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

Optimal. Leaf size=157 \[ \frac {5 c \left (b x+c x^2\right )^{3/2} (4 A c+3 b B)}{6 b x}+\frac {5}{4} c \sqrt {b x+c x^2} (4 A c+3 b B)+\frac {5}{4} b \sqrt {c} (4 A c+3 b B) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )-\frac {2 \left (b x+c x^2\right )^{5/2} (4 A c+3 b B)}{3 b x^3}-\frac {2 A \left (b x+c x^2\right )^{7/2}}{3 b x^5} \]

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Rubi [A]  time = 0.16, antiderivative size = 157, 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 = {792, 662, 664, 620, 206} \begin {gather*} -\frac {2 \left (b x+c x^2\right )^{5/2} (4 A c+3 b B)}{3 b x^3}+\frac {5 c \left (b x+c x^2\right )^{3/2} (4 A c+3 b B)}{6 b x}+\frac {5}{4} c \sqrt {b x+c x^2} (4 A c+3 b B)+\frac {5}{4} b \sqrt {c} (4 A c+3 b B) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )-\frac {2 A \left (b x+c x^2\right )^{7/2}}{3 b x^5} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [C]  time = 0.05, size = 86, normalized size = 0.55 \begin {gather*} -\frac {2 \sqrt {x (b+c x)} \left (b^2 x (4 A c+3 b B) \, _2F_1\left (-\frac {5}{2},-\frac {1}{2};\frac {1}{2};-\frac {c x}{b}\right )+A \sqrt {\frac {c x}{b}+1} (b+c x)^3\right )}{3 b x^2 \sqrt {\frac {c x}{b}+1}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sqrt[x*(b + c*x)]*(A*(b + c*x)^3*Sqrt[1 + (c*x)/b] + b^2*(3*b*B + 4*A*c)*x*Hypergeometric2F1[-5/2, -1/2, 1
/2, -((c*x)/b)]))/(3*b*x^2*Sqrt[1 + (c*x)/b])

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IntegrateAlgebraic [A]  time = 0.54, size = 119, normalized size = 0.76 \begin {gather*} \frac {\sqrt {b x+c x^2} \left (-8 A b^2-56 A b c x+12 A c^2 x^2-24 b^2 B x+27 b B c x^2+6 B c^2 x^3\right )}{12 x^2}-\frac {5}{8} \left (4 A b c^{3/2}+3 b^2 B \sqrt {c}\right ) \log \left (-2 \sqrt {c} \sqrt {b x+c x^2}+b+2 c x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[b*x + c*x^2]*(-8*A*b^2 - 24*b^2*B*x - 56*A*b*c*x + 27*b*B*c*x^2 + 12*A*c^2*x^2 + 6*B*c^2*x^3))/(12*x^2)
- (5*(3*b^2*B*Sqrt[c] + 4*A*b*c^(3/2))*Log[b + 2*c*x - 2*Sqrt[c]*Sqrt[b*x + c*x^2]])/8

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fricas [A]  time = 0.41, size = 224, normalized size = 1.43 \begin {gather*} \left [\frac {15 \, {\left (3 \, B b^{2} + 4 \, A b c\right )} \sqrt {c} x^{2} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) + 2 \, {\left (6 \, B c^{2} x^{3} - 8 \, A b^{2} + 3 \, {\left (9 \, B b c + 4 \, A c^{2}\right )} x^{2} - 8 \, {\left (3 \, B b^{2} + 7 \, A b c\right )} x\right )} \sqrt {c x^{2} + b x}}{24 \, x^{2}}, -\frac {15 \, {\left (3 \, B b^{2} + 4 \, A b c\right )} \sqrt {-c} x^{2} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x}\right ) - {\left (6 \, B c^{2} x^{3} - 8 \, A b^{2} + 3 \, {\left (9 \, B b c + 4 \, A c^{2}\right )} x^{2} - 8 \, {\left (3 \, B b^{2} + 7 \, A b c\right )} x\right )} \sqrt {c x^{2} + b x}}{12 \, x^{2}}\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^5,x, algorithm="fricas")

[Out]

[1/24*(15*(3*B*b^2 + 4*A*b*c)*sqrt(c)*x^2*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c)) + 2*(6*B*c^2*x^3 - 8*A*
b^2 + 3*(9*B*b*c + 4*A*c^2)*x^2 - 8*(3*B*b^2 + 7*A*b*c)*x)*sqrt(c*x^2 + b*x))/x^2, -1/12*(15*(3*B*b^2 + 4*A*b*
c)*sqrt(-c)*x^2*arctan(sqrt(c*x^2 + b*x)*sqrt(-c)/(c*x)) - (6*B*c^2*x^3 - 8*A*b^2 + 3*(9*B*b*c + 4*A*c^2)*x^2
- 8*(3*B*b^2 + 7*A*b*c)*x)*sqrt(c*x^2 + b*x))/x^2]

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giac [A]  time = 0.28, size = 211, normalized size = 1.34 \begin {gather*} \frac {1}{4} \, {\left (2 \, B c^{2} x + \frac {9 \, B b c^{2} + 4 \, A c^{3}}{c}\right )} \sqrt {c x^{2} + b x} - \frac {5 \, {\left (3 \, B b^{2} c + 4 \, A b c^{2}\right )} \log \left ({\left | -2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} \sqrt {c} - b \right |}\right )}{8 \, \sqrt {c}} + \frac {2 \, {\left (3 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{2} B b^{3} \sqrt {c} + 9 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{2} A b^{2} c^{\frac {3}{2}} + 3 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} A b^{3} c + A b^{4} \sqrt {c}\right )}}{3 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{3} \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^5,x, algorithm="giac")

[Out]

1/4*(2*B*c^2*x + (9*B*b*c^2 + 4*A*c^3)/c)*sqrt(c*x^2 + b*x) - 5/8*(3*B*b^2*c + 4*A*b*c^2)*log(abs(-2*(sqrt(c)*
x - sqrt(c*x^2 + b*x))*sqrt(c) - b))/sqrt(c) + 2/3*(3*(sqrt(c)*x - sqrt(c*x^2 + b*x))^2*B*b^3*sqrt(c) + 9*(sqr
t(c)*x - sqrt(c*x^2 + b*x))^2*A*b^2*c^(3/2) + 3*(sqrt(c)*x - sqrt(c*x^2 + b*x))*A*b^3*c + A*b^4*sqrt(c))/((sqr
t(c)*x - sqrt(c*x^2 + b*x))^3*sqrt(c))

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maple [B]  time = 0.05, size = 411, normalized size = 2.62 \begin {gather*} \frac {5 A b \,c^{\frac {3}{2}} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{2}+\frac {15 B \,b^{2} \sqrt {c}\, \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{8}-\frac {10 \sqrt {c \,x^{2}+b x}\, A \,c^{3} x}{b}-\frac {15 \sqrt {c \,x^{2}+b x}\, B \,c^{2} x}{2}+\frac {80 \left (c \,x^{2}+b x \right )^{\frac {3}{2}} A \,c^{4} x}{3 b^{3}}-5 \sqrt {c \,x^{2}+b x}\, A \,c^{2}-\frac {15 \sqrt {c \,x^{2}+b x}\, B b c}{4}+\frac {20 \left (c \,x^{2}+b x \right )^{\frac {3}{2}} B \,c^{3} x}{b^{2}}+\frac {40 \left (c \,x^{2}+b x \right )^{\frac {3}{2}} A \,c^{3}}{3 b^{2}}+\frac {10 \left (c \,x^{2}+b x \right )^{\frac {3}{2}} B \,c^{2}}{b}+\frac {128 \left (c \,x^{2}+b x \right )^{\frac {5}{2}} A \,c^{4}}{3 b^{4}}+\frac {32 \left (c \,x^{2}+b x \right )^{\frac {5}{2}} B \,c^{3}}{b^{3}}-\frac {128 \left (c \,x^{2}+b x \right )^{\frac {7}{2}} A \,c^{3}}{3 b^{4} x^{2}}-\frac {32 \left (c \,x^{2}+b x \right )^{\frac {7}{2}} B \,c^{2}}{b^{3} x^{2}}+\frac {16 \left (c \,x^{2}+b x \right )^{\frac {7}{2}} A \,c^{2}}{b^{3} x^{3}}+\frac {12 \left (c \,x^{2}+b x \right )^{\frac {7}{2}} B c}{b^{2} x^{3}}-\frac {8 \left (c \,x^{2}+b x \right )^{\frac {7}{2}} A c}{3 b^{2} x^{4}}-\frac {2 \left (c \,x^{2}+b x \right )^{\frac {7}{2}} B}{b \,x^{4}}-\frac {2 \left (c \,x^{2}+b x \right )^{\frac {7}{2}} A}{3 b \,x^{5}} \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^5,x)

[Out]

-2/3*A*(c*x^2+b*x)^(7/2)/b/x^5-8/3*A/b^2*c/x^4*(c*x^2+b*x)^(7/2)+16*A/b^3*c^2/x^3*(c*x^2+b*x)^(7/2)-128/3*A/b^
4*c^3/x^2*(c*x^2+b*x)^(7/2)+128/3*A/b^4*c^4*(c*x^2+b*x)^(5/2)+80/3*A/b^3*c^4*(c*x^2+b*x)^(3/2)*x+40/3*A/b^2*c^
3*(c*x^2+b*x)^(3/2)-10*A/b*c^3*(c*x^2+b*x)^(1/2)*x-5*A*c^2*(c*x^2+b*x)^(1/2)+5/2*A*b*c^(3/2)*ln((c*x+1/2*b)/c^
(1/2)+(c*x^2+b*x)^(1/2))-2*B/b/x^4*(c*x^2+b*x)^(7/2)+12*B/b^2*c/x^3*(c*x^2+b*x)^(7/2)-32*B/b^3*c^2/x^2*(c*x^2+
b*x)^(7/2)+32*B/b^3*c^3*(c*x^2+b*x)^(5/2)+20*B/b^2*c^3*(c*x^2+b*x)^(3/2)*x+10*B/b*c^2*(c*x^2+b*x)^(3/2)-15/2*B
*c^2*(c*x^2+b*x)^(1/2)*x-15/4*B*b*c*(c*x^2+b*x)^(1/2)+15/8*B*b^2*c^(1/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x)^(1
/2))

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maxima [A]  time = 0.96, size = 191, normalized size = 1.22 \begin {gather*} \frac {15}{8} \, B b^{2} \sqrt {c} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) + \frac {5}{2} \, A b c^{\frac {3}{2}} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) - \frac {15 \, \sqrt {c x^{2} + b x} B b^{2}}{4 \, x} - \frac {35 \, \sqrt {c x^{2} + b x} A b c}{6 \, x} + \frac {5 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} B b}{4 \, x^{2}} - \frac {5 \, \sqrt {c x^{2} + b x} A b^{2}}{6 \, x^{2}} + \frac {{\left (c x^{2} + b x\right )}^{\frac {5}{2}} B}{2 \, x^{3}} - \frac {5 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} A b}{6 \, x^{3}} + \frac {{\left (c x^{2} + b x\right )}^{\frac {5}{2}} A}{x^{4}} \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^5,x, algorithm="maxima")

[Out]

15/8*B*b^2*sqrt(c)*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c)) + 5/2*A*b*c^(3/2)*log(2*c*x + b + 2*sqrt(c*x^2
+ b*x)*sqrt(c)) - 15/4*sqrt(c*x^2 + b*x)*B*b^2/x - 35/6*sqrt(c*x^2 + b*x)*A*b*c/x + 5/4*(c*x^2 + b*x)^(3/2)*B
*b/x^2 - 5/6*sqrt(c*x^2 + b*x)*A*b^2/x^2 + 1/2*(c*x^2 + b*x)^(5/2)*B/x^3 - 5/6*(c*x^2 + b*x)^(3/2)*A*b/x^3 + (
c*x^2 + b*x)^(5/2)*A/x^4

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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^5} \,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^5,x)

[Out]

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

[Out]

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

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