3.3.82 \(\int (d+e x) (b x+c x^2)^{5/2} \, dx\)

Optimal. Leaf size=175 \[ -\frac {5 b^6 (2 c d-b e) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{1024 c^{9/2}}+\frac {5 b^4 (b+2 c x) \sqrt {b x+c x^2} (2 c d-b e)}{1024 c^4}-\frac {5 b^2 (b+2 c x) \left (b x+c x^2\right )^{3/2} (2 c d-b e)}{384 c^3}+\frac {(b+2 c x) \left (b x+c x^2\right )^{5/2} (2 c d-b e)}{24 c^2}+\frac {e \left (b x+c x^2\right )^{7/2}}{7 c} \]

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Rubi [A]  time = 0.07, antiderivative size = 175, normalized size of antiderivative = 1.00, number of steps used = 6, 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 {5 b^4 (b+2 c x) \sqrt {b x+c x^2} (2 c d-b e)}{1024 c^4}-\frac {5 b^2 (b+2 c x) \left (b x+c x^2\right )^{3/2} (2 c d-b e)}{384 c^3}-\frac {5 b^6 (2 c d-b e) \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} (2 c d-b e)}{24 c^2}+\frac {e \left (b x+c x^2\right )^{7/2}}{7 c} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(d + e*x)*(b*x + c*x^2)^(5/2),x]

[Out]

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

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

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Mathematica [A]  time = 0.29, size = 171, normalized size = 0.98 \begin {gather*} \frac {(x (b+c x))^{7/2} \left (\frac {49 (2 c d-b e) \left (\sqrt {c} \sqrt {x} \sqrt {\frac {c x}{b}+1} \left (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\right )-15 b^{11/2} \sinh ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )\right )}{3072 c^{7/2} x^{7/2} \sqrt {\frac {c x}{b}+1}}+7 e (b+c x)^3\right )}{49 c (b+c x)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x)*(b*x + c*x^2)^(5/2),x]

[Out]

((x*(b + c*x))^(7/2)*(7*e*(b + c*x)^3 + (49*(2*c*d - b*e)*(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])/S
qrt[b]]))/(3072*c^(7/2)*x^(7/2)*Sqrt[1 + (c*x)/b])))/(49*c*(b + c*x)^3)

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IntegrateAlgebraic [A]  time = 0.90, size = 200, normalized size = 1.14 \begin {gather*} \frac {\sqrt {b x+c x^2} \left (-105 b^6 e+210 b^5 c d+70 b^5 c e x-140 b^4 c^2 d x-56 b^4 c^2 e x^2+112 b^3 c^3 d x^2+48 b^3 c^3 e x^3+6048 b^2 c^4 d x^3+4736 b^2 c^4 e x^4+8960 b c^5 d x^4+7424 b c^5 e x^5+3584 c^6 d x^5+3072 c^6 e x^6\right )}{21504 c^4}-\frac {5 \left (b^7 e-2 b^6 c d\right ) \log \left (-2 \sqrt {c} \sqrt {b x+c x^2}+b+2 c x\right )}{2048 c^{9/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(d + e*x)*(b*x + c*x^2)^(5/2),x]

[Out]

(Sqrt[b*x + c*x^2]*(210*b^5*c*d - 105*b^6*e - 140*b^4*c^2*d*x + 70*b^5*c*e*x + 112*b^3*c^3*d*x^2 - 56*b^4*c^2*
e*x^2 + 6048*b^2*c^4*d*x^3 + 48*b^3*c^3*e*x^3 + 8960*b*c^5*d*x^4 + 4736*b^2*c^4*e*x^4 + 3584*c^6*d*x^5 + 7424*
b*c^5*e*x^5 + 3072*c^6*e*x^6))/(21504*c^4) - (5*(-2*b^6*c*d + b^7*e)*Log[b + 2*c*x - 2*Sqrt[c]*Sqrt[b*x + c*x^
2]])/(2048*c^(9/2))

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fricas [A]  time = 0.42, size = 397, normalized size = 2.27 \begin {gather*} \left [-\frac {105 \, {\left (2 \, b^{6} c d - b^{7} e\right )} \sqrt {c} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) - 2 \, {\left (3072 \, c^{7} e x^{6} + 210 \, b^{5} c^{2} d - 105 \, b^{6} c e + 256 \, {\left (14 \, c^{7} d + 29 \, b c^{6} e\right )} x^{5} + 128 \, {\left (70 \, b c^{6} d + 37 \, b^{2} c^{5} e\right )} x^{4} + 48 \, {\left (126 \, b^{2} c^{5} d + b^{3} c^{4} e\right )} x^{3} + 56 \, {\left (2 \, b^{3} c^{4} d - b^{4} c^{3} e\right )} x^{2} - 70 \, {\left (2 \, b^{4} c^{3} d - b^{5} c^{2} e\right )} x\right )} \sqrt {c x^{2} + b x}}{43008 \, c^{5}}, \frac {105 \, {\left (2 \, b^{6} c d - b^{7} e\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x}\right ) + {\left (3072 \, c^{7} e x^{6} + 210 \, b^{5} c^{2} d - 105 \, b^{6} c e + 256 \, {\left (14 \, c^{7} d + 29 \, b c^{6} e\right )} x^{5} + 128 \, {\left (70 \, b c^{6} d + 37 \, b^{2} c^{5} e\right )} x^{4} + 48 \, {\left (126 \, b^{2} c^{5} d + b^{3} c^{4} e\right )} x^{3} + 56 \, {\left (2 \, b^{3} c^{4} d - b^{4} c^{3} e\right )} x^{2} - 70 \, {\left (2 \, b^{4} c^{3} d - b^{5} c^{2} e\right )} x\right )} \sqrt {c x^{2} + b x}}{21504 \, c^{5}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.30, size = 233, normalized size = 1.33 \begin {gather*} \frac {1}{21504} \, \sqrt {c x^{2} + b x} {\left (2 \, {\left (4 \, {\left (2 \, {\left (8 \, {\left (2 \, {\left (12 \, c^{2} x e + \frac {14 \, c^{8} d + 29 \, b c^{7} e}{c^{6}}\right )} x + \frac {70 \, b c^{7} d + 37 \, b^{2} c^{6} e}{c^{6}}\right )} x + \frac {3 \, {\left (126 \, b^{2} c^{6} d + b^{3} c^{5} e\right )}}{c^{6}}\right )} x + \frac {7 \, {\left (2 \, b^{3} c^{5} d - b^{4} c^{4} e\right )}}{c^{6}}\right )} x - \frac {35 \, {\left (2 \, b^{4} c^{4} d - b^{5} c^{3} e\right )}}{c^{6}}\right )} x + \frac {105 \, {\left (2 \, b^{5} c^{3} d - b^{6} c^{2} e\right )}}{c^{6}}\right )} + \frac {5 \, {\left (2 \, b^{6} c d - b^{7} e\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 {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)*(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*c^2*x*e + (14*c^8*d + 29*b*c^7*e)/c^6)*x + (70*b*c^7*d + 37*b^2*c
^6*e)/c^6)*x + 3*(126*b^2*c^6*d + b^3*c^5*e)/c^6)*x + 7*(2*b^3*c^5*d - b^4*c^4*e)/c^6)*x - 35*(2*b^4*c^4*d - b
^5*c^3*e)/c^6)*x + 105*(2*b^5*c^3*d - b^6*c^2*e)/c^6) + 5/2048*(2*b^6*c*d - b^7*e)*log(abs(-2*(sqrt(c)*x - sqr
t(c*x^2 + b*x))*sqrt(c) - b))/c^(9/2)

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maple [B]  time = 0.04, size = 321, normalized size = 1.83 \begin {gather*} \frac {5 b^{7} e \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{2048 c^{\frac {9}{2}}}-\frac {5 b^{6} d \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{1024 c^{\frac {7}{2}}}-\frac {5 \sqrt {c \,x^{2}+b x}\, b^{5} e x}{512 c^{3}}+\frac {5 \sqrt {c \,x^{2}+b x}\, b^{4} d x}{256 c^{2}}-\frac {5 \sqrt {c \,x^{2}+b x}\, b^{6} e}{1024 c^{4}}+\frac {5 \sqrt {c \,x^{2}+b x}\, b^{5} d}{512 c^{3}}+\frac {5 \left (c \,x^{2}+b x \right )^{\frac {3}{2}} b^{3} e x}{192 c^{2}}-\frac {5 \left (c \,x^{2}+b x \right )^{\frac {3}{2}} b^{2} d x}{96 c}+\frac {5 \left (c \,x^{2}+b x \right )^{\frac {3}{2}} b^{4} e}{384 c^{3}}-\frac {5 \left (c \,x^{2}+b x \right )^{\frac {3}{2}} b^{3} d}{192 c^{2}}-\frac {\left (c \,x^{2}+b x \right )^{\frac {5}{2}} b e x}{12 c}+\frac {\left (c \,x^{2}+b x \right )^{\frac {5}{2}} d x}{6}-\frac {\left (c \,x^{2}+b x \right )^{\frac {5}{2}} b^{2} e}{24 c^{2}}+\frac {\left (c \,x^{2}+b x \right )^{\frac {5}{2}} b d}{12 c}+\frac {\left (c \,x^{2}+b x \right )^{\frac {7}{2}} e}{7 c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)*(c*x^2+b*x)^(5/2),x)

[Out]

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

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maxima [B]  time = 1.45, size = 318, normalized size = 1.82 \begin {gather*} \frac {1}{6} \, {\left (c x^{2} + b x\right )}^{\frac {5}{2}} d x + \frac {5 \, \sqrt {c x^{2} + b x} b^{4} d x}{256 \, c^{2}} - \frac {5 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} b^{2} d x}{96 \, c} - \frac {5 \, \sqrt {c x^{2} + b x} b^{5} e x}{512 \, c^{3}} + \frac {5 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} b^{3} e x}{192 \, c^{2}} - \frac {{\left (c x^{2} + b x\right )}^{\frac {5}{2}} b e x}{12 \, c} - \frac {5 \, b^{6} d \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{1024 \, c^{\frac {7}{2}}} + \frac {5 \, b^{7} e \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{2048 \, c^{\frac {9}{2}}} + \frac {5 \, \sqrt {c x^{2} + b x} b^{5} d}{512 \, c^{3}} - \frac {5 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} b^{3} d}{192 \, c^{2}} + \frac {{\left (c x^{2} + b x\right )}^{\frac {5}{2}} b d}{12 \, c} - \frac {5 \, \sqrt {c x^{2} + b x} b^{6} e}{1024 \, c^{4}} + \frac {5 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} b^{4} e}{384 \, c^{3}} - \frac {{\left (c x^{2} + b x\right )}^{\frac {5}{2}} b^{2} e}{24 \, c^{2}} + \frac {{\left (c x^{2} + b x\right )}^{\frac {7}{2}} e}{7 \, c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)*(c*x^2+b*x)^(5/2),x, algorithm="maxima")

[Out]

1/6*(c*x^2 + b*x)^(5/2)*d*x + 5/256*sqrt(c*x^2 + b*x)*b^4*d*x/c^2 - 5/96*(c*x^2 + b*x)^(3/2)*b^2*d*x/c - 5/512
*sqrt(c*x^2 + b*x)*b^5*e*x/c^3 + 5/192*(c*x^2 + b*x)^(3/2)*b^3*e*x/c^2 - 1/12*(c*x^2 + b*x)^(5/2)*b*e*x/c - 5/
1024*b^6*d*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c))/c^(7/2) + 5/2048*b^7*e*log(2*c*x + b + 2*sqrt(c*x^2 +
b*x)*sqrt(c))/c^(9/2) + 5/512*sqrt(c*x^2 + b*x)*b^5*d/c^3 - 5/192*(c*x^2 + b*x)^(3/2)*b^3*d/c^2 + 1/12*(c*x^2
+ b*x)^(5/2)*b*d/c - 5/1024*sqrt(c*x^2 + b*x)*b^6*e/c^4 + 5/384*(c*x^2 + b*x)^(3/2)*b^4*e/c^3 - 1/24*(c*x^2 +
b*x)^(5/2)*b^2*e/c^2 + 1/7*(c*x^2 + b*x)^(7/2)*e/c

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\left (c\,x^2+b\,x\right )}^{5/2}\,\left (d+e\,x\right ) \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x + c*x^2)^(5/2)*(d + e*x),x)

[Out]

int((b*x + c*x^2)^(5/2)*(d + e*x), x)

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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 {5}{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)**(5/2),x)

[Out]

Integral((x*(b + c*x))**(5/2)*(d + e*x), x)

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