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

Optimal. Leaf size=139 \[ \frac{3 c^3 \sqrt{b x+c x^2}}{64 b^2 x^{3/2}}-\frac{3 c^4 \tanh ^{-1}\left (\frac{\sqrt{b x+c x^2}}{\sqrt{b} \sqrt{x}}\right )}{64 b^{5/2}}-\frac{c^2 \sqrt{b x+c x^2}}{32 b x^{5/2}}-\frac{c \sqrt{b x+c x^2}}{8 x^{7/2}}-\frac{\left (b x+c x^2\right )^{3/2}}{4 x^{11/2}} \]

[Out]

-(c*Sqrt[b*x + c*x^2])/(8*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))

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Rubi [A]  time = 0.0646788, antiderivative size = 139, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, Rules used = {662, 672, 660, 207} \[ \frac{3 c^3 \sqrt{b x+c x^2}}{64 b^2 x^{3/2}}-\frac{3 c^4 \tanh ^{-1}\left (\frac{\sqrt{b x+c x^2}}{\sqrt{b} \sqrt{x}}\right )}{64 b^{5/2}}-\frac{c^2 \sqrt{b x+c x^2}}{32 b x^{5/2}}-\frac{c \sqrt{b x+c x^2}}{8 x^{7/2}}-\frac{\left (b x+c x^2\right )^{3/2}}{4 x^{11/2}} \]

Antiderivative was successfully verified.

[In]

Int[(b*x + c*x^2)^(3/2)/x^(13/2),x]

[Out]

-(c*Sqrt[b*x + c*x^2])/(8*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 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 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 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])

Rubi steps

\begin{align*} \int \frac{\left (b x+c x^2\right )^{3/2}}{x^{13/2}} \, dx &=-\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 ) \operatorname{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 [C]  time = 0.0176101, size = 42, normalized size = 0.3 \[ -\frac{2 c^4 (x (b+c x))^{5/2} \, _2F_1\left (\frac{5}{2},5;\frac{7}{2};\frac{c x}{b}+1\right )}{5 b^5 x^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*c^4*(x*(b + c*x))^(5/2)*Hypergeometric2F1[5/2, 5, 7/2, 1 + (c*x)/b])/(5*b^5*x^(5/2))

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Maple [A]  time = 0.186, size = 108, normalized size = 0.8 \begin{align*} -{\frac{1}{64}\sqrt{x \left ( cx+b \right ) } \left ( 3\,{\it Artanh} \left ({\frac{\sqrt{cx+b}}{\sqrt{b}}} \right ){x}^{4}{c}^{4}-3\,{x}^{3}{c}^{3}\sqrt{cx+b}\sqrt{b}+2\,{x}^{2}{b}^{3/2}{c}^{2}\sqrt{cx+b}+24\,x{b}^{5/2}c\sqrt{cx+b}+16\,{b}^{7/2}\sqrt{cx+b} \right ){b}^{-{\frac{5}{2}}}{x}^{-{\frac{9}{2}}}{\frac{1}{\sqrt{cx+b}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/64*(x*(c*x+b))^(1/2)/b^(5/2)*(3*arctanh((c*x+b)^(1/2)/b^(1/2))*x^4*c^4-3*x^3*c^3*(c*x+b)^(1/2)*b^(1/2)+2*x^
2*b^(3/2)*c^2*(c*x+b)^(1/2)+24*x*b^(5/2)*c*(c*x+b)^(1/2)+16*b^(7/2)*(c*x+b)^(1/2))/x^(9/2)/(c*x+b)^(1/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (c x^{2} + b x\right )}^{\frac{3}{2}}}{x^{\frac{13}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [A]  time = 2.09697, size = 468, normalized size = 3.37 \begin{align*} \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 ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]  time = 1.3455, size = 113, normalized size = 0.81 \begin{align*} \frac{1}{64} \, c^{4}{\left (\frac{3 \, \arctan \left (\frac{\sqrt{c x + b}}{\sqrt{-b}}\right )}{\sqrt{-b} b^{2}} + \frac{3 \,{\left (c x + b\right )}^{\frac{7}{2}} - 11 \,{\left (c x + b\right )}^{\frac{5}{2}} b - 11 \,{\left (c x + b\right )}^{\frac{3}{2}} b^{2} + 3 \, \sqrt{c x + b} b^{3}}{b^{2} c^{4} x^{4}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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