3.1.89 \(\int \frac {(b x+c x^2)^{3/2}}{x^{7/2}} \, dx\)

Optimal. Leaf size=75 \[ \frac {3 c \sqrt {b x+c x^2}}{\sqrt {x}}-3 \sqrt {b} c \tanh ^{-1}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right )-\frac {\left (b x+c x^2\right )^{3/2}}{x^{5/2}} \]

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Rubi [A]  time = 0.03, antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {662, 664, 660, 207} \begin {gather*} -\frac {\left (b x+c x^2\right )^{3/2}}{x^{5/2}}+\frac {3 c \sqrt {b x+c x^2}}{\sqrt {x}}-3 \sqrt {b} c \tanh ^{-1}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(b*x + c*x^2)^(3/2)/x^(7/2),x]

[Out]

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

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 664

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 + 2*p + 1)), x] - Dist[(p*(2*c*d - b*e))/(e^2*(m + 2*p + 1)), Int[(d + e*x)^(m + 1)*
(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] && (LeQ[-2, m, 0] || EqQ[m + p + 1, 0]) && NeQ[m + 2*p + 1, 0] && IntegerQ[2*p]

Rubi steps

\begin {align*} \int \frac {\left (b x+c x^2\right )^{3/2}}{x^{7/2}} \, dx &=-\frac {\left (b x+c x^2\right )^{3/2}}{x^{5/2}}+\frac {1}{2} (3 c) \int \frac {\sqrt {b x+c x^2}}{x^{3/2}} \, dx\\ &=\frac {3 c \sqrt {b x+c x^2}}{\sqrt {x}}-\frac {\left (b x+c x^2\right )^{3/2}}{x^{5/2}}+\frac {1}{2} (3 b c) \int \frac {1}{\sqrt {x} \sqrt {b x+c x^2}} \, dx\\ &=\frac {3 c \sqrt {b x+c x^2}}{\sqrt {x}}-\frac {\left (b x+c x^2\right )^{3/2}}{x^{5/2}}+(3 b c) \operatorname {Subst}\left (\int \frac {1}{-b+x^2} \, dx,x,\frac {\sqrt {b x+c x^2}}{\sqrt {x}}\right )\\ &=\frac {3 c \sqrt {b x+c x^2}}{\sqrt {x}}-\frac {\left (b x+c x^2\right )^{3/2}}{x^{5/2}}-3 \sqrt {b} c \tanh ^{-1}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right )\\ \end {align*}

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Mathematica [C]  time = 0.02, size = 40, normalized size = 0.53 \begin {gather*} \frac {2 c (x (b+c x))^{5/2} \, _2F_1\left (2,\frac {5}{2};\frac {7}{2};\frac {c x}{b}+1\right )}{5 b^2 x^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(b*x + c*x^2)^(3/2)/x^(7/2),x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.52, size = 61, normalized size = 0.81 \begin {gather*} \frac {(2 c x-b) \sqrt {b x+c x^2}}{x^{3/2}}-3 \sqrt {b} c \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {b x+c x^2}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(b*x + c*x^2)^(3/2)/x^(7/2),x]

[Out]

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

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fricas [A]  time = 0.42, size = 135, normalized size = 1.80 \begin {gather*} \left [\frac {3 \, \sqrt {b} c x^{2} \log \left (-\frac {c x^{2} + 2 \, b x - 2 \, \sqrt {c x^{2} + b x} \sqrt {b} \sqrt {x}}{x^{2}}\right ) + 2 \, \sqrt {c x^{2} + b x} {\left (2 \, c x - b\right )} \sqrt {x}}{2 \, x^{2}}, \frac {3 \, \sqrt {-b} c x^{2} \arctan \left (\frac {\sqrt {-b} \sqrt {x}}{\sqrt {c x^{2} + b x}}\right ) + \sqrt {c x^{2} + b x} {\left (2 \, c x - b\right )} \sqrt {x}}{x^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x)^(3/2)/x^(7/2),x, algorithm="fricas")

[Out]

[1/2*(3*sqrt(b)*c*x^2*log(-(c*x^2 + 2*b*x - 2*sqrt(c*x^2 + b*x)*sqrt(b)*sqrt(x))/x^2) + 2*sqrt(c*x^2 + b*x)*(2
*c*x - b)*sqrt(x))/x^2, (3*sqrt(-b)*c*x^2*arctan(sqrt(-b)*sqrt(x)/sqrt(c*x^2 + b*x)) + sqrt(c*x^2 + b*x)*(2*c*
x - b)*sqrt(x))/x^2]

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giac [A]  time = 0.23, size = 56, normalized size = 0.75 \begin {gather*} \frac {\frac {3 \, b c^{2} \arctan \left (\frac {\sqrt {c x + b}}{\sqrt {-b}}\right )}{\sqrt {-b}} + 2 \, \sqrt {c x + b} c^{2} - \frac {\sqrt {c x + b} b c}{x}}{c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x)^(3/2)/x^(7/2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.06, size = 68, normalized size = 0.91 \begin {gather*} \frac {\left (-3 b c x \arctanh \left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right )+2 \sqrt {c x +b}\, \sqrt {b}\, c x -\sqrt {c x +b}\, b^{\frac {3}{2}}\right ) \sqrt {\left (c x +b \right ) x}}{\sqrt {c x +b}\, \sqrt {b}\, x^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^2+b*x)^(3/2)/x^(7/2),x)

[Out]

(2*(c*x+b)^(1/2)*b^(1/2)*c*x-3*arctanh((c*x+b)^(1/2)/b^(1/2))*x*b*c-(c*x+b)^(1/2)*b^(3/2))*((c*x+b)*x)^(1/2)/x
^(3/2)/(c*x+b)^(1/2)/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 {7}{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^(7/2),x, algorithm="maxima")

[Out]

integrate((c*x^2 + b*x)^(3/2)/x^(7/2), 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^{7/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x + c*x^2)^(3/2)/x^(7/2),x)

[Out]

int((b*x + c*x^2)^(3/2)/x^(7/2), 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 {7}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**2+b*x)**(3/2)/x**(7/2),x)

[Out]

Integral((x*(b + c*x))**(3/2)/x**(7/2), x)

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