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

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

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Rubi [A]  time = 0.05, antiderivative size = 137, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {640, 612, 620, 206} \begin {gather*} -\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} \end {gather*}

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 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])

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 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]

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*}

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Mathematica [A]  time = 0.25, size = 146, normalized size = 1.07 \begin {gather*} \frac {\sqrt {x (b+c x)} \left (\sqrt {c} \left (15 b^4 e-10 b^3 c (3 d+e x)+4 b^2 c^2 x (5 d+2 e x)+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}} \end {gather*}

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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IntegrateAlgebraic [A]  time = 0.69, size = 152, normalized size = 1.11 \begin {gather*} \frac {3 \left (b^5 e-2 b^4 c d\right ) \log \left (-2 \sqrt {c} \sqrt {b x+c x^2}+b+2 c x\right )}{256 c^{7/2}}+\frac {\sqrt {b x+c x^2} \left (15 b^4 e-30 b^3 c d-10 b^3 c e x+20 b^2 c^2 d x+8 b^2 c^2 e x^2+240 b c^3 d x^2+176 b c^3 e x^3+160 c^4 d x^3+128 c^4 e x^4\right )}{640 c^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[b*x + c*x^2]*(-30*b^3*c*d + 15*b^4*e + 20*b^2*c^2*d*x - 10*b^3*c*e*x + 240*b*c^3*d*x^2 + 8*b^2*c^2*e*x^2
+ 160*c^4*d*x^3 + 176*b*c^3*e*x^3 + 128*c^4*e*x^4))/(640*c^3) + (3*(-2*b^4*c*d + b^5*e)*Log[b + 2*c*x - 2*Sqr
t[c]*Sqrt[b*x + c*x^2]])/(256*c^(7/2))

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fricas [A]  time = 0.43, size = 302, normalized size = 2.20 \begin {gather*} \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 {gather*}

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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giac [A]  time = 0.22, size = 171, normalized size = 1.25 \begin {gather*} \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 {gather*}

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)

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maple [B]  time = 0.05, size = 239, normalized size = 1.74 \begin {gather*} -\frac {3 b^{5} e \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{256 c^{\frac {7}{2}}}+\frac {3 b^{4} d \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{128 c^{\frac {5}{2}}}+\frac {3 \sqrt {c \,x^{2}+b x}\, b^{3} e x}{64 c^{2}}-\frac {3 \sqrt {c \,x^{2}+b x}\, b^{2} d x}{32 c}+\frac {3 \sqrt {c \,x^{2}+b x}\, b^{4} e}{128 c^{3}}-\frac {3 \sqrt {c \,x^{2}+b x}\, b^{3} d}{64 c^{2}}-\frac {\left (c \,x^{2}+b x \right )^{\frac {3}{2}} b e x}{8 c}+\frac {\left (c \,x^{2}+b x \right )^{\frac {3}{2}} d x}{4}-\frac {\left (c \,x^{2}+b x \right )^{\frac {3}{2}} b^{2} e}{16 c^{2}}+\frac {\left (c \,x^{2}+b x \right )^{\frac {3}{2}} b d}{8 c}+\frac {\left (c \,x^{2}+b x \right )^{\frac {5}{2}} e}{5 c} \end {gather*}

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((c*x+1/2*b)/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((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x)^(1/2))

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maxima [B]  time = 1.41, size = 236, normalized size = 1.72 \begin {gather*} \frac {1}{4} \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} d x - \frac {3 \, \sqrt {c x^{2} + b x} b^{2} d x}{32 \, c} + \frac {3 \, \sqrt {c x^{2} + b x} b^{3} e x}{64 \, c^{2}} - \frac {{\left (c x^{2} + b x\right )}^{\frac {3}{2}} b e x}{8 \, c} + \frac {3 \, b^{4} d \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{128 \, c^{\frac {5}{2}}} - \frac {3 \, b^{5} e \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{256 \, c^{\frac {7}{2}}} - \frac {3 \, \sqrt {c x^{2} + b x} b^{3} d}{64 \, c^{2}} + \frac {{\left (c x^{2} + b x\right )}^{\frac {3}{2}} b d}{8 \, c} + \frac {3 \, \sqrt {c x^{2} + b x} b^{4} e}{128 \, c^{3}} - \frac {{\left (c x^{2} + b x\right )}^{\frac {3}{2}} b^{2} e}{16 \, c^{2}} + \frac {{\left (c x^{2} + b x\right )}^{\frac {5}{2}} e}{5 \, c} \end {gather*}

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]

1/4*(c*x^2 + b*x)^(3/2)*d*x - 3/32*sqrt(c*x^2 + b*x)*b^2*d*x/c + 3/64*sqrt(c*x^2 + b*x)*b^3*e*x/c^2 - 1/8*(c*x
^2 + b*x)^(3/2)*b*e*x/c + 3/128*b^4*d*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c))/c^(5/2) - 3/256*b^5*e*log(2
*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c))/c^(7/2) - 3/64*sqrt(c*x^2 + b*x)*b^3*d/c^2 + 1/8*(c*x^2 + b*x)^(3/2)*b
*d/c + 3/128*sqrt(c*x^2 + b*x)*b^4*e/c^3 - 1/16*(c*x^2 + b*x)^(3/2)*b^2*e/c^2 + 1/5*(c*x^2 + b*x)^(5/2)*e/c

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mupad [B]  time = 0.67, size = 208, normalized size = 1.52 \begin {gather*} \frac {e\,{\left (c\,x^2+b\,x\right )}^{5/2}}{5\,c}-\frac {3\,b^2\,d\,\left (\sqrt {c\,x^2+b\,x}\,\left (\frac {x}{2}+\frac {b}{4\,c}\right )-\frac {b^2\,\ln \left (\frac {\frac {b}{2}+c\,x}{\sqrt {c}}+\sqrt {c\,x^2+b\,x}\right )}{8\,c^{3/2}}\right )}{16\,c}+\frac {d\,{\left (c\,x^2+b\,x\right )}^{3/2}\,\left (\frac {b}{2}+c\,x\right )}{4\,c}-\frac {b\,e\,\left (\frac {x\,{\left (c\,x^2+b\,x\right )}^{3/2}}{4}+\frac {b\,{\left (c\,x^2+b\,x\right )}^{3/2}}{8\,c}-\frac {3\,b^2\,\left (\frac {\sqrt {c\,x^2+b\,x}\,\left (b+2\,c\,x\right )}{4\,c}-\frac {b^2\,\ln \left (\frac {\frac {b}{2}+c\,x}{\sqrt {c}}+\sqrt {c\,x^2+b\,x}\right )}{8\,c^{3/2}}\right )}{16\,c}\right )}{2\,c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x + c*x^2)^(3/2)*(d + e*x),x)

[Out]

(e*(b*x + c*x^2)^(5/2))/(5*c) - (3*b^2*d*((b*x + c*x^2)^(1/2)*(x/2 + b/(4*c)) - (b^2*log((b/2 + c*x)/c^(1/2) +
(b*x + c*x^2)^(1/2)))/(8*c^(3/2))))/(16*c) + (d*(b*x + c*x^2)^(3/2)*(b/2 + c*x))/(4*c) - (b*e*((x*(b*x + c*x^
2)^(3/2))/4 + (b*(b*x + c*x^2)^(3/2))/(8*c) - (3*b^2*(((b*x + c*x^2)^(1/2)*(b + 2*c*x))/(4*c) - (b^2*log((b/2
+ c*x)/c^(1/2) + (b*x + c*x^2)^(1/2)))/(8*c^(3/2))))/(16*c)))/(2*c)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (x \left (b + c x\right )\right )^{\frac {3}{2}} \left (d + e x\right )\, dx \end {gather*}

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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