3.21.92 \(\int \frac {(a+b x)^{5/2} (A+B x)}{(d+e x)^{11/2}} \, dx\)

Optimal. Leaf size=95 \[ \frac {2 (a+b x)^{7/2} (-9 a B e+2 A b e+7 b B d)}{63 e (d+e x)^{7/2} (b d-a e)^2}-\frac {2 (a+b x)^{7/2} (B d-A e)}{9 e (d+e x)^{9/2} (b d-a e)} \]

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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)^{7/2} (-9 a B e+2 A b e+7 b B d)}{63 e (d+e x)^{7/2} (b d-a e)^2}-\frac {2 (a+b x)^{7/2} (B d-A e)}{9 e (d+e x)^{9/2} (b d-a e)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((a + b*x)^(5/2)*(A + B*x))/(d + e*x)^(11/2),x]

[Out]

(-2*(B*d - A*e)*(a + b*x)^(7/2))/(9*e*(b*d - a*e)*(d + e*x)^(9/2)) + (2*(7*b*B*d + 2*A*b*e - 9*a*B*e)*(a + b*x
)^(7/2))/(63*e*(b*d - a*e)^2*(d + e*x)^(7/2))

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)^{5/2} (A+B x)}{(d+e x)^{11/2}} \, dx &=-\frac {2 (B d-A e) (a+b x)^{7/2}}{9 e (b d-a e) (d+e x)^{9/2}}+\frac {(7 b B d+2 A b e-9 a B e) \int \frac {(a+b x)^{5/2}}{(d+e x)^{9/2}} \, dx}{9 e (b d-a e)}\\ &=-\frac {2 (B d-A e) (a+b x)^{7/2}}{9 e (b d-a e) (d+e x)^{9/2}}+\frac {2 (7 b B d+2 A b e-9 a B e) (a+b x)^{7/2}}{63 e (b d-a e)^2 (d+e x)^{7/2}}\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 66, normalized size = 0.69 \begin {gather*} \frac {2 (a+b x)^{7/2} (A (-7 a e+9 b d+2 b e x)+B (-2 a d-9 a e x+7 b d x))}{63 (d+e x)^{9/2} (b d-a e)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x)^(5/2)*(A + B*x))/(d + e*x)^(11/2),x]

[Out]

(2*(a + b*x)^(7/2)*(B*(-2*a*d + 7*b*d*x - 9*a*e*x) + A*(9*b*d - 7*a*e + 2*b*e*x)))/(63*(b*d - a*e)^2*(d + e*x)
^(9/2))

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IntegrateAlgebraic [A]  time = 0.17, size = 73, normalized size = 0.77 \begin {gather*} -\frac {2 (a+b x)^{9/2} \left (-\frac {9 A b (d+e x)}{a+b x}+\frac {9 a B (d+e x)}{a+b x}+7 A e-7 B d\right )}{63 (d+e x)^{9/2} (b d-a e)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((a + b*x)^(5/2)*(A + B*x))/(d + e*x)^(11/2),x]

[Out]

(-2*(a + b*x)^(9/2)*(-7*B*d + 7*A*e - (9*A*b*(d + e*x))/(a + b*x) + (9*a*B*(d + e*x))/(a + b*x)))/(63*(b*d - a
*e)^2*(d + e*x)^(9/2))

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fricas [B]  time = 152.73, size = 391, normalized size = 4.12 \begin {gather*} -\frac {2 \, {\left (7 \, A a^{4} e - {\left (7 \, B b^{4} d - {\left (9 \, B a b^{3} - 2 \, A b^{4}\right )} e\right )} x^{4} - {\left ({\left (19 \, B a b^{3} + 9 \, A b^{4}\right )} d - {\left (27 \, B a^{2} b^{2} + A a b^{3}\right )} e\right )} x^{3} - 3 \, {\left ({\left (5 \, B a^{2} b^{2} + 9 \, A a b^{3}\right )} d - {\left (9 \, B a^{3} b + 5 \, A a^{2} b^{2}\right )} e\right )} x^{2} + {\left (2 \, B a^{4} - 9 \, A a^{3} b\right )} d - {\left ({\left (B a^{3} b + 27 \, A a^{2} b^{2}\right )} d - {\left (9 \, B a^{4} + 19 \, A a^{3} b\right )} e\right )} x\right )} \sqrt {b x + a} \sqrt {e x + d}}{63 \, {\left (b^{2} d^{7} - 2 \, a b d^{6} e + a^{2} d^{5} e^{2} + {\left (b^{2} d^{2} e^{5} - 2 \, a b d e^{6} + a^{2} e^{7}\right )} x^{5} + 5 \, {\left (b^{2} d^{3} e^{4} - 2 \, a b d^{2} e^{5} + a^{2} d e^{6}\right )} x^{4} + 10 \, {\left (b^{2} d^{4} e^{3} - 2 \, a b d^{3} e^{4} + a^{2} d^{2} e^{5}\right )} x^{3} + 10 \, {\left (b^{2} d^{5} e^{2} - 2 \, a b d^{4} e^{3} + a^{2} d^{3} e^{4}\right )} x^{2} + 5 \, {\left (b^{2} d^{6} e - 2 \, a b d^{5} e^{2} + a^{2} d^{4} e^{3}\right )} x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(5/2)*(B*x+A)/(e*x+d)^(11/2),x, algorithm="fricas")

[Out]

-2/63*(7*A*a^4*e - (7*B*b^4*d - (9*B*a*b^3 - 2*A*b^4)*e)*x^4 - ((19*B*a*b^3 + 9*A*b^4)*d - (27*B*a^2*b^2 + A*a
*b^3)*e)*x^3 - 3*((5*B*a^2*b^2 + 9*A*a*b^3)*d - (9*B*a^3*b + 5*A*a^2*b^2)*e)*x^2 + (2*B*a^4 - 9*A*a^3*b)*d - (
(B*a^3*b + 27*A*a^2*b^2)*d - (9*B*a^4 + 19*A*a^3*b)*e)*x)*sqrt(b*x + a)*sqrt(e*x + d)/(b^2*d^7 - 2*a*b*d^6*e +
a^2*d^5*e^2 + (b^2*d^2*e^5 - 2*a*b*d*e^6 + a^2*e^7)*x^5 + 5*(b^2*d^3*e^4 - 2*a*b*d^2*e^5 + a^2*d*e^6)*x^4 + 1
0*(b^2*d^4*e^3 - 2*a*b*d^3*e^4 + a^2*d^2*e^5)*x^3 + 10*(b^2*d^5*e^2 - 2*a*b*d^4*e^3 + a^2*d^3*e^4)*x^2 + 5*(b^
2*d^6*e - 2*a*b*d^5*e^2 + a^2*d^4*e^3)*x)

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giac [B]  time = 5.77, size = 356, normalized size = 3.75 \begin {gather*} \frac {2 \, {\left (b x + a\right )}^{\frac {7}{2}} {\left (\frac {{\left (7 \, B b^{12} d^{3} {\left | b \right |} e^{4} - 23 \, B a b^{11} d^{2} {\left | b \right |} e^{5} + 2 \, A b^{12} d^{2} {\left | b \right |} e^{5} + 25 \, B a^{2} b^{10} d {\left | b \right |} e^{6} - 4 \, A a b^{11} d {\left | b \right |} e^{6} - 9 \, B a^{3} b^{9} {\left | b \right |} e^{7} + 2 \, A a^{2} b^{10} {\left | b \right |} e^{7}\right )} {\left (b x + a\right )}}{b^{6} d^{4} e^{4} - 4 \, a b^{5} d^{3} e^{5} + 6 \, a^{2} b^{4} d^{2} e^{6} - 4 \, a^{3} b^{3} d e^{7} + a^{4} b^{2} e^{8}} - \frac {9 \, {\left (B a b^{12} d^{3} {\left | b \right |} e^{4} - A b^{13} d^{3} {\left | b \right |} e^{4} - 3 \, B a^{2} b^{11} d^{2} {\left | b \right |} e^{5} + 3 \, A a b^{12} d^{2} {\left | b \right |} e^{5} + 3 \, B a^{3} b^{10} d {\left | b \right |} e^{6} - 3 \, A a^{2} b^{11} d {\left | b \right |} e^{6} - B a^{4} b^{9} {\left | b \right |} e^{7} + A a^{3} b^{10} {\left | b \right |} e^{7}\right )}}{b^{6} d^{4} e^{4} - 4 \, a b^{5} d^{3} e^{5} + 6 \, a^{2} b^{4} d^{2} e^{6} - 4 \, a^{3} b^{3} d e^{7} + a^{4} b^{2} e^{8}}\right )}}{63 \, {\left (b^{2} d + {\left (b x + a\right )} b e - a b e\right )}^{\frac {9}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(5/2)*(B*x+A)/(e*x+d)^(11/2),x, algorithm="giac")

[Out]

2/63*(b*x + a)^(7/2)*((7*B*b^12*d^3*abs(b)*e^4 - 23*B*a*b^11*d^2*abs(b)*e^5 + 2*A*b^12*d^2*abs(b)*e^5 + 25*B*a
^2*b^10*d*abs(b)*e^6 - 4*A*a*b^11*d*abs(b)*e^6 - 9*B*a^3*b^9*abs(b)*e^7 + 2*A*a^2*b^10*abs(b)*e^7)*(b*x + a)/(
b^6*d^4*e^4 - 4*a*b^5*d^3*e^5 + 6*a^2*b^4*d^2*e^6 - 4*a^3*b^3*d*e^7 + a^4*b^2*e^8) - 9*(B*a*b^12*d^3*abs(b)*e^
4 - A*b^13*d^3*abs(b)*e^4 - 3*B*a^2*b^11*d^2*abs(b)*e^5 + 3*A*a*b^12*d^2*abs(b)*e^5 + 3*B*a^3*b^10*d*abs(b)*e^
6 - 3*A*a^2*b^11*d*abs(b)*e^6 - B*a^4*b^9*abs(b)*e^7 + A*a^3*b^10*abs(b)*e^7)/(b^6*d^4*e^4 - 4*a*b^5*d^3*e^5 +
6*a^2*b^4*d^2*e^6 - 4*a^3*b^3*d*e^7 + a^4*b^2*e^8))/(b^2*d + (b*x + a)*b*e - a*b*e)^(9/2)

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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 {7}{2}} \left (-2 A b e x +9 B a e x -7 B b d x +7 A a e -9 A b d +2 B a d \right )}{63 \left (e x +d \right )^{\frac {9}{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)^(5/2)*(B*x+A)/(e*x+d)^(11/2),x)

[Out]

-2/63*(b*x+a)^(7/2)*(-2*A*b*e*x+9*B*a*e*x-7*B*b*d*x+7*A*a*e-9*A*b*d+2*B*a*d)/(e*x+d)^(9/2)/(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)^(5/2)*(B*x+A)/(e*x+d)^(11/2),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 = 2.40, size = 325, normalized size = 3.42 \begin {gather*} \frac {\sqrt {d+e\,x}\,\left (\frac {x^3\,\sqrt {a+b\,x}\,\left (18\,A\,b^4\,d-2\,A\,a\,b^3\,e+38\,B\,a\,b^3\,d-54\,B\,a^2\,b^2\,e\right )}{63\,e^5\,{\left (a\,e-b\,d\right )}^2}-\frac {\sqrt {a+b\,x}\,\left (14\,A\,a^4\,e+4\,B\,a^4\,d-18\,A\,a^3\,b\,d\right )}{63\,e^5\,{\left (a\,e-b\,d\right )}^2}+\frac {x^4\,\sqrt {a+b\,x}\,\left (4\,A\,b^4\,e+14\,B\,b^4\,d-18\,B\,a\,b^3\,e\right )}{63\,e^5\,{\left (a\,e-b\,d\right )}^2}-\frac {x\,\sqrt {a+b\,x}\,\left (18\,B\,a^4\,e+38\,A\,a^3\,b\,e-2\,B\,a^3\,b\,d-54\,A\,a^2\,b^2\,d\right )}{63\,e^5\,{\left (a\,e-b\,d\right )}^2}+\frac {2\,a\,b\,x^2\,\sqrt {a+b\,x}\,\left (9\,A\,b^2\,d-9\,B\,a^2\,e-5\,A\,a\,b\,e+5\,B\,a\,b\,d\right )}{21\,e^5\,{\left (a\,e-b\,d\right )}^2}\right )}{x^5+\frac {d^5}{e^5}+\frac {5\,d\,x^4}{e}+\frac {5\,d^4\,x}{e^4}+\frac {10\,d^2\,x^3}{e^2}+\frac {10\,d^3\,x^2}{e^3}} \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)^(11/2),x)

[Out]

((d + e*x)^(1/2)*((x^3*(a + b*x)^(1/2)*(18*A*b^4*d - 2*A*a*b^3*e + 38*B*a*b^3*d - 54*B*a^2*b^2*e))/(63*e^5*(a*
e - b*d)^2) - ((a + b*x)^(1/2)*(14*A*a^4*e + 4*B*a^4*d - 18*A*a^3*b*d))/(63*e^5*(a*e - b*d)^2) + (x^4*(a + b*x
)^(1/2)*(4*A*b^4*e + 14*B*b^4*d - 18*B*a*b^3*e))/(63*e^5*(a*e - b*d)^2) - (x*(a + b*x)^(1/2)*(18*B*a^4*e + 38*
A*a^3*b*e - 2*B*a^3*b*d - 54*A*a^2*b^2*d))/(63*e^5*(a*e - b*d)^2) + (2*a*b*x^2*(a + b*x)^(1/2)*(9*A*b^2*d - 9*
B*a^2*e - 5*A*a*b*e + 5*B*a*b*d))/(21*e^5*(a*e - b*d)^2)))/(x^5 + d^5/e^5 + (5*d*x^4)/e + (5*d^4*x)/e^4 + (10*
d^2*x^3)/e^2 + (10*d^3*x^2)/e^3)

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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)**(5/2)*(B*x+A)/(e*x+d)**(11/2),x)

[Out]

Timed out

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