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

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

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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 (A b-a B)}{b \sqrt {a+b x} (d+e x)^{3/2} (b d-a e)}+\frac {4 \sqrt {a+b x} (3 a B e-4 A b e+b B d)}{3 \sqrt {d+e x} (b d-a e)^3}+\frac {2 \sqrt {a+b x} (3 a B e-4 A b e+b B d)}{3 b (d+e x)^{3/2} (b d-a e)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.07, size = 133, normalized size = 0.96 \begin {gather*} \frac {2 \left (a^2 A e^2+a^2 B e (2 d+3 e x)-2 a A b e (3 d+2 e x)+2 a b B \left (3 d^2+5 d e x+3 e^2 x^2\right )-A b^2 \left (3 d^2+12 d e x+8 e^2 x^2\right )+b^2 B d x (3 d+2 e x)\right )}{3 \sqrt {a+b x} (d+e x)^{3/2} (b d-a e)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(a^2*A*e^2 - 2*a*A*b*e*(3*d + 2*e*x) + b^2*B*d*x*(3*d + 2*e*x) + a^2*B*e*(2*d + 3*e*x) + 2*a*b*B*(3*d^2 + 5
*d*e*x + 3*e^2*x^2) - A*b^2*(3*d^2 + 12*d*e*x + 8*e^2*x^2)))/(3*(b*d - a*e)^3*Sqrt[a + b*x]*(d + e*x)^(3/2))

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IntegrateAlgebraic [A]  time = 0.17, size = 133, normalized size = 0.96 \begin {gather*} \frac {2 (a+b x)^{3/2} \left (-\frac {3 A b^2 (d+e x)^2}{(a+b x)^2}-\frac {6 A b e (d+e x)}{a+b x}+\frac {3 a B e (d+e x)}{a+b x}+\frac {3 a b B (d+e x)^2}{(a+b x)^2}+\frac {3 b B d (d+e x)}{a+b x}+A e^2-B d e\right )}{3 (d+e x)^{3/2} (b d-a e)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 8.39, size = 336, normalized size = 2.42 \begin {gather*} \frac {2 \, {\left (A a^{2} e^{2} + 3 \, {\left (2 \, B a b - A b^{2}\right )} d^{2} + 2 \, {\left (B a^{2} - 3 \, A a b\right )} d e + 2 \, {\left (B b^{2} d e + {\left (3 \, 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 - 6 \, A b^{2}\right )} d e + {\left (3 \, B a^{2} - 4 \, A a b\right )} e^{2}\right )} x\right )} \sqrt {b x + a} \sqrt {e x + d}}{3 \, {\left (a b^{3} d^{5} - 3 \, a^{2} b^{2} d^{4} e + 3 \, a^{3} b d^{3} e^{2} - a^{4} d^{2} e^{3} + {\left (b^{4} d^{3} e^{2} - 3 \, a b^{3} d^{2} e^{3} + 3 \, a^{2} b^{2} d e^{4} - a^{3} b e^{5}\right )} x^{3} + {\left (2 \, b^{4} d^{4} e - 5 \, a b^{3} d^{3} e^{2} + 3 \, a^{2} b^{2} d^{2} e^{3} + a^{3} b d e^{4} - a^{4} e^{5}\right )} x^{2} + {\left (b^{4} d^{5} - a b^{3} d^{4} e - 3 \, a^{2} b^{2} d^{3} e^{2} + 5 \, a^{3} b d^{2} e^{3} - 2 \, a^{4} d 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)^(3/2)/(e*x+d)^(5/2),x, algorithm="fricas")

[Out]

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

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giac [B]  time = 1.98, size = 584, normalized size = 4.20 \begin {gather*} \frac {2 \, \sqrt {b x + a} {\left (\frac {{\left (2 \, B b^{6} d^{3} {\left | b \right |} e^{2} - B a b^{5} d^{2} {\left | b \right |} e^{3} - 5 \, A b^{6} d^{2} {\left | b \right |} e^{3} - 4 \, B a^{2} b^{4} d {\left | b \right |} e^{4} + 10 \, A a b^{5} d {\left | b \right |} e^{4} + 3 \, B a^{3} b^{3} {\left | b \right |} e^{5} - 5 \, A a^{2} b^{4} {\left | b \right |} e^{5}\right )} {\left (b x + a\right )}}{b^{7} d^{5} e - 5 \, a b^{6} d^{4} e^{2} + 10 \, a^{2} b^{5} d^{3} e^{3} - 10 \, a^{3} b^{4} d^{2} e^{4} + 5 \, a^{4} b^{3} d e^{5} - a^{5} b^{2} e^{6}} + \frac {3 \, {\left (B b^{7} d^{4} {\left | b \right |} e - 2 \, B a b^{6} d^{3} {\left | b \right |} e^{2} - 2 \, A b^{7} d^{3} {\left | b \right |} e^{2} + 6 \, A a b^{6} d^{2} {\left | b \right |} e^{3} + 2 \, B a^{3} b^{4} d {\left | b \right |} e^{4} - 6 \, A a^{2} b^{5} d {\left | b \right |} e^{4} - B a^{4} b^{3} {\left | b \right |} e^{5} + 2 \, A a^{3} b^{4} {\left | b \right |} e^{5}\right )}}{b^{7} d^{5} e - 5 \, a b^{6} d^{4} e^{2} + 10 \, a^{2} b^{5} d^{3} e^{3} - 10 \, a^{3} b^{4} d^{2} e^{4} + 5 \, a^{4} b^{3} d e^{5} - a^{5} b^{2} e^{6}}\right )}}{3 \, {\left (b^{2} d + {\left (b x + a\right )} b e - a b e\right )}^{\frac {3}{2}}} + \frac {4 \, {\left (B^{2} a^{2} b^{5} e - 2 \, A B a b^{6} e + A^{2} b^{7} e\right )}}{{\left (B a b^{\frac {9}{2}} d e^{\frac {1}{2}} - A b^{\frac {11}{2}} d e^{\frac {1}{2}} - B a^{2} b^{\frac {7}{2}} e^{\frac {3}{2}} + A a b^{\frac {9}{2}} e^{\frac {3}{2}} - {\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 b^{\frac {5}{2}} e^{\frac {1}{2}} + {\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 {7}{2}} e^{\frac {1}{2}}\right )} {\left (b^{2} d^{2} {\left | b \right |} - 2 \, a b d {\left | b \right |} e + a^{2} {\left | b \right |} e^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 176, normalized size = 1.27 \begin {gather*} -\frac {2 \left (-8 A \,b^{2} e^{2} x^{2}+6 B a b \,e^{2} x^{2}+2 B \,b^{2} d e \,x^{2}-4 A a b \,e^{2} x -12 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 +A \,a^{2} e^{2}-6 A a b d e -3 A \,b^{2} d^{2}+2 B \,a^{2} d e +6 B a b \,d^{2}\right )}{3 \sqrt {b x +a}\, \left (e x +d \right )^{\frac {3}{2}} \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)^(3/2)/(e*x+d)^(5/2),x)

[Out]

-2/3*(-8*A*b^2*e^2*x^2+6*B*a*b*e^2*x^2+2*B*b^2*d*e*x^2-4*A*a*b*e^2*x-12*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+A*a^2*e^2-6*A*a*b*d*e-3*A*b^2*d^2+2*B*a^2*d*e+6*B*a*b*d^2)/(b*x+a)^(1/2)/(e*x+d)^(3/2)/(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)^(3/2)/(e*x+d)^(5/2),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 = 2.20, size = 182, normalized size = 1.31 \begin {gather*} -\frac {\sqrt {d+e\,x}\,\left (\frac {4\,B\,a^2\,d\,e+2\,A\,a^2\,e^2+12\,B\,a\,b\,d^2-12\,A\,a\,b\,d\,e-6\,A\,b^2\,d^2}{3\,e^2\,{\left (a\,e-b\,d\right )}^3}+\frac {2\,x\,\left (a\,e+3\,b\,d\right )\,\left (3\,B\,a\,e-4\,A\,b\,e+B\,b\,d\right )}{3\,e^2\,{\left (a\,e-b\,d\right )}^3}+\frac {4\,b\,x^2\,\left (3\,B\,a\,e-4\,A\,b\,e+B\,b\,d\right )}{3\,e\,{\left (a\,e-b\,d\right )}^3}\right )}{x^2\,\sqrt {a+b\,x}+\frac {d^2\,\sqrt {a+b\,x}}{e^2}+\frac {2\,d\,x\,\sqrt {a+b\,x}}{e}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((d + e*x)^(1/2)*((2*A*a^2*e^2 - 6*A*b^2*d^2 + 12*B*a*b*d^2 + 4*B*a^2*d*e - 12*A*a*b*d*e)/(3*e^2*(a*e - b*d)^
3) + (2*x*(a*e + 3*b*d)*(3*B*a*e - 4*A*b*e + B*b*d))/(3*e^2*(a*e - b*d)^3) + (4*b*x^2*(3*B*a*e - 4*A*b*e + B*b
*d))/(3*e*(a*e - b*d)^3)))/(x^2*(a + b*x)^(1/2) + (d^2*(a + b*x)^(1/2))/e^2 + (2*d*x*(a + b*x)^(1/2))/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)**(3/2)/(e*x+d)**(5/2),x)

[Out]

Timed out

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