3.22.19 \(\int \frac {A+B x}{(a+b x)^{5/2} (d+e x)^{3/2}} \, dx\)

Optimal. Leaf size=139 \[ -\frac {2 (B d-A e)}{e (a+b x)^{3/2} \sqrt {d+e x} (b d-a e)}-\frac {4 \sqrt {d+e x} (a B e-4 A b e+3 b B d)}{3 \sqrt {a+b x} (b d-a e)^3}+\frac {2 \sqrt {d+e x} (a B e-4 A b e+3 b B d)}{3 e (a+b x)^{3/2} (b d-a e)^2} \]

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Rubi [A]  time = 0.09, antiderivative size = 139, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {78, 45, 37} \begin {gather*} -\frac {2 (B d-A e)}{e (a+b x)^{3/2} \sqrt {d+e x} (b d-a e)}-\frac {4 \sqrt {d+e x} (a B e-4 A b e+3 b B d)}{3 \sqrt {a+b x} (b d-a e)^3}+\frac {2 \sqrt {d+e x} (a B e-4 A b e+3 b B d)}{3 e (a+b x)^{3/2} (b d-a e)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(A + B*x)/((a + b*x)^(5/2)*(d + e*x)^(3/2)),x]

[Out]

(-2*(B*d - A*e))/(e*(b*d - a*e)*(a + b*x)^(3/2)*Sqrt[d + e*x]) + (2*(3*b*B*d - 4*A*b*e + a*B*e)*Sqrt[d + e*x])
/(3*e*(b*d - a*e)^2*(a + b*x)^(3/2)) - (4*(3*b*B*d - 4*A*b*e + a*B*e)*Sqrt[d + e*x])/(3*(b*d - a*e)^3*Sqrt[a +
b*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 45

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 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]))))

Rubi steps

\begin {align*} \int \frac {A+B x}{(a+b x)^{5/2} (d+e x)^{3/2}} \, dx &=-\frac {2 (B d-A e)}{e (b d-a e) (a+b x)^{3/2} \sqrt {d+e x}}-\frac {(3 b B d-4 A b e+a B e) \int \frac {1}{(a+b x)^{5/2} \sqrt {d+e x}} \, dx}{e (b d-a e)}\\ &=-\frac {2 (B d-A e)}{e (b d-a e) (a+b x)^{3/2} \sqrt {d+e x}}+\frac {2 (3 b B d-4 A b e+a B e) \sqrt {d+e x}}{3 e (b d-a e)^2 (a+b x)^{3/2}}+\frac {(2 (3 b B d-4 A b e+a B e)) \int \frac {1}{(a+b x)^{3/2} \sqrt {d+e x}} \, dx}{3 (b d-a e)^2}\\ &=-\frac {2 (B d-A e)}{e (b d-a e) (a+b x)^{3/2} \sqrt {d+e x}}+\frac {2 (3 b B d-4 A b e+a B e) \sqrt {d+e x}}{3 e (b d-a e)^2 (a+b x)^{3/2}}-\frac {4 (3 b B d-4 A b e+a B e) \sqrt {d+e x}}{3 (b d-a e)^3 \sqrt {a+b x}}\\ \end {align*}

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Mathematica [A]  time = 0.07, size = 125, normalized size = 0.90 \begin {gather*} -\frac {2 \left (A \left (-3 a^2 e^2-6 a b e (d+2 e x)+b^2 \left (d^2-4 d e x-8 e^2 x^2\right )\right )+B \left (3 a^2 e (2 d+e x)+2 a b \left (d^2+5 d e x+e^2 x^2\right )+3 b^2 d x (d+2 e x)\right )\right )}{3 (a+b x)^{3/2} \sqrt {d+e x} (b d-a e)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x)/((a + b*x)^(5/2)*(d + e*x)^(3/2)),x]

[Out]

(-2*(A*(-3*a^2*e^2 - 6*a*b*e*(d + 2*e*x) + b^2*(d^2 - 4*d*e*x - 8*e^2*x^2)) + B*(3*a^2*e*(2*d + e*x) + 3*b^2*d
*x*(d + 2*e*x) + 2*a*b*(d^2 + 5*d*e*x + e^2*x^2))))/(3*(b*d - a*e)^3*(a + b*x)^(3/2)*Sqrt[d + e*x])

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IntegrateAlgebraic [A]  time = 0.16, size = 133, normalized size = 0.96 \begin {gather*} \frac {2 (d+e x)^{3/2} \left (\frac {3 A e^2 (a+b x)^2}{(d+e x)^2}+\frac {6 A b e (a+b x)}{d+e x}-\frac {3 b B d (a+b x)}{d+e x}-\frac {3 a B e (a+b x)}{d+e x}-\frac {3 B d e (a+b x)^2}{(d+e x)^2}+a b B-A b^2\right )}{3 (a+b x)^{3/2} (b d-a e)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(A + B*x)/((a + b*x)^(5/2)*(d + e*x)^(3/2)),x]

[Out]

(2*(d + e*x)^(3/2)*(-(A*b^2) + a*b*B - (3*B*d*e*(a + b*x)^2)/(d + e*x)^2 + (3*A*e^2*(a + b*x)^2)/(d + e*x)^2 -
(3*b*B*d*(a + b*x))/(d + e*x) + (6*A*b*e*(a + b*x))/(d + e*x) - (3*a*B*e*(a + b*x))/(d + e*x)))/(3*(b*d - a*e
)^3*(a + b*x)^(3/2))

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fricas [B]  time = 6.14, size = 337, normalized size = 2.42 \begin {gather*} \frac {2 \, {\left (3 \, A a^{2} e^{2} - {\left (2 \, B a b + A b^{2}\right )} d^{2} - 6 \, {\left (B a^{2} - A a b\right )} d e - 2 \, {\left (3 \, B b^{2} d e + {\left (B a b - 4 \, A b^{2}\right )} e^{2}\right )} x^{2} - {\left (3 \, B b^{2} d^{2} + 2 \, {\left (5 \, B a b - 2 \, A b^{2}\right )} d e + 3 \, {\left (B a^{2} - 4 \, A a b\right )} e^{2}\right )} x\right )} \sqrt {b x + a} \sqrt {e x + d}}{3 \, {\left (a^{2} b^{3} d^{4} - 3 \, a^{3} b^{2} d^{3} e + 3 \, a^{4} b d^{2} e^{2} - a^{5} d e^{3} + {\left (b^{5} d^{3} e - 3 \, a b^{4} d^{2} e^{2} + 3 \, a^{2} b^{3} d e^{3} - a^{3} b^{2} e^{4}\right )} x^{3} + {\left (b^{5} d^{4} - a b^{4} d^{3} e - 3 \, a^{2} b^{3} d^{2} e^{2} + 5 \, a^{3} b^{2} d e^{3} - 2 \, a^{4} b e^{4}\right )} x^{2} + {\left (2 \, a b^{4} d^{4} - 5 \, a^{2} b^{3} d^{3} e + 3 \, a^{3} b^{2} d^{2} e^{2} + a^{4} b d e^{3} - a^{5} e^{4}\right )} x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/(b*x+a)^(5/2)/(e*x+d)^(3/2),x, algorithm="fricas")

[Out]

2/3*(3*A*a^2*e^2 - (2*B*a*b + A*b^2)*d^2 - 6*(B*a^2 - A*a*b)*d*e - 2*(3*B*b^2*d*e + (B*a*b - 4*A*b^2)*e^2)*x^2
- (3*B*b^2*d^2 + 2*(5*B*a*b - 2*A*b^2)*d*e + 3*(B*a^2 - 4*A*a*b)*e^2)*x)*sqrt(b*x + a)*sqrt(e*x + d)/(a^2*b^3
*d^4 - 3*a^3*b^2*d^3*e + 3*a^4*b*d^2*e^2 - a^5*d*e^3 + (b^5*d^3*e - 3*a*b^4*d^2*e^2 + 3*a^2*b^3*d*e^3 - a^3*b^
2*e^4)*x^3 + (b^5*d^4 - a*b^4*d^3*e - 3*a^2*b^3*d^2*e^2 + 5*a^3*b^2*d*e^3 - 2*a^4*b*e^4)*x^2 + (2*a*b^4*d^4 -
5*a^2*b^3*d^3*e + 3*a^3*b^2*d^2*e^2 + a^4*b*d*e^3 - a^5*e^4)*x)

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giac [B]  time = 2.51, size = 565, normalized size = 4.06 \begin {gather*} -\frac {2 \, {\left (B b^{2} d e - A b^{2} e^{2}\right )} \sqrt {b x + a}}{{\left (b^{3} d^{3} {\left | b \right |} - 3 \, a b^{2} d^{2} {\left | b \right |} e + 3 \, a^{2} b d {\left | b \right |} e^{2} - a^{3} {\left | b \right |} e^{3}\right )} \sqrt {b^{2} d + {\left (b x + a\right )} b e - a b e}} - \frac {4 \, {\left (3 \, B b^{\frac {13}{2}} d^{3} e^{\frac {1}{2}} - 4 \, B a b^{\frac {11}{2}} d^{2} e^{\frac {3}{2}} - 5 \, A b^{\frac {13}{2}} d^{2} e^{\frac {3}{2}} - 6 \, {\left (\sqrt {b x + a} \sqrt {b} e^{\frac {1}{2}} - \sqrt {b^{2} d + {\left (b x + a\right )} b e - a b e}\right )}^{2} B b^{\frac {9}{2}} d^{2} e^{\frac {1}{2}} - B a^{2} b^{\frac {9}{2}} d e^{\frac {5}{2}} + 10 \, A a b^{\frac {11}{2}} d e^{\frac {5}{2}} + 12 \, {\left (\sqrt {b x + a} \sqrt {b} e^{\frac {1}{2}} - \sqrt {b^{2} d + {\left (b x + a\right )} b e - a b e}\right )}^{2} A b^{\frac {9}{2}} d e^{\frac {3}{2}} + 3 \, {\left (\sqrt {b x + a} \sqrt {b} e^{\frac {1}{2}} - \sqrt {b^{2} d + {\left (b x + a\right )} b e - a b e}\right )}^{4} B b^{\frac {5}{2}} d e^{\frac {1}{2}} + 2 \, B a^{3} b^{\frac {7}{2}} e^{\frac {7}{2}} - 5 \, A a^{2} b^{\frac {9}{2}} e^{\frac {7}{2}} + 6 \, {\left (\sqrt {b x + a} \sqrt {b} e^{\frac {1}{2}} - \sqrt {b^{2} d + {\left (b x + a\right )} b e - a b e}\right )}^{2} B a^{2} b^{\frac {5}{2}} e^{\frac {5}{2}} - 12 \, {\left (\sqrt {b x + a} \sqrt {b} e^{\frac {1}{2}} - \sqrt {b^{2} d + {\left (b x + a\right )} b e - a b e}\right )}^{2} A a b^{\frac {7}{2}} e^{\frac {5}{2}} - 3 \, {\left (\sqrt {b x + a} \sqrt {b} e^{\frac {1}{2}} - \sqrt {b^{2} d + {\left (b x + a\right )} b e - a b e}\right )}^{4} A b^{\frac {5}{2}} e^{\frac {3}{2}}\right )}}{3 \, {\left (b^{2} d^{2} {\left | b \right |} - 2 \, a b d {\left | b \right |} e + a^{2} {\left | b \right |} e^{2}\right )} {\left (b^{2} d - a b e - {\left (\sqrt {b x + a} \sqrt {b} e^{\frac {1}{2}} - \sqrt {b^{2} d + {\left (b x + a\right )} b e - a b e}\right )}^{2}\right )}^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/(b*x+a)^(5/2)/(e*x+d)^(3/2),x, algorithm="giac")

[Out]

-2*(B*b^2*d*e - A*b^2*e^2)*sqrt(b*x + a)/((b^3*d^3*abs(b) - 3*a*b^2*d^2*abs(b)*e + 3*a^2*b*d*abs(b)*e^2 - a^3*
abs(b)*e^3)*sqrt(b^2*d + (b*x + a)*b*e - a*b*e)) - 4/3*(3*B*b^(13/2)*d^3*e^(1/2) - 4*B*a*b^(11/2)*d^2*e^(3/2)
- 5*A*b^(13/2)*d^2*e^(3/2) - 6*(sqrt(b*x + a)*sqrt(b)*e^(1/2) - sqrt(b^2*d + (b*x + a)*b*e - a*b*e))^2*B*b^(9/
2)*d^2*e^(1/2) - B*a^2*b^(9/2)*d*e^(5/2) + 10*A*a*b^(11/2)*d*e^(5/2) + 12*(sqrt(b*x + a)*sqrt(b)*e^(1/2) - sqr
t(b^2*d + (b*x + a)*b*e - a*b*e))^2*A*b^(9/2)*d*e^(3/2) + 3*(sqrt(b*x + a)*sqrt(b)*e^(1/2) - sqrt(b^2*d + (b*x
+ a)*b*e - a*b*e))^4*B*b^(5/2)*d*e^(1/2) + 2*B*a^3*b^(7/2)*e^(7/2) - 5*A*a^2*b^(9/2)*e^(7/2) + 6*(sqrt(b*x +
a)*sqrt(b)*e^(1/2) - sqrt(b^2*d + (b*x + a)*b*e - a*b*e))^2*B*a^2*b^(5/2)*e^(5/2) - 12*(sqrt(b*x + a)*sqrt(b)*
e^(1/2) - sqrt(b^2*d + (b*x + a)*b*e - a*b*e))^2*A*a*b^(7/2)*e^(5/2) - 3*(sqrt(b*x + a)*sqrt(b)*e^(1/2) - sqrt
(b^2*d + (b*x + a)*b*e - a*b*e))^4*A*b^(5/2)*e^(3/2))/((b^2*d^2*abs(b) - 2*a*b*d*abs(b)*e + a^2*abs(b)*e^2)*(b
^2*d - a*b*e - (sqrt(b*x + a)*sqrt(b)*e^(1/2) - sqrt(b^2*d + (b*x + a)*b*e - a*b*e))^2)^3)

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maple [A]  time = 0.01, size = 177, normalized size = 1.27 \begin {gather*} -\frac {2 \left (8 A \,b^{2} e^{2} x^{2}-2 B a b \,e^{2} x^{2}-6 B \,b^{2} d e \,x^{2}+12 A a b \,e^{2} x +4 A \,b^{2} d e x -3 B \,a^{2} e^{2} x -10 B a b d e x -3 B \,b^{2} d^{2} x +3 A \,a^{2} e^{2}+6 A a b d e -A \,b^{2} d^{2}-6 B \,a^{2} d e -2 B a b \,d^{2}\right )}{3 \left (b x +a \right )^{\frac {3}{2}} \sqrt {e x +d}\, \left (a^{3} e^{3}-3 a^{2} b d \,e^{2}+3 a \,b^{2} d^{2} e -b^{3} d^{3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)/(b*x+a)^(5/2)/(e*x+d)^(3/2),x)

[Out]

-2/3*(8*A*b^2*e^2*x^2-2*B*a*b*e^2*x^2-6*B*b^2*d*e*x^2+12*A*a*b*e^2*x+4*A*b^2*d*e*x-3*B*a^2*e^2*x-10*B*a*b*d*e*
x-3*B*b^2*d^2*x+3*A*a^2*e^2+6*A*a*b*d*e-A*b^2*d^2-6*B*a^2*d*e-2*B*a*b*d^2)/(b*x+a)^(3/2)/(e*x+d)^(1/2)/(a^3*e^
3-3*a^2*b*d*e^2+3*a*b^2*d^2*e-b^3*d^3)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/(b*x+a)^(5/2)/(e*x+d)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(a*e-b*d>0)', see `assume?` for
more details)Is a*e-b*d zero or nonzero?

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mupad [B]  time = 2.26, size = 193, normalized size = 1.39 \begin {gather*} \frac {\sqrt {d+e\,x}\,\left (\frac {4\,x^2\,\left (B\,a\,e-4\,A\,b\,e+3\,B\,b\,d\right )}{3\,{\left (a\,e-b\,d\right )}^3}+\frac {12\,B\,a^2\,d\,e-6\,A\,a^2\,e^2+4\,B\,a\,b\,d^2-12\,A\,a\,b\,d\,e+2\,A\,b^2\,d^2}{3\,b\,e\,{\left (a\,e-b\,d\right )}^3}+\frac {2\,x\,\left (3\,a\,e+b\,d\right )\,\left (B\,a\,e-4\,A\,b\,e+3\,B\,b\,d\right )}{3\,b\,e\,{\left (a\,e-b\,d\right )}^3}\right )}{x^2\,\sqrt {a+b\,x}+\frac {a\,d\,\sqrt {a+b\,x}}{b\,e}+\frac {x\,\left (a\,e+b\,d\right )\,\sqrt {a+b\,x}}{b\,e}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*x)/((a + b*x)^(5/2)*(d + e*x)^(3/2)),x)

[Out]

((d + e*x)^(1/2)*((4*x^2*(B*a*e - 4*A*b*e + 3*B*b*d))/(3*(a*e - b*d)^3) + (2*A*b^2*d^2 - 6*A*a^2*e^2 + 4*B*a*b
*d^2 + 12*B*a^2*d*e - 12*A*a*b*d*e)/(3*b*e*(a*e - b*d)^3) + (2*x*(3*a*e + b*d)*(B*a*e - 4*A*b*e + 3*B*b*d))/(3
*b*e*(a*e - b*d)^3)))/(x^2*(a + b*x)^(1/2) + (a*d*(a + b*x)^(1/2))/(b*e) + (x*(a*e + b*d)*(a + b*x)^(1/2))/(b*
e))

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sympy [F(-1)]  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)/(b*x+a)**(5/2)/(e*x+d)**(3/2),x)

[Out]

Timed out

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