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

Optimal. Leaf size=145 \[ -\frac {2 \sqrt {a+b x} (B d-A e)}{5 e (d+e x)^{5/2} (b d-a e)}+\frac {4 b \sqrt {a+b x} (-5 a B e+4 A b e+b B d)}{15 e \sqrt {d+e x} (b d-a e)^3}+\frac {2 \sqrt {a+b x} (-5 a B e+4 A b e+b B d)}{15 e (d+e x)^{3/2} (b d-a e)^2} \]

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Rubi [A]  time = 0.09, antiderivative size = 145, 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 \sqrt {a+b x} (B d-A e)}{5 e (d+e x)^{5/2} (b d-a e)}+\frac {4 b \sqrt {a+b x} (-5 a B e+4 A b e+b B d)}{15 e \sqrt {d+e x} (b d-a e)^3}+\frac {2 \sqrt {a+b x} (-5 a B e+4 A b e+b B d)}{15 e (d+e x)^{3/2} (b d-a e)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(B*d - A*e)*Sqrt[a + b*x])/(5*e*(b*d - a*e)*(d + e*x)^(5/2)) + (2*(b*B*d + 4*A*b*e - 5*a*B*e)*Sqrt[a + b*x
])/(15*e*(b*d - a*e)^2*(d + e*x)^(3/2)) + (4*b*(b*B*d + 4*A*b*e - 5*a*B*e)*Sqrt[a + b*x])/(15*e*(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}{\sqrt {a+b x} (d+e x)^{7/2}} \, dx &=-\frac {2 (B d-A e) \sqrt {a+b x}}{5 e (b d-a e) (d+e x)^{5/2}}+\frac {(b B d+4 A b e-5 a B e) \int \frac {1}{\sqrt {a+b x} (d+e x)^{5/2}} \, dx}{5 e (b d-a e)}\\ &=-\frac {2 (B d-A e) \sqrt {a+b x}}{5 e (b d-a e) (d+e x)^{5/2}}+\frac {2 (b B d+4 A b e-5 a B e) \sqrt {a+b x}}{15 e (b d-a e)^2 (d+e x)^{3/2}}+\frac {(2 b (b B d+4 A b e-5 a B e)) \int \frac {1}{\sqrt {a+b x} (d+e x)^{3/2}} \, dx}{15 e (b d-a e)^2}\\ &=-\frac {2 (B d-A e) \sqrt {a+b x}}{5 e (b d-a e) (d+e x)^{5/2}}+\frac {2 (b B d+4 A b e-5 a B e) \sqrt {a+b x}}{15 e (b d-a e)^2 (d+e x)^{3/2}}+\frac {4 b (b B d+4 A b e-5 a B e) \sqrt {a+b x}}{15 e (b d-a e)^3 \sqrt {d+e x}}\\ \end {align*}

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Mathematica [A]  time = 0.11, size = 133, normalized size = 0.92 \begin {gather*} \frac {2 \sqrt {a+b x} \left (A \left (3 a^2 e^2-2 a b e (5 d+2 e x)+b^2 \left (15 d^2+20 d e x+8 e^2 x^2\right )\right )+B \left (a^2 e (2 d+5 e x)-2 a b \left (5 d^2+13 d e x+5 e^2 x^2\right )+b^2 d x (5 d+2 e x)\right )\right )}{15 (d+e x)^{5/2} (b d-a e)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[a + b*x]*(B*(b^2*d*x*(5*d + 2*e*x) + a^2*e*(2*d + 5*e*x) - 2*a*b*(5*d^2 + 13*d*e*x + 5*e^2*x^2)) + A*(
3*a^2*e^2 - 2*a*b*e*(5*d + 2*e*x) + b^2*(15*d^2 + 20*d*e*x + 8*e^2*x^2))))/(15*(b*d - a*e)^3*(d + e*x)^(5/2))

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IntegrateAlgebraic [A]  time = 0.14, size = 134, normalized size = 0.92 \begin {gather*} \frac {2 \sqrt {a+b x} \left (\frac {3 A e^2 (a+b x)^2}{(d+e x)^2}-\frac {10 A b e (a+b x)}{d+e x}+\frac {5 b B d (a+b x)}{d+e x}+\frac {5 a B e (a+b x)}{d+e x}-\frac {3 B d e (a+b x)^2}{(d+e x)^2}-15 a b B+15 A b^2\right )}{15 \sqrt {d+e x} (b d-a e)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[a + b*x]*(15*A*b^2 - 15*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
+ (5*b*B*d*(a + b*x))/(d + e*x) - (10*A*b*e*(a + b*x))/(d + e*x) + (5*a*B*e*(a + b*x))/(d + e*x)))/(15*(b*d -
a*e)^3*Sqrt[d + e*x])

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fricas [B]  time = 12.19, size = 316, normalized size = 2.18 \begin {gather*} \frac {2 \, {\left (3 \, A a^{2} e^{2} - 5 \, {\left (2 \, B a b - 3 \, A b^{2}\right )} d^{2} + 2 \, {\left (B a^{2} - 5 \, A a b\right )} d e + 2 \, {\left (B b^{2} d e - {\left (5 \, B a b - 4 \, A b^{2}\right )} e^{2}\right )} x^{2} + {\left (5 \, B b^{2} d^{2} - 2 \, {\left (13 \, B a b - 10 \, A b^{2}\right )} d e + {\left (5 \, B a^{2} - 4 \, A a b\right )} e^{2}\right )} x\right )} \sqrt {b x + a} \sqrt {e x + d}}{15 \, {\left (b^{3} d^{6} - 3 \, a b^{2} d^{5} e + 3 \, a^{2} b d^{4} e^{2} - a^{3} d^{3} e^{3} + {\left (b^{3} d^{3} e^{3} - 3 \, a b^{2} d^{2} e^{4} + 3 \, a^{2} b d e^{5} - a^{3} e^{6}\right )} x^{3} + 3 \, {\left (b^{3} d^{4} e^{2} - 3 \, a b^{2} d^{3} e^{3} + 3 \, a^{2} b d^{2} e^{4} - a^{3} d e^{5}\right )} x^{2} + 3 \, {\left (b^{3} d^{5} e - 3 \, a b^{2} d^{4} e^{2} + 3 \, a^{2} b d^{3} e^{3} - a^{3} d^{2} e^{4}\right )} x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*(3*A*a^2*e^2 - 5*(2*B*a*b - 3*A*b^2)*d^2 + 2*(B*a^2 - 5*A*a*b)*d*e + 2*(B*b^2*d*e - (5*B*a*b - 4*A*b^2)*e
^2)*x^2 + (5*B*b^2*d^2 - 2*(13*B*a*b - 10*A*b^2)*d*e + (5*B*a^2 - 4*A*a*b)*e^2)*x)*sqrt(b*x + a)*sqrt(e*x + d)
/(b^3*d^6 - 3*a*b^2*d^5*e + 3*a^2*b*d^4*e^2 - a^3*d^3*e^3 + (b^3*d^3*e^3 - 3*a*b^2*d^2*e^4 + 3*a^2*b*d*e^5 - a
^3*e^6)*x^3 + 3*(b^3*d^4*e^2 - 3*a*b^2*d^3*e^3 + 3*a^2*b*d^2*e^4 - a^3*d*e^5)*x^2 + 3*(b^3*d^5*e - 3*a*b^2*d^4
*e^2 + 3*a^2*b*d^3*e^3 - a^3*d^2*e^4)*x)

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giac [B]  time = 1.36, size = 353, normalized size = 2.43 \begin {gather*} \frac {2 \, {\left ({\left (b x + a\right )} {\left (\frac {2 \, {\left (B b^{6} d {\left | b \right |} e^{3} - 5 \, B a b^{5} {\left | b \right |} e^{4} + 4 \, A b^{6} {\left | b \right |} e^{4}\right )} {\left (b x + a\right )}}{b^{5} d^{3} e^{2} - 3 \, a b^{4} d^{2} e^{3} + 3 \, a^{2} b^{3} d e^{4} - a^{3} b^{2} e^{5}} + \frac {5 \, {\left (B b^{7} d^{2} {\left | b \right |} e^{2} - 6 \, B a b^{6} d {\left | b \right |} e^{3} + 4 \, A b^{7} d {\left | b \right |} e^{3} + 5 \, B a^{2} b^{5} {\left | b \right |} e^{4} - 4 \, A a b^{6} {\left | b \right |} e^{4}\right )}}{b^{5} d^{3} e^{2} - 3 \, a b^{4} d^{2} e^{3} + 3 \, a^{2} b^{3} d e^{4} - a^{3} b^{2} e^{5}}\right )} - \frac {15 \, {\left (B a b^{7} d^{2} {\left | b \right |} e^{2} - A b^{8} d^{2} {\left | b \right |} e^{2} - 2 \, B a^{2} b^{6} d {\left | b \right |} e^{3} + 2 \, A a b^{7} d {\left | b \right |} e^{3} + B a^{3} b^{5} {\left | b \right |} e^{4} - A a^{2} b^{6} {\left | b \right |} e^{4}\right )}}{b^{5} d^{3} e^{2} - 3 \, a b^{4} d^{2} e^{3} + 3 \, a^{2} b^{3} d e^{4} - a^{3} b^{2} e^{5}}\right )} \sqrt {b x + a}}{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)/(e*x+d)^(7/2)/(b*x+a)^(1/2),x, algorithm="giac")

[Out]

2/15*((b*x + a)*(2*(B*b^6*d*abs(b)*e^3 - 5*B*a*b^5*abs(b)*e^4 + 4*A*b^6*abs(b)*e^4)*(b*x + a)/(b^5*d^3*e^2 - 3
*a*b^4*d^2*e^3 + 3*a^2*b^3*d*e^4 - a^3*b^2*e^5) + 5*(B*b^7*d^2*abs(b)*e^2 - 6*B*a*b^6*d*abs(b)*e^3 + 4*A*b^7*d
*abs(b)*e^3 + 5*B*a^2*b^5*abs(b)*e^4 - 4*A*a*b^6*abs(b)*e^4)/(b^5*d^3*e^2 - 3*a*b^4*d^2*e^3 + 3*a^2*b^3*d*e^4
- a^3*b^2*e^5)) - 15*(B*a*b^7*d^2*abs(b)*e^2 - A*b^8*d^2*abs(b)*e^2 - 2*B*a^2*b^6*d*abs(b)*e^3 + 2*A*a*b^7*d*a
bs(b)*e^3 + B*a^3*b^5*abs(b)*e^4 - A*a^2*b^6*abs(b)*e^4)/(b^5*d^3*e^2 - 3*a*b^4*d^2*e^3 + 3*a^2*b^3*d*e^4 - a^
3*b^2*e^5))*sqrt(b*x + a)/(b^2*d + (b*x + a)*b*e - a*b*e)^(5/2)

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maple [A]  time = 0.01, size = 177, normalized size = 1.22 \begin {gather*} -\frac {2 \sqrt {b x +a}\, \left (8 A \,b^{2} e^{2} x^{2}-10 B a b \,e^{2} x^{2}+2 B \,b^{2} d e \,x^{2}-4 A a b \,e^{2} x +20 A \,b^{2} d e x +5 B \,a^{2} e^{2} x -26 B a b d e x +5 B \,b^{2} d^{2} x +3 A \,a^{2} e^{2}-10 A a b d e +15 A \,b^{2} d^{2}+2 B \,a^{2} d e -10 B a b \,d^{2}\right )}{15 \left (e x +d \right )^{\frac {5}{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)/(e*x+d)^(7/2)/(b*x+a)^(1/2),x)

[Out]

-2/15*(b*x+a)^(1/2)*(8*A*b^2*e^2*x^2-10*B*a*b*e^2*x^2+2*B*b^2*d*e*x^2-4*A*a*b*e^2*x+20*A*b^2*d*e*x+5*B*a^2*e^2
*x-26*B*a*b*d*e*x+5*B*b^2*d^2*x+3*A*a^2*e^2-10*A*a*b*d*e+15*A*b^2*d^2+2*B*a^2*d*e-10*B*a*b*d^2)/(e*x+d)^(5/2)/
(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)/(e*x+d)^(7/2)/(b*x+a)^(1/2),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 = 2.25, size = 283, normalized size = 1.95 \begin {gather*} -\frac {\sqrt {d+e\,x}\,\left (\frac {4\,B\,a^3\,d\,e+6\,A\,a^3\,e^2-20\,B\,a^2\,b\,d^2-20\,A\,a^2\,b\,d\,e+30\,A\,a\,b^2\,d^2}{15\,e^3\,{\left (a\,e-b\,d\right )}^3}+\frac {x\,\left (10\,B\,a^3\,e^2-48\,B\,a^2\,b\,d\,e-2\,A\,a^2\,b\,e^2-10\,B\,a\,b^2\,d^2+20\,A\,a\,b^2\,d\,e+30\,A\,b^3\,d^2\right )}{15\,e^3\,{\left (a\,e-b\,d\right )}^3}+\frac {4\,b^2\,x^3\,\left (4\,A\,b\,e-5\,B\,a\,e+B\,b\,d\right )}{15\,e^2\,{\left (a\,e-b\,d\right )}^3}+\frac {2\,b\,x^2\,\left (a\,e+5\,b\,d\right )\,\left (4\,A\,b\,e-5\,B\,a\,e+B\,b\,d\right )}{15\,e^3\,{\left (a\,e-b\,d\right )}^3}\right )}{x^3\,\sqrt {a+b\,x}+\frac {d^3\,\sqrt {a+b\,x}}{e^3}+\frac {3\,d\,x^2\,\sqrt {a+b\,x}}{e}+\frac {3\,d^2\,x\,\sqrt {a+b\,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)*((6*A*a^3*e^2 + 4*B*a^3*d*e + 30*A*a*b^2*d^2 - 20*B*a^2*b*d^2 - 20*A*a^2*b*d*e)/(15*e^3*(a*e
- b*d)^3) + (x*(30*A*b^3*d^2 + 10*B*a^3*e^2 - 2*A*a^2*b*e^2 - 10*B*a*b^2*d^2 + 20*A*a*b^2*d*e - 48*B*a^2*b*d*
e))/(15*e^3*(a*e - b*d)^3) + (4*b^2*x^3*(4*A*b*e - 5*B*a*e + B*b*d))/(15*e^2*(a*e - b*d)^3) + (2*b*x^2*(a*e +
5*b*d)*(4*A*b*e - 5*B*a*e + B*b*d))/(15*e^3*(a*e - b*d)^3)))/(x^3*(a + b*x)^(1/2) + (d^3*(a + b*x)^(1/2))/e^3
+ (3*d*x^2*(a + b*x)^(1/2))/e + (3*d^2*x*(a + b*x)^(1/2))/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)/(e*x+d)**(7/2)/(b*x+a)**(1/2),x)

[Out]

Timed out

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