\(\int \frac {(b x+c x^2)^{3/2}}{x^{13/2}} \, dx\) [92]

Optimal result

Integrand size = 19, antiderivative size = 139 \[ \int \frac {\left (b x+c x^2\right )^{3/2}}{x^{13/2}} \, dx=-\frac {c \sqrt {b x+c x^2}}{8 x^{7/2}}-\frac {c^2 \sqrt {b x+c x^2}}{32 b x^{5/2}}+\frac {3 c^3 \sqrt {b x+c x^2}}{64 b^2 x^{3/2}}-\frac {\left (b x+c x^2\right )^{3/2}}{4 x^{11/2}}-\frac {3 c^4 \text {arctanh}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right )}{64 b^{5/2}} \]

[Out]

-1/4*(c*x^2+b*x)^(3/2)/x^(11/2)-3/64*c^4*arctanh((c*x^2+b*x)^(1/2)/b^(1/2)/x^(1/2))/b^(5/2)-1/8*c*(c*x^2+b*x)^
(1/2)/x^(7/2)-1/32*c^2*(c*x^2+b*x)^(1/2)/b/x^(5/2)+3/64*c^3*(c*x^2+b*x)^(1/2)/b^2/x^(3/2)

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {676, 686, 674, 213} \[ \int \frac {\left (b x+c x^2\right )^{3/2}}{x^{13/2}} \, dx=-\frac {3 c^4 \text {arctanh}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right )}{64 b^{5/2}}+\frac {3 c^3 \sqrt {b x+c x^2}}{64 b^2 x^{3/2}}-\frac {c^2 \sqrt {b x+c x^2}}{32 b x^{5/2}}-\frac {c \sqrt {b x+c x^2}}{8 x^{7/2}}-\frac {\left (b x+c x^2\right )^{3/2}}{4 x^{11/2}} \]

[In]

Int[(b*x + c*x^2)^(3/2)/x^(13/2),x]

[Out]

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

Mathematica [A] (verified)

Time = 0.22 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.75 \[ \int \frac {\left (b x+c x^2\right )^{3/2}}{x^{13/2}} \, dx=-\frac {\sqrt {x (b+c x)} \left (\sqrt {b} \sqrt {b+c x} \left (16 b^3+24 b^2 c x+2 b c^2 x^2-3 c^3 x^3\right )+3 c^4 x^4 \text {arctanh}\left (\frac {\sqrt {b+c x}}{\sqrt {b}}\right )\right )}{64 b^{5/2} x^{9/2} \sqrt {b+c x}} \]

[In]

Integrate[(b*x + c*x^2)^(3/2)/x^(13/2),x]

[Out]

-1/64*(Sqrt[x*(b + c*x)]*(Sqrt[b]*Sqrt[b + c*x]*(16*b^3 + 24*b^2*c*x + 2*b*c^2*x^2 - 3*c^3*x^3) + 3*c^4*x^4*Ar
cTanh[Sqrt[b + c*x]/Sqrt[b]]))/(b^(5/2)*x^(9/2)*Sqrt[b + c*x])

Maple [A] (verified)

Time = 2.04 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.67

method	result	size
risch	\(-\frac {\left (c x +b \right ) \left (-3 c^{3} x^{3}+2 b \,c^{2} x^{2}+24 b^{2} c x +16 b^{3}\right )}{64 x^{\frac {7}{2}} b^{2} \sqrt {x \left (c x +b \right )}}-\frac {3 c^{4} \operatorname {arctanh}\left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right ) \sqrt {c x +b}\, \sqrt {x}}{64 b^{\frac {5}{2}} \sqrt {x \left (c x +b \right )}}\)	\(93\)
default	\(-\frac {\sqrt {x \left (c x +b \right )}\, \left (3 \,\operatorname {arctanh}\left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right ) c^{4} x^{4}-3 c^{3} x^{3} \sqrt {c x +b}\, \sqrt {b}+2 b^{\frac {3}{2}} c^{2} x^{2} \sqrt {c x +b}+24 b^{\frac {5}{2}} c x \sqrt {c x +b}+16 b^{\frac {7}{2}} \sqrt {c x +b}\right )}{64 b^{\frac {5}{2}} x^{\frac {9}{2}} \sqrt {c x +b}}\)	\(108\)

[In]

int((c*x^2+b*x)^(3/2)/x^(13/2),x,method=_RETURNVERBOSE)

[Out]

-1/64*(c*x+b)*(-3*c^3*x^3+2*b*c^2*x^2+24*b^2*c*x+16*b^3)/x^(7/2)/b^2/(x*(c*x+b))^(1/2)-3/64*c^4/b^(5/2)*arctan
h((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 = 0.27 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.41 \[ \int \frac {\left (b x+c x^2\right )^{3/2}}{x^{13/2}} \, dx=\left [\frac {3 \, \sqrt {b} c^{4} x^{5} \log \left (-\frac {c x^{2} + 2 \, b x - 2 \, \sqrt {c x^{2} + b x} \sqrt {b} \sqrt {x}}{x^{2}}\right ) + 2 \, {\left (3 \, b c^{3} x^{3} - 2 \, b^{2} c^{2} x^{2} - 24 \, b^{3} c x - 16 \, b^{4}\right )} \sqrt {c x^{2} + b x} \sqrt {x}}{128 \, b^{3} x^{5}}, \frac {3 \, \sqrt {-b} c^{4} x^{5} \arctan \left (\frac {\sqrt {-b} \sqrt {x}}{\sqrt {c x^{2} + b x}}\right ) + {\left (3 \, b c^{3} x^{3} - 2 \, b^{2} c^{2} x^{2} - 24 \, b^{3} c x - 16 \, b^{4}\right )} \sqrt {c x^{2} + b x} \sqrt {x}}{64 \, b^{3} x^{5}}\right ] \]

[In]

integrate((c*x^2+b*x)^(3/2)/x^(13/2),x, algorithm="fricas")

[Out]

[1/128*(3*sqrt(b)*c^4*x^5*log(-(c*x^2 + 2*b*x - 2*sqrt(c*x^2 + b*x)*sqrt(b)*sqrt(x))/x^2) + 2*(3*b*c^3*x^3 - 2
*b^2*c^2*x^2 - 24*b^3*c*x - 16*b^4)*sqrt(c*x^2 + b*x)*sqrt(x))/(b^3*x^5), 1/64*(3*sqrt(-b)*c^4*x^5*arctan(sqrt
(-b)*sqrt(x)/sqrt(c*x^2 + b*x)) + (3*b*c^3*x^3 - 2*b^2*c^2*x^2 - 24*b^3*c*x - 16*b^4)*sqrt(c*x^2 + b*x)*sqrt(x
))/(b^3*x^5)]

Sympy [F]

\[ \int \frac {\left (b x+c x^2\right )^{3/2}}{x^{13/2}} \, dx=\int \frac {\left (x \left (b + c x\right )\right )^{\frac {3}{2}}}{x^{\frac {13}{2}}}\, dx \]

[In]

integrate((c*x**2+b*x)**(3/2)/x**(13/2),x)

[Out]

Integral((x*(b + c*x))**(3/2)/x**(13/2), x)

Maxima [F]

\[ \int \frac {\left (b x+c x^2\right )^{3/2}}{x^{13/2}} \, dx=\int { \frac {{\left (c x^{2} + b x\right )}^{\frac {3}{2}}}{x^{\frac {13}{2}}} \,d x } \]

[In]

integrate((c*x^2+b*x)^(3/2)/x^(13/2),x, algorithm="maxima")

[Out]

integrate((c*x^2 + b*x)^(3/2)/x^(13/2), x)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.71 \[ \int \frac {\left (b x+c x^2\right )^{3/2}}{x^{13/2}} \, dx=\frac {\frac {3 \, c^{5} \arctan \left (\frac {\sqrt {c x + b}}{\sqrt {-b}}\right )}{\sqrt {-b} b^{2}} + \frac {3 \, {\left (c x + b\right )}^{\frac {7}{2}} c^{5} - 11 \, {\left (c x + b\right )}^{\frac {5}{2}} b c^{5} - 11 \, {\left (c x + b\right )}^{\frac {3}{2}} b^{2} c^{5} + 3 \, \sqrt {c x + b} b^{3} c^{5}}{b^{2} c^{4} x^{4}}}{64 \, c} \]

[In]

integrate((c*x^2+b*x)^(3/2)/x^(13/2),x, algorithm="giac")

[Out]

1/64*(3*c^5*arctan(sqrt(c*x + b)/sqrt(-b))/(sqrt(-b)*b^2) + (3*(c*x + b)^(7/2)*c^5 - 11*(c*x + b)^(5/2)*b*c^5
- 11*(c*x + b)^(3/2)*b^2*c^5 + 3*sqrt(c*x + b)*b^3*c^5)/(b^2*c^4*x^4))/c

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (b x+c x^2\right )^{3/2}}{x^{13/2}} \, dx=\int \frac {{\left (c\,x^2+b\,x\right )}^{3/2}}{x^{13/2}} \,d x \]

[In]

int((b*x + c*x^2)^(3/2)/x^(13/2),x)

[Out]

int((b*x + c*x^2)^(3/2)/x^(13/2), x)