3.21.79 \(\int \frac {(a+b x)^{3/2} (A+B x)}{(d+e x)^{9/2}} \, dx\)

Optimal. Leaf size=95 \[ \frac {2 (a+b x)^{5/2} (-7 a B e+2 A b e+5 b B d)}{35 e (d+e x)^{5/2} (b d-a e)^2}-\frac {2 (a+b x)^{5/2} (B d-A e)}{7 e (d+e x)^{7/2} (b d-a e)} \]

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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)^{5/2} (-7 a B e+2 A b e+5 b B d)}{35 e (d+e x)^{5/2} (b d-a e)^2}-\frac {2 (a+b x)^{5/2} (B d-A e)}{7 e (d+e x)^{7/2} (b d-a e)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((a + b*x)^(3/2)*(A + B*x))/(d + e*x)^(9/2),x]

[Out]

(-2*(B*d - A*e)*(a + b*x)^(5/2))/(7*e*(b*d - a*e)*(d + e*x)^(7/2)) + (2*(5*b*B*d + 2*A*b*e - 7*a*B*e)*(a + b*x
)^(5/2))/(35*e*(b*d - a*e)^2*(d + e*x)^(5/2))

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 {(a+b x)^{3/2} (A+B x)}{(d+e x)^{9/2}} \, dx &=-\frac {2 (B d-A e) (a+b x)^{5/2}}{7 e (b d-a e) (d+e x)^{7/2}}+\frac {(5 b B d+2 A b e-7 a B e) \int \frac {(a+b x)^{3/2}}{(d+e x)^{7/2}} \, dx}{7 e (b d-a e)}\\ &=-\frac {2 (B d-A e) (a+b x)^{5/2}}{7 e (b d-a e) (d+e x)^{7/2}}+\frac {2 (5 b B d+2 A b e-7 a B e) (a+b x)^{5/2}}{35 e (b d-a e)^2 (d+e x)^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 0.07, size = 66, normalized size = 0.69 \begin {gather*} \frac {2 (a+b x)^{5/2} (A (-5 a e+7 b d+2 b e x)+B (-2 a d-7 a e x+5 b d x))}{35 (d+e x)^{7/2} (b d-a e)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x)^(3/2)*(A + B*x))/(d + e*x)^(9/2),x]

[Out]

(2*(a + b*x)^(5/2)*(B*(-2*a*d + 5*b*d*x - 7*a*e*x) + A*(7*b*d - 5*a*e + 2*b*e*x)))/(35*(b*d - a*e)^2*(d + e*x)
^(7/2))

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IntegrateAlgebraic [A]  time = 0.17, size = 73, normalized size = 0.77 \begin {gather*} -\frac {2 (a+b x)^{7/2} \left (-\frac {7 A b (d+e x)}{a+b x}+\frac {7 a B (d+e x)}{a+b x}+5 A e-5 B d\right )}{35 (d+e x)^{7/2} (b d-a e)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((a + b*x)^(3/2)*(A + B*x))/(d + e*x)^(9/2),x]

[Out]

(-2*(a + b*x)^(7/2)*(-5*B*d + 5*A*e - (7*A*b*(d + e*x))/(a + b*x) + (7*a*B*(d + e*x))/(a + b*x)))/(35*(b*d - a
*e)^2*(d + e*x)^(7/2))

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fricas [B]  time = 35.59, size = 306, normalized size = 3.22 \begin {gather*} -\frac {2 \, {\left (5 \, A a^{3} e - {\left (5 \, B b^{3} d - {\left (7 \, B a b^{2} - 2 \, A b^{3}\right )} e\right )} x^{3} - {\left ({\left (8 \, B a b^{2} + 7 \, A b^{3}\right )} d - {\left (14 \, B a^{2} b + A a b^{2}\right )} e\right )} x^{2} + {\left (2 \, B a^{3} - 7 \, A a^{2} b\right )} d - {\left ({\left (B a^{2} b + 14 \, A a b^{2}\right )} d - {\left (7 \, B a^{3} + 8 \, A a^{2} b\right )} e\right )} x\right )} \sqrt {b x + a} \sqrt {e x + d}}{35 \, {\left (b^{2} d^{6} - 2 \, a b d^{5} e + a^{2} d^{4} e^{2} + {\left (b^{2} d^{2} e^{4} - 2 \, a b d e^{5} + a^{2} e^{6}\right )} x^{4} + 4 \, {\left (b^{2} d^{3} e^{3} - 2 \, a b d^{2} e^{4} + a^{2} d e^{5}\right )} x^{3} + 6 \, {\left (b^{2} d^{4} e^{2} - 2 \, a b d^{3} e^{3} + a^{2} d^{2} e^{4}\right )} x^{2} + 4 \, {\left (b^{2} d^{5} e - 2 \, a b d^{4} e^{2} + a^{2} d^{3} e^{3}\right )} x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(3/2)*(B*x+A)/(e*x+d)^(9/2),x, algorithm="fricas")

[Out]

-2/35*(5*A*a^3*e - (5*B*b^3*d - (7*B*a*b^2 - 2*A*b^3)*e)*x^3 - ((8*B*a*b^2 + 7*A*b^3)*d - (14*B*a^2*b + A*a*b^
2)*e)*x^2 + (2*B*a^3 - 7*A*a^2*b)*d - ((B*a^2*b + 14*A*a*b^2)*d - (7*B*a^3 + 8*A*a^2*b)*e)*x)*sqrt(b*x + a)*sq
rt(e*x + d)/(b^2*d^6 - 2*a*b*d^5*e + a^2*d^4*e^2 + (b^2*d^2*e^4 - 2*a*b*d*e^5 + a^2*e^6)*x^4 + 4*(b^2*d^3*e^3
- 2*a*b*d^2*e^4 + a^2*d*e^5)*x^3 + 6*(b^2*d^4*e^2 - 2*a*b*d^3*e^3 + a^2*d^2*e^4)*x^2 + 4*(b^2*d^5*e - 2*a*b*d^
4*e^2 + a^2*d^3*e^3)*x)

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giac [B]  time = 3.21, size = 268, normalized size = 2.82 \begin {gather*} \frac {2 \, {\left (b x + a\right )}^{\frac {5}{2}} {\left (\frac {{\left (5 \, B b^{9} d^{2} {\left | b \right |} e^{3} - 12 \, B a b^{8} d {\left | b \right |} e^{4} + 2 \, A b^{9} d {\left | b \right |} e^{4} + 7 \, B a^{2} b^{7} {\left | b \right |} e^{5} - 2 \, A a b^{8} {\left | b \right |} e^{5}\right )} {\left (b x + a\right )}}{b^{5} d^{3} e^{3} - 3 \, a b^{4} d^{2} e^{4} + 3 \, a^{2} b^{3} d e^{5} - a^{3} b^{2} e^{6}} - \frac {7 \, {\left (B a b^{9} d^{2} {\left | b \right |} e^{3} - A b^{10} d^{2} {\left | b \right |} e^{3} - 2 \, B a^{2} b^{8} d {\left | b \right |} e^{4} + 2 \, A a b^{9} d {\left | b \right |} e^{4} + B a^{3} b^{7} {\left | b \right |} e^{5} - A a^{2} b^{8} {\left | b \right |} e^{5}\right )}}{b^{5} d^{3} e^{3} - 3 \, a b^{4} d^{2} e^{4} + 3 \, a^{2} b^{3} d e^{5} - a^{3} b^{2} e^{6}}\right )}}{35 \, {\left (b^{2} d + {\left (b x + a\right )} b e - a b e\right )}^{\frac {7}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(3/2)*(B*x+A)/(e*x+d)^(9/2),x, algorithm="giac")

[Out]

2/35*(b*x + a)^(5/2)*((5*B*b^9*d^2*abs(b)*e^3 - 12*B*a*b^8*d*abs(b)*e^4 + 2*A*b^9*d*abs(b)*e^4 + 7*B*a^2*b^7*a
bs(b)*e^5 - 2*A*a*b^8*abs(b)*e^5)*(b*x + a)/(b^5*d^3*e^3 - 3*a*b^4*d^2*e^4 + 3*a^2*b^3*d*e^5 - a^3*b^2*e^6) -
7*(B*a*b^9*d^2*abs(b)*e^3 - A*b^10*d^2*abs(b)*e^3 - 2*B*a^2*b^8*d*abs(b)*e^4 + 2*A*a*b^9*d*abs(b)*e^4 + B*a^3*
b^7*abs(b)*e^5 - A*a^2*b^8*abs(b)*e^5)/(b^5*d^3*e^3 - 3*a*b^4*d^2*e^4 + 3*a^2*b^3*d*e^5 - a^3*b^2*e^6))/(b^2*d
+ (b*x + a)*b*e - a*b*e)^(7/2)

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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 {5}{2}} \left (-2 A b e x +7 B a e x -5 B b d x +5 A a e -7 A b d +2 B a d \right )}{35 \left (e x +d \right )^{\frac {7}{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)^(3/2)*(B*x+A)/(e*x+d)^(9/2),x)

[Out]

-2/35*(b*x+a)^(5/2)*(-2*A*b*e*x+7*B*a*e*x-5*B*b*d*x+5*A*a*e-7*A*b*d+2*B*a*d)/(e*x+d)^(7/2)/(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)^(3/2)*(B*x+A)/(e*x+d)^(9/2),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 = 2.14, size = 257, normalized size = 2.71 \begin {gather*} -\frac {\sqrt {d+e\,x}\,\left (\frac {\sqrt {a+b\,x}\,\left (10\,A\,a^3\,e+4\,B\,a^3\,d-14\,A\,a^2\,b\,d\right )}{35\,e^4\,{\left (a\,e-b\,d\right )}^2}-\frac {x^3\,\sqrt {a+b\,x}\,\left (4\,A\,b^3\,e+10\,B\,b^3\,d-14\,B\,a\,b^2\,e\right )}{35\,e^4\,{\left (a\,e-b\,d\right )}^2}+\frac {x\,\sqrt {a+b\,x}\,\left (14\,B\,a^3\,e-28\,A\,a\,b^2\,d+16\,A\,a^2\,b\,e-2\,B\,a^2\,b\,d\right )}{35\,e^4\,{\left (a\,e-b\,d\right )}^2}-\frac {x^2\,\sqrt {a+b\,x}\,\left (14\,A\,b^3\,d-2\,A\,a\,b^2\,e+16\,B\,a\,b^2\,d-28\,B\,a^2\,b\,e\right )}{35\,e^4\,{\left (a\,e-b\,d\right )}^2}\right )}{x^4+\frac {d^4}{e^4}+\frac {4\,d\,x^3}{e}+\frac {4\,d^3\,x}{e^3}+\frac {6\,d^2\,x^2}{e^2}} \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)^(9/2),x)

[Out]

-((d + e*x)^(1/2)*(((a + b*x)^(1/2)*(10*A*a^3*e + 4*B*a^3*d - 14*A*a^2*b*d))/(35*e^4*(a*e - b*d)^2) - (x^3*(a
+ b*x)^(1/2)*(4*A*b^3*e + 10*B*b^3*d - 14*B*a*b^2*e))/(35*e^4*(a*e - b*d)^2) + (x*(a + b*x)^(1/2)*(14*B*a^3*e
- 28*A*a*b^2*d + 16*A*a^2*b*e - 2*B*a^2*b*d))/(35*e^4*(a*e - b*d)^2) - (x^2*(a + b*x)^(1/2)*(14*A*b^3*d - 2*A*
a*b^2*e + 16*B*a*b^2*d - 28*B*a^2*b*e))/(35*e^4*(a*e - b*d)^2)))/(x^4 + d^4/e^4 + (4*d*x^3)/e + (4*d^3*x)/e^3
+ (6*d^2*x^2)/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)**(3/2)*(B*x+A)/(e*x+d)**(9/2),x)

[Out]

Timed out

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