3.1.100 \(\int \frac {(A+B x) (b x+c x^2)^{5/2}}{x^4} \, dx\)

Optimal. Leaf size=153 \[ \frac {5 b^2 (6 A c+b B) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{8 \sqrt {c}}+\frac {2 \left (b x+c x^2\right )^{5/2} (6 A c+b B)}{b x^2}-\frac {5 c \left (b x+c x^2\right )^{3/2} (6 A c+b B)}{3 b}-\frac {5}{8} (b+2 c x) \sqrt {b x+c x^2} (6 A c+b B)-\frac {2 A \left (b x+c x^2\right )^{7/2}}{b x^4} \]

________________________________________________________________________________________

Rubi [A]  time = 0.17, antiderivative size = 153, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {792, 662, 664, 612, 620, 206} \begin {gather*} \frac {5 b^2 (6 A c+b B) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{8 \sqrt {c}}+\frac {2 \left (b x+c x^2\right )^{5/2} (6 A c+b B)}{b x^2}-\frac {5 c \left (b x+c x^2\right )^{3/2} (6 A c+b B)}{3 b}-\frac {5}{8} (b+2 c x) \sqrt {b x+c x^2} (6 A c+b B)-\frac {2 A \left (b x+c x^2\right )^{7/2}}{b x^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(b*x + c*x^2)^(5/2))/x^4,x]

[Out]

(-5*(b*B + 6*A*c)*(b + 2*c*x)*Sqrt[b*x + c*x^2])/8 - (5*c*(b*B + 6*A*c)*(b*x + c*x^2)^(3/2))/(3*b) + (2*(b*B +
6*A*c)*(b*x + c*x^2)^(5/2))/(b*x^2) - (2*A*(b*x + c*x^2)^(7/2))/(b*x^4) + (5*b^2*(b*B + 6*A*c)*ArcTanh[(Sqrt[
c]*x)/Sqrt[b*x + c*x^2]])/(8*Sqrt[c])

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 662

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 + p + 1)), x] - Dist[(c*p)/(e^2*(m + p + 1)), Int[(d + e*x)^(m + 2)*(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] && (LtQ[m, -2] || EqQ[m + 2*p + 1, 0]) && NeQ[m + p + 1, 0] && IntegerQ[2*p]

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 792

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp
[((d*g - e*f)*(d + e*x)^m*(a + b*x + c*x^2)^(p + 1))/((2*c*d - b*e)*(m + p + 1)), x] + Dist[(m*(g*(c*d - b*e)
+ c*e*f) + e*(p + 1)*(2*c*f - b*g))/(e*(2*c*d - b*e)*(m + p + 1)), Int[(d + e*x)^(m + 1)*(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] && ((L
tQ[m, -1] &&  !IGtQ[m + p + 1, 0]) || (LtQ[m, 0] && LtQ[p, -1]) || EqQ[m + 2*p + 2, 0]) && NeQ[m + p + 1, 0]

Rubi steps

\begin {align*} \int \frac {(A+B x) \left (b x+c x^2\right )^{5/2}}{x^4} \, dx &=-\frac {2 A \left (b x+c x^2\right )^{7/2}}{b x^4}+\frac {\left (2 \left (-4 (-b B+A c)+\frac {7}{2} (-b B+2 A c)\right )\right ) \int \frac {\left (b x+c x^2\right )^{5/2}}{x^3} \, dx}{b}\\ &=\frac {2 (b B+6 A c) \left (b x+c x^2\right )^{5/2}}{b x^2}-\frac {2 A \left (b x+c x^2\right )^{7/2}}{b x^4}-\frac {(5 c (b B+6 A c)) \int \frac {\left (b x+c x^2\right )^{3/2}}{x} \, dx}{b}\\ &=-\frac {5 c (b B+6 A c) \left (b x+c x^2\right )^{3/2}}{3 b}+\frac {2 (b B+6 A c) \left (b x+c x^2\right )^{5/2}}{b x^2}-\frac {2 A \left (b x+c x^2\right )^{7/2}}{b x^4}-\frac {1}{2} (5 c (b B+6 A c)) \int \sqrt {b x+c x^2} \, dx\\ &=-\frac {5}{8} (b B+6 A c) (b+2 c x) \sqrt {b x+c x^2}-\frac {5 c (b B+6 A c) \left (b x+c x^2\right )^{3/2}}{3 b}+\frac {2 (b B+6 A c) \left (b x+c x^2\right )^{5/2}}{b x^2}-\frac {2 A \left (b x+c x^2\right )^{7/2}}{b x^4}+\frac {1}{16} \left (5 b^2 (b B+6 A c)\right ) \int \frac {1}{\sqrt {b x+c x^2}} \, dx\\ &=-\frac {5}{8} (b B+6 A c) (b+2 c x) \sqrt {b x+c x^2}-\frac {5 c (b B+6 A c) \left (b x+c x^2\right )^{3/2}}{3 b}+\frac {2 (b B+6 A c) \left (b x+c x^2\right )^{5/2}}{b x^2}-\frac {2 A \left (b x+c x^2\right )^{7/2}}{b x^4}+\frac {1}{8} \left (5 b^2 (b B+6 A c)\right ) \operatorname {Subst}\left (\int \frac {1}{1-c x^2} \, dx,x,\frac {x}{\sqrt {b x+c x^2}}\right )\\ &=-\frac {5}{8} (b B+6 A c) (b+2 c x) \sqrt {b x+c x^2}-\frac {5 c (b B+6 A c) \left (b x+c x^2\right )^{3/2}}{3 b}+\frac {2 (b B+6 A c) \left (b x+c x^2\right )^{5/2}}{b x^2}-\frac {2 A \left (b x+c x^2\right )^{7/2}}{b x^4}+\frac {5 b^2 (b B+6 A c) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{8 \sqrt {c}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.21, size = 117, normalized size = 0.76 \begin {gather*} \frac {\sqrt {x (b+c x)} \left (\frac {15 b^{3/2} \sqrt {x} (6 A c+b B) \sinh ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{\sqrt {c} \sqrt {\frac {c x}{b}+1}}-6 A \left (8 b^2-9 b c x-2 c^2 x^2\right )+B x \left (33 b^2+26 b c x+8 c^2 x^2\right )\right )}{24 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(b*x + c*x^2)^(5/2))/x^4,x]

[Out]

(Sqrt[x*(b + c*x)]*(-6*A*(8*b^2 - 9*b*c*x - 2*c^2*x^2) + B*x*(33*b^2 + 26*b*c*x + 8*c^2*x^2) + (15*b^(3/2)*(b*
B + 6*A*c)*Sqrt[x]*ArcSinh[(Sqrt[c]*Sqrt[x])/Sqrt[b]])/(Sqrt[c]*Sqrt[1 + (c*x)/b])))/(24*x)

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.62, size = 116, normalized size = 0.76 \begin {gather*} \frac {\sqrt {b x+c x^2} \left (-48 A b^2+54 A b c x+12 A c^2 x^2+33 b^2 B x+26 b B c x^2+8 B c^2 x^3\right )}{24 x}-\frac {5 \left (6 A b^2 c+b^3 B\right ) \log \left (-2 \sqrt {c} \sqrt {b x+c x^2}+b+2 c x\right )}{16 \sqrt {c}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((A + B*x)*(b*x + c*x^2)^(5/2))/x^4,x]

[Out]

(Sqrt[b*x + c*x^2]*(-48*A*b^2 + 33*b^2*B*x + 54*A*b*c*x + 26*b*B*c*x^2 + 12*A*c^2*x^2 + 8*B*c^2*x^3))/(24*x) -
(5*(b^3*B + 6*A*b^2*c)*Log[b + 2*c*x - 2*Sqrt[c]*Sqrt[b*x + c*x^2]])/(16*Sqrt[c])

________________________________________________________________________________________

fricas [A]  time = 0.41, size = 240, normalized size = 1.57 \begin {gather*} \left [\frac {15 \, {\left (B b^{3} + 6 \, A b^{2} c\right )} \sqrt {c} x \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) + 2 \, {\left (8 \, B c^{3} x^{3} - 48 \, A b^{2} c + 2 \, {\left (13 \, B b c^{2} + 6 \, A c^{3}\right )} x^{2} + 3 \, {\left (11 \, B b^{2} c + 18 \, A b c^{2}\right )} x\right )} \sqrt {c x^{2} + b x}}{48 \, c x}, -\frac {15 \, {\left (B b^{3} + 6 \, A b^{2} c\right )} \sqrt {-c} x \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x}\right ) - {\left (8 \, B c^{3} x^{3} - 48 \, A b^{2} c + 2 \, {\left (13 \, B b c^{2} + 6 \, A c^{3}\right )} x^{2} + 3 \, {\left (11 \, B b^{2} c + 18 \, A b c^{2}\right )} x\right )} \sqrt {c x^{2} + b x}}{24 \, c x}\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^4,x, algorithm="fricas")

[Out]

[1/48*(15*(B*b^3 + 6*A*b^2*c)*sqrt(c)*x*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c)) + 2*(8*B*c^3*x^3 - 48*A*b
^2*c + 2*(13*B*b*c^2 + 6*A*c^3)*x^2 + 3*(11*B*b^2*c + 18*A*b*c^2)*x)*sqrt(c*x^2 + b*x))/(c*x), -1/24*(15*(B*b^
3 + 6*A*b^2*c)*sqrt(-c)*x*arctan(sqrt(c*x^2 + b*x)*sqrt(-c)/(c*x)) - (8*B*c^3*x^3 - 48*A*b^2*c + 2*(13*B*b*c^2
+ 6*A*c^3)*x^2 + 3*(11*B*b^2*c + 18*A*b*c^2)*x)*sqrt(c*x^2 + b*x))/(c*x)]

________________________________________________________________________________________

giac [A]  time = 0.24, size = 141, normalized size = 0.92 \begin {gather*} \frac {2 \, A b^{3}}{\sqrt {c} x - \sqrt {c x^{2} + b x}} + \frac {1}{24} \, \sqrt {c x^{2} + b x} {\left (2 \, {\left (4 \, B c^{2} x + \frac {13 \, B b c^{3} + 6 \, A c^{4}}{c^{2}}\right )} x + \frac {3 \, {\left (11 \, B b^{2} c^{2} + 18 \, A b c^{3}\right )}}{c^{2}}\right )} - \frac {5 \, {\left (B b^{3} + 6 \, A b^{2} c\right )} \log \left ({\left | -2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} \sqrt {c} - b \right |}\right )}{16 \, \sqrt {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^4,x, algorithm="giac")

[Out]

2*A*b^3/(sqrt(c)*x - sqrt(c*x^2 + b*x)) + 1/24*sqrt(c*x^2 + b*x)*(2*(4*B*c^2*x + (13*B*b*c^3 + 6*A*c^4)/c^2)*x
+ 3*(11*B*b^2*c^2 + 18*A*b*c^3)/c^2) - 5/16*(B*b^3 + 6*A*b^2*c)*log(abs(-2*(sqrt(c)*x - sqrt(c*x^2 + b*x))*sq
rt(c) - b))/sqrt(c)

________________________________________________________________________________________

maple [B]  time = 0.06, size = 358, normalized size = 2.34 \begin {gather*} \frac {15 A \,b^{2} \sqrt {c}\, \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{8}+\frac {5 B \,b^{3} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{16 \sqrt {c}}-\frac {15 \sqrt {c \,x^{2}+b x}\, A \,c^{2} x}{2}-\frac {5 \sqrt {c \,x^{2}+b x}\, B b c x}{4}-\frac {15 \sqrt {c \,x^{2}+b x}\, A b c}{4}+\frac {20 \left (c \,x^{2}+b x \right )^{\frac {3}{2}} A \,c^{3} x}{b^{2}}-\frac {5 \sqrt {c \,x^{2}+b x}\, B \,b^{2}}{8}+\frac {10 \left (c \,x^{2}+b x \right )^{\frac {3}{2}} B \,c^{2} x}{3 b}+\frac {10 \left (c \,x^{2}+b x \right )^{\frac {3}{2}} A \,c^{2}}{b}+\frac {5 \left (c \,x^{2}+b x \right )^{\frac {3}{2}} B c}{3}+\frac {32 \left (c \,x^{2}+b x \right )^{\frac {5}{2}} A \,c^{3}}{b^{3}}+\frac {16 \left (c \,x^{2}+b x \right )^{\frac {5}{2}} B \,c^{2}}{3 b^{2}}-\frac {32 \left (c \,x^{2}+b x \right )^{\frac {7}{2}} A \,c^{2}}{b^{3} x^{2}}-\frac {16 \left (c \,x^{2}+b x \right )^{\frac {7}{2}} B c}{3 b^{2} x^{2}}+\frac {12 \left (c \,x^{2}+b x \right )^{\frac {7}{2}} A c}{b^{2} x^{3}}+\frac {2 \left (c \,x^{2}+b x \right )^{\frac {7}{2}} B}{b \,x^{3}}-\frac {2 \left (c \,x^{2}+b x \right )^{\frac {7}{2}} A}{b \,x^{4}} \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^4,x)

[Out]

-2*A*(c*x^2+b*x)^(7/2)/b/x^4+12*A/b^2*c/x^3*(c*x^2+b*x)^(7/2)-32*A/b^3*c^2/x^2*(c*x^2+b*x)^(7/2)+32*A/b^3*c^3*
(c*x^2+b*x)^(5/2)+20*A/b^2*c^3*(c*x^2+b*x)^(3/2)*x+10*A/b*c^2*(c*x^2+b*x)^(3/2)-15/2*A*c^2*(c*x^2+b*x)^(1/2)*x
-15/4*A*b*c*(c*x^2+b*x)^(1/2)+15/8*A*b^2*c^(1/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x)^(1/2))+2*B/b/x^3*(c*x^2+b*
x)^(7/2)-16/3*B/b^2*c/x^2*(c*x^2+b*x)^(7/2)+16/3*B/b^2*c^2*(c*x^2+b*x)^(5/2)+10/3*B/b*c^2*(c*x^2+b*x)^(3/2)*x+
5/3*B*c*(c*x^2+b*x)^(3/2)-5/4*B*b*c*(c*x^2+b*x)^(1/2)*x-5/8*B*b^2*(c*x^2+b*x)^(1/2)+5/16*B*b^3/c^(1/2)*ln((c*x
+1/2*b)/c^(1/2)+(c*x^2+b*x)^(1/2))

________________________________________________________________________________________

maxima [A]  time = 0.91, size = 172, normalized size = 1.12 \begin {gather*} \frac {5 \, B b^{3} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{16 \, \sqrt {c}} + \frac {15}{8} \, A b^{2} \sqrt {c} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) + \frac {5}{8} \, \sqrt {c x^{2} + b x} B b^{2} + \frac {5 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} B b}{12 \, x} - \frac {15 \, \sqrt {c x^{2} + b x} A b^{2}}{4 \, x} + \frac {{\left (c x^{2} + b x\right )}^{\frac {5}{2}} B}{3 \, x^{2}} + \frac {5 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} A b}{4 \, x^{2}} + \frac {{\left (c x^{2} + b x\right )}^{\frac {5}{2}} A}{2 \, x^{3}} \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^4,x, algorithm="maxima")

[Out]

5/16*B*b^3*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c))/sqrt(c) + 15/8*A*b^2*sqrt(c)*log(2*c*x + b + 2*sqrt(c*
x^2 + b*x)*sqrt(c)) + 5/8*sqrt(c*x^2 + b*x)*B*b^2 + 5/12*(c*x^2 + b*x)^(3/2)*B*b/x - 15/4*sqrt(c*x^2 + b*x)*A*
b^2/x + 1/3*(c*x^2 + b*x)^(5/2)*B/x^2 + 5/4*(c*x^2 + b*x)^(3/2)*A*b/x^2 + 1/2*(c*x^2 + b*x)^(5/2)*A/x^3

________________________________________________________________________________________

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^4} \,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^4,x)

[Out]

int(((b*x + c*x^2)^(5/2)*(A + B*x))/x^4, x)

________________________________________________________________________________________

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^{4}}\, 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**4,x)

[Out]

Integral((x*(b + c*x))**(5/2)*(A + B*x)/x**4, x)

________________________________________________________________________________________