3.6.44 \(\int \frac {A+B x}{x^{7/2} (a+b x)^{5/2}} \, dx\) [544]

Optimal. Leaf size=144 \[ -\frac {2 A}{5 a x^{5/2} (a+b x)^{3/2}}-\frac {2 (8 A b-5 a B)}{15 a^2 x^{3/2} (a+b x)^{3/2}}-\frac {4 (8 A b-5 a B)}{5 a^3 x^{3/2} \sqrt {a+b x}}+\frac {16 (8 A b-5 a B) \sqrt {a+b x}}{15 a^4 x^{3/2}}-\frac {32 b (8 A b-5 a B) \sqrt {a+b x}}{15 a^5 \sqrt {x}} \]

[Out]

-2/5*A/a/x^(5/2)/(b*x+a)^(3/2)-2/15*(8*A*b-5*B*a)/a^2/x^(3/2)/(b*x+a)^(3/2)-4/5*(8*A*b-5*B*a)/a^3/x^(3/2)/(b*x
+a)^(1/2)+16/15*(8*A*b-5*B*a)*(b*x+a)^(1/2)/a^4/x^(3/2)-32/15*b*(8*A*b-5*B*a)*(b*x+a)^(1/2)/a^5/x^(1/2)

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Rubi [A]
time = 0.04, antiderivative size = 144, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {79, 47, 37} \begin {gather*} -\frac {32 b \sqrt {a+b x} (8 A b-5 a B)}{15 a^5 \sqrt {x}}+\frac {16 \sqrt {a+b x} (8 A b-5 a B)}{15 a^4 x^{3/2}}-\frac {4 (8 A b-5 a B)}{5 a^3 x^{3/2} \sqrt {a+b x}}-\frac {2 (8 A b-5 a B)}{15 a^2 x^{3/2} (a+b x)^{3/2}}-\frac {2 A}{5 a x^{5/2} (a+b x)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*A)/(5*a*x^(5/2)*(a + b*x)^(3/2)) - (2*(8*A*b - 5*a*B))/(15*a^2*x^(3/2)*(a + b*x)^(3/2)) - (4*(8*A*b - 5*a*
B))/(5*a^3*x^(3/2)*Sqrt[a + b*x]) + (16*(8*A*b - 5*a*B)*Sqrt[a + b*x])/(15*a^4*x^(3/2)) - (32*b*(8*A*b - 5*a*B
)*Sqrt[a + b*x])/(15*a^5*Sqrt[x])

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n +
1)/((b*c - a*d)*(m + 1))), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rule 47

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

Rule 79

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] || L
tQ[p, n]))))

Rubi steps

\begin {align*} \int \frac {A+B x}{x^{7/2} (a+b x)^{5/2}} \, dx &=-\frac {2 A}{5 a x^{5/2} (a+b x)^{3/2}}+\frac {\left (2 \left (-4 A b+\frac {5 a B}{2}\right )\right ) \int \frac {1}{x^{5/2} (a+b x)^{5/2}} \, dx}{5 a}\\ &=-\frac {2 A}{5 a x^{5/2} (a+b x)^{3/2}}-\frac {2 (8 A b-5 a B)}{15 a^2 x^{3/2} (a+b x)^{3/2}}-\frac {(2 (8 A b-5 a B)) \int \frac {1}{x^{5/2} (a+b x)^{3/2}} \, dx}{5 a^2}\\ &=-\frac {2 A}{5 a x^{5/2} (a+b x)^{3/2}}-\frac {2 (8 A b-5 a B)}{15 a^2 x^{3/2} (a+b x)^{3/2}}-\frac {4 (8 A b-5 a B)}{5 a^3 x^{3/2} \sqrt {a+b x}}-\frac {(8 (8 A b-5 a B)) \int \frac {1}{x^{5/2} \sqrt {a+b x}} \, dx}{5 a^3}\\ &=-\frac {2 A}{5 a x^{5/2} (a+b x)^{3/2}}-\frac {2 (8 A b-5 a B)}{15 a^2 x^{3/2} (a+b x)^{3/2}}-\frac {4 (8 A b-5 a B)}{5 a^3 x^{3/2} \sqrt {a+b x}}+\frac {16 (8 A b-5 a B) \sqrt {a+b x}}{15 a^4 x^{3/2}}+\frac {(16 b (8 A b-5 a B)) \int \frac {1}{x^{3/2} \sqrt {a+b x}} \, dx}{15 a^4}\\ &=-\frac {2 A}{5 a x^{5/2} (a+b x)^{3/2}}-\frac {2 (8 A b-5 a B)}{15 a^2 x^{3/2} (a+b x)^{3/2}}-\frac {4 (8 A b-5 a B)}{5 a^3 x^{3/2} \sqrt {a+b x}}+\frac {16 (8 A b-5 a B) \sqrt {a+b x}}{15 a^4 x^{3/2}}-\frac {32 b (8 A b-5 a B) \sqrt {a+b x}}{15 a^5 \sqrt {x}}\\ \end {align*}

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Mathematica [A]
time = 0.20, size = 94, normalized size = 0.65 \begin {gather*} -\frac {2 \left (128 A b^4 x^4+24 a^2 b^2 x^2 (2 A-5 B x)+16 a b^3 x^3 (12 A-5 B x)+a^4 (3 A+5 B x)-2 a^3 b x (4 A+15 B x)\right )}{15 a^5 x^{5/2} (a+b x)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(128*A*b^4*x^4 + 24*a^2*b^2*x^2*(2*A - 5*B*x) + 16*a*b^3*x^3*(12*A - 5*B*x) + a^4*(3*A + 5*B*x) - 2*a^3*b*
x*(4*A + 15*B*x)))/(15*a^5*x^(5/2)*(a + b*x)^(3/2))

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Maple [A]
time = 0.12, size = 101, normalized size = 0.70

method	result	size
risch	\(-\frac {2 \sqrt {b x +a}\, \left (73 A \,b^{2} x^{2}-40 B a b \,x^{2}-14 a A b x +5 a^{2} B x +3 a^{2} A \right )}{15 a^{5} x^{\frac {5}{2}}}-\frac {2 b^{2} \left (11 b^{2} A x -8 a b B x +12 a b A -9 a^{2} B \right ) \sqrt {x}}{3 \left (b x +a \right )^{\frac {3}{2}} a^{5}}\)	\(97\)
gosper	\(-\frac {2 \left (128 A \,b^{4} x^{4}-80 B a \,b^{3} x^{4}+192 A a \,b^{3} x^{3}-120 B \,a^{2} b^{2} x^{3}+48 A \,a^{2} b^{2} x^{2}-30 B \,a^{3} b \,x^{2}-8 A \,a^{3} b x +5 B \,a^{4} x +3 A \,a^{4}\right )}{15 \left (b x +a \right )^{\frac {3}{2}} x^{\frac {5}{2}} a^{5}}\)	\(101\)
default	\(-\frac {2 \left (128 A \,b^{4} x^{4}-80 B a \,b^{3} x^{4}+192 A a \,b^{3} x^{3}-120 B \,a^{2} b^{2} x^{3}+48 A \,a^{2} b^{2} x^{2}-30 B \,a^{3} b \,x^{2}-8 A \,a^{3} b x +5 B \,a^{4} x +3 A \,a^{4}\right )}{15 \left (b x +a \right )^{\frac {3}{2}} x^{\frac {5}{2}} a^{5}}\)	\(101\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/15*(128*A*b^4*x^4-80*B*a*b^3*x^4+192*A*a*b^3*x^3-120*B*a^2*b^2*x^3+48*A*a^2*b^2*x^2-30*B*a^3*b*x^2-8*A*a^3*
b*x+5*B*a^4*x+3*A*a^4)/(b*x+a)^(3/2)/x^(5/2)/a^5

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Maxima [A]
time = 0.27, size = 176, normalized size = 1.22 \begin {gather*} -\frac {4 \, B b x}{3 \, {\left (b x^{2} + a x\right )}^{\frac {3}{2}} a^{2}} + \frac {32 \, B b^{2} x}{3 \, \sqrt {b x^{2} + a x} a^{4}} + \frac {32 \, A b^{2} x}{15 \, {\left (b x^{2} + a x\right )}^{\frac {3}{2}} a^{3}} - \frac {256 \, A b^{3} x}{15 \, \sqrt {b x^{2} + a x} a^{5}} - \frac {2 \, B}{3 \, {\left (b x^{2} + a x\right )}^{\frac {3}{2}} a} + \frac {16 \, B b}{3 \, \sqrt {b x^{2} + a x} a^{3}} + \frac {16 \, A b}{15 \, {\left (b x^{2} + a x\right )}^{\frac {3}{2}} a^{2}} - \frac {128 \, A b^{2}}{15 \, \sqrt {b x^{2} + a x} a^{4}} - \frac {2 \, A}{5 \, {\left (b x^{2} + a x\right )}^{\frac {3}{2}} a x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4/3*B*b*x/((b*x^2 + a*x)^(3/2)*a^2) + 32/3*B*b^2*x/(sqrt(b*x^2 + a*x)*a^4) + 32/15*A*b^2*x/((b*x^2 + a*x)^(3/
2)*a^3) - 256/15*A*b^3*x/(sqrt(b*x^2 + a*x)*a^5) - 2/3*B/((b*x^2 + a*x)^(3/2)*a) + 16/3*B*b/(sqrt(b*x^2 + a*x)
*a^3) + 16/15*A*b/((b*x^2 + a*x)^(3/2)*a^2) - 128/15*A*b^2/(sqrt(b*x^2 + a*x)*a^4) - 2/5*A/((b*x^2 + a*x)^(3/2
)*a*x)

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Fricas [A]
time = 0.93, size = 127, normalized size = 0.88 \begin {gather*} -\frac {2 \, {\left (3 \, A a^{4} - 16 \, {\left (5 \, B a b^{3} - 8 \, A b^{4}\right )} x^{4} - 24 \, {\left (5 \, B a^{2} b^{2} - 8 \, A a b^{3}\right )} x^{3} - 6 \, {\left (5 \, B a^{3} b - 8 \, A a^{2} b^{2}\right )} x^{2} + {\left (5 \, B a^{4} - 8 \, A a^{3} b\right )} x\right )} \sqrt {b x + a} \sqrt {x}}{15 \, {\left (a^{5} b^{2} x^{5} + 2 \, a^{6} b x^{4} + a^{7} x^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/15*(3*A*a^4 - 16*(5*B*a*b^3 - 8*A*b^4)*x^4 - 24*(5*B*a^2*b^2 - 8*A*a*b^3)*x^3 - 6*(5*B*a^3*b - 8*A*a^2*b^2)
*x^2 + (5*B*a^4 - 8*A*a^3*b)*x)*sqrt(b*x + a)*sqrt(x)/(a^5*b^2*x^5 + 2*a^6*b*x^4 + a^7*x^3)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 341 vs. \(2 (114) = 228\).
time = 0.96, size = 341, normalized size = 2.37 \begin {gather*} \frac {2 \, \sqrt {b x + a} {\left ({\left (b x + a\right )} {\left (\frac {{\left (40 \, B a^{8} b^{7} - 73 \, A a^{7} b^{8}\right )} {\left (b x + a\right )}}{a^{12} b^{2} {\left | b \right |}} - \frac {5 \, {\left (17 \, B a^{9} b^{7} - 32 \, A a^{8} b^{8}\right )}}{a^{12} b^{2} {\left | b \right |}}\right )} + \frac {45 \, {\left (B a^{10} b^{7} - 2 \, A a^{9} b^{8}\right )}}{a^{12} b^{2} {\left | b \right |}}\right )}}{15 \, {\left ({\left (b x + a\right )} b - a b\right )}^{\frac {5}{2}}} + \frac {4 \, {\left (6 \, B a {\left (\sqrt {b x + a} \sqrt {b} - \sqrt {{\left (b x + a\right )} b - a b}\right )}^{4} b^{\frac {7}{2}} + 18 \, B a^{2} {\left (\sqrt {b x + a} \sqrt {b} - \sqrt {{\left (b x + a\right )} b - a b}\right )}^{2} b^{\frac {9}{2}} - 9 \, A {\left (\sqrt {b x + a} \sqrt {b} - \sqrt {{\left (b x + a\right )} b - a b}\right )}^{4} b^{\frac {9}{2}} + 8 \, B a^{3} b^{\frac {11}{2}} - 24 \, A a {\left (\sqrt {b x + a} \sqrt {b} - \sqrt {{\left (b x + a\right )} b - a b}\right )}^{2} b^{\frac {11}{2}} - 11 \, A a^{2} b^{\frac {13}{2}}\right )}}{3 \, {\left ({\left (\sqrt {b x + a} \sqrt {b} - \sqrt {{\left (b x + a\right )} b - a b}\right )}^{2} + a b\right )}^{3} a^{4} {\left | b \right |}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*sqrt(b*x + a)*((b*x + a)*((40*B*a^8*b^7 - 73*A*a^7*b^8)*(b*x + a)/(a^12*b^2*abs(b)) - 5*(17*B*a^9*b^7 - 3
2*A*a^8*b^8)/(a^12*b^2*abs(b))) + 45*(B*a^10*b^7 - 2*A*a^9*b^8)/(a^12*b^2*abs(b)))/((b*x + a)*b - a*b)^(5/2) +
4/3*(6*B*a*(sqrt(b*x + a)*sqrt(b) - sqrt((b*x + a)*b - a*b))^4*b^(7/2) + 18*B*a^2*(sqrt(b*x + a)*sqrt(b) - sq
rt((b*x + a)*b - a*b))^2*b^(9/2) - 9*A*(sqrt(b*x + a)*sqrt(b) - sqrt((b*x + a)*b - a*b))^4*b^(9/2) + 8*B*a^3*b
^(11/2) - 24*A*a*(sqrt(b*x + a)*sqrt(b) - sqrt((b*x + a)*b - a*b))^2*b^(11/2) - 11*A*a^2*b^(13/2))/(((sqrt(b*x
+ a)*sqrt(b) - sqrt((b*x + a)*b - a*b))^2 + a*b)^3*a^4*abs(b))

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Mupad [B]
time = 1.01, size = 129, normalized size = 0.90 \begin {gather*} -\frac {\sqrt {a+b\,x}\,\left (\frac {2\,A}{5\,a\,b^2}+\frac {16\,x^3\,\left (8\,A\,b-5\,B\,a\right )}{5\,a^4}+\frac {4\,x^2\,\left (8\,A\,b-5\,B\,a\right )}{5\,a^3\,b}+\frac {x^4\,\left (256\,A\,b^4-160\,B\,a\,b^3\right )}{15\,a^5\,b^2}+\frac {x\,\left (10\,B\,a^4-16\,A\,a^3\,b\right )}{15\,a^5\,b^2}\right )}{x^{9/2}+\frac {2\,a\,x^{7/2}}{b}+\frac {a^2\,x^{5/2}}{b^2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((a + b*x)^(1/2)*((2*A)/(5*a*b^2) + (16*x^3*(8*A*b - 5*B*a))/(5*a^4) + (4*x^2*(8*A*b - 5*B*a))/(5*a^3*b) + (x
^4*(256*A*b^4 - 160*B*a*b^3))/(15*a^5*b^2) + (x*(10*B*a^4 - 16*A*a^3*b))/(15*a^5*b^2)))/(x^(9/2) + (2*a*x^(7/2
))/b + (a^2*x^(5/2))/b^2)

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