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

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

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Rubi [A]  time = 0.05, 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 = {662, 672, 660, 207} \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]

-(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 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])

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 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]

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*}

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Mathematica [C]  time = 0.02, size = 42, normalized size = 0.38 \begin {gather*} \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}} \end {gather*}

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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IntegrateAlgebraic [A]  time = 0.61, size = 82, normalized size = 0.74 \begin {gather*} \frac {c^3 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {b x+c x^2}}\right )}{8 b^{3/2}}+\frac {\sqrt {b x+c x^2} \left (-8 b^2-14 b c x-3 c^2 x^2\right )}{24 b x^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.42, 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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giac [A]  time = 0.23, 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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maple [A]  time = 0.06, size = 90, normalized size = 0.81 \begin {gather*} \frac {\sqrt {\left (c x +b \right ) x}\, \left (3 c^{3} x^{3} \arctanh \left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right )-3 \sqrt {c x +b}\, \sqrt {b}\, c^{2} x^{2}-14 \sqrt {c x +b}\, b^{\frac {3}{2}} c x -8 \sqrt {c x +b}\, b^{\frac {5}{2}}\right )}{24 \sqrt {c x +b}\, b^{\frac {3}{2}} x^{\frac {7}{2}}} \end {gather*}

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*((c*x+b)*x)^(1/2)/b^(3/2)*(3*arctanh((c*x+b)^(1/2)/b^(1/2))*c^3*x^3-3*(c*x+b)^(1/2)*b^(1/2)*c^2*x^2-14*(c
*x+b)^(1/2)*b^(3/2)*c*x-8*(c*x+b)^(1/2)*b^(5/2))/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*} \int \frac {{\left (c x^{2} + b x\right )}^{\frac {3}{2}}}{x^{\frac {11}{2}}}\,{d x} \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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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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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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