3.297 $$\int (d+e x) (b x+c x^2)^{3/2} \, dx$$

Optimal. Leaf size=137 $-\frac{3 b^2 (b+2 c x) \sqrt{b x+c x^2} (2 c d-b e)}{128 c^3}+\frac{3 b^4 (2 c d-b e) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{128 c^{7/2}}+\frac{(b+2 c x) \left (b x+c x^2\right )^{3/2} (2 c d-b e)}{16 c^2}+\frac{e \left (b x+c x^2\right )^{5/2}}{5 c}$

[Out]

(-3*b^2*(2*c*d - b*e)*(b + 2*c*x)*Sqrt[b*x + c*x^2])/(128*c^3) + ((2*c*d - b*e)*(b + 2*c*x)*(b*x + c*x^2)^(3/2
))/(16*c^2) + (e*(b*x + c*x^2)^(5/2))/(5*c) + (3*b^4*(2*c*d - b*e)*ArcTanh[(Sqrt[c]*x)/Sqrt[b*x + c*x^2]])/(12
8*c^(7/2))

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Rubi [A]  time = 0.0526371, antiderivative size = 137, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 19, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.21, Rules used = {640, 612, 620, 206} $-\frac{3 b^2 (b+2 c x) \sqrt{b x+c x^2} (2 c d-b e)}{128 c^3}+\frac{3 b^4 (2 c d-b e) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{128 c^{7/2}}+\frac{(b+2 c x) \left (b x+c x^2\right )^{3/2} (2 c d-b e)}{16 c^2}+\frac{e \left (b x+c x^2\right )^{5/2}}{5 c}$

Antiderivative was successfully veriﬁed.

[In]

Int[(d + e*x)*(b*x + c*x^2)^(3/2),x]

[Out]

(-3*b^2*(2*c*d - b*e)*(b + 2*c*x)*Sqrt[b*x + c*x^2])/(128*c^3) + ((2*c*d - b*e)*(b + 2*c*x)*(b*x + c*x^2)^(3/2
))/(16*c^2) + (e*(b*x + c*x^2)^(5/2))/(5*c) + (3*b^4*(2*c*d - b*e)*ArcTanh[(Sqrt[c]*x)/Sqrt[b*x + c*x^2]])/(12
8*c^(7/2))

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 (d+e x) \left (b x+c x^2\right )^{3/2} \, dx &=\frac{e \left (b x+c x^2\right )^{5/2}}{5 c}+\frac{(2 c d-b e) \int \left (b x+c x^2\right )^{3/2} \, dx}{2 c}\\ &=\frac{(2 c d-b e) (b+2 c x) \left (b x+c x^2\right )^{3/2}}{16 c^2}+\frac{e \left (b x+c x^2\right )^{5/2}}{5 c}-\frac{\left (3 b^2 (2 c d-b e)\right ) \int \sqrt{b x+c x^2} \, dx}{32 c^2}\\ &=-\frac{3 b^2 (2 c d-b e) (b+2 c x) \sqrt{b x+c x^2}}{128 c^3}+\frac{(2 c d-b e) (b+2 c x) \left (b x+c x^2\right )^{3/2}}{16 c^2}+\frac{e \left (b x+c x^2\right )^{5/2}}{5 c}+\frac{\left (3 b^4 (2 c d-b e)\right ) \int \frac{1}{\sqrt{b x+c x^2}} \, dx}{256 c^3}\\ &=-\frac{3 b^2 (2 c d-b e) (b+2 c x) \sqrt{b x+c x^2}}{128 c^3}+\frac{(2 c d-b e) (b+2 c x) \left (b x+c x^2\right )^{3/2}}{16 c^2}+\frac{e \left (b x+c x^2\right )^{5/2}}{5 c}+\frac{\left (3 b^4 (2 c d-b e)\right ) \operatorname{Subst}\left (\int \frac{1}{1-c x^2} \, dx,x,\frac{x}{\sqrt{b x+c x^2}}\right )}{128 c^3}\\ &=-\frac{3 b^2 (2 c d-b e) (b+2 c x) \sqrt{b x+c x^2}}{128 c^3}+\frac{(2 c d-b e) (b+2 c x) \left (b x+c x^2\right )^{3/2}}{16 c^2}+\frac{e \left (b x+c x^2\right )^{5/2}}{5 c}+\frac{3 b^4 (2 c d-b e) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{128 c^{7/2}}\\ \end{align*}

Mathematica [A]  time = 0.265041, size = 146, normalized size = 1.07 $\frac{\sqrt{x (b+c x)} \left (\sqrt{c} \left (4 b^2 c^2 x (5 d+2 e x)-10 b^3 c (3 d+e x)+15 b^4 e+16 b c^3 x^2 (15 d+11 e x)+32 c^4 x^3 (5 d+4 e x)\right )-\frac{15 b^{7/2} (b e-2 c d) \sinh ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{b}}\right )}{\sqrt{x} \sqrt{\frac{c x}{b}+1}}\right )}{640 c^{7/2}}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(d + e*x)*(b*x + c*x^2)^(3/2),x]

[Out]

(Sqrt[x*(b + c*x)]*(Sqrt[c]*(15*b^4*e - 10*b^3*c*(3*d + e*x) + 4*b^2*c^2*x*(5*d + 2*e*x) + 32*c^4*x^3*(5*d + 4
*e*x) + 16*b*c^3*x^2*(15*d + 11*e*x)) - (15*b^(7/2)*(-2*c*d + b*e)*ArcSinh[(Sqrt[c]*Sqrt[x])/Sqrt[b]])/(Sqrt[x
]*Sqrt[1 + (c*x)/b])))/(640*c^(7/2))

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Maple [B]  time = 0.049, size = 239, normalized size = 1.7 \begin{align*}{\frac{e}{5\,c} \left ( c{x}^{2}+bx \right ) ^{{\frac{5}{2}}}}-{\frac{bxe}{8\,c} \left ( c{x}^{2}+bx \right ) ^{{\frac{3}{2}}}}-{\frac{{b}^{2}e}{16\,{c}^{2}} \left ( c{x}^{2}+bx \right ) ^{{\frac{3}{2}}}}+{\frac{3\,e{b}^{3}x}{64\,{c}^{2}}\sqrt{c{x}^{2}+bx}}+{\frac{3\,e{b}^{4}}{128\,{c}^{3}}\sqrt{c{x}^{2}+bx}}-{\frac{3\,e{b}^{5}}{256}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx} \right ){c}^{-{\frac{7}{2}}}}+{\frac{dx}{4} \left ( c{x}^{2}+bx \right ) ^{{\frac{3}{2}}}}+{\frac{bd}{8\,c} \left ( c{x}^{2}+bx \right ) ^{{\frac{3}{2}}}}-{\frac{3\,{b}^{2}dx}{32\,c}\sqrt{c{x}^{2}+bx}}-{\frac{3\,d{b}^{3}}{64\,{c}^{2}}\sqrt{c{x}^{2}+bx}}+{\frac{3\,d{b}^{4}}{128}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx} \right ){c}^{-{\frac{5}{2}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)*(c*x^2+b*x)^(3/2),x)

[Out]

1/5*e*(c*x^2+b*x)^(5/2)/c-1/8*e*b/c*x*(c*x^2+b*x)^(3/2)-1/16*e*b^2/c^2*(c*x^2+b*x)^(3/2)+3/64*e*b^3/c^2*(c*x^2
+b*x)^(1/2)*x+3/128*e*b^4/c^3*(c*x^2+b*x)^(1/2)-3/256*e*b^5/c^(7/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x)^(1/2))+
1/4*d*x*(c*x^2+b*x)^(3/2)+1/8*d/c*(c*x^2+b*x)^(3/2)*b-3/32*d*b^2/c*(c*x^2+b*x)^(1/2)*x-3/64*d*b^3/c^2*(c*x^2+b
*x)^(1/2)+3/128*d*b^4/c^(5/2)*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((e*x+d)*(c*x^2+b*x)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.35333, size = 689, normalized size = 5.03 \begin{align*} \left [-\frac{15 \,{\left (2 \, b^{4} c d - b^{5} e\right )} \sqrt{c} \log \left (2 \, c x + b - 2 \, \sqrt{c x^{2} + b x} \sqrt{c}\right ) - 2 \,{\left (128 \, c^{5} e x^{4} - 30 \, b^{3} c^{2} d + 15 \, b^{4} c e + 16 \,{\left (10 \, c^{5} d + 11 \, b c^{4} e\right )} x^{3} + 8 \,{\left (30 \, b c^{4} d + b^{2} c^{3} e\right )} x^{2} + 10 \,{\left (2 \, b^{2} c^{3} d - b^{3} c^{2} e\right )} x\right )} \sqrt{c x^{2} + b x}}{1280 \, c^{4}}, -\frac{15 \,{\left (2 \, b^{4} c d - b^{5} e\right )} \sqrt{-c} \arctan \left (\frac{\sqrt{c x^{2} + b x} \sqrt{-c}}{c x}\right ) -{\left (128 \, c^{5} e x^{4} - 30 \, b^{3} c^{2} d + 15 \, b^{4} c e + 16 \,{\left (10 \, c^{5} d + 11 \, b c^{4} e\right )} x^{3} + 8 \,{\left (30 \, b c^{4} d + b^{2} c^{3} e\right )} x^{2} + 10 \,{\left (2 \, b^{2} c^{3} d - b^{3} c^{2} e\right )} x\right )} \sqrt{c x^{2} + b x}}{640 \, c^{4}}\right ] \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)*(c*x^2+b*x)^(3/2),x, algorithm="fricas")

[Out]

[-1/1280*(15*(2*b^4*c*d - b^5*e)*sqrt(c)*log(2*c*x + b - 2*sqrt(c*x^2 + b*x)*sqrt(c)) - 2*(128*c^5*e*x^4 - 30*
b^3*c^2*d + 15*b^4*c*e + 16*(10*c^5*d + 11*b*c^4*e)*x^3 + 8*(30*b*c^4*d + b^2*c^3*e)*x^2 + 10*(2*b^2*c^3*d - b
^3*c^2*e)*x)*sqrt(c*x^2 + b*x))/c^4, -1/640*(15*(2*b^4*c*d - b^5*e)*sqrt(-c)*arctan(sqrt(c*x^2 + b*x)*sqrt(-c)
/(c*x)) - (128*c^5*e*x^4 - 30*b^3*c^2*d + 15*b^4*c*e + 16*(10*c^5*d + 11*b*c^4*e)*x^3 + 8*(30*b*c^4*d + b^2*c^
3*e)*x^2 + 10*(2*b^2*c^3*d - b^3*c^2*e)*x)*sqrt(c*x^2 + b*x))/c^4]

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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{3}{2}} \left (d + e x\right )\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)*(c*x**2+b*x)**(3/2),x)

[Out]

Integral((x*(b + c*x))**(3/2)*(d + e*x), x)

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Giac [A]  time = 1.27819, size = 231, normalized size = 1.69 \begin{align*} \frac{1}{640} \, \sqrt{c x^{2} + b x}{\left (2 \,{\left (4 \,{\left (2 \,{\left (8 \, c x e + \frac{10 \, c^{5} d + 11 \, b c^{4} e}{c^{4}}\right )} x + \frac{30 \, b c^{4} d + b^{2} c^{3} e}{c^{4}}\right )} x + \frac{5 \,{\left (2 \, b^{2} c^{3} d - b^{3} c^{2} e\right )}}{c^{4}}\right )} x - \frac{15 \,{\left (2 \, b^{3} c^{2} d - b^{4} c e\right )}}{c^{4}}\right )} - \frac{3 \,{\left (2 \, b^{4} c d - b^{5} e\right )} \log \left ({\left | -2 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )} \sqrt{c} - b \right |}\right )}{256 \, c^{\frac{7}{2}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)*(c*x^2+b*x)^(3/2),x, algorithm="giac")

[Out]

1/640*sqrt(c*x^2 + b*x)*(2*(4*(2*(8*c*x*e + (10*c^5*d + 11*b*c^4*e)/c^4)*x + (30*b*c^4*d + b^2*c^3*e)/c^4)*x +
5*(2*b^2*c^3*d - b^3*c^2*e)/c^4)*x - 15*(2*b^3*c^2*d - b^4*c*e)/c^4) - 3/256*(2*b^4*c*d - b^5*e)*log(abs(-2*(
sqrt(c)*x - sqrt(c*x^2 + b*x))*sqrt(c) - b))/c^(7/2)