3.3.22 \(\int \frac {(A+B x) (b x+c x^2)^{5/2}}{x^{13/2}} \, dx\)

Optimal. Leaf size=175 \[ \frac {5 c^2 \sqrt {b x+c x^2} (A c+6 b B)}{8 b \sqrt {x}}-\frac {5 c^2 (A c+6 b B) \tanh ^{-1}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right )}{8 \sqrt {b}}-\frac {\left (b x+c x^2\right )^{5/2} (A c+6 b B)}{12 b x^{9/2}}-\frac {5 c \left (b x+c x^2\right )^{3/2} (A c+6 b B)}{24 b x^{5/2}}-\frac {A \left (b x+c x^2\right )^{7/2}}{3 b x^{13/2}} \]

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Rubi [A]  time = 0.17, antiderivative size = 175, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {792, 662, 664, 660, 207} \begin {gather*} \frac {5 c^2 \sqrt {b x+c x^2} (A c+6 b B)}{8 b \sqrt {x}}-\frac {5 c^2 (A c+6 b B) \tanh ^{-1}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right )}{8 \sqrt {b}}-\frac {\left (b x+c x^2\right )^{5/2} (A c+6 b B)}{12 b x^{9/2}}-\frac {5 c \left (b x+c x^2\right )^{3/2} (A c+6 b B)}{24 b x^{5/2}}-\frac {A \left (b x+c x^2\right )^{7/2}}{3 b x^{13/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(b*x + c*x^2)^(5/2))/x^(13/2),x]

[Out]

(5*c^2*(6*b*B + A*c)*Sqrt[b*x + c*x^2])/(8*b*Sqrt[x]) - (5*c*(6*b*B + A*c)*(b*x + c*x^2)^(3/2))/(24*b*x^(5/2))
- ((6*b*B + A*c)*(b*x + c*x^2)^(5/2))/(12*b*x^(9/2)) - (A*(b*x + c*x^2)^(7/2))/(3*b*x^(13/2)) - (5*c^2*(6*b*B
+ A*c)*ArcTanh[Sqrt[b*x + c*x^2]/(Sqrt[b]*Sqrt[x])])/(8*Sqrt[b])

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 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^{13/2}} \, dx &=-\frac {A \left (b x+c x^2\right )^{7/2}}{3 b x^{13/2}}+\frac {\left (-\frac {13}{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^{11/2}} \, dx}{3 b}\\ &=-\frac {(6 b B+A c) \left (b x+c x^2\right )^{5/2}}{12 b x^{9/2}}-\frac {A \left (b x+c x^2\right )^{7/2}}{3 b x^{13/2}}+\frac {(5 c (6 b B+A c)) \int \frac {\left (b x+c x^2\right )^{3/2}}{x^{7/2}} \, dx}{24 b}\\ &=-\frac {5 c (6 b B+A c) \left (b x+c x^2\right )^{3/2}}{24 b x^{5/2}}-\frac {(6 b B+A c) \left (b x+c x^2\right )^{5/2}}{12 b x^{9/2}}-\frac {A \left (b x+c x^2\right )^{7/2}}{3 b x^{13/2}}+\frac {\left (5 c^2 (6 b B+A c)\right ) \int \frac {\sqrt {b x+c x^2}}{x^{3/2}} \, dx}{16 b}\\ &=\frac {5 c^2 (6 b B+A c) \sqrt {b x+c x^2}}{8 b \sqrt {x}}-\frac {5 c (6 b B+A c) \left (b x+c x^2\right )^{3/2}}{24 b x^{5/2}}-\frac {(6 b B+A c) \left (b x+c x^2\right )^{5/2}}{12 b x^{9/2}}-\frac {A \left (b x+c x^2\right )^{7/2}}{3 b x^{13/2}}+\frac {1}{16} \left (5 c^2 (6 b B+A c)\right ) \int \frac {1}{\sqrt {x} \sqrt {b x+c x^2}} \, dx\\ &=\frac {5 c^2 (6 b B+A c) \sqrt {b x+c x^2}}{8 b \sqrt {x}}-\frac {5 c (6 b B+A c) \left (b x+c x^2\right )^{3/2}}{24 b x^{5/2}}-\frac {(6 b B+A c) \left (b x+c x^2\right )^{5/2}}{12 b x^{9/2}}-\frac {A \left (b x+c x^2\right )^{7/2}}{3 b x^{13/2}}+\frac {1}{8} \left (5 c^2 (6 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 )\\ &=\frac {5 c^2 (6 b B+A c) \sqrt {b x+c x^2}}{8 b \sqrt {x}}-\frac {5 c (6 b B+A c) \left (b x+c x^2\right )^{3/2}}{24 b x^{5/2}}-\frac {(6 b B+A c) \left (b x+c x^2\right )^{5/2}}{12 b x^{9/2}}-\frac {A \left (b x+c x^2\right )^{7/2}}{3 b x^{13/2}}-\frac {5 c^2 (6 b B+A c) \tanh ^{-1}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right )}{8 \sqrt {b}}\\ \end {align*}

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Mathematica [C]  time = 0.03, size = 68, normalized size = 0.39 \begin {gather*} -\frac {(b+c x)^3 \sqrt {x (b+c x)} \left (7 A b^3+c^2 x^3 (A c+6 b B) \, _2F_1\left (3,\frac {7}{2};\frac {9}{2};\frac {c x}{b}+1\right )\right )}{21 b^4 x^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(b*x + c*x^2)^(5/2))/x^(13/2),x]

[Out]

-1/21*((b + c*x)^3*Sqrt[x*(b + c*x)]*(7*A*b^3 + c^2*(6*b*B + A*c)*x^3*Hypergeometric2F1[3, 7/2, 9/2, 1 + (c*x)
/b]))/(b^4*x^(7/2))

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IntegrateAlgebraic [A]  time = 1.04, size = 116, normalized size = 0.66 \begin {gather*} \frac {\sqrt {b x+c x^2} \left (-8 A b^2-26 A b c x-33 A c^2 x^2-12 b^2 B x-54 b B c x^2+48 B c^2 x^3\right )}{24 x^{7/2}}-\frac {5 \left (A c^3+6 b B c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {b x+c x^2}}\right )}{8 \sqrt {b}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((A + B*x)*(b*x + c*x^2)^(5/2))/x^(13/2),x]

[Out]

(Sqrt[b*x + c*x^2]*(-8*A*b^2 - 12*b^2*B*x - 26*A*b*c*x - 54*b*B*c*x^2 - 33*A*c^2*x^2 + 48*B*c^2*x^3))/(24*x^(7
/2)) - (5*(6*b*B*c^2 + A*c^3)*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[b*x + c*x^2]])/(8*Sqrt[b])

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fricas [A]  time = 0.42, size = 258, normalized size = 1.47 \begin {gather*} \left [\frac {15 \, {\left (6 \, B b c^{2} + A c^{3}\right )} \sqrt {b} x^{4} \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 \, B b c^{2} x^{3} - 8 \, A b^{3} - 3 \, {\left (18 \, B b^{2} c + 11 \, A b c^{2}\right )} x^{2} - 2 \, {\left (6 \, B b^{3} + 13 \, A b^{2} c\right )} x\right )} \sqrt {c x^{2} + b x} \sqrt {x}}{48 \, b x^{4}}, \frac {15 \, {\left (6 \, B b c^{2} + A c^{3}\right )} \sqrt {-b} x^{4} \arctan \left (\frac {\sqrt {-b} \sqrt {x}}{\sqrt {c x^{2} + b x}}\right ) + {\left (48 \, B b c^{2} x^{3} - 8 \, A b^{3} - 3 \, {\left (18 \, B b^{2} c + 11 \, A b c^{2}\right )} x^{2} - 2 \, {\left (6 \, B b^{3} + 13 \, A b^{2} c\right )} x\right )} \sqrt {c x^{2} + b x} \sqrt {x}}{24 \, b x^{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^(13/2),x, algorithm="fricas")

[Out]

[1/48*(15*(6*B*b*c^2 + A*c^3)*sqrt(b)*x^4*log(-(c*x^2 + 2*b*x - 2*sqrt(c*x^2 + b*x)*sqrt(b)*sqrt(x))/x^2) + 2*
(48*B*b*c^2*x^3 - 8*A*b^3 - 3*(18*B*b^2*c + 11*A*b*c^2)*x^2 - 2*(6*B*b^3 + 13*A*b^2*c)*x)*sqrt(c*x^2 + b*x)*sq
rt(x))/(b*x^4), 1/24*(15*(6*B*b*c^2 + A*c^3)*sqrt(-b)*x^4*arctan(sqrt(-b)*sqrt(x)/sqrt(c*x^2 + b*x)) + (48*B*b
*c^2*x^3 - 8*A*b^3 - 3*(18*B*b^2*c + 11*A*b*c^2)*x^2 - 2*(6*B*b^3 + 13*A*b^2*c)*x)*sqrt(c*x^2 + b*x)*sqrt(x))/
(b*x^4)]

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giac [A]  time = 0.35, size = 151, normalized size = 0.86 \begin {gather*} \frac {48 \, \sqrt {c x + b} B c^{3} + \frac {15 \, {\left (6 \, B b c^{3} + A c^{4}\right )} \arctan \left (\frac {\sqrt {c x + b}}{\sqrt {-b}}\right )}{\sqrt {-b}} - \frac {54 \, {\left (c x + b\right )}^{\frac {5}{2}} B b c^{3} - 96 \, {\left (c x + b\right )}^{\frac {3}{2}} B b^{2} c^{3} + 42 \, \sqrt {c x + b} B b^{3} c^{3} + 33 \, {\left (c x + b\right )}^{\frac {5}{2}} A c^{4} - 40 \, {\left (c x + b\right )}^{\frac {3}{2}} A b c^{4} + 15 \, \sqrt {c x + b} A b^{2} c^{4}}{c^{3} x^{3}}}{24 \, 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^(13/2),x, algorithm="giac")

[Out]

1/24*(48*sqrt(c*x + b)*B*c^3 + 15*(6*B*b*c^3 + A*c^4)*arctan(sqrt(c*x + b)/sqrt(-b))/sqrt(-b) - (54*(c*x + b)^
(5/2)*B*b*c^3 - 96*(c*x + b)^(3/2)*B*b^2*c^3 + 42*sqrt(c*x + b)*B*b^3*c^3 + 33*(c*x + b)^(5/2)*A*c^4 - 40*(c*x
+ b)^(3/2)*A*b*c^4 + 15*sqrt(c*x + b)*A*b^2*c^4)/(c^3*x^3))/c

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maple [A]  time = 0.07, size = 166, normalized size = 0.95 \begin {gather*} -\frac {\sqrt {\left (c x +b \right ) x}\, \left (15 A \,c^{3} x^{3} \arctanh \left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right )+90 B b \,c^{2} x^{3} \arctanh \left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right )-48 \sqrt {c x +b}\, B \sqrt {b}\, c^{2} x^{3}+33 \sqrt {c x +b}\, A \sqrt {b}\, c^{2} x^{2}+54 \sqrt {c x +b}\, B \,b^{\frac {3}{2}} c \,x^{2}+26 \sqrt {c x +b}\, A \,b^{\frac {3}{2}} c x +12 \sqrt {c x +b}\, B \,b^{\frac {5}{2}} x +8 \sqrt {c x +b}\, A \,b^{\frac {5}{2}}\right )}{24 \sqrt {c x +b}\, \sqrt {b}\, x^{\frac {7}{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^(13/2),x)

[Out]

-1/24*((c*x+b)*x)^(1/2)*(15*A*arctanh((c*x+b)^(1/2)/b^(1/2))*x^3*c^3+90*B*arctanh((c*x+b)^(1/2)/b^(1/2))*x^3*b
*c^2-48*(c*x+b)^(1/2)*B*b^(1/2)*c^2*x^3+33*(c*x+b)^(1/2)*A*b^(1/2)*c^2*x^2+54*(c*x+b)^(1/2)*B*b^(3/2)*c*x^2+26
*(c*x+b)^(1/2)*A*b^(3/2)*c*x+12*(c*x+b)^(1/2)*B*b^(5/2)*x+8*(c*x+b)^(1/2)*A*b^(5/2))/x^(7/2)/(c*x+b)^(1/2)/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 {13}{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^(13/2),x, algorithm="maxima")

[Out]

integrate((c*x^2 + b*x)^(5/2)*(B*x + A)/x^(13/2), 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^{13/2}} \,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^(13/2),x)

[Out]

int(((b*x + c*x^2)^(5/2)*(A + B*x))/x^(13/2), 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**(13/2),x)

[Out]

Timed out

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