3.1.91 \(\int \frac {(b x+c x^2)^{3/2}}{x^{11/2}} \, dx\) [91]

Optimal. Leaf size=111 \[ -\frac {c \sqrt {b x+c x^2}}{4 x^{5/2}}-\frac {c^2 \sqrt {b x+c x^2}}{8 b x^{3/2}}-\frac {\left (b x+c x^2\right )^{3/2}}{3 x^{9/2}}+\frac {c^3 \tanh ^{-1}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right )}{8 b^{3/2}} \]

[Out]

-1/3*(c*x^2+b*x)^(3/2)/x^(9/2)+1/8*c^3*arctanh((c*x^2+b*x)^(1/2)/b^(1/2)/x^(1/2))/b^(3/2)-1/4*c*(c*x^2+b*x)^(1
/2)/x^(5/2)-1/8*c^2*(c*x^2+b*x)^(1/2)/b/x^(3/2)

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Rubi [A]
time = 0.03, antiderivative size = 111, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {676, 686, 674, 213} \begin {gather*} \frac {c^3 \tanh ^{-1}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right )}{8 b^{3/2}}-\frac {c^2 \sqrt {b x+c x^2}}{8 b x^{3/2}}-\frac {c \sqrt {b x+c x^2}}{4 x^{5/2}}-\frac {\left (b x+c x^2\right )^{3/2}}{3 x^{9/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(b*x + c*x^2)^(3/2)/x^(11/2),x]

[Out]

-1/4*(c*Sqrt[b*x + c*x^2])/x^(5/2) - (c^2*Sqrt[b*x + c*x^2])/(8*b*x^(3/2)) - (b*x + c*x^2)^(3/2)/(3*x^(9/2)) +
(c^3*ArcTanh[Sqrt[b*x + c*x^2]/(Sqrt[b]*Sqrt[x])])/(8*b^(3/2))

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

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Mathematica [A]
time = 0.13, size = 94, normalized size = 0.85 \begin {gather*} \frac {\sqrt {x (b+c x)} \left (-\sqrt {b} \sqrt {b+c x} \left (8 b^2+14 b c x+3 c^2 x^2\right )+3 c^3 x^3 \tanh ^{-1}\left (\frac {\sqrt {b+c x}}{\sqrt {b}}\right )\right )}{24 b^{3/2} x^{7/2} \sqrt {b+c x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(b*x + c*x^2)^(3/2)/x^(11/2),x]

[Out]

(Sqrt[x*(b + c*x)]*(-(Sqrt[b]*Sqrt[b + c*x]*(8*b^2 + 14*b*c*x + 3*c^2*x^2)) + 3*c^3*x^3*ArcTanh[Sqrt[b + c*x]/
Sqrt[b]]))/(24*b^(3/2)*x^(7/2)*Sqrt[b + c*x])

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Maple [A]
time = 0.43, size = 90, normalized size = 0.81

method	result	size
risch	\(-\frac {\left (c x +b \right ) \left (3 c^{2} x^{2}+14 b c x +8 b^{2}\right )}{24 x^{\frac {5}{2}} b \sqrt {x \left (c x +b \right )}}+\frac {c^{3} \arctanh \left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right ) \sqrt {c x +b}\, \sqrt {x}}{8 b^{\frac {3}{2}} \sqrt {x \left (c x +b \right )}}\)	\(82\)
default	\(\frac {\sqrt {x \left (c x +b \right )}\, \left (3 \arctanh \left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right ) c^{3} x^{3}-3 c^{2} x^{2} \sqrt {b}\, \sqrt {c x +b}-14 b^{\frac {3}{2}} c x \sqrt {c x +b}-8 b^{\frac {5}{2}} \sqrt {c x +b}\right )}{24 b^{\frac {3}{2}} x^{\frac {7}{2}} \sqrt {c x +b}}\)	\(90\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^2+b*x)^(3/2)/x^(11/2),x,method=_RETURNVERBOSE)

[Out]

1/24*(x*(c*x+b))^(1/2)/b^(3/2)*(3*arctanh((c*x+b)^(1/2)/b^(1/2))*c^3*x^3-3*c^2*x^2*b^(1/2)*(c*x+b)^(1/2)-14*b^
(3/2)*c*x*(c*x+b)^(1/2)-8*b^(5/2)*(c*x+b)^(1/2))/x^(7/2)/(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((c*x^2+b*x)^(3/2)/x^(11/2),x, algorithm="maxima")

[Out]

integrate((c*x^2 + b*x)^(3/2)/x^(11/2), x)

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Fricas [A]
time = 1.45, size = 174, normalized size = 1.57 \begin {gather*} \left [\frac {3 \, \sqrt {b} c^{3} 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 (3 \, b c^{2} x^{2} + 14 \, b^{2} c x + 8 \, b^{3}\right )} \sqrt {c x^{2} + b x} \sqrt {x}}{48 \, b^{2} x^{4}}, -\frac {3 \, \sqrt {-b} c^{3} x^{4} \arctan \left (\frac {\sqrt {-b} \sqrt {x}}{\sqrt {c x^{2} + b x}}\right ) + {\left (3 \, b c^{2} x^{2} + 14 \, b^{2} c x + 8 \, b^{3}\right )} \sqrt {c x^{2} + b x} \sqrt {x}}{24 \, b^{2} x^{4}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x)^(3/2)/x^(11/2),x, algorithm="fricas")

[Out]

[1/48*(3*sqrt(b)*c^3*x^4*log(-(c*x^2 + 2*b*x + 2*sqrt(c*x^2 + b*x)*sqrt(b)*sqrt(x))/x^2) - 2*(3*b*c^2*x^2 + 14
*b^2*c*x + 8*b^3)*sqrt(c*x^2 + b*x)*sqrt(x))/(b^2*x^4), -1/24*(3*sqrt(-b)*c^3*x^4*arctan(sqrt(-b)*sqrt(x)/sqrt
(c*x^2 + b*x)) + (3*b*c^2*x^2 + 14*b^2*c*x + 8*b^3)*sqrt(c*x^2 + b*x)*sqrt(x))/(b^2*x^4)]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (x \left (b + c x\right )\right )^{\frac {3}{2}}}{x^{\frac {11}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**2+b*x)**(3/2)/x**(11/2),x)

[Out]

Integral((x*(b + c*x))**(3/2)/x**(11/2), x)

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Giac [A]
time = 0.87, size = 84, normalized size = 0.76 \begin {gather*} -\frac {\frac {3 \, c^{4} \arctan \left (\frac {\sqrt {c x + b}}{\sqrt {-b}}\right )}{\sqrt {-b} b} + \frac {3 \, {\left (c x + b\right )}^{\frac {5}{2}} c^{4} + 8 \, {\left (c x + b\right )}^{\frac {3}{2}} b c^{4} - 3 \, \sqrt {c x + b} b^{2} c^{4}}{b c^{3} x^{3}}}{24 \, c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x)^(3/2)/x^(11/2),x, algorithm="giac")

[Out]

-1/24*(3*c^4*arctan(sqrt(c*x + b)/sqrt(-b))/(sqrt(-b)*b) + (3*(c*x + b)^(5/2)*c^4 + 8*(c*x + b)^(3/2)*b*c^4 -
3*sqrt(c*x + b)*b^2*c^4)/(b*c^3*x^3))/c

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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 )}^{3/2}}{x^{11/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x + c*x^2)^(3/2)/x^(11/2),x)

[Out]

int((b*x + c*x^2)^(3/2)/x^(11/2), x)

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