 48*(B*b^
3*c^4 + 126*A*b^2*c^5)*x^3 - 56*(B*b^4*c^3 - 2*A*b^3*c^4)*x^2 + 70*(B*b^5*c^2 - 2*A*b^4*c^3)*x)*sqrt(c*x^2 + b
*x))/c^5, -1/21504*(105*(B*b^7 - 2*A*b^6*c)*sqrt(-c)*arctan(sqrt(c*x^2 + b*x)*sqrt(-c)/(c*x)) - (3072*B*c^7*x^
6 - 105*B*b^6*c + 210*A*b^5*c^2 + 256*(29*B*b*c^6 + 14*A*c^7)*x^5 + 128*(37*B*b^2*c^5 + 70*A*b*c^6)*x^4 + 48*(
B*b^3*c^4 + 126*A*b^2*c^5)*x^3 - 56*(B*b^4*c^3 - 2*A*b^3*c^4)*x^2 + 70*(B*b^5*c^2 - 2*A*b^4*c^3)*x)*sqrt(c*x^2
+ b*x))/c^5]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.30025, size = 298, normalized size = 1.74 \begin{align*} \frac{1}{21504} \, \sqrt{c x^{2} + b x}{\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} + 70 \, A b c^{7}}{c^{6}}\right )} x + \frac{3 \,{\left (B b^{3} c^{5} + 126 \, A b^{2} c^{6}\right )}}{c^{6}}\right )} x - \frac{7 \,{\left (B b^{4} c^{4} - 2 \, A b^{3} c^{5}\right )}}{c^{6}}\right )} x + \frac{35 \,{\left (B b^{5} c^{3} - 2 \, A b^{4} c^{4}\right )}}{c^{6}}\right )} x - \frac{105 \,{\left (B b^{6} c^{2} - 2 \, A b^{5} c^{3}\right )}}{c^{6}}\right )} - \frac{5 \,{\left (B b^{7} - 2 \, A b^{6} c\right )} \log \left ({\left | -2 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\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)^(5/2),x, algorithm="giac")

[Out]

1/21504*sqrt(c*x^2 + b*x)*(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 + 70*A*b
*c^7)/c^6)*x + 3*(B*b^3*c^5 + 126*A*b^2*c^6)/c^6)*x - 7*(B*b^4*c^4 - 2*A*b^3*c^5)/c^6)*x + 35*(B*b^5*c^3 - 2*A
*b^4*c^4)/c^6)*x - 105*(B*b^6*c^2 - 2*A*b^5*c^3)/c^6) - 5/2048*(B*b^7 - 2*A*b^6*c)*log(abs(-2*(sqrt(c)*x - sqr
t(c*x^2 + b*x))*sqrt(c) - b))/c^(9/2)