\(\int \frac {(b x+c x^2)^{3/2}}{x^{15/2}} \, dx\) [93]

Optimal result

Integrand size = 19, antiderivative size = 167 \[ \int \frac {\left (b x+c x^2\right )^{3/2}}{x^{15/2}} \, dx=-\frac {3 c \sqrt {b x+c x^2}}{40 x^{9/2}}-\frac {c^2 \sqrt {b x+c x^2}}{80 b x^{7/2}}+\frac {c^3 \sqrt {b x+c x^2}}{64 b^2 x^{5/2}}-\frac {3 c^4 \sqrt {b x+c x^2}}{128 b^3 x^{3/2}}-\frac {\left (b x+c x^2\right )^{3/2}}{5 x^{13/2}}+\frac {3 c^5 \text {arctanh}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right )}{128 b^{7/2}} \]

[Out]

-1/5*(c*x^2+b*x)^(3/2)/x^(13/2)+3/128*c^5*arctanh((c*x^2+b*x)^(1/2)/b^(1/2)/x^(1/2))/b^(7/2)-3/40*c*(c*x^2+b*x
)^(1/2)/x^(9/2)-1/80*c^2*(c*x^2+b*x)^(1/2)/b/x^(7/2)+1/64*c^3*(c*x^2+b*x)^(1/2)/b^2/x^(5/2)-3/128*c^4*(c*x^2+b
*x)^(1/2)/b^3/x^(3/2)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {676, 686, 674, 213} \[ \int \frac {\left (b x+c x^2\right )^{3/2}}{x^{15/2}} \, dx=\frac {3 c^5 \text {arctanh}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right )}{128 b^{7/2}}-\frac {3 c^4 \sqrt {b x+c x^2}}{128 b^3 x^{3/2}}+\frac {c^3 \sqrt {b x+c x^2}}{64 b^2 x^{5/2}}-\frac {c^2 \sqrt {b x+c x^2}}{80 b x^{7/2}}-\frac {3 c \sqrt {b x+c x^2}}{40 x^{9/2}}-\frac {\left (b x+c x^2\right )^{3/2}}{5 x^{13/2}} \]

[In]

Int[(b*x + c*x^2)^(3/2)/x^(15/2),x]

[Out]

(-3*c*Sqrt[b*x + c*x^2])/(40*x^(9/2)) - (c^2*Sqrt[b*x + c*x^2])/(80*b*x^(7/2)) + (c^3*Sqrt[b*x + c*x^2])/(64*b
^2*x^(5/2)) - (3*c^4*Sqrt[b*x + c*x^2])/(128*b^3*x^(3/2)) - (b*x + c*x^2)^(3/2)/(5*x^(13/2)) + (3*c^5*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]

Rubi steps \begin{align*} \text {integral}& = -\frac {\left (b x+c x^2\right )^{3/2}}{5 x^{13/2}}+\frac {1}{10} (3 c) \int \frac {\sqrt {b x+c x^2}}{x^{11/2}} \, dx \\ & = -\frac {3 c \sqrt {b x+c x^2}}{40 x^{9/2}}-\frac {\left (b x+c x^2\right )^{3/2}}{5 x^{13/2}}+\frac {1}{80} \left (3 c^2\right ) \int \frac {1}{x^{7/2} \sqrt {b x+c x^2}} \, dx \\ & = -\frac {3 c \sqrt {b x+c x^2}}{40 x^{9/2}}-\frac {c^2 \sqrt {b x+c x^2}}{80 b x^{7/2}}-\frac {\left (b x+c x^2\right )^{3/2}}{5 x^{13/2}}-\frac {c^3 \int \frac {1}{x^{5/2} \sqrt {b x+c x^2}} \, dx}{32 b} \\ & = -\frac {3 c \sqrt {b x+c x^2}}{40 x^{9/2}}-\frac {c^2 \sqrt {b x+c x^2}}{80 b x^{7/2}}+\frac {c^3 \sqrt {b x+c x^2}}{64 b^2 x^{5/2}}-\frac {\left (b x+c x^2\right )^{3/2}}{5 x^{13/2}}+\frac {\left (3 c^4\right ) \int \frac {1}{x^{3/2} \sqrt {b x+c x^2}} \, dx}{128 b^2} \\ & = -\frac {3 c \sqrt {b x+c x^2}}{40 x^{9/2}}-\frac {c^2 \sqrt {b x+c x^2}}{80 b x^{7/2}}+\frac {c^3 \sqrt {b x+c x^2}}{64 b^2 x^{5/2}}-\frac {3 c^4 \sqrt {b x+c x^2}}{128 b^3 x^{3/2}}-\frac {\left (b x+c x^2\right )^{3/2}}{5 x^{13/2}}-\frac {\left (3 c^5\right ) \int \frac {1}{\sqrt {x} \sqrt {b x+c x^2}} \, dx}{256 b^3} \\ & = -\frac {3 c \sqrt {b x+c x^2}}{40 x^{9/2}}-\frac {c^2 \sqrt {b x+c x^2}}{80 b x^{7/2}}+\frac {c^3 \sqrt {b x+c x^2}}{64 b^2 x^{5/2}}-\frac {3 c^4 \sqrt {b x+c x^2}}{128 b^3 x^{3/2}}-\frac {\left (b x+c x^2\right )^{3/2}}{5 x^{13/2}}-\frac {\left (3 c^5\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 {3 c \sqrt {b x+c x^2}}{40 x^{9/2}}-\frac {c^2 \sqrt {b x+c x^2}}{80 b x^{7/2}}+\frac {c^3 \sqrt {b x+c x^2}}{64 b^2 x^{5/2}}-\frac {3 c^4 \sqrt {b x+c x^2}}{128 b^3 x^{3/2}}-\frac {\left (b x+c x^2\right )^{3/2}}{5 x^{13/2}}+\frac {3 c^5 \tanh ^{-1}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right )}{128 b^{7/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.27 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.69 \[ \int \frac {\left (b x+c x^2\right )^{3/2}}{x^{15/2}} \, dx=\frac {\sqrt {b+c x} \left (-\sqrt {b} \sqrt {b+c x} \left (128 b^4+176 b^3 c x+8 b^2 c^2 x^2-10 b c^3 x^3+15 c^4 x^4\right )+15 c^5 x^5 \text {arctanh}\left (\frac {\sqrt {b+c x}}{\sqrt {b}}\right )\right )}{640 b^{7/2} x^{9/2} \sqrt {x (b+c x)}} \]

[In]

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

[Out]

(Sqrt[b + c*x]*(-(Sqrt[b]*Sqrt[b + c*x]*(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^5*x^5*ArcTanh[Sqrt[b + c*x]/Sqrt[b]]))/(640*b^(7/2)*x^(9/2)*Sqrt[x*(b + c*x)])

Maple [A] (verified)

Time = 2.04 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.62

method	result	size
risch	\(-\frac {\left (c x +b \right ) \left (15 c^{4} x^{4}-10 b \,c^{3} x^{3}+8 b^{2} c^{2} x^{2}+176 b^{3} c x +128 b^{4}\right )}{640 x^{\frac {9}{2}} b^{3} \sqrt {x \left (c x +b \right )}}+\frac {3 c^{5} \operatorname {arctanh}\left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right ) \sqrt {c x +b}\, \sqrt {x}}{128 b^{\frac {7}{2}} \sqrt {x \left (c x +b \right )}}\)	\(104\)
default	\(\frac {\sqrt {x \left (c x +b \right )}\, \left (15 \,\operatorname {arctanh}\left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right ) x^{5} c^{5}-15 c^{4} x^{4} \sqrt {c x +b}\, \sqrt {b}+10 b^{\frac {3}{2}} c^{3} x^{3} \sqrt {c x +b}-8 b^{\frac {5}{2}} c^{2} x^{2} \sqrt {c x +b}-176 b^{\frac {7}{2}} c x \sqrt {c x +b}-128 b^{\frac {9}{2}} \sqrt {c x +b}\right )}{640 b^{\frac {7}{2}} x^{\frac {11}{2}} \sqrt {c x +b}}\)	\(126\)

[In]

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

[Out]

-1/640*(c*x+b)*(15*c^4*x^4-10*b*c^3*x^3+8*b^2*c^2*x^2+176*b^3*c*x+128*b^4)/x^(9/2)/b^3/(x*(c*x+b))^(1/2)+3/128
*c^5/b^(7/2)*arctanh((c*x+b)^(1/2)/b^(1/2))*(c*x+b)^(1/2)*x^(1/2)/(x*(c*x+b))^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.31 \[ \int \frac {\left (b x+c x^2\right )^{3/2}}{x^{15/2}} \, dx=\left [\frac {15 \, \sqrt {b} c^{5} x^{6} \log \left (-\frac {c x^{2} + 2 \, b x + 2 \, \sqrt {c x^{2} + b x} \sqrt {b} \sqrt {x}}{x^{2}}\right ) - 2 \, {\left (15 \, b c^{4} x^{4} - 10 \, b^{2} c^{3} x^{3} + 8 \, b^{3} c^{2} x^{2} + 176 \, b^{4} c x + 128 \, b^{5}\right )} \sqrt {c x^{2} + b x} \sqrt {x}}{1280 \, b^{4} x^{6}}, -\frac {15 \, \sqrt {-b} c^{5} x^{6} \arctan \left (\frac {\sqrt {-b} \sqrt {x}}{\sqrt {c x^{2} + b x}}\right ) + {\left (15 \, b c^{4} x^{4} - 10 \, b^{2} c^{3} x^{3} + 8 \, b^{3} c^{2} x^{2} + 176 \, b^{4} c x + 128 \, b^{5}\right )} \sqrt {c x^{2} + b x} \sqrt {x}}{640 \, b^{4} x^{6}}\right ] \]

[In]

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

[Out]

[1/1280*(15*sqrt(b)*c^5*x^6*log(-(c*x^2 + 2*b*x + 2*sqrt(c*x^2 + b*x)*sqrt(b)*sqrt(x))/x^2) - 2*(15*b*c^4*x^4
- 10*b^2*c^3*x^3 + 8*b^3*c^2*x^2 + 176*b^4*c*x + 128*b^5)*sqrt(c*x^2 + b*x)*sqrt(x))/(b^4*x^6), -1/640*(15*sqr
t(-b)*c^5*x^6*arctan(sqrt(-b)*sqrt(x)/sqrt(c*x^2 + b*x)) + (15*b*c^4*x^4 - 10*b^2*c^3*x^3 + 8*b^3*c^2*x^2 + 17
6*b^4*c*x + 128*b^5)*sqrt(c*x^2 + b*x)*sqrt(x))/(b^4*x^6)]

Sympy [F(-1)]

Timed out. \[ \int \frac {\left (b x+c x^2\right )^{3/2}}{x^{15/2}} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F]

\[ \int \frac {\left (b x+c x^2\right )^{3/2}}{x^{15/2}} \, dx=\int { \frac {{\left (c x^{2} + b x\right )}^{\frac {3}{2}}}{x^{\frac {15}{2}}} \,d x } \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.68 \[ \int \frac {\left (b x+c x^2\right )^{3/2}}{x^{15/2}} \, dx=-\frac {\frac {15 \, c^{6} \arctan \left (\frac {\sqrt {c x + b}}{\sqrt {-b}}\right )}{\sqrt {-b} b^{3}} + \frac {15 \, {\left (c x + b\right )}^{\frac {9}{2}} c^{6} - 70 \, {\left (c x + b\right )}^{\frac {7}{2}} b c^{6} + 128 \, {\left (c x + b\right )}^{\frac {5}{2}} b^{2} c^{6} + 70 \, {\left (c x + b\right )}^{\frac {3}{2}} b^{3} c^{6} - 15 \, \sqrt {c x + b} b^{4} c^{6}}{b^{3} c^{5} x^{5}}}{640 \, c} \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (b x+c x^2\right )^{3/2}}{x^{15/2}} \, dx=\int \frac {{\left (c\,x^2+b\,x\right )}^{3/2}}{x^{15/2}} \,d x \]

[In]

int((b*x + c*x^2)^(3/2)/x^(15/2),x)

[Out]

int((b*x + c*x^2)^(3/2)/x^(15/2), x)