3.6.15 \(\int \frac {x^{5/2} (A+B x)}{\sqrt {a+b x}} \, dx\) [515]

Optimal. Leaf size=159 \[ \frac {5 a^2 (8 A b-7 a B) \sqrt {x} \sqrt {a+b x}}{64 b^4}-\frac {5 a (8 A b-7 a B) x^{3/2} \sqrt {a+b x}}{96 b^3}+\frac {(8 A b-7 a B) x^{5/2} \sqrt {a+b x}}{24 b^2}+\frac {B x^{7/2} \sqrt {a+b x}}{4 b}-\frac {5 a^3 (8 A b-7 a B) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{64 b^{9/2}} \]

[Out]

-5/64*a^3*(8*A*b-7*B*a)*arctanh(b^(1/2)*x^(1/2)/(b*x+a)^(1/2))/b^(9/2)-5/96*a*(8*A*b-7*B*a)*x^(3/2)*(b*x+a)^(1
/2)/b^3+1/24*(8*A*b-7*B*a)*x^(5/2)*(b*x+a)^(1/2)/b^2+1/4*B*x^(7/2)*(b*x+a)^(1/2)/b+5/64*a^2*(8*A*b-7*B*a)*x^(1
/2)*(b*x+a)^(1/2)/b^4

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Rubi [A]
time = 0.05, antiderivative size = 159, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {81, 52, 65, 223, 212} \begin {gather*} -\frac {5 a^3 (8 A b-7 a B) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{64 b^{9/2}}+\frac {5 a^2 \sqrt {x} \sqrt {a+b x} (8 A b-7 a B)}{64 b^4}-\frac {5 a x^{3/2} \sqrt {a+b x} (8 A b-7 a B)}{96 b^3}+\frac {x^{5/2} \sqrt {a+b x} (8 A b-7 a B)}{24 b^2}+\frac {B x^{7/2} \sqrt {a+b x}}{4 b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(5*a^2*(8*A*b - 7*a*B)*Sqrt[x]*Sqrt[a + b*x])/(64*b^4) - (5*a*(8*A*b - 7*a*B)*x^(3/2)*Sqrt[a + b*x])/(96*b^3)
+ ((8*A*b - 7*a*B)*x^(5/2)*Sqrt[a + b*x])/(24*b^2) + (B*x^(7/2)*Sqrt[a + b*x])/(4*b) - (5*a^3*(8*A*b - 7*a*B)*
ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[a + b*x]])/(64*b^(9/2))

Rule 52

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + n + 1))), x] + Dist[n*((b*c - a*d)/(b*(m + n + 1))), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 81

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[b*(c + d*x)^
(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), x] + Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(
d*f*(n + p + 2)), Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2,
0]

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 223

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rubi steps

\begin {align*} \int \frac {x^{5/2} (A+B x)}{\sqrt {a+b x}} \, dx &=\frac {B x^{7/2} \sqrt {a+b x}}{4 b}+\frac {\left (4 A b-\frac {7 a B}{2}\right ) \int \frac {x^{5/2}}{\sqrt {a+b x}} \, dx}{4 b}\\ &=\frac {(8 A b-7 a B) x^{5/2} \sqrt {a+b x}}{24 b^2}+\frac {B x^{7/2} \sqrt {a+b x}}{4 b}-\frac {(5 a (8 A b-7 a B)) \int \frac {x^{3/2}}{\sqrt {a+b x}} \, dx}{48 b^2}\\ &=-\frac {5 a (8 A b-7 a B) x^{3/2} \sqrt {a+b x}}{96 b^3}+\frac {(8 A b-7 a B) x^{5/2} \sqrt {a+b x}}{24 b^2}+\frac {B x^{7/2} \sqrt {a+b x}}{4 b}+\frac {\left (5 a^2 (8 A b-7 a B)\right ) \int \frac {\sqrt {x}}{\sqrt {a+b x}} \, dx}{64 b^3}\\ &=\frac {5 a^2 (8 A b-7 a B) \sqrt {x} \sqrt {a+b x}}{64 b^4}-\frac {5 a (8 A b-7 a B) x^{3/2} \sqrt {a+b x}}{96 b^3}+\frac {(8 A b-7 a B) x^{5/2} \sqrt {a+b x}}{24 b^2}+\frac {B x^{7/2} \sqrt {a+b x}}{4 b}-\frac {\left (5 a^3 (8 A b-7 a B)\right ) \int \frac {1}{\sqrt {x} \sqrt {a+b x}} \, dx}{128 b^4}\\ &=\frac {5 a^2 (8 A b-7 a B) \sqrt {x} \sqrt {a+b x}}{64 b^4}-\frac {5 a (8 A b-7 a B) x^{3/2} \sqrt {a+b x}}{96 b^3}+\frac {(8 A b-7 a B) x^{5/2} \sqrt {a+b x}}{24 b^2}+\frac {B x^{7/2} \sqrt {a+b x}}{4 b}-\frac {\left (5 a^3 (8 A b-7 a B)\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a+b x^2}} \, dx,x,\sqrt {x}\right )}{64 b^4}\\ &=\frac {5 a^2 (8 A b-7 a B) \sqrt {x} \sqrt {a+b x}}{64 b^4}-\frac {5 a (8 A b-7 a B) x^{3/2} \sqrt {a+b x}}{96 b^3}+\frac {(8 A b-7 a B) x^{5/2} \sqrt {a+b x}}{24 b^2}+\frac {B x^{7/2} \sqrt {a+b x}}{4 b}-\frac {\left (5 a^3 (8 A b-7 a B)\right ) \text {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {\sqrt {x}}{\sqrt {a+b x}}\right )}{64 b^4}\\ &=\frac {5 a^2 (8 A b-7 a B) \sqrt {x} \sqrt {a+b x}}{64 b^4}-\frac {5 a (8 A b-7 a B) x^{3/2} \sqrt {a+b x}}{96 b^3}+\frac {(8 A b-7 a B) x^{5/2} \sqrt {a+b x}}{24 b^2}+\frac {B x^{7/2} \sqrt {a+b x}}{4 b}-\frac {5 a^3 (8 A b-7 a B) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{64 b^{9/2}}\\ \end {align*}

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Mathematica [A]
time = 0.22, size = 119, normalized size = 0.75 \begin {gather*} \frac {\sqrt {b} \sqrt {x} \sqrt {a+b x} \left (-105 a^3 B+16 b^3 x^2 (4 A+3 B x)-8 a b^2 x (10 A+7 B x)+10 a^2 b (12 A+7 B x)\right )-15 a^3 (-8 A b+7 a B) \log \left (-\sqrt {b} \sqrt {x}+\sqrt {a+b x}\right )}{192 b^{9/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[b]*Sqrt[x]*Sqrt[a + b*x]*(-105*a^3*B + 16*b^3*x^2*(4*A + 3*B*x) - 8*a*b^2*x*(10*A + 7*B*x) + 10*a^2*b*(1
2*A + 7*B*x)) - 15*a^3*(-8*A*b + 7*a*B)*Log[-(Sqrt[b]*Sqrt[x]) + Sqrt[a + b*x]])/(192*b^(9/2))

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Maple [A]
time = 0.08, size = 218, normalized size = 1.37

method	result	size
risch	\(\frac {\left (48 b^{3} B \,x^{3}+64 A \,b^{3} x^{2}-56 B a \,b^{2} x^{2}-80 a \,b^{2} A x +70 a^{2} b B x +120 a^{2} b A -105 a^{3} B \right ) \sqrt {b x +a}\, \sqrt {x}}{192 b^{4}}+\frac {\left (-\frac {5 a^{3} \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right ) A}{16 b^{\frac {7}{2}}}+\frac {35 a^{4} \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right ) B}{128 b^{\frac {9}{2}}}\right ) \sqrt {\left (b x +a \right ) x}}{\sqrt {b x +a}\, \sqrt {x}}\)	\(162\)
default	\(-\frac {\sqrt {x}\, \sqrt {b x +a}\, \left (-96 B \,b^{\frac {7}{2}} x^{3} \sqrt {\left (b x +a \right ) x}-128 A \,b^{\frac {7}{2}} x^{2} \sqrt {\left (b x +a \right ) x}+112 B a \,b^{\frac {5}{2}} x^{2} \sqrt {\left (b x +a \right ) x}+160 A \,b^{\frac {5}{2}} \sqrt {\left (b x +a \right ) x}\, a x -140 B \,b^{\frac {3}{2}} \sqrt {\left (b x +a \right ) x}\, a^{2} x +120 A \ln \left (\frac {2 \sqrt {\left (b x +a \right ) x}\, \sqrt {b}+2 b x +a}{2 \sqrt {b}}\right ) a^{3} b -240 A \,b^{\frac {3}{2}} \sqrt {\left (b x +a \right ) x}\, a^{2}-105 B \ln \left (\frac {2 \sqrt {\left (b x +a \right ) x}\, \sqrt {b}+2 b x +a}{2 \sqrt {b}}\right ) a^{4}+210 B \sqrt {b}\, \sqrt {\left (b x +a \right ) x}\, a^{3}\right )}{384 b^{\frac {9}{2}} \sqrt {\left (b x +a \right ) x}}\)	\(218\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(5/2)*(B*x+A)/(b*x+a)^(1/2),x,method=_RETURNVERBOSE)

[Out]

-1/384*x^(1/2)*(b*x+a)^(1/2)/b^(9/2)*(-96*B*b^(7/2)*x^3*((b*x+a)*x)^(1/2)-128*A*b^(7/2)*x^2*((b*x+a)*x)^(1/2)+
112*B*a*b^(5/2)*x^2*((b*x+a)*x)^(1/2)+160*A*b^(5/2)*((b*x+a)*x)^(1/2)*a*x-140*B*b^(3/2)*((b*x+a)*x)^(1/2)*a^2*
x+120*A*ln(1/2*(2*((b*x+a)*x)^(1/2)*b^(1/2)+2*b*x+a)/b^(1/2))*a^3*b-240*A*b^(3/2)*((b*x+a)*x)^(1/2)*a^2-105*B*
ln(1/2*(2*((b*x+a)*x)^(1/2)*b^(1/2)+2*b*x+a)/b^(1/2))*a^4+210*B*b^(1/2)*((b*x+a)*x)^(1/2)*a^3)/((b*x+a)*x)^(1/
2)

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Maxima [A]
time = 0.27, size = 206, normalized size = 1.30 \begin {gather*} \frac {\sqrt {b x^{2} + a x} B x^{3}}{4 \, b} - \frac {7 \, \sqrt {b x^{2} + a x} B a x^{2}}{24 \, b^{2}} + \frac {\sqrt {b x^{2} + a x} A x^{2}}{3 \, b} + \frac {35 \, \sqrt {b x^{2} + a x} B a^{2} x}{96 \, b^{3}} - \frac {5 \, \sqrt {b x^{2} + a x} A a x}{12 \, b^{2}} + \frac {35 \, B a^{4} \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right )}{128 \, b^{\frac {9}{2}}} - \frac {5 \, A a^{3} \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right )}{16 \, b^{\frac {7}{2}}} - \frac {35 \, \sqrt {b x^{2} + a x} B a^{3}}{64 \, b^{4}} + \frac {5 \, \sqrt {b x^{2} + a x} A a^{2}}{8 \, b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)*(B*x+A)/(b*x+a)^(1/2),x, algorithm="maxima")

[Out]

1/4*sqrt(b*x^2 + a*x)*B*x^3/b - 7/24*sqrt(b*x^2 + a*x)*B*a*x^2/b^2 + 1/3*sqrt(b*x^2 + a*x)*A*x^2/b + 35/96*sqr
t(b*x^2 + a*x)*B*a^2*x/b^3 - 5/12*sqrt(b*x^2 + a*x)*A*a*x/b^2 + 35/128*B*a^4*log(2*b*x + a + 2*sqrt(b*x^2 + a*
x)*sqrt(b))/b^(9/2) - 5/16*A*a^3*log(2*b*x + a + 2*sqrt(b*x^2 + a*x)*sqrt(b))/b^(7/2) - 35/64*sqrt(b*x^2 + a*x
)*B*a^3/b^4 + 5/8*sqrt(b*x^2 + a*x)*A*a^2/b^3

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Fricas [A]
time = 1.25, size = 249, normalized size = 1.57 \begin {gather*} \left [-\frac {15 \, {\left (7 \, B a^{4} - 8 \, A a^{3} b\right )} \sqrt {b} \log \left (2 \, b x - 2 \, \sqrt {b x + a} \sqrt {b} \sqrt {x} + a\right ) - 2 \, {\left (48 \, B b^{4} x^{3} - 105 \, B a^{3} b + 120 \, A a^{2} b^{2} - 8 \, {\left (7 \, B a b^{3} - 8 \, A b^{4}\right )} x^{2} + 10 \, {\left (7 \, B a^{2} b^{2} - 8 \, A a b^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {x}}{384 \, b^{5}}, -\frac {15 \, {\left (7 \, B a^{4} - 8 \, A a^{3} b\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-b}}{b \sqrt {x}}\right ) - {\left (48 \, B b^{4} x^{3} - 105 \, B a^{3} b + 120 \, A a^{2} b^{2} - 8 \, {\left (7 \, B a b^{3} - 8 \, A b^{4}\right )} x^{2} + 10 \, {\left (7 \, B a^{2} b^{2} - 8 \, A a b^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {x}}{192 \, b^{5}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)*(B*x+A)/(b*x+a)^(1/2),x, algorithm="fricas")

[Out]

[-1/384*(15*(7*B*a^4 - 8*A*a^3*b)*sqrt(b)*log(2*b*x - 2*sqrt(b*x + a)*sqrt(b)*sqrt(x) + a) - 2*(48*B*b^4*x^3 -
105*B*a^3*b + 120*A*a^2*b^2 - 8*(7*B*a*b^3 - 8*A*b^4)*x^2 + 10*(7*B*a^2*b^2 - 8*A*a*b^3)*x)*sqrt(b*x + a)*sqr
t(x))/b^5, -1/192*(15*(7*B*a^4 - 8*A*a^3*b)*sqrt(-b)*arctan(sqrt(b*x + a)*sqrt(-b)/(b*sqrt(x))) - (48*B*b^4*x^
3 - 105*B*a^3*b + 120*A*a^2*b^2 - 8*(7*B*a*b^3 - 8*A*b^4)*x^2 + 10*(7*B*a^2*b^2 - 8*A*a*b^3)*x)*sqrt(b*x + a)*
sqrt(x))/b^5]

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Sympy [A]
time = 68.45, size = 303, normalized size = 1.91 \begin {gather*} \frac {5 A a^{\frac {5}{2}} \sqrt {x}}{8 b^{3} \sqrt {1 + \frac {b x}{a}}} + \frac {5 A a^{\frac {3}{2}} x^{\frac {3}{2}}}{24 b^{2} \sqrt {1 + \frac {b x}{a}}} - \frac {A \sqrt {a} x^{\frac {5}{2}}}{12 b \sqrt {1 + \frac {b x}{a}}} - \frac {5 A a^{3} \operatorname {asinh}{\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}} \right )}}{8 b^{\frac {7}{2}}} + \frac {A x^{\frac {7}{2}}}{3 \sqrt {a} \sqrt {1 + \frac {b x}{a}}} - \frac {35 B a^{\frac {7}{2}} \sqrt {x}}{64 b^{4} \sqrt {1 + \frac {b x}{a}}} - \frac {35 B a^{\frac {5}{2}} x^{\frac {3}{2}}}{192 b^{3} \sqrt {1 + \frac {b x}{a}}} + \frac {7 B a^{\frac {3}{2}} x^{\frac {5}{2}}}{96 b^{2} \sqrt {1 + \frac {b x}{a}}} - \frac {B \sqrt {a} x^{\frac {7}{2}}}{24 b \sqrt {1 + \frac {b x}{a}}} + \frac {35 B a^{4} \operatorname {asinh}{\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}} \right )}}{64 b^{\frac {9}{2}}} + \frac {B x^{\frac {9}{2}}}{4 \sqrt {a} \sqrt {1 + \frac {b x}{a}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(5/2)*(B*x+A)/(b*x+a)**(1/2),x)

[Out]

5*A*a**(5/2)*sqrt(x)/(8*b**3*sqrt(1 + b*x/a)) + 5*A*a**(3/2)*x**(3/2)/(24*b**2*sqrt(1 + b*x/a)) - A*sqrt(a)*x*
*(5/2)/(12*b*sqrt(1 + b*x/a)) - 5*A*a**3*asinh(sqrt(b)*sqrt(x)/sqrt(a))/(8*b**(7/2)) + A*x**(7/2)/(3*sqrt(a)*s
qrt(1 + b*x/a)) - 35*B*a**(7/2)*sqrt(x)/(64*b**4*sqrt(1 + b*x/a)) - 35*B*a**(5/2)*x**(3/2)/(192*b**3*sqrt(1 +
b*x/a)) + 7*B*a**(3/2)*x**(5/2)/(96*b**2*sqrt(1 + b*x/a)) - B*sqrt(a)*x**(7/2)/(24*b*sqrt(1 + b*x/a)) + 35*B*a
**4*asinh(sqrt(b)*sqrt(x)/sqrt(a))/(64*b**(9/2)) + B*x**(9/2)/(4*sqrt(a)*sqrt(1 + b*x/a))

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: NotImplementedError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)*(B*x+A)/(b*x+a)^(1/2),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: Warning, choosing root of [1,0,%%%{-4,[1
,0,0]%%%}+%%%{-4,[0,1,1]%%%}+%%%{-4,[0,1,0]%%%}+%%%{-4,[0,0,1]%%%},0,%%%{6,[2,0,0]%%%}+%%%{12,[1,1,1]%%%}+%%%{
4,[1,1,0]%%%}+%%%{4,[

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^{5/2}\,\left (A+B\,x\right )}{\sqrt {a+b\,x}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^(5/2)*(A + B*x))/(a + b*x)^(1/2),x)

[Out]

int((x^(5/2)*(A + B*x))/(a + b*x)^(1/2), x)

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