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

Optimal. Leaf size=150 \[ b^{3/2} (5 a B+2 A b) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )+\frac {b^2 \sqrt {x} \sqrt {a+b x} (5 a B+2 A b)}{a}-\frac {2 (a+b x)^{5/2} (5 a B+2 A b)}{15 a x^{3/2}}-\frac {2 b (a+b x)^{3/2} (5 a B+2 A b)}{3 a \sqrt {x}}-\frac {2 A (a+b x)^{7/2}}{5 a x^{5/2}} \]

[Out]

-2/15*(2*A*b+5*B*a)*(b*x+a)^(5/2)/a/x^(3/2)-2/5*A*(b*x+a)^(7/2)/a/x^(5/2)+b^(3/2)*(2*A*b+5*B*a)*arctanh(b^(1/2
)*x^(1/2)/(b*x+a)^(1/2))-2/3*b*(2*A*b+5*B*a)*(b*x+a)^(3/2)/a/x^(1/2)+b^2*(2*A*b+5*B*a)*x^(1/2)*(b*x+a)^(1/2)/a

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Rubi [A]  time = 0.07, antiderivative size = 150, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {78, 47, 50, 63, 217, 206} \[ \frac {b^2 \sqrt {x} \sqrt {a+b x} (5 a B+2 A b)}{a}+b^{3/2} (5 a B+2 A b) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )-\frac {2 (a+b x)^{5/2} (5 a B+2 A b)}{15 a x^{3/2}}-\frac {2 b (a+b x)^{3/2} (5 a B+2 A b)}{3 a \sqrt {x}}-\frac {2 A (a+b x)^{7/2}}{5 a x^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b^2*(2*A*b + 5*a*B)*Sqrt[x]*Sqrt[a + b*x])/a - (2*b*(2*A*b + 5*a*B)*(a + b*x)^(3/2))/(3*a*Sqrt[x]) - (2*(2*A*
b + 5*a*B)*(a + b*x)^(5/2))/(15*a*x^(3/2)) - (2*A*(a + b*x)^(7/2))/(5*a*x^(5/2)) + b^(3/2)*(2*A*b + 5*a*B)*Arc
Tanh[(Sqrt[b]*Sqrt[x])/Sqrt[a + b*x]]

Rule 47

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

Rule 50

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 63

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 78

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

Rule 206

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

Rule 217

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 {(a+b x)^{5/2} (A+B x)}{x^{7/2}} \, dx &=-\frac {2 A (a+b x)^{7/2}}{5 a x^{5/2}}+\frac {\left (2 \left (A b+\frac {5 a B}{2}\right )\right ) \int \frac {(a+b x)^{5/2}}{x^{5/2}} \, dx}{5 a}\\ &=-\frac {2 (2 A b+5 a B) (a+b x)^{5/2}}{15 a x^{3/2}}-\frac {2 A (a+b x)^{7/2}}{5 a x^{5/2}}+\frac {(b (2 A b+5 a B)) \int \frac {(a+b x)^{3/2}}{x^{3/2}} \, dx}{3 a}\\ &=-\frac {2 b (2 A b+5 a B) (a+b x)^{3/2}}{3 a \sqrt {x}}-\frac {2 (2 A b+5 a B) (a+b x)^{5/2}}{15 a x^{3/2}}-\frac {2 A (a+b x)^{7/2}}{5 a x^{5/2}}+\frac {\left (b^2 (2 A b+5 a B)\right ) \int \frac {\sqrt {a+b x}}{\sqrt {x}} \, dx}{a}\\ &=\frac {b^2 (2 A b+5 a B) \sqrt {x} \sqrt {a+b x}}{a}-\frac {2 b (2 A b+5 a B) (a+b x)^{3/2}}{3 a \sqrt {x}}-\frac {2 (2 A b+5 a B) (a+b x)^{5/2}}{15 a x^{3/2}}-\frac {2 A (a+b x)^{7/2}}{5 a x^{5/2}}+\frac {1}{2} \left (b^2 (2 A b+5 a B)\right ) \int \frac {1}{\sqrt {x} \sqrt {a+b x}} \, dx\\ &=\frac {b^2 (2 A b+5 a B) \sqrt {x} \sqrt {a+b x}}{a}-\frac {2 b (2 A b+5 a B) (a+b x)^{3/2}}{3 a \sqrt {x}}-\frac {2 (2 A b+5 a B) (a+b x)^{5/2}}{15 a x^{3/2}}-\frac {2 A (a+b x)^{7/2}}{5 a x^{5/2}}+\left (b^2 (2 A b+5 a B)\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+b x^2}} \, dx,x,\sqrt {x}\right )\\ &=\frac {b^2 (2 A b+5 a B) \sqrt {x} \sqrt {a+b x}}{a}-\frac {2 b (2 A b+5 a B) (a+b x)^{3/2}}{3 a \sqrt {x}}-\frac {2 (2 A b+5 a B) (a+b x)^{5/2}}{15 a x^{3/2}}-\frac {2 A (a+b x)^{7/2}}{5 a x^{5/2}}+\left (b^2 (2 A b+5 a B)\right ) \operatorname {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {\sqrt {x}}{\sqrt {a+b x}}\right )\\ &=\frac {b^2 (2 A b+5 a B) \sqrt {x} \sqrt {a+b x}}{a}-\frac {2 b (2 A b+5 a B) (a+b x)^{3/2}}{3 a \sqrt {x}}-\frac {2 (2 A b+5 a B) (a+b x)^{5/2}}{15 a x^{3/2}}-\frac {2 A (a+b x)^{7/2}}{5 a x^{5/2}}+b^{3/2} (2 A b+5 a B) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )\\ \end {align*}

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Mathematica [C]  time = 0.05, size = 76, normalized size = 0.51 \[ \frac {2 \sqrt {a+b x} \left (-\frac {a^2 x (5 a B+2 A b) \, _2F_1\left (-\frac {5}{2},-\frac {3}{2};-\frac {1}{2};-\frac {b x}{a}\right )}{\sqrt {\frac {b x}{a}+1}}-3 A (a+b x)^3\right )}{15 a x^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[a + b*x]*(-3*A*(a + b*x)^3 - (a^2*(2*A*b + 5*a*B)*x*Hypergeometric2F1[-5/2, -3/2, -1/2, -((b*x)/a)])/S
qrt[1 + (b*x)/a]))/(15*a*x^(5/2))

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fricas [A]  time = 0.66, size = 217, normalized size = 1.45 \[ \left [\frac {15 \, {\left (5 \, B a b + 2 \, A b^{2}\right )} \sqrt {b} x^{3} \log \left (2 \, b x + 2 \, \sqrt {b x + a} \sqrt {b} \sqrt {x} + a\right ) + 2 \, {\left (15 \, B b^{2} x^{3} - 6 \, A a^{2} - 2 \, {\left (35 \, B a b + 23 \, A b^{2}\right )} x^{2} - 2 \, {\left (5 \, B a^{2} + 11 \, A a b\right )} x\right )} \sqrt {b x + a} \sqrt {x}}{30 \, x^{3}}, -\frac {15 \, {\left (5 \, B a b + 2 \, A b^{2}\right )} \sqrt {-b} x^{3} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-b}}{b \sqrt {x}}\right ) - {\left (15 \, B b^{2} x^{3} - 6 \, A a^{2} - 2 \, {\left (35 \, B a b + 23 \, A b^{2}\right )} x^{2} - 2 \, {\left (5 \, B a^{2} + 11 \, A a b\right )} x\right )} \sqrt {b x + a} \sqrt {x}}{15 \, x^{3}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/30*(15*(5*B*a*b + 2*A*b^2)*sqrt(b)*x^3*log(2*b*x + 2*sqrt(b*x + a)*sqrt(b)*sqrt(x) + a) + 2*(15*B*b^2*x^3 -
6*A*a^2 - 2*(35*B*a*b + 23*A*b^2)*x^2 - 2*(5*B*a^2 + 11*A*a*b)*x)*sqrt(b*x + a)*sqrt(x))/x^3, -1/15*(15*(5*B*
a*b + 2*A*b^2)*sqrt(-b)*x^3*arctan(sqrt(b*x + a)*sqrt(-b)/(b*sqrt(x))) - (15*B*b^2*x^3 - 6*A*a^2 - 2*(35*B*a*b
+ 23*A*b^2)*x^2 - 2*(5*B*a^2 + 11*A*a*b)*x)*sqrt(b*x + a)*sqrt(x))/x^3]

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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maple [A]  time = 0.01, size = 206, normalized size = 1.37 \[ \frac {\sqrt {b x +a}\, \left (30 A \,b^{3} x^{3} \ln \left (\frac {2 b x +a +2 \sqrt {\left (b x +a \right ) x}\, \sqrt {b}}{2 \sqrt {b}}\right )+75 B a \,b^{2} x^{3} \ln \left (\frac {2 b x +a +2 \sqrt {\left (b x +a \right ) x}\, \sqrt {b}}{2 \sqrt {b}}\right )+30 \sqrt {\left (b x +a \right ) x}\, B \,b^{\frac {5}{2}} x^{3}-92 \sqrt {\left (b x +a \right ) x}\, A \,b^{\frac {5}{2}} x^{2}-140 \sqrt {\left (b x +a \right ) x}\, B a \,b^{\frac {3}{2}} x^{2}-44 \sqrt {\left (b x +a \right ) x}\, A a \,b^{\frac {3}{2}} x -20 \sqrt {\left (b x +a \right ) x}\, B \,a^{2} \sqrt {b}\, x -12 \sqrt {\left (b x +a \right ) x}\, A \,a^{2} \sqrt {b}\right )}{30 \sqrt {\left (b x +a \right ) x}\, \sqrt {b}\, x^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/30*(b*x+a)^(1/2)/x^(5/2)*(30*A*ln(1/2*(2*b*x+a+2*((b*x+a)*x)^(1/2)*b^(1/2))/b^(1/2))*x^3*b^3+75*B*a*b^2*x^3*
ln(1/2*(2*b*x+a+2*((b*x+a)*x)^(1/2)*b^(1/2))/b^(1/2))+30*((b*x+a)*x)^(1/2)*B*b^(5/2)*x^3-92*((b*x+a)*x)^(1/2)*
A*b^(5/2)*x^2-140*((b*x+a)*x)^(1/2)*B*a*b^(3/2)*x^2-44*((b*x+a)*x)^(1/2)*A*a*b^(3/2)*x-20*((b*x+a)*x)^(1/2)*B*
a^2*b^(1/2)*x-12*((b*x+a)*x)^(1/2)*A*a^2*b^(1/2))/((b*x+a)*x)^(1/2)/b^(1/2)

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maxima [B]  time = 0.91, size = 244, normalized size = 1.63 \[ \frac {5}{2} \, B a b^{\frac {3}{2}} \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right ) + A b^{\frac {5}{2}} \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right ) - \frac {35 \, \sqrt {b x^{2} + a x} B a b}{6 \, x} - \frac {38 \, \sqrt {b x^{2} + a x} A b^{2}}{15 \, x} - \frac {5 \, \sqrt {b x^{2} + a x} B a^{2}}{6 \, x^{2}} - \frac {7 \, \sqrt {b x^{2} + a x} A a b}{30 \, x^{2}} - \frac {5 \, {\left (b x^{2} + a x\right )}^{\frac {3}{2}} B a}{6 \, x^{3}} + \frac {3 \, \sqrt {b x^{2} + a x} A a^{2}}{10 \, x^{3}} - \frac {{\left (b x^{2} + a x\right )}^{\frac {3}{2}} A b}{3 \, x^{3}} + \frac {{\left (b x^{2} + a x\right )}^{\frac {5}{2}} B}{x^{4}} - \frac {{\left (b x^{2} + a x\right )}^{\frac {3}{2}} A a}{2 \, x^{4}} - \frac {{\left (b x^{2} + a x\right )}^{\frac {5}{2}} A}{5 \, x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/2*B*a*b^(3/2)*log(2*b*x + a + 2*sqrt(b*x^2 + a*x)*sqrt(b)) + A*b^(5/2)*log(2*b*x + a + 2*sqrt(b*x^2 + a*x)*s
qrt(b)) - 35/6*sqrt(b*x^2 + a*x)*B*a*b/x - 38/15*sqrt(b*x^2 + a*x)*A*b^2/x - 5/6*sqrt(b*x^2 + a*x)*B*a^2/x^2 -
7/30*sqrt(b*x^2 + a*x)*A*a*b/x^2 - 5/6*(b*x^2 + a*x)^(3/2)*B*a/x^3 + 3/10*sqrt(b*x^2 + a*x)*A*a^2/x^3 - 1/3*(
b*x^2 + a*x)^(3/2)*A*b/x^3 + (b*x^2 + a*x)^(5/2)*B/x^4 - 1/2*(b*x^2 + a*x)^(3/2)*A*a/x^4 - 1/5*(b*x^2 + a*x)^(
5/2)*A/x^5

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\left (A+B\,x\right )\,{\left (a+b\,x\right )}^{5/2}}{x^{7/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 92.56, size = 201, normalized size = 1.34 \[ A \left (- \frac {2 a^{2} \sqrt {b} \sqrt {\frac {a}{b x} + 1}}{5 x^{2}} - \frac {22 a b^{\frac {3}{2}} \sqrt {\frac {a}{b x} + 1}}{15 x} - \frac {46 b^{\frac {5}{2}} \sqrt {\frac {a}{b x} + 1}}{15} - b^{\frac {5}{2}} \log {\left (\frac {a}{b x} \right )} + 2 b^{\frac {5}{2}} \log {\left (\sqrt {\frac {a}{b x} + 1} + 1 \right )}\right ) + B \left (- \frac {2 a^{2} \sqrt {b} \sqrt {\frac {a}{b x} + 1}}{3 x} - \frac {14 a b^{\frac {3}{2}} \sqrt {\frac {a}{b x} + 1}}{3} - \frac {5 a b^{\frac {3}{2}} \log {\left (\frac {a}{b x} \right )}}{2} + 5 a b^{\frac {3}{2}} \log {\left (\sqrt {\frac {a}{b x} + 1} + 1 \right )} + b^{\frac {5}{2}} x \sqrt {\frac {a}{b x} + 1}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

A*(-2*a**2*sqrt(b)*sqrt(a/(b*x) + 1)/(5*x**2) - 22*a*b**(3/2)*sqrt(a/(b*x) + 1)/(15*x) - 46*b**(5/2)*sqrt(a/(b
*x) + 1)/15 - b**(5/2)*log(a/(b*x)) + 2*b**(5/2)*log(sqrt(a/(b*x) + 1) + 1)) + B*(-2*a**2*sqrt(b)*sqrt(a/(b*x)
+ 1)/(3*x) - 14*a*b**(3/2)*sqrt(a/(b*x) + 1)/3 - 5*a*b**(3/2)*log(a/(b*x))/2 + 5*a*b**(3/2)*log(sqrt(a/(b*x)
+ 1) + 1) + b**(5/2)*x*sqrt(a/(b*x) + 1))

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