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

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

[Out]

5/384*b^2*(-2*A*c+B*b)*(2*c*x+b)*(c*x^2+b*x)^(3/2)/c^3-1/24*(-2*A*c+B*b)*(2*c*x+b)*(c*x^2+b*x)^(5/2)/c^2+1/7*B
*(c*x^2+b*x)^(7/2)/c+5/1024*b^6*(-2*A*c+B*b)*arctanh(x*c^(1/2)/(c*x^2+b*x)^(1/2))/c^(9/2)-5/1024*b^4*(-2*A*c+B
*b)*(2*c*x+b)*(c*x^2+b*x)^(1/2)/c^4

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Rubi [A]  time = 0.07, antiderivative size = 171, 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} \[ -\frac {5 b^4 (b+2 c x) \sqrt {b x+c x^2} (b B-2 A c)}{1024 c^4}+\frac {5 b^2 (b+2 c x) \left (b x+c x^2\right )^{3/2} (b B-2 A c)}{384 c^3}+\frac {5 b^6 (b B-2 A c) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{1024 c^{9/2}}-\frac {(b+2 c x) \left (b x+c x^2\right )^{5/2} (b B-2 A c)}{24 c^2}+\frac {B \left (b x+c x^2\right )^{7/2}}{7 c} \]

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 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 (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*}

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Mathematica [A]  time = 0.34, size = 171, normalized size = 1.00 \[ \frac {(x (b+c x))^{7/2} \left (\frac {49 (b B-2 A c) \left (15 b^{11/2} \sinh ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )-\sqrt {c} \sqrt {x} \sqrt {\frac {c x}{b}+1} \left (15 b^5-10 b^4 c x+8 b^3 c^2 x^2+432 b^2 c^3 x^3+640 b c^4 x^4+256 c^5 x^5\right )\right )}{3072 c^{7/2} x^{7/2} \sqrt {\frac {c x}{b}+1}}+7 B (b+c x)^3\right )}{49 c (b+c x)^3} \]

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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fricas [A]  time = 0.64, size = 392, normalized size = 2.29 \[ \left [-\frac {105 \, {\left (B b^{7} - 2 \, A b^{6} c\right )} \sqrt {c} \log \left (2 \, c x + b - 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) - 2 \, {\left (3072 \, B c^{7} x^{6} - 105 \, B b^{6} c + 210 \, A b^{5} c^{2} + 256 \, {\left (29 \, B b c^{6} + 14 \, A c^{7}\right )} x^{5} + 128 \, {\left (37 \, B b^{2} c^{5} + 70 \, A b c^{6}\right )} x^{4} + 48 \, {\left (B b^{3} c^{4} + 126 \, A b^{2} c^{5}\right )} x^{3} - 56 \, {\left (B b^{4} c^{3} - 2 \, A b^{3} c^{4}\right )} x^{2} + 70 \, {\left (B b^{5} c^{2} - 2 \, A b^{4} c^{3}\right )} x\right )} \sqrt {c x^{2} + b x}}{43008 \, c^{5}}, -\frac {105 \, {\left (B b^{7} - 2 \, A b^{6} c\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x}\right ) - {\left (3072 \, B c^{7} x^{6} - 105 \, B b^{6} c + 210 \, A b^{5} c^{2} + 256 \, {\left (29 \, B b c^{6} + 14 \, A c^{7}\right )} x^{5} + 128 \, {\left (37 \, B b^{2} c^{5} + 70 \, A b c^{6}\right )} x^{4} + 48 \, {\left (B b^{3} c^{4} + 126 \, A b^{2} c^{5}\right )} x^{3} - 56 \, {\left (B b^{4} c^{3} - 2 \, A b^{3} c^{4}\right )} x^{2} + 70 \, {\left (B b^{5} c^{2} - 2 \, A b^{4} c^{3}\right )} x\right )} \sqrt {c x^{2} + b x}}{21504 \, c^{5}}\right ] \]

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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giac [A]  time = 0.26, size = 221, normalized size = 1.29 \[ \frac {1}{21504} \, \sqrt {c x^{2} + b x} {\left (2 \, {\left (4 \, {\left (2 \, {\left (8 \, {\left (2 \, {\left (12 \, B c^{2} x + \frac {29 \, B b c^{7} + 14 \, A c^{8}}{c^{6}}\right )} x + \frac {37 \, B b^{2} c^{6} + 70 \, A b c^{7}}{c^{6}}\right )} x + \frac {3 \, {\left (B b^{3} c^{5} + 126 \, A b^{2} c^{6}\right )}}{c^{6}}\right )} x - \frac {7 \, {\left (B b^{4} c^{4} - 2 \, A b^{3} c^{5}\right )}}{c^{6}}\right )} x + \frac {35 \, {\left (B b^{5} c^{3} - 2 \, A b^{4} c^{4}\right )}}{c^{6}}\right )} x - \frac {105 \, {\left (B b^{6} c^{2} - 2 \, A b^{5} c^{3}\right )}}{c^{6}}\right )} - \frac {5 \, {\left (B b^{7} - 2 \, A b^{6} c\right )} \log \left ({\left | -2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} \sqrt {c} - b \right |}\right )}{2048 \, c^{\frac {9}{2}}} \]

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)

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maple [B]  time = 0.05, size = 321, normalized size = 1.88 \[ -\frac {5 A \,b^{6} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{1024 c^{\frac {7}{2}}}+\frac {5 B \,b^{7} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{2048 c^{\frac {9}{2}}}+\frac {5 \sqrt {c \,x^{2}+b x}\, A \,b^{4} x}{256 c^{2}}-\frac {5 \sqrt {c \,x^{2}+b x}\, B \,b^{5} x}{512 c^{3}}+\frac {5 \sqrt {c \,x^{2}+b x}\, A \,b^{5}}{512 c^{3}}-\frac {5 \left (c \,x^{2}+b x \right )^{\frac {3}{2}} A \,b^{2} x}{96 c}-\frac {5 \sqrt {c \,x^{2}+b x}\, B \,b^{6}}{1024 c^{4}}+\frac {5 \left (c \,x^{2}+b x \right )^{\frac {3}{2}} B \,b^{3} x}{192 c^{2}}-\frac {5 \left (c \,x^{2}+b x \right )^{\frac {3}{2}} A \,b^{3}}{192 c^{2}}+\frac {\left (c \,x^{2}+b x \right )^{\frac {5}{2}} A x}{6}+\frac {5 \left (c \,x^{2}+b x \right )^{\frac {3}{2}} B \,b^{4}}{384 c^{3}}-\frac {\left (c \,x^{2}+b x \right )^{\frac {5}{2}} B b x}{12 c}+\frac {\left (c \,x^{2}+b x \right )^{\frac {5}{2}} A b}{12 c}-\frac {\left (c \,x^{2}+b x \right )^{\frac {5}{2}} B \,b^{2}}{24 c^{2}}+\frac {\left (c \,x^{2}+b x \right )^{\frac {7}{2}} B}{7 c} \]

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*(c*x^2+b*x)^(7/2)*B/c-1/12*B*b/c*x*(c*x^2+b*x)^(5/2)-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((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x)^(1/2))+1/6*A*x*(c*x^2+b*x)^(5/2)+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((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x)^(1/2))

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maxima [B]  time = 0.62, size = 318, normalized size = 1.86 \[ \frac {1}{6} \, {\left (c x^{2} + b x\right )}^{\frac {5}{2}} A x - \frac {5 \, \sqrt {c x^{2} + b x} B b^{5} x}{512 \, c^{3}} + \frac {5 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} B b^{3} x}{192 \, c^{2}} + \frac {5 \, \sqrt {c x^{2} + b x} A b^{4} x}{256 \, c^{2}} - \frac {{\left (c x^{2} + b x\right )}^{\frac {5}{2}} B b x}{12 \, c} - \frac {5 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} A b^{2} x}{96 \, c} + \frac {5 \, B b^{7} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{2048 \, c^{\frac {9}{2}}} - \frac {5 \, A b^{6} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{1024 \, c^{\frac {7}{2}}} - \frac {5 \, \sqrt {c x^{2} + b x} B b^{6}}{1024 \, c^{4}} + \frac {5 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} B b^{4}}{384 \, c^{3}} + \frac {5 \, \sqrt {c x^{2} + b x} A b^{5}}{512 \, c^{3}} - \frac {{\left (c x^{2} + b x\right )}^{\frac {5}{2}} B b^{2}}{24 \, c^{2}} - \frac {5 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} A b^{3}}{192 \, c^{2}} + \frac {{\left (c x^{2} + b x\right )}^{\frac {7}{2}} B}{7 \, c} + \frac {{\left (c x^{2} + b x\right )}^{\frac {5}{2}} A b}{12 \, c} \]

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]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x + c*x^2)^(5/2)*(A + B*x),x)

[Out]

int((b*x + c*x^2)^(5/2)*(A + B*x), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (x \left (b + c x\right )\right )^{\frac {5}{2}} \left (A + B x\right )\, dx \]

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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