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

Optimal. Leaf size=170 \[ -\frac {b^5 (5 b B-12 A c) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{512 c^{7/2}}+\frac {b^3 (b+2 c x) \sqrt {b x+c x^2} (5 b B-12 A c)}{512 c^3}-\frac {b (b+2 c x) \left (b x+c x^2\right )^{3/2} (5 b B-12 A c)}{192 c^2}-\frac {\left (b x+c x^2\right )^{5/2} (5 b B-12 A c)}{60 c}+\frac {B \left (b x+c x^2\right )^{7/2}}{6 c x} \]

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Rubi [A]  time = 0.13, antiderivative size = 170, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {794, 664, 612, 620, 206} \begin {gather*} \frac {b^3 (b+2 c x) \sqrt {b x+c x^2} (5 b B-12 A c)}{512 c^3}-\frac {b^5 (5 b B-12 A c) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{512 c^{7/2}}-\frac {b (b+2 c x) \left (b x+c x^2\right )^{3/2} (5 b B-12 A c)}{192 c^2}-\frac {\left (b x+c x^2\right )^{5/2} (5 b B-12 A c)}{60 c}+\frac {B \left (b x+c x^2\right )^{7/2}}{6 c x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(b*x + c*x^2)^(5/2))/x,x]

[Out]

(b^3*(5*b*B - 12*A*c)*(b + 2*c*x)*Sqrt[b*x + c*x^2])/(512*c^3) - (b*(5*b*B - 12*A*c)*(b + 2*c*x)*(b*x + c*x^2)
^(3/2))/(192*c^2) - ((5*b*B - 12*A*c)*(b*x + c*x^2)^(5/2))/(60*c) + (B*(b*x + c*x^2)^(7/2))/(6*c*x) - (b^5*(5*
b*B - 12*A*c)*ArcTanh[(Sqrt[c]*x)/Sqrt[b*x + c*x^2]])/(512*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 664

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m + 1)*(
a + b*x + c*x^2)^p)/(e*(m + 2*p + 1)), x] - Dist[(p*(2*c*d - b*e))/(e^2*(m + 2*p + 1)), Int[(d + e*x)^(m + 1)*
(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a
*e^2, 0] && GtQ[p, 0] && (LeQ[-2, m, 0] || EqQ[m + p + 1, 0]) && NeQ[m + 2*p + 1, 0] && IntegerQ[2*p]

Rule 794

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp
[(g*(d + e*x)^m*(a + b*x + c*x^2)^(p + 1))/(c*(m + 2*p + 2)), x] + Dist[(m*(g*(c*d - b*e) + c*e*f) + e*(p + 1)
*(2*c*f - b*g))/(c*e*(m + 2*p + 2)), Int[(d + e*x)^m*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g
, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ[m + 2*p + 2, 0] && (NeQ[m, 2] || Eq
Q[d, 0])

Rubi steps

\begin {align*} \int \frac {(A+B x) \left (b x+c x^2\right )^{5/2}}{x} \, dx &=\frac {B \left (b x+c x^2\right )^{7/2}}{6 c x}+\frac {\left (b B-A c+\frac {7}{2} (-b B+2 A c)\right ) \int \frac {\left (b x+c x^2\right )^{5/2}}{x} \, dx}{6 c}\\ &=-\frac {(5 b B-12 A c) \left (b x+c x^2\right )^{5/2}}{60 c}+\frac {B \left (b x+c x^2\right )^{7/2}}{6 c x}-\frac {(b (5 b B-12 A c)) \int \left (b x+c x^2\right )^{3/2} \, dx}{24 c}\\ &=-\frac {b (5 b B-12 A c) (b+2 c x) \left (b x+c x^2\right )^{3/2}}{192 c^2}-\frac {(5 b B-12 A c) \left (b x+c x^2\right )^{5/2}}{60 c}+\frac {B \left (b x+c x^2\right )^{7/2}}{6 c x}+\frac {\left (b^3 (5 b B-12 A c)\right ) \int \sqrt {b x+c x^2} \, dx}{128 c^2}\\ &=\frac {b^3 (5 b B-12 A c) (b+2 c x) \sqrt {b x+c x^2}}{512 c^3}-\frac {b (5 b B-12 A c) (b+2 c x) \left (b x+c x^2\right )^{3/2}}{192 c^2}-\frac {(5 b B-12 A c) \left (b x+c x^2\right )^{5/2}}{60 c}+\frac {B \left (b x+c x^2\right )^{7/2}}{6 c x}-\frac {\left (b^5 (5 b B-12 A c)\right ) \int \frac {1}{\sqrt {b x+c x^2}} \, dx}{1024 c^3}\\ &=\frac {b^3 (5 b B-12 A c) (b+2 c x) \sqrt {b x+c x^2}}{512 c^3}-\frac {b (5 b B-12 A c) (b+2 c x) \left (b x+c x^2\right )^{3/2}}{192 c^2}-\frac {(5 b B-12 A c) \left (b x+c x^2\right )^{5/2}}{60 c}+\frac {B \left (b x+c x^2\right )^{7/2}}{6 c x}-\frac {\left (b^5 (5 b B-12 A c)\right ) \operatorname {Subst}\left (\int \frac {1}{1-c x^2} \, dx,x,\frac {x}{\sqrt {b x+c x^2}}\right )}{512 c^3}\\ &=\frac {b^3 (5 b B-12 A c) (b+2 c x) \sqrt {b x+c x^2}}{512 c^3}-\frac {b (5 b B-12 A c) (b+2 c x) \left (b x+c x^2\right )^{3/2}}{192 c^2}-\frac {(5 b B-12 A c) \left (b x+c x^2\right )^{5/2}}{60 c}+\frac {B \left (b x+c x^2\right )^{7/2}}{6 c x}-\frac {b^5 (5 b B-12 A c) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{512 c^{7/2}}\\ \end {align*}

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Mathematica [A]  time = 0.29, size = 166, normalized size = 0.98 \begin {gather*} \frac {\sqrt {x (b+c x)} \left (\sqrt {c} \left (-10 b^4 c (18 A+5 B x)+40 b^3 c^2 x (3 A+B x)+48 b^2 c^3 x^2 (62 A+45 B x)+64 b c^4 x^3 (63 A+50 B x)+256 c^5 x^4 (6 A+5 B x)+75 b^5 B\right )-\frac {15 b^{9/2} (5 b B-12 A c) \sinh ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{\sqrt {x} \sqrt {\frac {c x}{b}+1}}\right )}{7680 c^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(b*x + c*x^2)^(5/2))/x,x]

[Out]

(Sqrt[x*(b + c*x)]*(Sqrt[c]*(75*b^5*B + 40*b^3*c^2*x*(3*A + B*x) + 256*c^5*x^4*(6*A + 5*B*x) - 10*b^4*c*(18*A
+ 5*B*x) + 48*b^2*c^3*x^2*(62*A + 45*B*x) + 64*b*c^4*x^3*(63*A + 50*B*x)) - (15*b^(9/2)*(5*b*B - 12*A*c)*ArcSi
nh[(Sqrt[c]*Sqrt[x])/Sqrt[b]])/(Sqrt[x]*Sqrt[1 + (c*x)/b])))/(7680*c^(7/2))

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IntegrateAlgebraic [A]  time = 0.89, size = 177, normalized size = 1.04 \begin {gather*} \frac {\left (5 b^6 B-12 A b^5 c\right ) \log \left (-2 \sqrt {c} \sqrt {b x+c x^2}+b+2 c x\right )}{1024 c^{7/2}}+\frac {\sqrt {b x+c x^2} \left (-180 A b^4 c+120 A b^3 c^2 x+2976 A b^2 c^3 x^2+4032 A b c^4 x^3+1536 A c^5 x^4+75 b^5 B-50 b^4 B c x+40 b^3 B c^2 x^2+2160 b^2 B c^3 x^3+3200 b B c^4 x^4+1280 B c^5 x^5\right )}{7680 c^3} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((A + B*x)*(b*x + c*x^2)^(5/2))/x,x]

[Out]

(Sqrt[b*x + c*x^2]*(75*b^5*B - 180*A*b^4*c - 50*b^4*B*c*x + 120*A*b^3*c^2*x + 40*b^3*B*c^2*x^2 + 2976*A*b^2*c^
3*x^2 + 2160*b^2*B*c^3*x^3 + 4032*A*b*c^4*x^3 + 3200*b*B*c^4*x^4 + 1536*A*c^5*x^4 + 1280*B*c^5*x^5))/(7680*c^3
) + ((5*b^6*B - 12*A*b^5*c)*Log[b + 2*c*x - 2*Sqrt[c]*Sqrt[b*x + c*x^2]])/(1024*c^(7/2))

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fricas [A]  time = 0.45, size = 351, normalized size = 2.06 \begin {gather*} \left [-\frac {15 \, {\left (5 \, B b^{6} - 12 \, A b^{5} c\right )} \sqrt {c} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) - 2 \, {\left (1280 \, B c^{6} x^{5} + 75 \, B b^{5} c - 180 \, A b^{4} c^{2} + 128 \, {\left (25 \, B b c^{5} + 12 \, A c^{6}\right )} x^{4} + 144 \, {\left (15 \, B b^{2} c^{4} + 28 \, A b c^{5}\right )} x^{3} + 8 \, {\left (5 \, B b^{3} c^{3} + 372 \, A b^{2} c^{4}\right )} x^{2} - 10 \, {\left (5 \, B b^{4} c^{2} - 12 \, A b^{3} c^{3}\right )} x\right )} \sqrt {c x^{2} + b x}}{15360 \, c^{4}}, \frac {15 \, {\left (5 \, B b^{6} - 12 \, A b^{5} c\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x}\right ) + {\left (1280 \, B c^{6} x^{5} + 75 \, B b^{5} c - 180 \, A b^{4} c^{2} + 128 \, {\left (25 \, B b c^{5} + 12 \, A c^{6}\right )} x^{4} + 144 \, {\left (15 \, B b^{2} c^{4} + 28 \, A b c^{5}\right )} x^{3} + 8 \, {\left (5 \, B b^{3} c^{3} + 372 \, A b^{2} c^{4}\right )} x^{2} - 10 \, {\left (5 \, B b^{4} c^{2} - 12 \, A b^{3} c^{3}\right )} x\right )} \sqrt {c x^{2} + b x}}{7680 \, c^{4}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/15360*(15*(5*B*b^6 - 12*A*b^5*c)*sqrt(c)*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c)) - 2*(1280*B*c^6*x^5
+ 75*B*b^5*c - 180*A*b^4*c^2 + 128*(25*B*b*c^5 + 12*A*c^6)*x^4 + 144*(15*B*b^2*c^4 + 28*A*b*c^5)*x^3 + 8*(5*B*
b^3*c^3 + 372*A*b^2*c^4)*x^2 - 10*(5*B*b^4*c^2 - 12*A*b^3*c^3)*x)*sqrt(c*x^2 + b*x))/c^4, 1/7680*(15*(5*B*b^6
- 12*A*b^5*c)*sqrt(-c)*arctan(sqrt(c*x^2 + b*x)*sqrt(-c)/(c*x)) + (1280*B*c^6*x^5 + 75*B*b^5*c - 180*A*b^4*c^2
+ 128*(25*B*b*c^5 + 12*A*c^6)*x^4 + 144*(15*B*b^2*c^4 + 28*A*b*c^5)*x^3 + 8*(5*B*b^3*c^3 + 372*A*b^2*c^4)*x^2
- 10*(5*B*b^4*c^2 - 12*A*b^3*c^3)*x)*sqrt(c*x^2 + b*x))/c^4]

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giac [A]  time = 0.23, size = 198, normalized size = 1.16 \begin {gather*} \frac {1}{7680} \, \sqrt {c x^{2} + b x} {\left (2 \, {\left (4 \, {\left (2 \, {\left (8 \, {\left (10 \, B c^{2} x + \frac {25 \, B b c^{6} + 12 \, A c^{7}}{c^{5}}\right )} x + \frac {9 \, {\left (15 \, B b^{2} c^{5} + 28 \, A b c^{6}\right )}}{c^{5}}\right )} x + \frac {5 \, B b^{3} c^{4} + 372 \, A b^{2} c^{5}}{c^{5}}\right )} x - \frac {5 \, {\left (5 \, B b^{4} c^{3} - 12 \, A b^{3} c^{4}\right )}}{c^{5}}\right )} x + \frac {15 \, {\left (5 \, B b^{5} c^{2} - 12 \, A b^{4} c^{3}\right )}}{c^{5}}\right )} + \frac {{\left (5 \, B b^{6} - 12 \, A b^{5} c\right )} \log \left ({\left | -2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} \sqrt {c} - b \right |}\right )}{1024 \, c^{\frac {7}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x)^(5/2)/x,x, algorithm="giac")

[Out]

1/7680*sqrt(c*x^2 + b*x)*(2*(4*(2*(8*(10*B*c^2*x + (25*B*b*c^6 + 12*A*c^7)/c^5)*x + 9*(15*B*b^2*c^5 + 28*A*b*c
^6)/c^5)*x + (5*B*b^3*c^4 + 372*A*b^2*c^5)/c^5)*x - 5*(5*B*b^4*c^3 - 12*A*b^3*c^4)/c^5)*x + 15*(5*B*b^5*c^2 -
12*A*b^4*c^3)/c^5) + 1/1024*(5*B*b^6 - 12*A*b^5*c)*log(abs(-2*(sqrt(c)*x - sqrt(c*x^2 + b*x))*sqrt(c) - b))/c^
(7/2)

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maple [A]  time = 0.05, size = 274, normalized size = 1.61 \begin {gather*} \frac {3 A \,b^{5} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{256 c^{\frac {5}{2}}}-\frac {5 B \,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 {3 \sqrt {c \,x^{2}+b x}\, A \,b^{3} x}{64 c}+\frac {5 \sqrt {c \,x^{2}+b x}\, B \,b^{4} x}{256 c^{2}}-\frac {3 \sqrt {c \,x^{2}+b x}\, A \,b^{4}}{128 c^{2}}+\frac {\left (c \,x^{2}+b x \right )^{\frac {3}{2}} A b x}{8}+\frac {5 \sqrt {c \,x^{2}+b x}\, B \,b^{5}}{512 c^{3}}-\frac {5 \left (c \,x^{2}+b x \right )^{\frac {3}{2}} B \,b^{2} x}{96 c}+\frac {\left (c \,x^{2}+b x \right )^{\frac {3}{2}} A \,b^{2}}{16 c}-\frac {5 \left (c \,x^{2}+b x \right )^{\frac {3}{2}} B \,b^{3}}{192 c^{2}}+\frac {\left (c \,x^{2}+b x \right )^{\frac {5}{2}} B x}{6}+\frac {\left (c \,x^{2}+b x \right )^{\frac {5}{2}} A}{5}+\frac {\left (c \,x^{2}+b x \right )^{\frac {5}{2}} B b}{12 c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(c*x^2+b*x)^(5/2)/x,x)

[Out]

1/6*B*x*(c*x^2+b*x)^(5/2)+1/12*B/c*(c*x^2+b*x)^(5/2)*b-5/96*B*b^2/c*(c*x^2+b*x)^(3/2)*x-5/192*B*b^3/c^2*(c*x^2
+b*x)^(3/2)+5/256*B*b^4/c^2*(c*x^2+b*x)^(1/2)*x+5/512*B*b^5/c^3*(c*x^2+b*x)^(1/2)-5/1024*B*b^6/c^(7/2)*ln((c*x
+1/2*b)/c^(1/2)+(c*x^2+b*x)^(1/2))+1/5*A*(c*x^2+b*x)^(5/2)+1/8*A*b*(c*x^2+b*x)^(3/2)*x+1/16*A/c*(c*x^2+b*x)^(3
/2)*b^2-3/64*A*b^3/c*(c*x^2+b*x)^(1/2)*x-3/128*A*b^4/c^2*(c*x^2+b*x)^(1/2)+3/256*A*b^5/c^(5/2)*ln((c*x+1/2*b)/
c^(1/2)+(c*x^2+b*x)^(1/2))

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maxima [A]  time = 0.97, size = 271, normalized size = 1.59 \begin {gather*} \frac {1}{6} \, {\left (c x^{2} + b x\right )}^{\frac {5}{2}} B x + \frac {1}{8} \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} A b x + \frac {5 \, \sqrt {c x^{2} + b x} B b^{4} x}{256 \, c^{2}} - \frac {5 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} B b^{2} x}{96 \, c} - \frac {3 \, \sqrt {c x^{2} + b x} A b^{3} x}{64 \, c} - \frac {5 \, B b^{6} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{1024 \, c^{\frac {7}{2}}} + \frac {3 \, A b^{5} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{256 \, c^{\frac {5}{2}}} + \frac {1}{5} \, {\left (c x^{2} + b x\right )}^{\frac {5}{2}} A + \frac {5 \, \sqrt {c x^{2} + b x} B b^{5}}{512 \, c^{3}} - \frac {5 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} B b^{3}}{192 \, c^{2}} - \frac {3 \, \sqrt {c x^{2} + b x} A b^{4}}{128 \, c^{2}} + \frac {{\left (c x^{2} + b x\right )}^{\frac {5}{2}} B b}{12 \, c} + \frac {{\left (c x^{2} + b x\right )}^{\frac {3}{2}} A b^{2}}{16 \, c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(c*x^2 + b*x)^(5/2)*B*x + 1/8*(c*x^2 + b*x)^(3/2)*A*b*x + 5/256*sqrt(c*x^2 + b*x)*B*b^4*x/c^2 - 5/96*(c*x^
2 + b*x)^(3/2)*B*b^2*x/c - 3/64*sqrt(c*x^2 + b*x)*A*b^3*x/c - 5/1024*B*b^6*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)
*sqrt(c))/c^(7/2) + 3/256*A*b^5*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c))/c^(5/2) + 1/5*(c*x^2 + b*x)^(5/2)
*A + 5/512*sqrt(c*x^2 + b*x)*B*b^5/c^3 - 5/192*(c*x^2 + b*x)^(3/2)*B*b^3/c^2 - 3/128*sqrt(c*x^2 + b*x)*A*b^4/c
^2 + 1/12*(c*x^2 + b*x)^(5/2)*B*b/c + 1/16*(c*x^2 + b*x)^(3/2)*A*b^2/c

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((b*x + c*x^2)^(5/2)*(A + B*x))/x,x)

[Out]

int(((b*x + c*x^2)^(5/2)*(A + B*x))/x, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x**2+b*x)**(5/2)/x,x)

[Out]

Integral((x*(b + c*x))**(5/2)*(A + B*x)/x, x)

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