3.3.19 \(\int \frac {(A+B x) (b x+c x^2)^{5/2}}{x^{7/2}} \, dx\)

Optimal. Leaf size=131 \[ -2 A b^{5/2} \tanh ^{-1}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right )+\frac {2 A b^2 \sqrt {b x+c x^2}}{\sqrt {x}}+\frac {2 A \left (b x+c x^2\right )^{5/2}}{5 x^{5/2}}+\frac {2 A b \left (b x+c x^2\right )^{3/2}}{3 x^{3/2}}+\frac {2 B \left (b x+c x^2\right )^{7/2}}{7 c x^{7/2}} \]

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Rubi [A]  time = 0.12, antiderivative size = 131, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {794, 664, 660, 207} \begin {gather*} \frac {2 A b^2 \sqrt {b x+c x^2}}{\sqrt {x}}-2 A b^{5/2} \tanh ^{-1}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right )+\frac {2 A \left (b x+c x^2\right )^{5/2}}{5 x^{5/2}}+\frac {2 A b \left (b x+c x^2\right )^{3/2}}{3 x^{3/2}}+\frac {2 B \left (b x+c x^2\right )^{7/2}}{7 c x^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(b*x + c*x^2)^(5/2))/x^(7/2),x]

[Out]

(2*A*b^2*Sqrt[b*x + c*x^2])/Sqrt[x] + (2*A*b*(b*x + c*x^2)^(3/2))/(3*x^(3/2)) + (2*A*(b*x + c*x^2)^(5/2))/(5*x
^(5/2)) + (2*B*(b*x + c*x^2)^(7/2))/(7*c*x^(7/2)) - 2*A*b^(5/2)*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 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]

Rule 794

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp
[(g*(d + e*x)^m*(a + b*x + c*x^2)^(p + 1))/(c*(m + 2*p + 2)), x] + Dist[(m*(g*(c*d - b*e) + c*e*f) + e*(p + 1)
*(2*c*f - b*g))/(c*e*(m + 2*p + 2)), Int[(d + e*x)^m*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g
, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ[m + 2*p + 2, 0] && (NeQ[m, 2] || Eq
Q[d, 0])

Rubi steps

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

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Mathematica [A]  time = 0.21, size = 95, normalized size = 0.73 \begin {gather*} \frac {(x (b+c x))^{5/2} \left (-\frac {14 A b^{5/2} \tanh ^{-1}\left (\frac {\sqrt {b+c x}}{\sqrt {b}}\right )}{(b+c x)^{5/2}}+A \left (\frac {14 b^2}{(b+c x)^2}+\frac {14 b}{3 (b+c x)}+\frac {14}{5}\right )+\frac {2 B (b+c x)}{c}\right )}{7 x^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(b*x + c*x^2)^(5/2))/x^(7/2),x]

[Out]

((x*(b + c*x))^(5/2)*((2*B*(b + c*x))/c + A*(14/5 + (14*b^2)/(b + c*x)^2 + (14*b)/(3*(b + c*x))) - (14*A*b^(5/
2)*ArcTanh[Sqrt[b + c*x]/Sqrt[b]])/(b + c*x)^(5/2)))/(7*x^(5/2))

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IntegrateAlgebraic [A]  time = 0.75, size = 117, normalized size = 0.89 \begin {gather*} \frac {2 \sqrt {b x+c x^2} \left (161 A b^2 c+77 A b c^2 x+21 A c^3 x^2+15 b^3 B+45 b^2 B c x+45 b B c^2 x^2+15 B c^3 x^3\right )}{105 c \sqrt {x}}-2 A b^{5/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {b x+c x^2}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((A + B*x)*(b*x + c*x^2)^(5/2))/x^(7/2),x]

[Out]

(2*Sqrt[b*x + c*x^2]*(15*b^3*B + 161*A*b^2*c + 45*b^2*B*c*x + 77*A*b*c^2*x + 45*b*B*c^2*x^2 + 21*A*c^3*x^2 + 1
5*B*c^3*x^3))/(105*c*Sqrt[x]) - 2*A*b^(5/2)*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[b*x + c*x^2]]

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fricas [A]  time = 0.42, size = 245, normalized size = 1.87 \begin {gather*} \left [\frac {105 \, A b^{\frac {5}{2}} c x \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 (15 \, B c^{3} x^{3} + 15 \, B b^{3} + 161 \, A b^{2} c + 3 \, {\left (15 \, B b c^{2} + 7 \, A c^{3}\right )} x^{2} + {\left (45 \, B b^{2} c + 77 \, A b c^{2}\right )} x\right )} \sqrt {c x^{2} + b x} \sqrt {x}}{105 \, c x}, \frac {2 \, {\left (105 \, A \sqrt {-b} b^{2} c x \arctan \left (\frac {\sqrt {-b} \sqrt {x}}{\sqrt {c x^{2} + b x}}\right ) + {\left (15 \, B c^{3} x^{3} + 15 \, B b^{3} + 161 \, A b^{2} c + 3 \, {\left (15 \, B b c^{2} + 7 \, A c^{3}\right )} x^{2} + {\left (45 \, B b^{2} c + 77 \, A b c^{2}\right )} x\right )} \sqrt {c x^{2} + b x} \sqrt {x}\right )}}{105 \, c x}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x)^(5/2)/x^(7/2),x, algorithm="fricas")

[Out]

[1/105*(105*A*b^(5/2)*c*x*log(-(c*x^2 + 2*b*x - 2*sqrt(c*x^2 + b*x)*sqrt(b)*sqrt(x))/x^2) + 2*(15*B*c^3*x^3 +
15*B*b^3 + 161*A*b^2*c + 3*(15*B*b*c^2 + 7*A*c^3)*x^2 + (45*B*b^2*c + 77*A*b*c^2)*x)*sqrt(c*x^2 + b*x)*sqrt(x)
)/(c*x), 2/105*(105*A*sqrt(-b)*b^2*c*x*arctan(sqrt(-b)*sqrt(x)/sqrt(c*x^2 + b*x)) + (15*B*c^3*x^3 + 15*B*b^3 +
161*A*b^2*c + 3*(15*B*b*c^2 + 7*A*c^3)*x^2 + (45*B*b^2*c + 77*A*b*c^2)*x)*sqrt(c*x^2 + b*x)*sqrt(x))/(c*x)]

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giac [A]  time = 0.22, size = 139, normalized size = 1.06 \begin {gather*} \frac {2 \, A b^{3} \arctan \left (\frac {\sqrt {c x + b}}{\sqrt {-b}}\right )}{\sqrt {-b}} - \frac {2 \, {\left (105 \, A b^{3} c \arctan \left (\frac {\sqrt {b}}{\sqrt {-b}}\right ) + 15 \, B \sqrt {-b} b^{\frac {7}{2}} + 161 \, A \sqrt {-b} b^{\frac {5}{2}} c\right )}}{105 \, \sqrt {-b} c} + \frac {2 \, {\left (15 \, {\left (c x + b\right )}^{\frac {7}{2}} B c^{6} + 21 \, {\left (c x + b\right )}^{\frac {5}{2}} A c^{7} + 35 \, {\left (c x + b\right )}^{\frac {3}{2}} A b c^{7} + 105 \, \sqrt {c x + b} A b^{2} c^{7}\right )}}{105 \, c^{7}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x)^(5/2)/x^(7/2),x, algorithm="giac")

[Out]

2*A*b^3*arctan(sqrt(c*x + b)/sqrt(-b))/sqrt(-b) - 2/105*(105*A*b^3*c*arctan(sqrt(b)/sqrt(-b)) + 15*B*sqrt(-b)*
b^(7/2) + 161*A*sqrt(-b)*b^(5/2)*c)/(sqrt(-b)*c) + 2/105*(15*(c*x + b)^(7/2)*B*c^6 + 21*(c*x + b)^(5/2)*A*c^7
+ 35*(c*x + b)^(3/2)*A*b*c^7 + 105*sqrt(c*x + b)*A*b^2*c^7)/c^7

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maple [A]  time = 0.06, size = 151, normalized size = 1.15 \begin {gather*} -\frac {2 \sqrt {\left (c x +b \right ) x}\, \left (-15 \sqrt {c x +b}\, B \,c^{3} x^{3}-21 \sqrt {c x +b}\, A \,c^{3} x^{2}-45 \sqrt {c x +b}\, B b \,c^{2} x^{2}+105 A \,b^{\frac {5}{2}} c \arctanh \left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right )-77 \sqrt {c x +b}\, A b \,c^{2} x -45 \sqrt {c x +b}\, B \,b^{2} c x -161 \sqrt {c x +b}\, A \,b^{2} c -15 \sqrt {c x +b}\, B \,b^{3}\right )}{105 \sqrt {c x +b}\, c \sqrt {x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(c*x^2+b*x)^(5/2)/x^(7/2),x)

[Out]

-2/105*((c*x+b)*x)^(1/2)*(-15*B*x^3*c^3*(c*x+b)^(1/2)-21*A*x^2*c^3*(c*x+b)^(1/2)-45*B*x^2*b*c^2*(c*x+b)^(1/2)+
105*A*b^(5/2)*c*arctanh((c*x+b)^(1/2)/b^(1/2))-77*A*x*b*c^2*(c*x+b)^(1/2)-45*B*x*b^2*c*(c*x+b)^(1/2)-161*A*b^2
*c*(c*x+b)^(1/2)-15*B*b^3*(c*x+b)^(1/2))/x^(1/2)/(c*x+b)^(1/2)/c

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} A b^{2} \int \frac {\sqrt {c x + b}}{x}\,{d x} + \frac {2 \, {\left (35 \, {\left (B b^{2} c + 2 \, A b c^{2}\right )} x^{3} + {\left (15 \, B c^{3} x^{3} + 3 \, B b c^{2} x^{2} - 4 \, B b^{2} c x + 8 \, B b^{3}\right )} x^{2} + 35 \, {\left (B b^{3} + 2 \, A b^{2} c\right )} x^{2} + 7 \, {\left (3 \, {\left (2 \, B b c^{2} + A c^{3}\right )} x^{3} + {\left (2 \, B b^{2} c + A b c^{2}\right )} x^{2} - 2 \, {\left (2 \, B b^{3} + A b^{2} c\right )} x\right )} x\right )} \sqrt {c x + b}}{105 \, c x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x)^(5/2)/x^(7/2),x, algorithm="maxima")

[Out]

A*b^2*integrate(sqrt(c*x + b)/x, x) + 2/105*(35*(B*b^2*c + 2*A*b*c^2)*x^3 + (15*B*c^3*x^3 + 3*B*b*c^2*x^2 - 4*
B*b^2*c*x + 8*B*b^3)*x^2 + 35*(B*b^3 + 2*A*b^2*c)*x^2 + 7*(3*(2*B*b*c^2 + A*c^3)*x^3 + (2*B*b^2*c + A*b*c^2)*x
^2 - 2*(2*B*b^3 + A*b^2*c)*x)*x)*sqrt(c*x + b)/(c*x^2)

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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 )}^{5/2}\,\left (A+B\,x\right )}{x^{7/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((b*x + c*x^2)^(5/2)*(A + B*x))/x^(7/2),x)

[Out]

int(((b*x + c*x^2)^(5/2)*(A + B*x))/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 {5}{2}} \left (A + B x\right )}{x^{\frac {7}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x**2+b*x)**(5/2)/x**(7/2),x)

[Out]

Integral((x*(b + c*x))**(5/2)*(A + B*x)/x**(7/2), x)

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