3.21.67 \(\int \frac {\sqrt {a+b x} (A+B x)}{(d+e x)^{7/2}} \, dx\)

Optimal. Leaf size=95 \[ \frac {2 (a+b x)^{3/2} (-5 a B e+2 A b e+3 b B d)}{15 e (d+e x)^{3/2} (b d-a e)^2}-\frac {2 (a+b x)^{3/2} (B d-A e)}{5 e (d+e x)^{5/2} (b d-a e)} \]

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Rubi [A]  time = 0.05, antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {78, 37} \begin {gather*} \frac {2 (a+b x)^{3/2} (-5 a B e+2 A b e+3 b B d)}{15 e (d+e x)^{3/2} (b d-a e)^2}-\frac {2 (a+b x)^{3/2} (B d-A e)}{5 e (d+e x)^{5/2} (b d-a e)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(B*d - A*e)*(a + b*x)^(3/2))/(5*e*(b*d - a*e)*(d + e*x)^(5/2)) + (2*(3*b*B*d + 2*A*b*e - 5*a*B*e)*(a + b*x
)^(3/2))/(15*e*(b*d - a*e)^2*(d + e*x)^(3/2))

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 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 {\sqrt {a+b x} (A+B x)}{(d+e x)^{7/2}} \, dx &=-\frac {2 (B d-A e) (a+b x)^{3/2}}{5 e (b d-a e) (d+e x)^{5/2}}+\frac {(3 b B d+2 A b e-5 a B e) \int \frac {\sqrt {a+b x}}{(d+e x)^{5/2}} \, dx}{5 e (b d-a e)}\\ &=-\frac {2 (B d-A e) (a+b x)^{3/2}}{5 e (b d-a e) (d+e x)^{5/2}}+\frac {2 (3 b B d+2 A b e-5 a B e) (a+b x)^{3/2}}{15 e (b d-a e)^2 (d+e x)^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 66, normalized size = 0.69 \begin {gather*} \frac {2 (a+b x)^{3/2} (A (-3 a e+5 b d+2 b e x)+B (-2 a d-5 a e x+3 b d x))}{15 (d+e x)^{5/2} (b d-a e)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(a + b*x)^(3/2)*(B*(-2*a*d + 3*b*d*x - 5*a*e*x) + A*(5*b*d - 3*a*e + 2*b*e*x)))/(15*(b*d - a*e)^2*(d + e*x)
^(5/2))

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IntegrateAlgebraic [A]  time = 0.13, size = 73, normalized size = 0.77 \begin {gather*} \frac {2 (a+b x)^{3/2} \left (-\frac {3 A e (a+b x)}{d+e x}+\frac {3 B d (a+b x)}{d+e x}-5 a B+5 A b\right )}{15 (d+e x)^{3/2} (b d-a e)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(Sqrt[a + b*x]*(A + B*x))/(d + e*x)^(7/2),x]

[Out]

(2*(a + b*x)^(3/2)*(5*A*b - 5*a*B + (3*B*d*(a + b*x))/(d + e*x) - (3*A*e*(a + b*x))/(d + e*x)))/(15*(b*d - a*e
)^2*(d + e*x)^(3/2))

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fricas [B]  time = 12.80, size = 221, normalized size = 2.33 \begin {gather*} -\frac {2 \, {\left (3 \, A a^{2} e - {\left (3 \, B b^{2} d - {\left (5 \, B a b - 2 \, A b^{2}\right )} e\right )} x^{2} + {\left (2 \, B a^{2} - 5 \, A a b\right )} d - {\left ({\left (B a b + 5 \, A b^{2}\right )} d - {\left (5 \, B a^{2} + A a b\right )} e\right )} x\right )} \sqrt {b x + a} \sqrt {e x + d}}{15 \, {\left (b^{2} d^{5} - 2 \, a b d^{4} e + a^{2} d^{3} e^{2} + {\left (b^{2} d^{2} e^{3} - 2 \, a b d e^{4} + a^{2} e^{5}\right )} x^{3} + 3 \, {\left (b^{2} d^{3} e^{2} - 2 \, a b d^{2} e^{3} + a^{2} d e^{4}\right )} x^{2} + 3 \, {\left (b^{2} d^{4} e - 2 \, a b d^{3} e^{2} + a^{2} d^{2} e^{3}\right )} x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(b*x+a)^(1/2)/(e*x+d)^(7/2),x, algorithm="fricas")

[Out]

-2/15*(3*A*a^2*e - (3*B*b^2*d - (5*B*a*b - 2*A*b^2)*e)*x^2 + (2*B*a^2 - 5*A*a*b)*d - ((B*a*b + 5*A*b^2)*d - (5
*B*a^2 + A*a*b)*e)*x)*sqrt(b*x + a)*sqrt(e*x + d)/(b^2*d^5 - 2*a*b*d^4*e + a^2*d^3*e^2 + (b^2*d^2*e^3 - 2*a*b*
d*e^4 + a^2*e^5)*x^3 + 3*(b^2*d^3*e^2 - 2*a*b*d^2*e^3 + a^2*d*e^4)*x^2 + 3*(b^2*d^4*e - 2*a*b*d^3*e^2 + a^2*d^
2*e^3)*x)

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giac [B]  time = 2.12, size = 180, normalized size = 1.89 \begin {gather*} \frac {2 \, {\left (b x + a\right )}^{\frac {3}{2}} {\left (\frac {{\left (3 \, B b^{6} d {\left | b \right |} e^{2} - 5 \, B a b^{5} {\left | b \right |} e^{3} + 2 \, A b^{6} {\left | b \right |} e^{3}\right )} {\left (b x + a\right )}}{b^{4} d^{2} e^{2} - 2 \, a b^{3} d e^{3} + a^{2} b^{2} e^{4}} - \frac {5 \, {\left (B a b^{6} d {\left | b \right |} e^{2} - A b^{7} d {\left | b \right |} e^{2} - B a^{2} b^{5} {\left | b \right |} e^{3} + A a b^{6} {\left | b \right |} e^{3}\right )}}{b^{4} d^{2} e^{2} - 2 \, a b^{3} d e^{3} + a^{2} b^{2} e^{4}}\right )}}{15 \, {\left (b^{2} d + {\left (b x + a\right )} b e - a b e\right )}^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(b*x+a)^(1/2)/(e*x+d)^(7/2),x, algorithm="giac")

[Out]

2/15*(b*x + a)^(3/2)*((3*B*b^6*d*abs(b)*e^2 - 5*B*a*b^5*abs(b)*e^3 + 2*A*b^6*abs(b)*e^3)*(b*x + a)/(b^4*d^2*e^
2 - 2*a*b^3*d*e^3 + a^2*b^2*e^4) - 5*(B*a*b^6*d*abs(b)*e^2 - A*b^7*d*abs(b)*e^2 - B*a^2*b^5*abs(b)*e^3 + A*a*b
^6*abs(b)*e^3)/(b^4*d^2*e^2 - 2*a*b^3*d*e^3 + a^2*b^2*e^4))/(b^2*d + (b*x + a)*b*e - a*b*e)^(5/2)

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maple [A]  time = 0.01, size = 74, normalized size = 0.78 \begin {gather*} -\frac {2 \left (b x +a \right )^{\frac {3}{2}} \left (-2 A b e x +5 B a e x -3 B b d x +3 A a e -5 A b d +2 B a d \right )}{15 \left (e x +d \right )^{\frac {5}{2}} \left (a^{2} e^{2}-2 b e a d +b^{2} d^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(b*x+a)^(1/2)/(e*x+d)^(7/2),x)

[Out]

-2/15*(b*x+a)^(3/2)*(-2*A*b*e*x+5*B*a*e*x-3*B*b*d*x+3*A*a*e-5*A*b*d+2*B*a*d)/(e*x+d)^(5/2)/(a^2*e^2-2*a*b*d*e+
b^2*d^2)

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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)^(1/2)/(e*x+d)^(7/2),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 = 1.90, size = 179, normalized size = 1.88 \begin {gather*} \frac {\sqrt {d+e\,x}\,\left (\frac {x^2\,\sqrt {a+b\,x}\,\left (4\,A\,b^2\,e+6\,B\,b^2\,d-10\,B\,a\,b\,e\right )}{15\,e^3\,{\left (a\,e-b\,d\right )}^2}-\frac {\sqrt {a+b\,x}\,\left (6\,A\,a^2\,e+4\,B\,a^2\,d-10\,A\,a\,b\,d\right )}{15\,e^3\,{\left (a\,e-b\,d\right )}^2}+\frac {x\,\sqrt {a+b\,x}\,\left (10\,A\,b^2\,d-10\,B\,a^2\,e-2\,A\,a\,b\,e+2\,B\,a\,b\,d\right )}{15\,e^3\,{\left (a\,e-b\,d\right )}^2}\right )}{x^3+\frac {d^3}{e^3}+\frac {3\,d\,x^2}{e}+\frac {3\,d^2\,x}{e^2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*x)*(a + b*x)^(1/2))/(d + e*x)^(7/2),x)

[Out]

((d + e*x)^(1/2)*((x^2*(a + b*x)^(1/2)*(4*A*b^2*e + 6*B*b^2*d - 10*B*a*b*e))/(15*e^3*(a*e - b*d)^2) - ((a + b*
x)^(1/2)*(6*A*a^2*e + 4*B*a^2*d - 10*A*a*b*d))/(15*e^3*(a*e - b*d)^2) + (x*(a + b*x)^(1/2)*(10*A*b^2*d - 10*B*
a^2*e - 2*A*a*b*e + 2*B*a*b*d))/(15*e^3*(a*e - b*d)^2)))/(x^3 + d^3/e^3 + (3*d*x^2)/e + (3*d^2*x)/e^2)

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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)**(1/2)/(e*x+d)**(7/2),x)

[Out]

Timed out

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