3.1772 \(\int \frac{a+b x}{(c+d x) (e+f x)^{7/2}} \, dx\)

Optimal. Leaf size=151 \[ \frac{2 d^{3/2} (b c-a d) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{e+f x}}{\sqrt{d e-c f}}\right )}{(d e-c f)^{7/2}}-\frac{2 d (b c-a d)}{\sqrt{e+f x} (d e-c f)^3}-\frac{2 (b c-a d)}{3 (e+f x)^{3/2} (d e-c f)^2}-\frac{2 (b e-a f)}{5 f (e+f x)^{5/2} (d e-c f)} \]

[Out]

(-2*(b*e - a*f))/(5*f*(d*e - c*f)*(e + f*x)^(5/2)) - (2*(b*c - a*d))/(3*(d*e - c*f)^2*(e + f*x)^(3/2)) - (2*d*
(b*c - a*d))/((d*e - c*f)^3*Sqrt[e + f*x]) + (2*d^(3/2)*(b*c - a*d)*ArcTanh[(Sqrt[d]*Sqrt[e + f*x])/Sqrt[d*e -
c*f]])/(d*e - c*f)^(7/2)

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Rubi [A]  time = 0.139259, antiderivative size = 151, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {78, 51, 63, 208} \[ \frac{2 d^{3/2} (b c-a d) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{e+f x}}{\sqrt{d e-c f}}\right )}{(d e-c f)^{7/2}}-\frac{2 d (b c-a d)}{\sqrt{e+f x} (d e-c f)^3}-\frac{2 (b c-a d)}{3 (e+f x)^{3/2} (d e-c f)^2}-\frac{2 (b e-a f)}{5 f (e+f x)^{5/2} (d e-c f)} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x)/((c + d*x)*(e + f*x)^(7/2)),x]

[Out]

(-2*(b*e - a*f))/(5*f*(d*e - c*f)*(e + f*x)^(5/2)) - (2*(b*c - a*d))/(3*(d*e - c*f)^2*(e + f*x)^(3/2)) - (2*d*
(b*c - a*d))/((d*e - c*f)^3*Sqrt[e + f*x]) + (2*d^(3/2)*(b*c - a*d)*ArcTanh[(Sqrt[d]*Sqrt[e + f*x])/Sqrt[d*e -
c*f]])/(d*e - c*f)^(7/2)

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)^{7/2}} \, dx &=-\frac{2 (b e-a f)}{5 f (d e-c f) (e+f x)^{5/2}}-\frac{(b c-a d) \int \frac{1}{(c+d x) (e+f x)^{5/2}} \, dx}{d e-c f}\\ &=-\frac{2 (b e-a f)}{5 f (d e-c f) (e+f x)^{5/2}}-\frac{2 (b c-a d)}{3 (d e-c f)^2 (e+f x)^{3/2}}-\frac{(d (b c-a d)) \int \frac{1}{(c+d x) (e+f x)^{3/2}} \, dx}{(d e-c f)^2}\\ &=-\frac{2 (b e-a f)}{5 f (d e-c f) (e+f x)^{5/2}}-\frac{2 (b c-a d)}{3 (d e-c f)^2 (e+f x)^{3/2}}-\frac{2 d (b c-a d)}{(d e-c f)^3 \sqrt{e+f x}}-\frac{\left (d^2 (b c-a d)\right ) \int \frac{1}{(c+d x) \sqrt{e+f x}} \, dx}{(d e-c f)^3}\\ &=-\frac{2 (b e-a f)}{5 f (d e-c f) (e+f x)^{5/2}}-\frac{2 (b c-a d)}{3 (d e-c f)^2 (e+f x)^{3/2}}-\frac{2 d (b c-a d)}{(d e-c f)^3 \sqrt{e+f x}}-\frac{\left (2 d^2 (b c-a d)\right ) \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)^3}\\ &=-\frac{2 (b e-a f)}{5 f (d e-c f) (e+f x)^{5/2}}-\frac{2 (b c-a d)}{3 (d e-c f)^2 (e+f x)^{3/2}}-\frac{2 d (b c-a d)}{(d e-c f)^3 \sqrt{e+f x}}+\frac{2 d^{3/2} (b c-a d) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{e+f x}}{\sqrt{d e-c f}}\right )}{(d e-c f)^{7/2}}\\ \end{align*}

Mathematica [C]  time = 0.0385567, size = 86, normalized size = 0.57 \[ -\frac{2 \left (5 f (e+f x) (b c-a d) \, _2F_1\left (-\frac{3}{2},1;-\frac{1}{2};\frac{d (e+f x)}{d e-c f}\right )+3 (b e-a f) (d e-c f)\right )}{15 f (e+f x)^{5/2} (d e-c f)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)/((c + d*x)*(e + f*x)^(7/2)),x]

[Out]

(-2*(3*(b*e - a*f)*(d*e - c*f) + 5*(b*c - a*d)*f*(e + f*x)*Hypergeometric2F1[-3/2, 1, -1/2, (d*(e + f*x))/(d*e
- c*f)]))/(15*f*(d*e - c*f)^2*(e + f*x)^(5/2))

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Maple [A]  time = 0.014, size = 234, normalized size = 1.6 \begin{align*} -{\frac{2\,a}{5\,cf-5\,de} \left ( fx+e \right ) ^{-{\frac{5}{2}}}}+{\frac{2\,be}{5\, \left ( cf-de \right ) f} \left ( fx+e \right ) ^{-{\frac{5}{2}}}}-2\,{\frac{a{d}^{2}}{ \left ( cf-de \right ) ^{3}\sqrt{fx+e}}}+2\,{\frac{bdc}{ \left ( cf-de \right ) ^{3}\sqrt{fx+e}}}+{\frac{2\,ad}{3\, \left ( cf-de \right ) ^{2}} \left ( fx+e \right ) ^{-{\frac{3}{2}}}}-{\frac{2\,bc}{3\, \left ( cf-de \right ) ^{2}} \left ( fx+e \right ) ^{-{\frac{3}{2}}}}-2\,{\frac{{d}^{3}a}{ \left ( cf-de \right ) ^{3}\sqrt{ \left ( cf-de \right ) d}}\arctan \left ({\frac{\sqrt{fx+e}d}{\sqrt{ \left ( cf-de \right ) d}}} \right ) }+2\,{\frac{{d}^{2}bc}{ \left ( cf-de \right ) ^{3}\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)^(7/2),x)

[Out]

-2/5/(c*f-d*e)/(f*x+e)^(5/2)*a+2/5/f/(c*f-d*e)/(f*x+e)^(5/2)*b*e-2/(c*f-d*e)^3*d^2/(f*x+e)^(1/2)*a+2/(c*f-d*e)
^3*d/(f*x+e)^(1/2)*b*c+2/3/(c*f-d*e)^2/(f*x+e)^(3/2)*a*d-2/3/(c*f-d*e)^2/(f*x+e)^(3/2)*b*c-2*d^3/(c*f-d*e)^3/(
(c*f-d*e)*d)^(1/2)*arctan((f*x+e)^(1/2)*d/((c*f-d*e)*d)^(1/2))*a+2*d^2/(c*f-d*e)^3/((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)^(7/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.46409, size = 1823, normalized size = 12.07 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)/(d*x+c)/(f*x+e)^(7/2),x, algorithm="fricas")

[Out]

[1/15*(15*((b*c*d - a*d^2)*f^4*x^3 + 3*(b*c*d - a*d^2)*e*f^3*x^2 + 3*(b*c*d - a*d^2)*e^2*f^2*x + (b*c*d - a*d^
2)*e^3*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*(3*b*d^2*e^3 - 3*a*c^2*f^3 + 15*(b*c*d - a*d^2)*f^3*x^2 + (14*b*c*d - 23*a*d^2)*e^2*f - (2*b*c^2 -
11*a*c*d)*e*f^2 + 5*(7*(b*c*d - a*d^2)*e*f^2 - (b*c^2 - a*c*d)*f^3)*x)*sqrt(f*x + e))/(d^3*e^6*f - 3*c*d^2*e^5
*f^2 + 3*c^2*d*e^4*f^3 - c^3*e^3*f^4 + (d^3*e^3*f^4 - 3*c*d^2*e^2*f^5 + 3*c^2*d*e*f^6 - c^3*f^7)*x^3 + 3*(d^3*
e^4*f^3 - 3*c*d^2*e^3*f^4 + 3*c^2*d*e^2*f^5 - c^3*e*f^6)*x^2 + 3*(d^3*e^5*f^2 - 3*c*d^2*e^4*f^3 + 3*c^2*d*e^3*
f^4 - c^3*e^2*f^5)*x), 2/15*(15*((b*c*d - a*d^2)*f^4*x^3 + 3*(b*c*d - a*d^2)*e*f^3*x^2 + 3*(b*c*d - a*d^2)*e^2
*f^2*x + (b*c*d - a*d^2)*e^3*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)) - (3*b*d^2*e^3 - 3*a*c^2*f^3 + 15*(b*c*d - a*d^2)*f^3*x^2 + (14*b*c*d - 23*a*d^2)*e^2*f - (2*b*c^
2 - 11*a*c*d)*e*f^2 + 5*(7*(b*c*d - a*d^2)*e*f^2 - (b*c^2 - a*c*d)*f^3)*x)*sqrt(f*x + e))/(d^3*e^6*f - 3*c*d^2
*e^5*f^2 + 3*c^2*d*e^4*f^3 - c^3*e^3*f^4 + (d^3*e^3*f^4 - 3*c*d^2*e^2*f^5 + 3*c^2*d*e*f^6 - c^3*f^7)*x^3 + 3*(
d^3*e^4*f^3 - 3*c*d^2*e^3*f^4 + 3*c^2*d*e^2*f^5 - c^3*e*f^6)*x^2 + 3*(d^3*e^5*f^2 - 3*c*d^2*e^4*f^3 + 3*c^2*d*
e^3*f^4 - c^3*e^2*f^5)*x)]

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Sympy [A]  time = 28.7082, size = 136, normalized size = 0.9 \begin{align*} - \frac{2 d \left (a d - b c\right )}{\sqrt{e + f x} \left (c f - d e\right )^{3}} - \frac{2 d \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 )^{3}} + \frac{2 \left (a d - b c\right )}{3 \left (e + f x\right )^{\frac{3}{2}} \left (c f - d e\right )^{2}} - \frac{2 \left (a f - b e\right )}{5 f \left (e + f x\right )^{\frac{5}{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)**(7/2),x)

[Out]

-2*d*(a*d - b*c)/(sqrt(e + f*x)*(c*f - d*e)**3) - 2*d*(a*d - b*c)*atan(sqrt(e + f*x)/sqrt((c*f - d*e)/d))/(sqr
t((c*f - d*e)/d)*(c*f - d*e)**3) + 2*(a*d - b*c)/(3*(e + f*x)**(3/2)*(c*f - d*e)**2) - 2*(a*f - b*e)/(5*f*(e +
f*x)**(5/2)*(c*f - d*e))

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Giac [B]  time = 2.44709, size = 385, normalized size = 2.55 \begin{align*} \frac{2 \,{\left (b c d^{2} - a d^{3}\right )} \arctan \left (\frac{\sqrt{f x + e} d}{\sqrt{c d f - d^{2} e}}\right )}{{\left (c^{3} f^{3} - 3 \, c^{2} d f^{2} e + 3 \, c d^{2} f e^{2} - d^{3} e^{3}\right )} \sqrt{c d f - d^{2} e}} + \frac{2 \,{\left (15 \,{\left (f x + e\right )}^{2} b c d f - 15 \,{\left (f x + e\right )}^{2} a d^{2} f - 5 \,{\left (f x + e\right )} b c^{2} f^{2} + 5 \,{\left (f x + e\right )} a c d f^{2} - 3 \, a c^{2} f^{3} + 5 \,{\left (f x + e\right )} b c d f e - 5 \,{\left (f x + e\right )} a d^{2} f e + 3 \, b c^{2} f^{2} e + 6 \, a c d f^{2} e - 6 \, b c d f e^{2} - 3 \, a d^{2} f e^{2} + 3 \, b d^{2} e^{3}\right )}}{15 \,{\left (c^{3} f^{4} - 3 \, c^{2} d f^{3} e + 3 \, c d^{2} f^{2} e^{2} - d^{3} f e^{3}\right )}{\left (f x + e\right )}^{\frac{5}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)/(d*x+c)/(f*x+e)^(7/2),x, algorithm="giac")

[Out]

2*(b*c*d^2 - a*d^3)*arctan(sqrt(f*x + e)*d/sqrt(c*d*f - d^2*e))/((c^3*f^3 - 3*c^2*d*f^2*e + 3*c*d^2*f*e^2 - d^
3*e^3)*sqrt(c*d*f - d^2*e)) + 2/15*(15*(f*x + e)^2*b*c*d*f - 15*(f*x + e)^2*a*d^2*f - 5*(f*x + e)*b*c^2*f^2 +
5*(f*x + e)*a*c*d*f^2 - 3*a*c^2*f^3 + 5*(f*x + e)*b*c*d*f*e - 5*(f*x + e)*a*d^2*f*e + 3*b*c^2*f^2*e + 6*a*c*d*
f^2*e - 6*b*c*d*f*e^2 - 3*a*d^2*f*e^2 + 3*b*d^2*e^3)/((c^3*f^4 - 3*c^2*d*f^3*e + 3*c*d^2*f^2*e^2 - d^3*f*e^3)*
(f*x + e)^(5/2))