3.91 \(\int \frac{(b x+c x^2)^{3/2}}{x^{11/2}} \, dx\)

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

[Out]

-(c*Sqrt[b*x + c*x^2])/(4*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))

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Rubi [A]  time = 0.0460828, antiderivative size = 111, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, Rules used = {662, 672, 660, 207} \[ \frac{c^3 \tanh ^{-1}\left (\frac{\sqrt{b x+c x^2}}{\sqrt{b} \sqrt{x}}\right )}{8 b^{3/2}}-\frac{c^2 \sqrt{b x+c x^2}}{8 b x^{3/2}}-\frac{c \sqrt{b x+c x^2}}{4 x^{5/2}}-\frac{\left (b x+c x^2\right )^{3/2}}{3 x^{9/2}} \]

Antiderivative was successfully verified.

[In]

Int[(b*x + c*x^2)^(3/2)/x^(11/2),x]

[Out]

-(c*Sqrt[b*x + c*x^2])/(4*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 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^{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 \operatorname{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*}

Mathematica [C]  time = 0.0185051, size = 42, normalized size = 0.38 \[ \frac{2 c^3 (x (b+c x))^{5/2} \, _2F_1\left (\frac{5}{2},4;\frac{7}{2};\frac{c x}{b}+1\right )}{5 b^4 x^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*c^3*(x*(b + c*x))^(5/2)*Hypergeometric2F1[5/2, 4, 7/2, 1 + (c*x)/b])/(5*b^4*x^(5/2))

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Maple [A]  time = 0.189, size = 90, normalized size = 0.8 \begin{align*}{\frac{1}{24}\sqrt{x \left ( cx+b \right ) } \left ( 3\,{\it Artanh} \left ({\frac{\sqrt{cx+b}}{\sqrt{b}}} \right ){x}^{3}{c}^{3}-3\,{x}^{2}{c}^{2}\sqrt{b}\sqrt{cx+b}-14\,x{b}^{3/2}c\sqrt{cx+b}-8\,{b}^{5/2}\sqrt{cx+b} \right ){b}^{-{\frac{3}{2}}}{x}^{-{\frac{7}{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^(11/2),x)

[Out]

1/24*(x*(c*x+b))^(1/2)/b^(3/2)*(3*arctanh((c*x+b)^(1/2)/b^(1/2))*x^3*c^3-3*x^2*c^2*b^(1/2)*(c*x+b)^(1/2)-14*x*
b^(3/2)*c*(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., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (c x^{2} + b x\right )}^{\frac{3}{2}}}{x^{\frac{11}{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^(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 = 2.15364, size = 423, normalized size = 3.81 \begin{align*} \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{align*}

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(-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**(11/2),x)

[Out]

Timed out

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Giac [A]  time = 1.33937, size = 97, normalized size = 0.87 \begin{align*} -\frac{1}{24} \, c^{3}{\left (\frac{3 \, \arctan \left (\frac{\sqrt{c x + b}}{\sqrt{-b}}\right )}{\sqrt{-b} b} + \frac{3 \,{\left (c x + b\right )}^{\frac{5}{2}} + 8 \,{\left (c x + b\right )}^{\frac{3}{2}} b - 3 \, \sqrt{c x + b} b^{2}}{b c^{3} x^{3}}\right )} \end{align*}

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