3.3.24 \(\int \frac {(A+B x) (b x+c x^2)^{5/2}}{x^{17/2}} \, dx\)

Optimal. Leaf size=216 \[ \frac {c^4 (10 b B-3 A c) \tanh ^{-1}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right )}{128 b^{5/2}}-\frac {c^3 \sqrt {b x+c x^2} (10 b B-3 A c)}{128 b^2 x^{3/2}}-\frac {c^2 \sqrt {b x+c x^2} (10 b B-3 A c)}{64 b x^{5/2}}-\frac {c \left (b x+c x^2\right )^{3/2} (10 b B-3 A c)}{48 b x^{9/2}}-\frac {\left (b x+c x^2\right )^{5/2} (10 b B-3 A c)}{40 b x^{13/2}}-\frac {A \left (b x+c x^2\right )^{7/2}}{5 b x^{17/2}} \]

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Rubi [A]  time = 0.21, antiderivative size = 216, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {792, 662, 672, 660, 207} \begin {gather*} -\frac {c^3 \sqrt {b x+c x^2} (10 b B-3 A c)}{128 b^2 x^{3/2}}+\frac {c^4 (10 b B-3 A c) \tanh ^{-1}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right )}{128 b^{5/2}}-\frac {c^2 \sqrt {b x+c x^2} (10 b B-3 A c)}{64 b x^{5/2}}-\frac {c \left (b x+c x^2\right )^{3/2} (10 b B-3 A c)}{48 b x^{9/2}}-\frac {\left (b x+c x^2\right )^{5/2} (10 b B-3 A c)}{40 b x^{13/2}}-\frac {A \left (b x+c x^2\right )^{7/2}}{5 b x^{17/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(b*x + c*x^2)^(5/2))/x^(17/2),x]

[Out]

-(c^2*(10*b*B - 3*A*c)*Sqrt[b*x + c*x^2])/(64*b*x^(5/2)) - (c^3*(10*b*B - 3*A*c)*Sqrt[b*x + c*x^2])/(128*b^2*x
^(3/2)) - (c*(10*b*B - 3*A*c)*(b*x + c*x^2)^(3/2))/(48*b*x^(9/2)) - ((10*b*B - 3*A*c)*(b*x + c*x^2)^(5/2))/(40
*b*x^(13/2)) - (A*(b*x + c*x^2)^(7/2))/(5*b*x^(17/2)) + (c^4*(10*b*B - 3*A*c)*ArcTanh[Sqrt[b*x + c*x^2]/(Sqrt[
b]*Sqrt[x])])/(128*b^(5/2))

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 672

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[(e*(d + e*x)^m*(a +
b*x + c*x^2)^(p + 1))/((m + p + 1)*(2*c*d - b*e)), x] + Dist[(c*(m + 2*p + 2))/((m + p + 1)*(2*c*d - b*e)), I
nt[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ
[c*d^2 - b*d*e + a*e^2, 0] && LtQ[m, 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^{17/2}} \, dx &=-\frac {A \left (b x+c x^2\right )^{7/2}}{5 b x^{17/2}}+\frac {\left (-\frac {17}{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^{15/2}} \, dx}{5 b}\\ &=-\frac {(10 b B-3 A c) \left (b x+c x^2\right )^{5/2}}{40 b x^{13/2}}-\frac {A \left (b x+c x^2\right )^{7/2}}{5 b x^{17/2}}+\frac {(c (10 b B-3 A c)) \int \frac {\left (b x+c x^2\right )^{3/2}}{x^{11/2}} \, dx}{16 b}\\ &=-\frac {c (10 b B-3 A c) \left (b x+c x^2\right )^{3/2}}{48 b x^{9/2}}-\frac {(10 b B-3 A c) \left (b x+c x^2\right )^{5/2}}{40 b x^{13/2}}-\frac {A \left (b x+c x^2\right )^{7/2}}{5 b x^{17/2}}+\frac {\left (c^2 (10 b B-3 A c)\right ) \int \frac {\sqrt {b x+c x^2}}{x^{7/2}} \, dx}{32 b}\\ &=-\frac {c^2 (10 b B-3 A c) \sqrt {b x+c x^2}}{64 b x^{5/2}}-\frac {c (10 b B-3 A c) \left (b x+c x^2\right )^{3/2}}{48 b x^{9/2}}-\frac {(10 b B-3 A c) \left (b x+c x^2\right )^{5/2}}{40 b x^{13/2}}-\frac {A \left (b x+c x^2\right )^{7/2}}{5 b x^{17/2}}+\frac {\left (c^3 (10 b B-3 A c)\right ) \int \frac {1}{x^{3/2} \sqrt {b x+c x^2}} \, dx}{128 b}\\ &=-\frac {c^2 (10 b B-3 A c) \sqrt {b x+c x^2}}{64 b x^{5/2}}-\frac {c^3 (10 b B-3 A c) \sqrt {b x+c x^2}}{128 b^2 x^{3/2}}-\frac {c (10 b B-3 A c) \left (b x+c x^2\right )^{3/2}}{48 b x^{9/2}}-\frac {(10 b B-3 A c) \left (b x+c x^2\right )^{5/2}}{40 b x^{13/2}}-\frac {A \left (b x+c x^2\right )^{7/2}}{5 b x^{17/2}}-\frac {\left (c^4 (10 b B-3 A c)\right ) \int \frac {1}{\sqrt {x} \sqrt {b x+c x^2}} \, dx}{256 b^2}\\ &=-\frac {c^2 (10 b B-3 A c) \sqrt {b x+c x^2}}{64 b x^{5/2}}-\frac {c^3 (10 b B-3 A c) \sqrt {b x+c x^2}}{128 b^2 x^{3/2}}-\frac {c (10 b B-3 A c) \left (b x+c x^2\right )^{3/2}}{48 b x^{9/2}}-\frac {(10 b B-3 A c) \left (b x+c x^2\right )^{5/2}}{40 b x^{13/2}}-\frac {A \left (b x+c x^2\right )^{7/2}}{5 b x^{17/2}}-\frac {\left (c^4 (10 b B-3 A c)\right ) \operatorname {Subst}\left (\int \frac {1}{-b+x^2} \, dx,x,\frac {\sqrt {b x+c x^2}}{\sqrt {x}}\right )}{128 b^2}\\ &=-\frac {c^2 (10 b B-3 A c) \sqrt {b x+c x^2}}{64 b x^{5/2}}-\frac {c^3 (10 b B-3 A c) \sqrt {b x+c x^2}}{128 b^2 x^{3/2}}-\frac {c (10 b B-3 A c) \left (b x+c x^2\right )^{3/2}}{48 b x^{9/2}}-\frac {(10 b B-3 A c) \left (b x+c x^2\right )^{5/2}}{40 b x^{13/2}}-\frac {A \left (b x+c x^2\right )^{7/2}}{5 b x^{17/2}}+\frac {c^4 (10 b B-3 A c) \tanh ^{-1}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right )}{128 b^{5/2}}\\ \end {align*}

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Mathematica [C]  time = 0.04, size = 69, normalized size = 0.32 \begin {gather*} -\frac {(b+c x)^3 \sqrt {x (b+c x)} \left (7 A b^5+c^4 x^5 (10 b B-3 A c) \, _2F_1\left (\frac {7}{2},5;\frac {9}{2};\frac {c x}{b}+1\right )\right )}{35 b^6 x^{11/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(b*x + c*x^2)^(5/2))/x^(17/2),x]

[Out]

-1/35*((b + c*x)^3*Sqrt[x*(b + c*x)]*(7*A*b^5 + c^4*(10*b*B - 3*A*c)*x^5*Hypergeometric2F1[7/2, 5, 9/2, 1 + (c
*x)/b]))/(b^6*x^(11/2))

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IntegrateAlgebraic [A]  time = 0.46, size = 199, normalized size = 0.92 \begin {gather*} \frac {(x (b+c x))^{5/2} \left (\frac {\left (10 b B c^4-3 A c^5\right ) \tanh ^{-1}\left (\frac {\sqrt {b+c x}}{\sqrt {b}}\right )}{128 b^{5/2}}+\frac {\sqrt {b+c x} \left (-45 A b^4 c+210 A b^3 c (b+c x)-384 A b^2 c (b+c x)^2-210 A b c (b+c x)^3+45 A c (b+c x)^4+150 b^5 B-700 b^4 B (b+c x)+1280 b^3 B (b+c x)^2-580 b^2 B (b+c x)^3-150 b B (b+c x)^4\right )}{1920 b^2 c x^5}\right )}{x^{5/2} (b+c x)^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((A + B*x)*(b*x + c*x^2)^(5/2))/x^(17/2),x]

[Out]

((x*(b + c*x))^(5/2)*((Sqrt[b + c*x]*(150*b^5*B - 45*A*b^4*c - 700*b^4*B*(b + c*x) + 210*A*b^3*c*(b + c*x) + 1
280*b^3*B*(b + c*x)^2 - 384*A*b^2*c*(b + c*x)^2 - 580*b^2*B*(b + c*x)^3 - 210*A*b*c*(b + c*x)^3 - 150*b*B*(b +
c*x)^4 + 45*A*c*(b + c*x)^4))/(1920*b^2*c*x^5) + ((10*b*B*c^4 - 3*A*c^5)*ArcTanh[Sqrt[b + c*x]/Sqrt[b]])/(128
*b^(5/2))))/(x^(5/2)*(b + c*x)^(5/2))

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fricas [A]  time = 0.43, size = 336, normalized size = 1.56 \begin {gather*} \left [-\frac {15 \, {\left (10 \, B b c^{4} - 3 \, A c^{5}\right )} \sqrt {b} x^{6} \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 (384 \, A b^{5} + 15 \, {\left (10 \, B b^{2} c^{3} - 3 \, A b c^{4}\right )} x^{4} + 10 \, {\left (118 \, B b^{3} c^{2} + 3 \, A b^{2} c^{3}\right )} x^{3} + 8 \, {\left (170 \, B b^{4} c + 93 \, A b^{3} c^{2}\right )} x^{2} + 48 \, {\left (10 \, B b^{5} + 21 \, A b^{4} c\right )} x\right )} \sqrt {c x^{2} + b x} \sqrt {x}}{3840 \, b^{3} x^{6}}, -\frac {15 \, {\left (10 \, B b c^{4} - 3 \, A c^{5}\right )} \sqrt {-b} x^{6} \arctan \left (\frac {\sqrt {-b} \sqrt {x}}{\sqrt {c x^{2} + b x}}\right ) + {\left (384 \, A b^{5} + 15 \, {\left (10 \, B b^{2} c^{3} - 3 \, A b c^{4}\right )} x^{4} + 10 \, {\left (118 \, B b^{3} c^{2} + 3 \, A b^{2} c^{3}\right )} x^{3} + 8 \, {\left (170 \, B b^{4} c + 93 \, A b^{3} c^{2}\right )} x^{2} + 48 \, {\left (10 \, B b^{5} + 21 \, A b^{4} c\right )} x\right )} \sqrt {c x^{2} + b x} \sqrt {x}}{1920 \, b^{3} x^{6}}\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^(17/2),x, algorithm="fricas")

[Out]

[-1/3840*(15*(10*B*b*c^4 - 3*A*c^5)*sqrt(b)*x^6*log(-(c*x^2 + 2*b*x - 2*sqrt(c*x^2 + b*x)*sqrt(b)*sqrt(x))/x^2
) + 2*(384*A*b^5 + 15*(10*B*b^2*c^3 - 3*A*b*c^4)*x^4 + 10*(118*B*b^3*c^2 + 3*A*b^2*c^3)*x^3 + 8*(170*B*b^4*c +
93*A*b^3*c^2)*x^2 + 48*(10*B*b^5 + 21*A*b^4*c)*x)*sqrt(c*x^2 + b*x)*sqrt(x))/(b^3*x^6), -1/1920*(15*(10*B*b*c
^4 - 3*A*c^5)*sqrt(-b)*x^6*arctan(sqrt(-b)*sqrt(x)/sqrt(c*x^2 + b*x)) + (384*A*b^5 + 15*(10*B*b^2*c^3 - 3*A*b*
c^4)*x^4 + 10*(118*B*b^3*c^2 + 3*A*b^2*c^3)*x^3 + 8*(170*B*b^4*c + 93*A*b^3*c^2)*x^2 + 48*(10*B*b^5 + 21*A*b^4
*c)*x)*sqrt(c*x^2 + b*x)*sqrt(x))/(b^3*x^6)]

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giac [A]  time = 0.44, size = 208, normalized size = 0.96 \begin {gather*} -\frac {\frac {15 \, {\left (10 \, B b c^{5} - 3 \, A c^{6}\right )} \arctan \left (\frac {\sqrt {c x + b}}{\sqrt {-b}}\right )}{\sqrt {-b} b^{2}} + \frac {150 \, {\left (c x + b\right )}^{\frac {9}{2}} B b c^{5} + 580 \, {\left (c x + b\right )}^{\frac {7}{2}} B b^{2} c^{5} - 1280 \, {\left (c x + b\right )}^{\frac {5}{2}} B b^{3} c^{5} + 700 \, {\left (c x + b\right )}^{\frac {3}{2}} B b^{4} c^{5} - 150 \, \sqrt {c x + b} B b^{5} c^{5} - 45 \, {\left (c x + b\right )}^{\frac {9}{2}} A c^{6} + 210 \, {\left (c x + b\right )}^{\frac {7}{2}} A b c^{6} + 384 \, {\left (c x + b\right )}^{\frac {5}{2}} A b^{2} c^{6} - 210 \, {\left (c x + b\right )}^{\frac {3}{2}} A b^{3} c^{6} + 45 \, \sqrt {c x + b} A b^{4} c^{6}}{b^{2} c^{5} x^{5}}}{1920 \, 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^(17/2),x, algorithm="giac")

[Out]

-1/1920*(15*(10*B*b*c^5 - 3*A*c^6)*arctan(sqrt(c*x + b)/sqrt(-b))/(sqrt(-b)*b^2) + (150*(c*x + b)^(9/2)*B*b*c^
5 + 580*(c*x + b)^(7/2)*B*b^2*c^5 - 1280*(c*x + b)^(5/2)*B*b^3*c^5 + 700*(c*x + b)^(3/2)*B*b^4*c^5 - 150*sqrt(
c*x + b)*B*b^5*c^5 - 45*(c*x + b)^(9/2)*A*c^6 + 210*(c*x + b)^(7/2)*A*b*c^6 + 384*(c*x + b)^(5/2)*A*b^2*c^6 -
210*(c*x + b)^(3/2)*A*b^3*c^6 + 45*sqrt(c*x + b)*A*b^4*c^6)/(b^2*c^5*x^5))/c

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maple [A]  time = 0.07, size = 223, normalized size = 1.03 \begin {gather*} -\frac {\sqrt {\left (c x +b \right ) x}\, \left (45 A \,c^{5} x^{5} \arctanh \left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right )-150 B b \,c^{4} x^{5} \arctanh \left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right )-45 \sqrt {c x +b}\, A \sqrt {b}\, c^{4} x^{4}+150 \sqrt {c x +b}\, B \,b^{\frac {3}{2}} c^{3} x^{4}+30 \sqrt {c x +b}\, A \,b^{\frac {3}{2}} c^{3} x^{3}+1180 \sqrt {c x +b}\, B \,b^{\frac {5}{2}} c^{2} x^{3}+744 \sqrt {c x +b}\, A \,b^{\frac {5}{2}} c^{2} x^{2}+1360 \sqrt {c x +b}\, B \,b^{\frac {7}{2}} c \,x^{2}+1008 \sqrt {c x +b}\, A \,b^{\frac {7}{2}} c x +480 \sqrt {c x +b}\, B \,b^{\frac {9}{2}} x +384 \sqrt {c x +b}\, A \,b^{\frac {9}{2}}\right )}{1920 \sqrt {c x +b}\, b^{\frac {5}{2}} x^{\frac {11}{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^(17/2),x)

[Out]

-1/1920*((c*x+b)*x)^(1/2)/b^(5/2)*(45*A*arctanh((c*x+b)^(1/2)/b^(1/2))*x^5*c^5-150*B*arctanh((c*x+b)^(1/2)/b^(
1/2))*x^5*b*c^4-45*(c*x+b)^(1/2)*A*b^(1/2)*c^4*x^4+150*(c*x+b)^(1/2)*B*b^(3/2)*c^3*x^4+30*(c*x+b)^(1/2)*A*b^(3
/2)*c^3*x^3+1180*(c*x+b)^(1/2)*B*b^(5/2)*c^2*x^3+744*(c*x+b)^(1/2)*A*b^(5/2)*c^2*x^2+1360*(c*x+b)^(1/2)*B*b^(7
/2)*c*x^2+1008*(c*x+b)^(1/2)*A*b^(7/2)*c*x+480*(c*x+b)^(1/2)*B*b^(9/2)*x+384*(c*x+b)^(1/2)*A*b^(9/2))/x^(11/2)
/(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 {17}{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^(17/2),x, algorithm="maxima")

[Out]

integrate((c*x^2 + b*x)^(5/2)*(B*x + A)/x^(17/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (c\,x^2+b\,x\right )}^{5/2}\,\left (A+B\,x\right )}{x^{17/2}} \,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^(17/2),x)

[Out]

int(((b*x + c*x^2)^(5/2)*(A + B*x))/x^(17/2), 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**(17/2),x)

[Out]

Timed out

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