3.3.97 \(\int (d+e x) (b x+c x^2)^{3/2} \, dx\) [297]

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

[Out]

1/16*(-b*e+2*c*d)*(2*c*x+b)*(c*x^2+b*x)^(3/2)/c^2+1/5*e*(c*x^2+b*x)^(5/2)/c+3/128*b^4*(-b*e+2*c*d)*arctanh(x*c
^(1/2)/(c*x^2+b*x)^(1/2))/c^(7/2)-3/128*b^2*(-b*e+2*c*d)*(2*c*x+b)*(c*x^2+b*x)^(1/2)/c^3

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Rubi [A]
time = 0.03, 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 = {654, 626, 634, 212} \begin {gather*} \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} \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 212

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

Rule 626

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 634

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 654

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 ) \text {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 = 0.23, size = 147, 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)+32 c^4 x^3 (5 d+4 e x)+16 b c^3 x^2 (15 d+11 e x)\right )+\frac {15 b^4 (-2 c d+b e) \log \left (-\sqrt {c} \sqrt {x}+\sqrt {b+c x}\right )}{\sqrt {x} \sqrt {b+c x}}\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^4*(-2*c*d + b*e)*Log[-(Sqrt[c]*Sqrt[x]) + Sqrt[b + c*x]])/(Sqrt[
x]*Sqrt[b + c*x])))/(640*c^(7/2))

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Maple [A]
time = 0.43, size = 201, normalized size = 1.47

method	result	size
risch	\(\frac {\left (128 c^{4} e \,x^{4}+176 b \,c^{3} e \,x^{3}+160 c^{4} d \,x^{3}+8 b^{2} c^{2} e \,x^{2}+240 b \,c^{3} d \,x^{2}-10 b^{3} c e x +20 x \,b^{2} c^{2} d +15 b^{4} e -30 b^{3} d c \right ) x \left (c x +b \right )}{640 c^{3} \sqrt {x \left (c x +b \right )}}-\frac {3 b^{5} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right ) e}{256 c^{\frac {7}{2}}}+\frac {3 b^{4} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right ) d}{128 c^{\frac {5}{2}}}\)	\(170\)
default	\(e \left (\frac {\left (c \,x^{2}+b x \right )^{\frac {5}{2}}}{5 c}-\frac {b \left (\frac {\left (2 c x +b \right ) \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{8 c}-\frac {3 b^{2} \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x}}{4 c}-\frac {b^{2} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{8 c^{\frac {3}{2}}}\right )}{16 c}\right )}{2 c}\right )+d \left (\frac {\left (2 c x +b \right ) \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{8 c}-\frac {3 b^{2} \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x}}{4 c}-\frac {b^{2} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{8 c^{\frac {3}{2}}}\right )}{16 c}\right )\)	\(201\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.31, size = 242, normalized size = 1.77 \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} x e}{64 \, c^{2}} - \frac {{\left (c x^{2} + b x\right )}^{\frac {3}{2}} b x e}{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*x*e/c^2 - 1/8*(c*x
^2 + b*x)^(3/2)*b*x*e/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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Fricas [A]
time = 1.49, size = 300, normalized size = 2.19 \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 (160 \, c^{5} d x^{3} + 240 \, b c^{4} d x^{2} + 20 \, b^{2} c^{3} d x - 30 \, b^{3} c^{2} d + {\left (128 \, c^{5} x^{4} + 176 \, b c^{4} x^{3} + 8 \, b^{2} c^{3} x^{2} - 10 \, b^{3} c^{2} x + 15 \, b^{4} c\right )} e\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 (160 \, c^{5} d x^{3} + 240 \, b c^{4} d x^{2} + 20 \, b^{2} c^{3} d x - 30 \, b^{3} c^{2} d + {\left (128 \, c^{5} x^{4} + 176 \, b c^{4} x^{3} + 8 \, b^{2} c^{3} x^{2} - 10 \, b^{3} c^{2} x + 15 \, b^{4} c\right )} e\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*(160*c^5*d*x^3 + 240
*b*c^4*d*x^2 + 20*b^2*c^3*d*x - 30*b^3*c^2*d + (128*c^5*x^4 + 176*b*c^4*x^3 + 8*b^2*c^3*x^2 - 10*b^3*c^2*x + 1
5*b^4*c)*e)*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)) - (160*c^5*d*x^3 + 240*b*c^4*d*x^2 + 20*b^2*c^3*d*x - 30*b^3*c^2*d + (128*c^5*x^4 + 176*b*c^4*x^3 + 8*b
^2*c^3*x^2 - 10*b^3*c^2*x + 15*b^4*c)*e)*sqrt(c*x^2 + b*x))/c^4]

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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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Giac [A]
time = 2.68, 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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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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