3.1771 \(\int \frac{a+b x}{(c+d x) (e+f x)^{5/2}} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.0980947, 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} \[ -\frac{2 (b c-a d)}{\sqrt{e+f x} (d e-c f)^2}-\frac{2 (b e-a f)}{3 f (e+f x)^{3/2} (d e-c f)}+\frac{2 \sqrt{d} (b c-a d) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{e+f x}}{\sqrt{d e-c f}}\right )}{(d e-c f)^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(b*e - a*f))/(3*f*(d*e - c*f)*(e + f*x)^(3/2)) - (2*(b*c - a*d))/((d*e - c*f)^2*Sqrt[e + f*x]) + (2*Sqrt[d
]*(b*c - a*d)*ArcTanh[(Sqrt[d]*Sqrt[e + f*x])/Sqrt[d*e - c*f]])/(d*e - c*f)^(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}{(c+d x) (e+f x)^{5/2}} \, dx &=-\frac{2 (b e-a f)}{3 f (d e-c f) (e+f x)^{3/2}}-\frac{(b c-a d) \int \frac{1}{(c+d x) (e+f x)^{3/2}} \, dx}{d e-c f}\\ &=-\frac{2 (b e-a f)}{3 f (d e-c f) (e+f x)^{3/2}}-\frac{2 (b c-a d)}{(d e-c f)^2 \sqrt{e+f x}}-\frac{(d (b c-a d)) \int \frac{1}{(c+d x) \sqrt{e+f x}} \, dx}{(d e-c f)^2}\\ &=-\frac{2 (b e-a f)}{3 f (d e-c f) (e+f x)^{3/2}}-\frac{2 (b c-a d)}{(d e-c f)^2 \sqrt{e+f x}}-\frac{(2 d (b c-a d)) \operatorname{Subst}\left (\int \frac{1}{c-\frac{d e}{f}+\frac{d x^2}{f}} \, dx,x,\sqrt{e+f x}\right )}{f (d e-c f)^2}\\ &=-\frac{2 (b e-a f)}{3 f (d e-c f) (e+f x)^{3/2}}-\frac{2 (b c-a d)}{(d e-c f)^2 \sqrt{e+f x}}+\frac{2 \sqrt{d} (b c-a d) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{e+f x}}{\sqrt{d e-c f}}\right )}{(d e-c f)^{5/2}}\\ \end{align*}

Mathematica [C]  time = 0.0383674, size = 86, normalized size = 0.72 \[ \frac{-6 f (e+f x) (b c-a d) \, _2F_1\left (-\frac{1}{2},1;\frac{1}{2};\frac{d (e+f x)}{d e-c f}\right )-2 (b e-a f) (d e-c f)}{3 f (e+f x)^{3/2} (d e-c f)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.011, size = 187, normalized size = 1.6 \begin{align*} -{\frac{2\,a}{3\,cf-3\,de} \left ( fx+e \right ) ^{-{\frac{3}{2}}}}+{\frac{2\,be}{3\, \left ( cf-de \right ) f} \left ( fx+e \right ) ^{-{\frac{3}{2}}}}+2\,{\frac{ad}{ \left ( cf-de \right ) ^{2}\sqrt{fx+e}}}-2\,{\frac{bc}{ \left ( cf-de \right ) ^{2}\sqrt{fx+e}}}+2\,{\frac{a{d}^{2}}{ \left ( cf-de \right ) ^{2}\sqrt{ \left ( cf-de \right ) d}}\arctan \left ({\frac{\sqrt{fx+e}d}{\sqrt{ \left ( cf-de \right ) d}}} \right ) }-2\,{\frac{bdc}{ \left ( cf-de \right ) ^{2}\sqrt{ \left ( cf-de \right ) d}}\arctan \left ({\frac{\sqrt{fx+e}d}{\sqrt{ \left ( cf-de \right ) d}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)/(d*x+c)/(f*x+e)^(5/2),x)

[Out]

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

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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)/(d*x+c)/(f*x+e)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.48643, size = 1057, normalized size = 8.88 \begin{align*} \left [-\frac{3 \,{\left ({\left (b c - a d\right )} f^{3} x^{2} + 2 \,{\left (b c - a d\right )} e f^{2} x +{\left (b c - a d\right )} e^{2} f\right )} \sqrt{\frac{d}{d e - c f}} \log \left (\frac{d f x + 2 \, d e - c f - 2 \,{\left (d e - c f\right )} \sqrt{f x + e} \sqrt{\frac{d}{d e - c f}}}{d x + c}\right ) + 2 \,{\left (b d e^{2} + a c f^{2} + 3 \,{\left (b c - a d\right )} f^{2} x + 2 \,{\left (b c - 2 \, a d\right )} e f\right )} \sqrt{f x + e}}{3 \,{\left (d^{2} e^{4} f - 2 \, c d e^{3} f^{2} + c^{2} e^{2} f^{3} +{\left (d^{2} e^{2} f^{3} - 2 \, c d e f^{4} + c^{2} f^{5}\right )} x^{2} + 2 \,{\left (d^{2} e^{3} f^{2} - 2 \, c d e^{2} f^{3} + c^{2} e f^{4}\right )} x\right )}}, \frac{2 \,{\left (3 \,{\left ({\left (b c - a d\right )} f^{3} x^{2} + 2 \,{\left (b c - a d\right )} e f^{2} x +{\left (b c - a d\right )} e^{2} f\right )} \sqrt{-\frac{d}{d e - c f}} \arctan \left (-\frac{{\left (d e - c f\right )} \sqrt{f x + e} \sqrt{-\frac{d}{d e - c f}}}{d f x + d e}\right ) -{\left (b d e^{2} + a c f^{2} + 3 \,{\left (b c - a d\right )} f^{2} x + 2 \,{\left (b c - 2 \, a d\right )} e f\right )} \sqrt{f x + e}\right )}}{3 \,{\left (d^{2} e^{4} f - 2 \, c d e^{3} f^{2} + c^{2} e^{2} f^{3} +{\left (d^{2} e^{2} f^{3} - 2 \, c d e f^{4} + c^{2} f^{5}\right )} x^{2} + 2 \,{\left (d^{2} e^{3} f^{2} - 2 \, c d e^{2} f^{3} + c^{2} e f^{4}\right )} x\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 21.3697, size = 105, normalized size = 0.88 \begin{align*} \frac{2 \left (a d - b c\right )}{\sqrt{e + f x} \left (c f - d e\right )^{2}} + \frac{2 \left (a d - b c\right ) \operatorname{atan}{\left (\frac{\sqrt{e + f x}}{\sqrt{\frac{c f - d e}{d}}} \right )}}{\sqrt{\frac{c f - d e}{d}} \left (c f - d e\right )^{2}} - \frac{2 \left (a f - b e\right )}{3 f \left (e + f x\right )^{\frac{3}{2}} \left (c f - d e\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)/(d*x+c)/(f*x+e)**(5/2),x)

[Out]

2*(a*d - b*c)/(sqrt(e + f*x)*(c*f - d*e)**2) + 2*(a*d - b*c)*atan(sqrt(e + f*x)/sqrt((c*f - d*e)/d))/(sqrt((c*
f - d*e)/d)*(c*f - d*e)**2) - 2*(a*f - b*e)/(3*f*(e + f*x)**(3/2)*(c*f - d*e))

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Giac [A]  time = 2.35905, size = 216, normalized size = 1.82 \begin{align*} -\frac{2 \,{\left (b c d - a d^{2}\right )} \arctan \left (\frac{\sqrt{f x + e} d}{\sqrt{c d f - d^{2} e}}\right )}{{\left (c^{2} f^{2} - 2 \, c d f e + d^{2} e^{2}\right )} \sqrt{c d f - d^{2} e}} - \frac{2 \,{\left (3 \,{\left (f x + e\right )} b c f - 3 \,{\left (f x + e\right )} a d f + a c f^{2} - b c f e - a d f e + b d e^{2}\right )}}{3 \,{\left (c^{2} f^{3} - 2 \, c d f^{2} e + d^{2} f e^{2}\right )}{\left (f x + e\right )}^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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