3.3.23 \(\int \frac {(A+B x) (b x+c x^2)^{5/2}}{x^{15/2}} \, dx\)

Optimal. Leaf size=179 \[ -\frac {5 c^3 (8 b B-A c) \tanh ^{-1}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right )}{64 b^{3/2}}-\frac {5 c^2 \sqrt {b x+c x^2} (8 b B-A c)}{64 b x^{3/2}}-\frac {\left (b x+c x^2\right )^{5/2} (8 b B-A c)}{24 b x^{11/2}}-\frac {5 c \left (b x+c x^2\right )^{3/2} (8 b B-A c)}{96 b x^{7/2}}-\frac {A \left (b x+c x^2\right )^{7/2}}{4 b x^{15/2}} \]

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Rubi [A]  time = 0.17, antiderivative size = 179, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {792, 662, 660, 207} \begin {gather*} -\frac {5 c^3 (8 b B-A c) \tanh ^{-1}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right )}{64 b^{3/2}}-\frac {5 c^2 \sqrt {b x+c x^2} (8 b B-A c)}{64 b x^{3/2}}-\frac {\left (b x+c x^2\right )^{5/2} (8 b B-A c)}{24 b x^{11/2}}-\frac {5 c \left (b x+c x^2\right )^{3/2} (8 b B-A c)}{96 b x^{7/2}}-\frac {A \left (b x+c x^2\right )^{7/2}}{4 b x^{15/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(b*x + c*x^2)^(5/2))/x^(15/2),x]

[Out]

(-5*c^2*(8*b*B - A*c)*Sqrt[b*x + c*x^2])/(64*b*x^(3/2)) - (5*c*(8*b*B - A*c)*(b*x + c*x^2)^(3/2))/(96*b*x^(7/2
)) - ((8*b*B - A*c)*(b*x + c*x^2)^(5/2))/(24*b*x^(11/2)) - (A*(b*x + c*x^2)^(7/2))/(4*b*x^(15/2)) - (5*c^3*(8*
b*B - A*c)*ArcTanh[Sqrt[b*x + c*x^2]/(Sqrt[b]*Sqrt[x])])/(64*b^(3/2))

Rule 207

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTanh[(Rt[b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rule 660

Int[1/(Sqrt[(d_.) + (e_.)*(x_)]*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[2*e, Subst[Int[1/(
2*c*d - b*e + e^2*x^2), x], x, Sqrt[a + b*x + c*x^2]/Sqrt[d + e*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]

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 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^{15/2}} \, dx &=-\frac {A \left (b x+c x^2\right )^{7/2}}{4 b x^{15/2}}+\frac {\left (-\frac {15}{2} (-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^{13/2}} \, dx}{4 b}\\ &=-\frac {(8 b B-A c) \left (b x+c x^2\right )^{5/2}}{24 b x^{11/2}}-\frac {A \left (b x+c x^2\right )^{7/2}}{4 b x^{15/2}}+\frac {(5 c (8 b B-A c)) \int \frac {\left (b x+c x^2\right )^{3/2}}{x^{9/2}} \, dx}{48 b}\\ &=-\frac {5 c (8 b B-A c) \left (b x+c x^2\right )^{3/2}}{96 b x^{7/2}}-\frac {(8 b B-A c) \left (b x+c x^2\right )^{5/2}}{24 b x^{11/2}}-\frac {A \left (b x+c x^2\right )^{7/2}}{4 b x^{15/2}}+\frac {\left (5 c^2 (8 b B-A c)\right ) \int \frac {\sqrt {b x+c x^2}}{x^{5/2}} \, dx}{64 b}\\ &=-\frac {5 c^2 (8 b B-A c) \sqrt {b x+c x^2}}{64 b x^{3/2}}-\frac {5 c (8 b B-A c) \left (b x+c x^2\right )^{3/2}}{96 b x^{7/2}}-\frac {(8 b B-A c) \left (b x+c x^2\right )^{5/2}}{24 b x^{11/2}}-\frac {A \left (b x+c x^2\right )^{7/2}}{4 b x^{15/2}}+\frac {\left (5 c^3 (8 b B-A c)\right ) \int \frac {1}{\sqrt {x} \sqrt {b x+c x^2}} \, dx}{128 b}\\ &=-\frac {5 c^2 (8 b B-A c) \sqrt {b x+c x^2}}{64 b x^{3/2}}-\frac {5 c (8 b B-A c) \left (b x+c x^2\right )^{3/2}}{96 b x^{7/2}}-\frac {(8 b B-A c) \left (b x+c x^2\right )^{5/2}}{24 b x^{11/2}}-\frac {A \left (b x+c x^2\right )^{7/2}}{4 b x^{15/2}}+\frac {\left (5 c^3 (8 b B-A c)\right ) \operatorname {Subst}\left (\int \frac {1}{-b+x^2} \, dx,x,\frac {\sqrt {b x+c x^2}}{\sqrt {x}}\right )}{64 b}\\ &=-\frac {5 c^2 (8 b B-A c) \sqrt {b x+c x^2}}{64 b x^{3/2}}-\frac {5 c (8 b B-A c) \left (b x+c x^2\right )^{3/2}}{96 b x^{7/2}}-\frac {(8 b B-A c) \left (b x+c x^2\right )^{5/2}}{24 b x^{11/2}}-\frac {A \left (b x+c x^2\right )^{7/2}}{4 b x^{15/2}}-\frac {5 c^3 (8 b B-A c) \tanh ^{-1}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right )}{64 b^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.14, size = 129, normalized size = 0.72 \begin {gather*} -\frac {(b+c x) \left (A \left (48 b^3+136 b^2 c x+118 b c^2 x^2+15 c^3 x^3\right )+8 b B x \left (8 b^2+26 b c x+33 c^2 x^2\right )\right )+15 c^3 x^4 \sqrt {\frac {c x}{b}+1} (8 b B-A c) \tanh ^{-1}\left (\sqrt {\frac {c x}{b}+1}\right )}{192 b x^{7/2} \sqrt {x (b+c x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(b*x + c*x^2)^(5/2))/x^(15/2),x]

[Out]

-1/192*((b + c*x)*(8*b*B*x*(8*b^2 + 26*b*c*x + 33*c^2*x^2) + A*(48*b^3 + 136*b^2*c*x + 118*b*c^2*x^2 + 15*c^3*
x^3)) + 15*c^3*(8*b*B - A*c)*x^4*Sqrt[1 + (c*x)/b]*ArcTanh[Sqrt[1 + (c*x)/b]])/(b*x^(7/2)*Sqrt[x*(b + c*x)])

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IntegrateAlgebraic [A]  time = 1.26, size = 135, normalized size = 0.75 \begin {gather*} \frac {\sqrt {b x+c x^2} \left (-48 A b^3-136 A b^2 c x-118 A b c^2 x^2-15 A c^3 x^3-64 b^3 B x-208 b^2 B c x^2-264 b B c^2 x^3\right )}{192 b x^{9/2}}-\frac {5 \left (8 b B c^3-A c^4\right ) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {b x+c x^2}}\right )}{64 b^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((A + B*x)*(b*x + c*x^2)^(5/2))/x^(15/2),x]

[Out]

(Sqrt[b*x + c*x^2]*(-48*A*b^3 - 64*b^3*B*x - 136*A*b^2*c*x - 208*b^2*B*c*x^2 - 118*A*b*c^2*x^2 - 264*b*B*c^2*x
^3 - 15*A*c^3*x^3))/(192*b*x^(9/2)) - (5*(8*b*B*c^3 - A*c^4)*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[b*x + c*x^2]])/(64
*b^(3/2))

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fricas [A]  time = 0.43, size = 289, normalized size = 1.61 \begin {gather*} \left [-\frac {15 \, {\left (8 \, B b c^{3} - A c^{4}\right )} \sqrt {b} x^{5} \log \left (-\frac {c x^{2} + 2 \, b x + 2 \, \sqrt {c x^{2} + b x} \sqrt {b} \sqrt {x}}{x^{2}}\right ) + 2 \, {\left (48 \, A b^{4} + 3 \, {\left (88 \, B b^{2} c^{2} + 5 \, A b c^{3}\right )} x^{3} + 2 \, {\left (104 \, B b^{3} c + 59 \, A b^{2} c^{2}\right )} x^{2} + 8 \, {\left (8 \, B b^{4} + 17 \, A b^{3} c\right )} x\right )} \sqrt {c x^{2} + b x} \sqrt {x}}{384 \, b^{2} x^{5}}, \frac {15 \, {\left (8 \, B b c^{3} - A c^{4}\right )} \sqrt {-b} x^{5} \arctan \left (\frac {\sqrt {-b} \sqrt {x}}{\sqrt {c x^{2} + b x}}\right ) - {\left (48 \, A b^{4} + 3 \, {\left (88 \, B b^{2} c^{2} + 5 \, A b c^{3}\right )} x^{3} + 2 \, {\left (104 \, B b^{3} c + 59 \, A b^{2} c^{2}\right )} x^{2} + 8 \, {\left (8 \, B b^{4} + 17 \, A b^{3} c\right )} x\right )} \sqrt {c x^{2} + b x} \sqrt {x}}{192 \, b^{2} x^{5}}\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^(15/2),x, algorithm="fricas")

[Out]

[-1/384*(15*(8*B*b*c^3 - A*c^4)*sqrt(b)*x^5*log(-(c*x^2 + 2*b*x + 2*sqrt(c*x^2 + b*x)*sqrt(b)*sqrt(x))/x^2) +
2*(48*A*b^4 + 3*(88*B*b^2*c^2 + 5*A*b*c^3)*x^3 + 2*(104*B*b^3*c + 59*A*b^2*c^2)*x^2 + 8*(8*B*b^4 + 17*A*b^3*c)
*x)*sqrt(c*x^2 + b*x)*sqrt(x))/(b^2*x^5), 1/192*(15*(8*B*b*c^3 - A*c^4)*sqrt(-b)*x^5*arctan(sqrt(-b)*sqrt(x)/s
qrt(c*x^2 + b*x)) - (48*A*b^4 + 3*(88*B*b^2*c^2 + 5*A*b*c^3)*x^3 + 2*(104*B*b^3*c + 59*A*b^2*c^2)*x^2 + 8*(8*B
*b^4 + 17*A*b^3*c)*x)*sqrt(c*x^2 + b*x)*sqrt(x))/(b^2*x^5)]

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giac [A]  time = 0.34, size = 177, normalized size = 0.99 \begin {gather*} \frac {\frac {15 \, {\left (8 \, B b c^{4} - A c^{5}\right )} \arctan \left (\frac {\sqrt {c x + b}}{\sqrt {-b}}\right )}{\sqrt {-b} b} - \frac {264 \, {\left (c x + b\right )}^{\frac {7}{2}} B b c^{4} - 584 \, {\left (c x + b\right )}^{\frac {5}{2}} B b^{2} c^{4} + 440 \, {\left (c x + b\right )}^{\frac {3}{2}} B b^{3} c^{4} - 120 \, \sqrt {c x + b} B b^{4} c^{4} + 15 \, {\left (c x + b\right )}^{\frac {7}{2}} A c^{5} + 73 \, {\left (c x + b\right )}^{\frac {5}{2}} A b c^{5} - 55 \, {\left (c x + b\right )}^{\frac {3}{2}} A b^{2} c^{5} + 15 \, \sqrt {c x + b} A b^{3} c^{5}}{b c^{4} x^{4}}}{192 \, 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^(15/2),x, algorithm="giac")

[Out]

1/192*(15*(8*B*b*c^4 - A*c^5)*arctan(sqrt(c*x + b)/sqrt(-b))/(sqrt(-b)*b) - (264*(c*x + b)^(7/2)*B*b*c^4 - 584
*(c*x + b)^(5/2)*B*b^2*c^4 + 440*(c*x + b)^(3/2)*B*b^3*c^4 - 120*sqrt(c*x + b)*B*b^4*c^4 + 15*(c*x + b)^(7/2)*
A*c^5 + 73*(c*x + b)^(5/2)*A*b*c^5 - 55*(c*x + b)^(3/2)*A*b^2*c^5 + 15*sqrt(c*x + b)*A*b^3*c^5)/(b*c^4*x^4))/c

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maple [A]  time = 0.07, size = 185, normalized size = 1.03 \begin {gather*} \frac {\sqrt {\left (c x +b \right ) x}\, \left (15 A \,c^{4} x^{4} \arctanh \left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right )-120 B b \,c^{3} x^{4} \arctanh \left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right )-15 \sqrt {c x +b}\, A \sqrt {b}\, c^{3} x^{3}-264 \sqrt {c x +b}\, B \,b^{\frac {3}{2}} c^{2} x^{3}-118 \sqrt {c x +b}\, A \,b^{\frac {3}{2}} c^{2} x^{2}-208 \sqrt {c x +b}\, B \,b^{\frac {5}{2}} c \,x^{2}-136 \sqrt {c x +b}\, A \,b^{\frac {5}{2}} c x -64 \sqrt {c x +b}\, B \,b^{\frac {7}{2}} x -48 \sqrt {c x +b}\, A \,b^{\frac {7}{2}}\right )}{192 \sqrt {c x +b}\, b^{\frac {3}{2}} x^{\frac {9}{2}}} \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^(15/2),x)

[Out]

1/192*((c*x+b)*x)^(1/2)/b^(3/2)*(15*A*arctanh((c*x+b)^(1/2)/b^(1/2))*x^4*c^4-120*B*arctanh((c*x+b)^(1/2)/b^(1/
2))*x^4*b*c^3-15*(c*x+b)^(1/2)*A*b^(1/2)*c^3*x^3-264*(c*x+b)^(1/2)*B*b^(3/2)*c^2*x^3-118*(c*x+b)^(1/2)*A*b^(3/
2)*c^2*x^2-208*(c*x+b)^(1/2)*B*b^(5/2)*c*x^2-136*(c*x+b)^(1/2)*A*b^(5/2)*c*x-64*(c*x+b)^(1/2)*B*b^(7/2)*x-48*(
c*x+b)^(1/2)*A*b^(7/2))/x^(9/2)/(c*x+b)^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (c x^{2} + b x\right )}^{\frac {5}{2}} {\left (B x + A\right )}}{x^{\frac {15}{2}}}\,{d x} \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^(15/2),x, algorithm="maxima")

[Out]

integrate((c*x^2 + b*x)^(5/2)*(B*x + A)/x^(15/2), x)

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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^{15/2}} \,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^(15/2),x)

[Out]

int(((b*x + c*x^2)^(5/2)*(A + B*x))/x^(15/2), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \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**(15/2),x)

[Out]

Timed out

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