3.1748 \(\int \frac{A+B x}{(a+b x) (d+e x)^{5/2}} \, dx\)

Optimal. Leaf size=119 \[ \frac{2 (A b-a B)}{\sqrt{d+e x} (b d-a e)^2}-\frac{2 (B d-A e)}{3 e (d+e x)^{3/2} (b d-a e)}-\frac{2 \sqrt{b} (A b-a B) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{(b d-a e)^{5/2}} \]

[Out]

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

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Rubi [A]  time = 0.0711895, antiderivative size = 119, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {78, 51, 63, 208} \[ \frac{2 (A b-a B)}{\sqrt{d+e x} (b d-a e)^2}-\frac{2 (B d-A e)}{3 e (d+e x)^{3/2} (b d-a e)}-\frac{2 \sqrt{b} (A b-a B) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{(b d-a e)^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]))))

Rule 51

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*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rubi steps

\begin{align*} \int \frac{A+B x}{(a+b x) (d+e x)^{5/2}} \, dx &=-\frac{2 (B d-A e)}{3 e (b d-a e) (d+e x)^{3/2}}+\frac{(A b-a B) \int \frac{1}{(a+b x) (d+e x)^{3/2}} \, dx}{b d-a e}\\ &=-\frac{2 (B d-A e)}{3 e (b d-a e) (d+e x)^{3/2}}+\frac{2 (A b-a B)}{(b d-a e)^2 \sqrt{d+e x}}+\frac{(b (A b-a B)) \int \frac{1}{(a+b x) \sqrt{d+e x}} \, dx}{(b d-a e)^2}\\ &=-\frac{2 (B d-A e)}{3 e (b d-a e) (d+e x)^{3/2}}+\frac{2 (A b-a B)}{(b d-a e)^2 \sqrt{d+e x}}+\frac{(2 b (A b-a B)) \operatorname{Subst}\left (\int \frac{1}{a-\frac{b d}{e}+\frac{b x^2}{e}} \, dx,x,\sqrt{d+e x}\right )}{e (b d-a e)^2}\\ &=-\frac{2 (B d-A e)}{3 e (b d-a e) (d+e x)^{3/2}}+\frac{2 (A b-a B)}{(b d-a e)^2 \sqrt{d+e x}}-\frac{2 \sqrt{b} (A b-a B) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{(b d-a e)^{5/2}}\\ \end{align*}

Mathematica [C]  time = 0.0412645, size = 86, normalized size = 0.72 \[ \frac{6 e (d+e x) (A b-a B) \, _2F_1\left (-\frac{1}{2},1;\frac{1}{2};\frac{b (d+e x)}{b d-a e}\right )-2 (b d-a e) (B d-A e)}{3 e (d+e x)^{3/2} (b d-a e)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(b*d - a*e)*(B*d - A*e) + 6*(A*b - a*B)*e*(d + e*x)*Hypergeometric2F1[-1/2, 1, 1/2, (b*(d + e*x))/(b*d - a
*e)])/(3*e*(b*d - a*e)^2*(d + e*x)^(3/2))

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Maple [A]  time = 0.012, size = 187, normalized size = 1.6 \begin{align*} -{\frac{2\,A}{3\,ae-3\,bd} \left ( ex+d \right ) ^{-{\frac{3}{2}}}}+{\frac{2\,Bd}{3\,e \left ( ae-bd \right ) } \left ( ex+d \right ) ^{-{\frac{3}{2}}}}+2\,{\frac{Ab}{ \left ( ae-bd \right ) ^{2}\sqrt{ex+d}}}-2\,{\frac{Ba}{ \left ( ae-bd \right ) ^{2}\sqrt{ex+d}}}+2\,{\frac{A{b}^{2}}{ \left ( ae-bd \right ) ^{2}\sqrt{ \left ( ae-bd \right ) b}}\arctan \left ({\frac{b\sqrt{ex+d}}{\sqrt{ \left ( ae-bd \right ) b}}} \right ) }-2\,{\frac{Bba}{ \left ( ae-bd \right ) ^{2}\sqrt{ \left ( ae-bd \right ) b}}\arctan \left ({\frac{b\sqrt{ex+d}}{\sqrt{ \left ( ae-bd \right ) b}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3/(a*e-b*d)/(e*x+d)^(3/2)*A+2/3/e/(a*e-b*d)/(e*x+d)^(3/2)*B*d+2/(a*e-b*d)^2/(e*x+d)^(1/2)*A*b-2/(a*e-b*d)^2
/(e*x+d)^(1/2)*B*a+2*b^2/(a*e-b*d)^2/((a*e-b*d)*b)^(1/2)*arctan(b*(e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2))*A-2*b/(a*
e-b*d)^2/((a*e-b*d)*b)^(1/2)*arctan(b*(e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2))*B*a

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.52437, size = 1057, normalized size = 8.88 \begin{align*} \left [-\frac{3 \,{\left ({\left (B a - A b\right )} e^{3} x^{2} + 2 \,{\left (B a - A b\right )} d e^{2} x +{\left (B a - A b\right )} d^{2} e\right )} \sqrt{\frac{b}{b d - a e}} \log \left (\frac{b e x + 2 \, b d - a e - 2 \,{\left (b d - a e\right )} \sqrt{e x + d} \sqrt{\frac{b}{b d - a e}}}{b x + a}\right ) + 2 \,{\left (B b d^{2} + A a e^{2} + 3 \,{\left (B a - A b\right )} e^{2} x + 2 \,{\left (B a - 2 \, A b\right )} d e\right )} \sqrt{e x + d}}{3 \,{\left (b^{2} d^{4} e - 2 \, a b d^{3} e^{2} + a^{2} d^{2} e^{3} +{\left (b^{2} d^{2} e^{3} - 2 \, a b d e^{4} + a^{2} e^{5}\right )} x^{2} + 2 \,{\left (b^{2} d^{3} e^{2} - 2 \, a b d^{2} e^{3} + a^{2} d e^{4}\right )} x\right )}}, \frac{2 \,{\left (3 \,{\left ({\left (B a - A b\right )} e^{3} x^{2} + 2 \,{\left (B a - A b\right )} d e^{2} x +{\left (B a - A b\right )} d^{2} e\right )} \sqrt{-\frac{b}{b d - a e}} \arctan \left (-\frac{{\left (b d - a e\right )} \sqrt{e x + d} \sqrt{-\frac{b}{b d - a e}}}{b e x + b d}\right ) -{\left (B b d^{2} + A a e^{2} + 3 \,{\left (B a - A b\right )} e^{2} x + 2 \,{\left (B a - 2 \, A b\right )} d e\right )} \sqrt{e x + d}\right )}}{3 \,{\left (b^{2} d^{4} e - 2 \, a b d^{3} e^{2} + a^{2} d^{2} e^{3} +{\left (b^{2} d^{2} e^{3} - 2 \, a b d e^{4} + a^{2} e^{5}\right )} x^{2} + 2 \,{\left (b^{2} d^{3} e^{2} - 2 \, a b d^{2} e^{3} + a^{2} d e^{4}\right )} x\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/3*(3*((B*a - A*b)*e^3*x^2 + 2*(B*a - A*b)*d*e^2*x + (B*a - A*b)*d^2*e)*sqrt(b/(b*d - a*e))*log((b*e*x + 2*
b*d - a*e - 2*(b*d - a*e)*sqrt(e*x + d)*sqrt(b/(b*d - a*e)))/(b*x + a)) + 2*(B*b*d^2 + A*a*e^2 + 3*(B*a - A*b)
*e^2*x + 2*(B*a - 2*A*b)*d*e)*sqrt(e*x + d))/(b^2*d^4*e - 2*a*b*d^3*e^2 + a^2*d^2*e^3 + (b^2*d^2*e^3 - 2*a*b*d
*e^4 + a^2*e^5)*x^2 + 2*(b^2*d^3*e^2 - 2*a*b*d^2*e^3 + a^2*d*e^4)*x), 2/3*(3*((B*a - A*b)*e^3*x^2 + 2*(B*a - A
*b)*d*e^2*x + (B*a - A*b)*d^2*e)*sqrt(-b/(b*d - a*e))*arctan(-(b*d - a*e)*sqrt(e*x + d)*sqrt(-b/(b*d - a*e))/(
b*e*x + b*d)) - (B*b*d^2 + A*a*e^2 + 3*(B*a - A*b)*e^2*x + 2*(B*a - 2*A*b)*d*e)*sqrt(e*x + d))/(b^2*d^4*e - 2*
a*b*d^3*e^2 + a^2*d^2*e^3 + (b^2*d^2*e^3 - 2*a*b*d*e^4 + a^2*e^5)*x^2 + 2*(b^2*d^3*e^2 - 2*a*b*d^2*e^3 + a^2*d
*e^4)*x)]

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Sympy [A]  time = 22.9697, size = 105, normalized size = 0.88 \begin{align*} - \frac{2 \left (- A b + B a\right )}{\sqrt{d + e x} \left (a e - b d\right )^{2}} - \frac{2 \left (- A b + B a\right ) \operatorname{atan}{\left (\frac{\sqrt{d + e x}}{\sqrt{\frac{a e - b d}{b}}} \right )}}{\sqrt{\frac{a e - b d}{b}} \left (a e - b d\right )^{2}} + \frac{2 \left (- A e + B d\right )}{3 e \left (d + e x\right )^{\frac{3}{2}} \left (a e - b d\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/(b*x+a)/(e*x+d)**(5/2),x)

[Out]

-2*(-A*b + B*a)/(sqrt(d + e*x)*(a*e - b*d)**2) - 2*(-A*b + B*a)*atan(sqrt(d + e*x)/sqrt((a*e - b*d)/b))/(sqrt(
(a*e - b*d)/b)*(a*e - b*d)**2) + 2*(-A*e + B*d)/(3*e*(d + e*x)**(3/2)*(a*e - b*d))

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Giac [A]  time = 2.07686, size = 217, normalized size = 1.82 \begin{align*} -\frac{2 \,{\left (B a b - A b^{2}\right )} \arctan \left (\frac{\sqrt{x e + d} b}{\sqrt{-b^{2} d + a b e}}\right )}{{\left (b^{2} d^{2} - 2 \, a b d e + a^{2} e^{2}\right )} \sqrt{-b^{2} d + a b e}} - \frac{2 \,{\left (B b d^{2} + 3 \,{\left (x e + d\right )} B a e - 3 \,{\left (x e + d\right )} A b e - B a d e - A b d e + A a e^{2}\right )}}{3 \,{\left (b^{2} d^{2} e - 2 \, a b d e^{2} + a^{2} e^{3}\right )}{\left (x e + d\right )}^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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