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

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

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Rubi [A]  time = 0.14, antiderivative size = 151, normalized size of antiderivative = 1.00, 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} \begin {gather*} \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)} \end {gather*}

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 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 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 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*}

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Mathematica [C]  time = 0.06, size = 86, normalized size = 0.57 \begin {gather*} -\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} \end {gather*}

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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IntegrateAlgebraic [A]  time = 0.27, size = 232, normalized size = 1.54 \begin {gather*} \frac {2 \left (a d^{5/2}-b c d^{3/2}\right ) \tan ^{-1}\left (\frac {\sqrt {d} \sqrt {e+f x} \sqrt {c f-d e}}{d e-c f}\right )}{(c f-d e)^{7/2}}-\frac {2 \left (3 a c^2 f^3-5 a c d f^2 (e+f x)-6 a c d e f^2+3 a d^2 e^2 f+5 a d^2 e f (e+f x)+15 a d^2 f (e+f x)^2+5 b c^2 f^2 (e+f x)-3 b c^2 e f^2+6 b c d e^2 f-5 b c d e f (e+f x)-15 b c d f (e+f x)^2-3 b d^2 e^3\right )}{15 f (e+f x)^{5/2} (c f-d e)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(-3*b*d^2*e^3 + 6*b*c*d*e^2*f + 3*a*d^2*e^2*f - 3*b*c^2*e*f^2 - 6*a*c*d*e*f^2 + 3*a*c^2*f^3 - 5*b*c*d*e*f*
(e + f*x) + 5*a*d^2*e*f*(e + f*x) + 5*b*c^2*f^2*(e + f*x) - 5*a*c*d*f^2*(e + f*x) - 15*b*c*d*f*(e + f*x)^2 + 1
5*a*d^2*f*(e + f*x)^2))/(15*f*(-(d*e) + c*f)^3*(e + f*x)^(5/2)) + (2*(-(b*c*d^(3/2)) + a*d^(5/2))*ArcTan[(Sqrt
[d]*Sqrt[-(d*e) + c*f]*Sqrt[e + f*x])/(d*e - c*f)])/(-(d*e) + c*f)^(7/2)

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fricas [B]  time = 1.45, size = 902, normalized size = 5.97 \begin {gather*} \left [\frac {15 \, {\left ({\left (b c d - a d^{2}\right )} f^{4} x^{3} + 3 \, {\left (b c d - a d^{2}\right )} e f^{3} x^{2} + 3 \, {\left (b c d - a d^{2}\right )} e^{2} f^{2} x + {\left (b c d - a d^{2}\right )} e^{3} 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 (3 \, b d^{2} e^{3} - 3 \, a c^{2} f^{3} + 15 \, {\left (b c d - a d^{2}\right )} f^{3} x^{2} + {\left (14 \, b c d - 23 \, a d^{2}\right )} e^{2} f - {\left (2 \, b c^{2} - 11 \, a c d\right )} e f^{2} + 5 \, {\left (7 \, {\left (b c d - a d^{2}\right )} e f^{2} - {\left (b c^{2} - a c d\right )} f^{3}\right )} x\right )} \sqrt {f x + e}}{15 \, {\left (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} + {\left (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}\right )} x^{3} + 3 \, {\left (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}\right )} x^{2} + 3 \, {\left (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}\right )} x\right )}}, \frac {2 \, {\left (15 \, {\left ({\left (b c d - a d^{2}\right )} f^{4} x^{3} + 3 \, {\left (b c d - a d^{2}\right )} e f^{3} x^{2} + 3 \, {\left (b c d - a d^{2}\right )} e^{2} f^{2} x + {\left (b c d - a d^{2}\right )} e^{3} 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 (3 \, b d^{2} e^{3} - 3 \, a c^{2} f^{3} + 15 \, {\left (b c d - a d^{2}\right )} f^{3} x^{2} + {\left (14 \, b c d - 23 \, a d^{2}\right )} e^{2} f - {\left (2 \, b c^{2} - 11 \, a c d\right )} e f^{2} + 5 \, {\left (7 \, {\left (b c d - a d^{2}\right )} e f^{2} - {\left (b c^{2} - a c d\right )} f^{3}\right )} x\right )} \sqrt {f x + e}\right )}}{15 \, {\left (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} + {\left (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}\right )} x^{3} + 3 \, {\left (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}\right )} x^{2} + 3 \, {\left (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}\right )} x\right )}}\right ] \end {gather*}

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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giac [B]  time = 1.27, size = 285, normalized size = 1.89 \begin {gather*} \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 {gather*}

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))

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maple [A]  time = 0.02, size = 234, normalized size = 1.55 \begin {gather*} -\frac {2 a \,d^{3} \arctan \left (\frac {\sqrt {f x +e}\, d}{\sqrt {\left (c f -d e \right ) d}}\right )}{\left (c f -d e \right )^{3} \sqrt {\left (c f -d e \right ) d}}+\frac {2 b c \,d^{2} \arctan \left (\frac {\sqrt {f x +e}\, d}{\sqrt {\left (c f -d e \right ) d}}\right )}{\left (c f -d e \right )^{3} \sqrt {\left (c f -d e \right ) d}}-\frac {2 a \,d^{2}}{\left (c f -d e \right )^{3} \sqrt {f x +e}}+\frac {2 b c d}{\left (c f -d e \right )^{3} \sqrt {f x +e}}+\frac {2 a d}{3 \left (c f -d e \right )^{2} \left (f x +e \right )^{\frac {3}{2}}}-\frac {2 b c}{3 \left (c f -d e \right )^{2} \left (f x +e \right )^{\frac {3}{2}}}-\frac {2 a}{5 \left (c f -d e \right ) \left (f x +e \right )^{\frac {5}{2}}}+\frac {2 b e}{5 \left (c f -d e \right ) \left (f x +e \right )^{\frac {5}{2}} f} \end {gather*}

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

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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)/(d*x+c)/(f*x+e)^(7/2),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(c*f-d*e>0)', see `assume?` for
more details)Is c*f-d*e positive or negative?

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mupad [B]  time = 1.36, size = 173, normalized size = 1.15 \begin {gather*} -\frac {\frac {2\,\left (a\,f-b\,e\right )}{5\,\left (c\,f-d\,e\right )}-\frac {2\,\left (e+f\,x\right )\,\left (a\,d\,f-b\,c\,f\right )}{3\,{\left (c\,f-d\,e\right )}^2}+\frac {2\,d\,{\left (e+f\,x\right )}^2\,\left (a\,d\,f-b\,c\,f\right )}{{\left (c\,f-d\,e\right )}^3}}{f\,{\left (e+f\,x\right )}^{5/2}}-\frac {2\,d^{3/2}\,\mathrm {atan}\left (\frac {\sqrt {d}\,\sqrt {e+f\,x}\,\left (c^3\,f^3-3\,c^2\,d\,e\,f^2+3\,c\,d^2\,e^2\,f-d^3\,e^3\right )}{{\left (c\,f-d\,e\right )}^{7/2}}\right )\,\left (a\,d-b\,c\right )}{{\left (c\,f-d\,e\right )}^{7/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- ((2*(a*f - b*e))/(5*(c*f - d*e)) - (2*(e + f*x)*(a*d*f - b*c*f))/(3*(c*f - d*e)^2) + (2*d*(e + f*x)^2*(a*d*f
- b*c*f))/(c*f - d*e)^3)/(f*(e + f*x)^(5/2)) - (2*d^(3/2)*atan((d^(1/2)*(e + f*x)^(1/2)*(c^3*f^3 - d^3*e^3 +
3*c*d^2*e^2*f - 3*c^2*d*e*f^2))/(c*f - d*e)^(7/2))*(a*d - b*c))/(c*f - d*e)^(7/2)

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sympy [A]  time = 49.92, size = 136, normalized size = 0.90 \begin {gather*} - \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 {gather*}

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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