3.3.13 \(\int \frac {(A+B x) (b x+c x^2)^{3/2}}{x^{15/2}} \, dx\) [213]

Optimal. Leaf size=216 \[ -\frac {c (2 b B-A c) \sqrt {b x+c x^2}}{16 b x^{7/2}}-\frac {c^2 (2 b B-A c) \sqrt {b x+c x^2}}{64 b^2 x^{5/2}}+\frac {3 c^3 (2 b B-A c) \sqrt {b x+c x^2}}{128 b^3 x^{3/2}}-\frac {(2 b B-A c) \left (b x+c x^2\right )^{3/2}}{8 b x^{11/2}}-\frac {A \left (b x+c x^2\right )^{5/2}}{5 b x^{15/2}}-\frac {3 c^4 (2 b B-A c) \tanh ^{-1}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right )}{128 b^{7/2}} \]

[Out]

-1/8*(-A*c+2*B*b)*(c*x^2+b*x)^(3/2)/b/x^(11/2)-1/5*A*(c*x^2+b*x)^(5/2)/b/x^(15/2)-3/128*c^4*(-A*c+2*B*b)*arcta
nh((c*x^2+b*x)^(1/2)/b^(1/2)/x^(1/2))/b^(7/2)-1/16*c*(-A*c+2*B*b)*(c*x^2+b*x)^(1/2)/b/x^(7/2)-1/64*c^2*(-A*c+2
*B*b)*(c*x^2+b*x)^(1/2)/b^2/x^(5/2)+3/128*c^3*(-A*c+2*B*b)*(c*x^2+b*x)^(1/2)/b^3/x^(3/2)

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Rubi [A]
time = 0.13, 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 = {806, 676, 686, 674, 213} \begin {gather*} -\frac {3 c^4 (2 b B-A c) \tanh ^{-1}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right )}{128 b^{7/2}}+\frac {3 c^3 \sqrt {b x+c x^2} (2 b B-A c)}{128 b^3 x^{3/2}}-\frac {c^2 \sqrt {b x+c x^2} (2 b B-A c)}{64 b^2 x^{5/2}}-\frac {c \sqrt {b x+c x^2} (2 b B-A c)}{16 b x^{7/2}}-\frac {\left (b x+c x^2\right )^{3/2} (2 b B-A c)}{8 b x^{11/2}}-\frac {A \left (b x+c x^2\right )^{5/2}}{5 b x^{15/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(b*x + c*x^2)^(3/2))/x^(15/2),x]

[Out]

-1/16*(c*(2*b*B - A*c)*Sqrt[b*x + c*x^2])/(b*x^(7/2)) - (c^2*(2*b*B - A*c)*Sqrt[b*x + c*x^2])/(64*b^2*x^(5/2))
+ (3*c^3*(2*b*B - A*c)*Sqrt[b*x + c*x^2])/(128*b^3*x^(3/2)) - ((2*b*B - A*c)*(b*x + c*x^2)^(3/2))/(8*b*x^(11/
2)) - (A*(b*x + c*x^2)^(5/2))/(5*b*x^(15/2)) - (3*c^4*(2*b*B - A*c)*ArcTanh[Sqrt[b*x + c*x^2]/(Sqrt[b]*Sqrt[x]
)])/(128*b^(7/2))

Rule 213

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

Rule 674

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 676

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 686

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))),
Int[(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] && E
qQ[c*d^2 - b*d*e + a*e^2, 0] && LtQ[m, 0] && NeQ[m + p + 1, 0] && IntegerQ[2*p]

Rule 806

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 )^{3/2}}{x^{15/2}} \, dx &=-\frac {A \left (b x+c x^2\right )^{5/2}}{5 b x^{15/2}}+\frac {\left (-\frac {15}{2} (-b B+A c)+\frac {5}{2} (-b B+2 A c)\right ) \int \frac {\left (b x+c x^2\right )^{3/2}}{x^{13/2}} \, dx}{5 b}\\ &=-\frac {(2 b B-A c) \left (b x+c x^2\right )^{3/2}}{8 b x^{11/2}}-\frac {A \left (b x+c x^2\right )^{5/2}}{5 b x^{15/2}}+\frac {(3 c (2 b B-A c)) \int \frac {\sqrt {b x+c x^2}}{x^{9/2}} \, dx}{16 b}\\ &=-\frac {c (2 b B-A c) \sqrt {b x+c x^2}}{16 b x^{7/2}}-\frac {(2 b B-A c) \left (b x+c x^2\right )^{3/2}}{8 b x^{11/2}}-\frac {A \left (b x+c x^2\right )^{5/2}}{5 b x^{15/2}}+\frac {\left (c^2 (2 b B-A c)\right ) \int \frac {1}{x^{5/2} \sqrt {b x+c x^2}} \, dx}{32 b}\\ &=-\frac {c (2 b B-A c) \sqrt {b x+c x^2}}{16 b x^{7/2}}-\frac {c^2 (2 b B-A c) \sqrt {b x+c x^2}}{64 b^2 x^{5/2}}-\frac {(2 b B-A c) \left (b x+c x^2\right )^{3/2}}{8 b x^{11/2}}-\frac {A \left (b x+c x^2\right )^{5/2}}{5 b x^{15/2}}-\frac {\left (3 c^3 (2 b B-A c)\right ) \int \frac {1}{x^{3/2} \sqrt {b x+c x^2}} \, dx}{128 b^2}\\ &=-\frac {c (2 b B-A c) \sqrt {b x+c x^2}}{16 b x^{7/2}}-\frac {c^2 (2 b B-A c) \sqrt {b x+c x^2}}{64 b^2 x^{5/2}}+\frac {3 c^3 (2 b B-A c) \sqrt {b x+c x^2}}{128 b^3 x^{3/2}}-\frac {(2 b B-A c) \left (b x+c x^2\right )^{3/2}}{8 b x^{11/2}}-\frac {A \left (b x+c x^2\right )^{5/2}}{5 b x^{15/2}}+\frac {\left (3 c^4 (2 b B-A c)\right ) \int \frac {1}{\sqrt {x} \sqrt {b x+c x^2}} \, dx}{256 b^3}\\ &=-\frac {c (2 b B-A c) \sqrt {b x+c x^2}}{16 b x^{7/2}}-\frac {c^2 (2 b B-A c) \sqrt {b x+c x^2}}{64 b^2 x^{5/2}}+\frac {3 c^3 (2 b B-A c) \sqrt {b x+c x^2}}{128 b^3 x^{3/2}}-\frac {(2 b B-A c) \left (b x+c x^2\right )^{3/2}}{8 b x^{11/2}}-\frac {A \left (b x+c x^2\right )^{5/2}}{5 b x^{15/2}}+\frac {\left (3 c^4 (2 b B-A c)\right ) \text {Subst}\left (\int \frac {1}{-b+x^2} \, dx,x,\frac {\sqrt {b x+c x^2}}{\sqrt {x}}\right )}{128 b^3}\\ &=-\frac {c (2 b B-A c) \sqrt {b x+c x^2}}{16 b x^{7/2}}-\frac {c^2 (2 b B-A c) \sqrt {b x+c x^2}}{64 b^2 x^{5/2}}+\frac {3 c^3 (2 b B-A c) \sqrt {b x+c x^2}}{128 b^3 x^{3/2}}-\frac {(2 b B-A c) \left (b x+c x^2\right )^{3/2}}{8 b x^{11/2}}-\frac {A \left (b x+c x^2\right )^{5/2}}{5 b x^{15/2}}-\frac {3 c^4 (2 b B-A c) \tanh ^{-1}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right )}{128 b^{7/2}}\\ \end {align*}

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Mathematica [A]
time = 0.34, size = 158, normalized size = 0.73 \begin {gather*} \frac {-\sqrt {b} (b+c x) \left (10 b B x \left (16 b^3+24 b^2 c x+2 b c^2 x^2-3 c^3 x^3\right )+A \left (128 b^4+176 b^3 c x+8 b^2 c^2 x^2-10 b c^3 x^3+15 c^4 x^4\right )\right )+15 c^4 (-2 b B+A c) x^5 \sqrt {b+c x} \tanh ^{-1}\left (\frac {\sqrt {b+c x}}{\sqrt {b}}\right )}{640 b^{7/2} x^{9/2} \sqrt {x (b+c x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(b*x + c*x^2)^(3/2))/x^(15/2),x]

[Out]

(-(Sqrt[b]*(b + c*x)*(10*b*B*x*(16*b^3 + 24*b^2*c*x + 2*b*c^2*x^2 - 3*c^3*x^3) + A*(128*b^4 + 176*b^3*c*x + 8*
b^2*c^2*x^2 - 10*b*c^3*x^3 + 15*c^4*x^4))) + 15*c^4*(-2*b*B + A*c)*x^5*Sqrt[b + c*x]*ArcTanh[Sqrt[b + c*x]/Sqr
t[b]])/(640*b^(7/2)*x^(9/2)*Sqrt[x*(b + c*x)])

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Maple [A]
time = 0.59, size = 223, normalized size = 1.03

method	result	size
risch	\(-\frac {\left (c x +b \right ) \left (15 A \,c^{4} x^{4}-30 B b \,c^{3} x^{4}-10 A b \,c^{3} x^{3}+20 B \,b^{2} c^{2} x^{3}+8 A \,b^{2} c^{2} x^{2}+240 B \,b^{3} c \,x^{2}+176 A \,b^{3} c x +160 B \,b^{4} x +128 A \,b^{4}\right )}{640 x^{\frac {9}{2}} b^{3} \sqrt {x \left (c x +b \right )}}+\frac {3 c^{4} \left (A c -2 B b \right ) \arctanh \left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right ) \sqrt {c x +b}\, \sqrt {x}}{128 b^{\frac {7}{2}} \sqrt {x \left (c x +b \right )}}\)	\(156\)
default	\(\frac {\sqrt {x \left (c x +b \right )}\, \left (15 A \arctanh \left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right ) c^{5} x^{5}-30 B \arctanh \left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right ) b \,c^{4} x^{5}-15 A \,c^{4} x^{4} \sqrt {b}\, \sqrt {c x +b}+30 B \,b^{\frac {3}{2}} c^{3} x^{4} \sqrt {c x +b}+10 A \,b^{\frac {3}{2}} c^{3} x^{3} \sqrt {c x +b}-20 B \,b^{\frac {5}{2}} c^{2} x^{3} \sqrt {c x +b}-8 A \,b^{\frac {5}{2}} c^{2} x^{2} \sqrt {c x +b}-240 B \,b^{\frac {7}{2}} c \,x^{2} \sqrt {c x +b}-176 A \,b^{\frac {7}{2}} c x \sqrt {c x +b}-160 B \,b^{\frac {9}{2}} x \sqrt {c x +b}-128 A \,b^{\frac {9}{2}} \sqrt {c x +b}\right )}{640 b^{\frac {7}{2}} x^{\frac {11}{2}} \sqrt {c x +b}}\)	\(223\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(c*x^2+b*x)^(3/2)/x^(15/2),x,method=_RETURNVERBOSE)

[Out]

1/640*(x*(c*x+b))^(1/2)/b^(7/2)*(15*A*arctanh((c*x+b)^(1/2)/b^(1/2))*c^5*x^5-30*B*arctanh((c*x+b)^(1/2)/b^(1/2
))*b*c^4*x^5-15*A*c^4*x^4*b^(1/2)*(c*x+b)^(1/2)+30*B*b^(3/2)*c^3*x^4*(c*x+b)^(1/2)+10*A*b^(3/2)*c^3*x^3*(c*x+b
)^(1/2)-20*B*b^(5/2)*c^2*x^3*(c*x+b)^(1/2)-8*A*b^(5/2)*c^2*x^2*(c*x+b)^(1/2)-240*B*b^(7/2)*c*x^2*(c*x+b)^(1/2)
-176*A*b^(7/2)*c*x*(c*x+b)^(1/2)-160*B*b^(9/2)*x*(c*x+b)^(1/2)-128*A*b^(9/2)*(c*x+b)^(1/2))/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*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x)^(3/2)/x^(15/2),x, algorithm="maxima")

[Out]

integrate((c*x^2 + b*x)^(3/2)*(B*x + A)/x^(15/2), x)

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Fricas [A]
time = 3.48, size = 335, normalized size = 1.55 \begin {gather*} \left [-\frac {15 \, {\left (2 \, B b c^{4} - 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 (128 \, A b^{5} - 15 \, {\left (2 \, B b^{2} c^{3} - A b c^{4}\right )} x^{4} + 10 \, {\left (2 \, B b^{3} c^{2} - A b^{2} c^{3}\right )} x^{3} + 8 \, {\left (30 \, B b^{4} c + A b^{3} c^{2}\right )} x^{2} + 16 \, {\left (10 \, B b^{5} + 11 \, A b^{4} c\right )} x\right )} \sqrt {c x^{2} + b x} \sqrt {x}}{1280 \, b^{4} x^{6}}, \frac {15 \, {\left (2 \, B b c^{4} - A c^{5}\right )} \sqrt {-b} x^{6} \arctan \left (\frac {\sqrt {-b} \sqrt {x}}{\sqrt {c x^{2} + b x}}\right ) - {\left (128 \, A b^{5} - 15 \, {\left (2 \, B b^{2} c^{3} - A b c^{4}\right )} x^{4} + 10 \, {\left (2 \, B b^{3} c^{2} - A b^{2} c^{3}\right )} x^{3} + 8 \, {\left (30 \, B b^{4} c + A b^{3} c^{2}\right )} x^{2} + 16 \, {\left (10 \, B b^{5} + 11 \, A b^{4} c\right )} x\right )} \sqrt {c x^{2} + b x} \sqrt {x}}{640 \, b^{4} 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)^(3/2)/x^(15/2),x, algorithm="fricas")

[Out]

[-1/1280*(15*(2*B*b*c^4 - 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*(128*A*b^5 - 15*(2*B*b^2*c^3 - A*b*c^4)*x^4 + 10*(2*B*b^3*c^2 - A*b^2*c^3)*x^3 + 8*(30*B*b^4*c + A*b^3*c^2)
*x^2 + 16*(10*B*b^5 + 11*A*b^4*c)*x)*sqrt(c*x^2 + b*x)*sqrt(x))/(b^4*x^6), 1/640*(15*(2*B*b*c^4 - A*c^5)*sqrt(
-b)*x^6*arctan(sqrt(-b)*sqrt(x)/sqrt(c*x^2 + b*x)) - (128*A*b^5 - 15*(2*B*b^2*c^3 - A*b*c^4)*x^4 + 10*(2*B*b^3
*c^2 - A*b^2*c^3)*x^3 + 8*(30*B*b^4*c + A*b^3*c^2)*x^2 + 16*(10*B*b^5 + 11*A*b^4*c)*x)*sqrt(c*x^2 + b*x)*sqrt(
x))/(b^4*x^6)]

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x**2+b*x)**(3/2)/x**(15/2),x)

[Out]

Exception raised: SystemError >> excessive stack use: stack is 4061 deep

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Giac [A]
time = 3.12, size = 192, normalized size = 0.89 \begin {gather*} \frac {\frac {15 \, {\left (2 \, B b c^{5} - A c^{6}\right )} \arctan \left (\frac {\sqrt {c x + b}}{\sqrt {-b}}\right )}{\sqrt {-b} b^{3}} + \frac {30 \, {\left (c x + b\right )}^{\frac {9}{2}} B b c^{5} - 140 \, {\left (c x + b\right )}^{\frac {7}{2}} B b^{2} c^{5} + 140 \, {\left (c x + b\right )}^{\frac {3}{2}} B b^{4} c^{5} - 30 \, \sqrt {c x + b} B b^{5} c^{5} - 15 \, {\left (c x + b\right )}^{\frac {9}{2}} A c^{6} + 70 \, {\left (c x + b\right )}^{\frac {7}{2}} A b c^{6} - 128 \, {\left (c x + b\right )}^{\frac {5}{2}} A b^{2} c^{6} - 70 \, {\left (c x + b\right )}^{\frac {3}{2}} A b^{3} c^{6} + 15 \, \sqrt {c x + b} A b^{4} c^{6}}{b^{3} c^{5} x^{5}}}{640 \, c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x)^(3/2)/x^(15/2),x, algorithm="giac")

[Out]

1/640*(15*(2*B*b*c^5 - A*c^6)*arctan(sqrt(c*x + b)/sqrt(-b))/(sqrt(-b)*b^3) + (30*(c*x + b)^(9/2)*B*b*c^5 - 14
0*(c*x + b)^(7/2)*B*b^2*c^5 + 140*(c*x + b)^(3/2)*B*b^4*c^5 - 30*sqrt(c*x + b)*B*b^5*c^5 - 15*(c*x + b)^(9/2)*
A*c^6 + 70*(c*x + b)^(7/2)*A*b*c^6 - 128*(c*x + b)^(5/2)*A*b^2*c^6 - 70*(c*x + b)^(3/2)*A*b^3*c^6 + 15*sqrt(c*
x + b)*A*b^4*c^6)/(b^3*c^5*x^5))/c

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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 )}^{3/2}\,\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)^(3/2)*(A + B*x))/x^(15/2),x)

[Out]

int(((b*x + c*x^2)^(3/2)*(A + B*x))/x^(15/2), x)

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