3.18.80 \(\int \frac {(a+b x)^2}{(c+d x) (e+f x)^{7/2}} \, dx\) [1780]

Optimal. Leaf size=173 \[ \frac {2 (b e-a f)^2}{5 f^2 (d e-c f) (e+f x)^{5/2}}-\frac {2 (b e-a f) (b d e-2 b c f+a d f)}{3 f^2 (d e-c f)^2 (e+f x)^{3/2}}+\frac {2 (b c-a d)^2}{(d e-c f)^3 \sqrt {e+f x}}-\frac {2 \sqrt {d} (b c-a d)^2 \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {e+f x}}{\sqrt {d e-c f}}\right )}{(d e-c f)^{7/2}} \]

[Out]

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

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Rubi [A]
time = 0.13, antiderivative size = 173, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {89, 65, 214} \begin {gather*} -\frac {2 (b e-a f) (a d f-2 b c f+b d e)}{3 f^2 (e+f x)^{3/2} (d e-c f)^2}+\frac {2 (b e-a f)^2}{5 f^2 (e+f x)^{5/2} (d e-c f)}+\frac {2 (b c-a d)^2}{\sqrt {e+f x} (d e-c f)^3}-\frac {2 \sqrt {d} (b c-a d)^2 \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {e+f x}}{\sqrt {d e-c f}}\right )}{(d e-c f)^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 65

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 89

Int[(((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_))/((a_.) + (b_.)*(x_)), x_Symbol] :> Int[ExpandIntegr
and[(e + f*x)^FractionalPart[p], (c + d*x)^n*((e + f*x)^IntegerPart[p]/(a + b*x)), x], x] /; FreeQ[{a, b, c, d
, e, f}, x] && IGtQ[n, 0] && LtQ[p, -1] && FractionQ[p]

Rule 214

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

Rubi steps

\begin {align*} \int \frac {(a+b x)^2}{(c+d x) (e+f x)^{7/2}} \, dx &=\int \left (\frac {(-b e+a f)^2}{f (-d e+c f) (e+f x)^{7/2}}+\frac {(-b e+a f) (-b d e+2 b c f-a d f)}{f (-d e+c f)^2 (e+f x)^{5/2}}+\frac {(b c-a d)^2 f}{(-d e+c f)^3 (e+f x)^{3/2}}+\frac {d (-b c+a d)^2}{(d e-c f)^3 (c+d x) \sqrt {e+f x}}\right ) \, dx\\ &=\frac {2 (b e-a f)^2}{5 f^2 (d e-c f) (e+f x)^{5/2}}-\frac {2 (b e-a f) (b d e-2 b c f+a d f)}{3 f^2 (d e-c f)^2 (e+f x)^{3/2}}+\frac {2 (b c-a d)^2}{(d e-c f)^3 \sqrt {e+f x}}+\frac {\left (d (b c-a d)^2\right ) \int \frac {1}{(c+d x) \sqrt {e+f x}} \, dx}{(d e-c f)^3}\\ &=\frac {2 (b e-a f)^2}{5 f^2 (d e-c f) (e+f x)^{5/2}}-\frac {2 (b e-a f) (b d e-2 b c f+a d f)}{3 f^2 (d e-c f)^2 (e+f x)^{3/2}}+\frac {2 (b c-a d)^2}{(d e-c f)^3 \sqrt {e+f x}}+\frac {\left (2 d (b c-a d)^2\right ) \text {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)^2}{5 f^2 (d e-c f) (e+f x)^{5/2}}-\frac {2 (b e-a f) (b d e-2 b c f+a d f)}{3 f^2 (d e-c f)^2 (e+f x)^{3/2}}+\frac {2 (b c-a d)^2}{(d e-c f)^3 \sqrt {e+f x}}-\frac {2 \sqrt {d} (b c-a d)^2 \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 [A]
time = 0.46, size = 253, normalized size = 1.46 \begin {gather*} -\frac {2 \left (b^2 \left (-d^2 e^3 (2 e+5 f x)+3 c d e^2 f (3 e+5 f x)+c^2 f^2 \left (8 e^2+20 e f x+15 f^2 x^2\right )\right )-2 a b f \left (3 d^2 e^3-c^2 f^2 (2 e+5 f x)+c d f \left (14 e^2+35 e f x+15 f^2 x^2\right )\right )+a^2 f^2 \left (3 c^2 f^2-c d f (11 e+5 f x)+d^2 \left (23 e^2+35 e f x+15 f^2 x^2\right )\right )\right )}{15 f^2 (-d e+c f)^3 (e+f x)^{5/2}}-\frac {2 \sqrt {d} (b c-a d)^2 \tan ^{-1}\left (\frac {\sqrt {d} \sqrt {e+f x}}{\sqrt {-d e+c f}}\right )}{(-d e+c f)^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(b^2*(-(d^2*e^3*(2*e + 5*f*x)) + 3*c*d*e^2*f*(3*e + 5*f*x) + c^2*f^2*(8*e^2 + 20*e*f*x + 15*f^2*x^2)) - 2*
a*b*f*(3*d^2*e^3 - c^2*f^2*(2*e + 5*f*x) + c*d*f*(14*e^2 + 35*e*f*x + 15*f^2*x^2)) + a^2*f^2*(3*c^2*f^2 - c*d*
f*(11*e + 5*f*x) + d^2*(23*e^2 + 35*e*f*x + 15*f^2*x^2))))/(15*f^2*(-(d*e) + c*f)^3*(e + f*x)^(5/2)) - (2*Sqrt
[d]*(b*c - a*d)^2*ArcTan[(Sqrt[d]*Sqrt[e + f*x])/Sqrt[-(d*e) + c*f]])/(-(d*e) + c*f)^(7/2)

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Maple [A]
time = 0.11, size = 214, normalized size = 1.24

method	result	size
derivativedivides	\(\frac {-\frac {2 d \,f^{2} \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) \arctan \left (\frac {d \sqrt {f x +e}}{\sqrt {\left (c f -d e \right ) d}}\right )}{\left (c f -d e \right )^{3} \sqrt {\left (c f -d e \right ) d}}-\frac {2 \left (a^{2} f^{2}-2 a b f e +b^{2} e^{2}\right )}{5 \left (c f -d e \right ) \left (f x +e \right )^{\frac {5}{2}}}-\frac {2 \left (-a^{2} d \,f^{2}+2 a b c \,f^{2}-2 b^{2} c e f +b^{2} d \,e^{2}\right )}{3 \left (c f -d e \right )^{2} \left (f x +e \right )^{\frac {3}{2}}}-\frac {2 f^{2} \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right )}{\left (c f -d e \right )^{3} \sqrt {f x +e}}}{f^{2}}\)	\(214\)
default	\(\frac {-\frac {2 d \,f^{2} \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) \arctan \left (\frac {d \sqrt {f x +e}}{\sqrt {\left (c f -d e \right ) d}}\right )}{\left (c f -d e \right )^{3} \sqrt {\left (c f -d e \right ) d}}-\frac {2 \left (a^{2} f^{2}-2 a b f e +b^{2} e^{2}\right )}{5 \left (c f -d e \right ) \left (f x +e \right )^{\frac {5}{2}}}-\frac {2 \left (-a^{2} d \,f^{2}+2 a b c \,f^{2}-2 b^{2} c e f +b^{2} d \,e^{2}\right )}{3 \left (c f -d e \right )^{2} \left (f x +e \right )^{\frac {3}{2}}}-\frac {2 f^{2} \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right )}{\left (c f -d e \right )^{3} \sqrt {f x +e}}}{f^{2}}\)	\(214\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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)^2/(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-%e*d>0)', see `assume?` fo
r more detai

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 576 vs. \(2 (167) = 334\).
time = 1.36, size = 1167, normalized size = 6.75 \begin {gather*} \left [-\frac {15 \, {\left ({\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} f^{5} x^{3} + 3 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} f^{4} x^{2} e + 3 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} f^{3} x e^{2} + {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} f^{2} e^{3}\right )} \sqrt {-\frac {d}{c f - d e}} \log \left (\frac {d f x - c f + 2 \, {\left (c f - d e\right )} \sqrt {f x + e} \sqrt {-\frac {d}{c f - d e}} + 2 \, d e}{d x + c}\right ) + 2 \, {\left (3 \, a^{2} c^{2} f^{4} + 15 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} f^{4} x^{2} + 5 \, {\left (2 \, a b c^{2} - a^{2} c d\right )} f^{4} x - 2 \, b^{2} d^{2} e^{4} - {\left (5 \, b^{2} d^{2} f x - 3 \, {\left (3 \, b^{2} c d - 2 \, a b d^{2}\right )} f\right )} e^{3} + {\left (15 \, b^{2} c d f^{2} x + {\left (8 \, b^{2} c^{2} - 28 \, a b c d + 23 \, a^{2} d^{2}\right )} f^{2}\right )} e^{2} + {\left (5 \, {\left (4 \, b^{2} c^{2} - 14 \, a b c d + 7 \, a^{2} d^{2}\right )} f^{3} x + {\left (4 \, a b c^{2} - 11 \, a^{2} c d\right )} f^{3}\right )} e\right )} \sqrt {f x + e}}{15 \, {\left (c^{3} f^{8} x^{3} - d^{3} f^{2} e^{6} - 3 \, {\left (d^{3} f^{3} x - c d^{2} f^{3}\right )} e^{5} - 3 \, {\left (d^{3} f^{4} x^{2} - 3 \, c d^{2} f^{4} x + c^{2} d f^{4}\right )} e^{4} - {\left (d^{3} f^{5} x^{3} - 9 \, c d^{2} f^{5} x^{2} + 9 \, c^{2} d f^{5} x - c^{3} f^{5}\right )} e^{3} + 3 \, {\left (c d^{2} f^{6} x^{3} - 3 \, c^{2} d f^{6} x^{2} + c^{3} f^{6} x\right )} e^{2} - 3 \, {\left (c^{2} d f^{7} x^{3} - c^{3} f^{7} x^{2}\right )} e\right )}}, -\frac {2 \, {\left (15 \, {\left ({\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} f^{5} x^{3} + 3 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} f^{4} x^{2} e + 3 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} f^{3} x e^{2} + {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} f^{2} e^{3}\right )} \sqrt {\frac {d}{c f - d e}} \arctan \left (-\frac {{\left (c f - d e\right )} \sqrt {f x + e} \sqrt {\frac {d}{c f - d e}}}{d f x + d e}\right ) + {\left (3 \, a^{2} c^{2} f^{4} + 15 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} f^{4} x^{2} + 5 \, {\left (2 \, a b c^{2} - a^{2} c d\right )} f^{4} x - 2 \, b^{2} d^{2} e^{4} - {\left (5 \, b^{2} d^{2} f x - 3 \, {\left (3 \, b^{2} c d - 2 \, a b d^{2}\right )} f\right )} e^{3} + {\left (15 \, b^{2} c d f^{2} x + {\left (8 \, b^{2} c^{2} - 28 \, a b c d + 23 \, a^{2} d^{2}\right )} f^{2}\right )} e^{2} + {\left (5 \, {\left (4 \, b^{2} c^{2} - 14 \, a b c d + 7 \, a^{2} d^{2}\right )} f^{3} x + {\left (4 \, a b c^{2} - 11 \, a^{2} c d\right )} f^{3}\right )} e\right )} \sqrt {f x + e}\right )}}{15 \, {\left (c^{3} f^{8} x^{3} - d^{3} f^{2} e^{6} - 3 \, {\left (d^{3} f^{3} x - c d^{2} f^{3}\right )} e^{5} - 3 \, {\left (d^{3} f^{4} x^{2} - 3 \, c d^{2} f^{4} x + c^{2} d f^{4}\right )} e^{4} - {\left (d^{3} f^{5} x^{3} - 9 \, c d^{2} f^{5} x^{2} + 9 \, c^{2} d f^{5} x - c^{3} f^{5}\right )} e^{3} + 3 \, {\left (c d^{2} f^{6} x^{3} - 3 \, c^{2} d f^{6} x^{2} + c^{3} f^{6} x\right )} e^{2} - 3 \, {\left (c^{2} d f^{7} x^{3} - c^{3} f^{7} x^{2}\right )} e\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/15*(15*((b^2*c^2 - 2*a*b*c*d + a^2*d^2)*f^5*x^3 + 3*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*f^4*x^2*e + 3*(b^2*c^2
- 2*a*b*c*d + a^2*d^2)*f^3*x*e^2 + (b^2*c^2 - 2*a*b*c*d + a^2*d^2)*f^2*e^3)*sqrt(-d/(c*f - d*e))*log((d*f*x -
c*f + 2*(c*f - d*e)*sqrt(f*x + e)*sqrt(-d/(c*f - d*e)) + 2*d*e)/(d*x + c)) + 2*(3*a^2*c^2*f^4 + 15*(b^2*c^2 -
2*a*b*c*d + a^2*d^2)*f^4*x^2 + 5*(2*a*b*c^2 - a^2*c*d)*f^4*x - 2*b^2*d^2*e^4 - (5*b^2*d^2*f*x - 3*(3*b^2*c*d
- 2*a*b*d^2)*f)*e^3 + (15*b^2*c*d*f^2*x + (8*b^2*c^2 - 28*a*b*c*d + 23*a^2*d^2)*f^2)*e^2 + (5*(4*b^2*c^2 - 14*
a*b*c*d + 7*a^2*d^2)*f^3*x + (4*a*b*c^2 - 11*a^2*c*d)*f^3)*e)*sqrt(f*x + e))/(c^3*f^8*x^3 - d^3*f^2*e^6 - 3*(d
^3*f^3*x - c*d^2*f^3)*e^5 - 3*(d^3*f^4*x^2 - 3*c*d^2*f^4*x + c^2*d*f^4)*e^4 - (d^3*f^5*x^3 - 9*c*d^2*f^5*x^2 +
9*c^2*d*f^5*x - c^3*f^5)*e^3 + 3*(c*d^2*f^6*x^3 - 3*c^2*d*f^6*x^2 + c^3*f^6*x)*e^2 - 3*(c^2*d*f^7*x^3 - c^3*f
^7*x^2)*e), -2/15*(15*((b^2*c^2 - 2*a*b*c*d + a^2*d^2)*f^5*x^3 + 3*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*f^4*x^2*e +
3*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*f^3*x*e^2 + (b^2*c^2 - 2*a*b*c*d + a^2*d^2)*f^2*e^3)*sqrt(d/(c*f - d*e))*ar
ctan(-(c*f - d*e)*sqrt(f*x + e)*sqrt(d/(c*f - d*e))/(d*f*x + d*e)) + (3*a^2*c^2*f^4 + 15*(b^2*c^2 - 2*a*b*c*d
+ a^2*d^2)*f^4*x^2 + 5*(2*a*b*c^2 - a^2*c*d)*f^4*x - 2*b^2*d^2*e^4 - (5*b^2*d^2*f*x - 3*(3*b^2*c*d - 2*a*b*d^2
)*f)*e^3 + (15*b^2*c*d*f^2*x + (8*b^2*c^2 - 28*a*b*c*d + 23*a^2*d^2)*f^2)*e^2 + (5*(4*b^2*c^2 - 14*a*b*c*d + 7
*a^2*d^2)*f^3*x + (4*a*b*c^2 - 11*a^2*c*d)*f^3)*e)*sqrt(f*x + e))/(c^3*f^8*x^3 - d^3*f^2*e^6 - 3*(d^3*f^3*x -
c*d^2*f^3)*e^5 - 3*(d^3*f^4*x^2 - 3*c*d^2*f^4*x + c^2*d*f^4)*e^4 - (d^3*f^5*x^3 - 9*c*d^2*f^5*x^2 + 9*c^2*d*f^
5*x - c^3*f^5)*e^3 + 3*(c*d^2*f^6*x^3 - 3*c^2*d*f^6*x^2 + c^3*f^6*x)*e^2 - 3*(c^2*d*f^7*x^3 - c^3*f^7*x^2)*e)]

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Sympy [A]
time = 134.76, size = 156, normalized size = 0.90 \begin {gather*} - \frac {2 \left (a d - b c\right )^{2}}{\sqrt {e + f x} \left (c f - d e\right )^{3}} - \frac {2 \left (a d - b c\right )^{2} \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 f - b e\right ) \left (a d f - 2 b c f + b d e\right )}{3 f^{2} \left (e + f x\right )^{\frac {3}{2}} \left (c f - d e\right )^{2}} - \frac {2 \left (a f - b e\right )^{2}}{5 f^{2} \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)**2/(d*x+c)/(f*x+e)**(7/2),x)

[Out]

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

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 432 vs. \(2 (167) = 334\).
time = 0.60, size = 432, normalized size = 2.50 \begin {gather*} -\frac {2 \, {\left (b^{2} c^{2} d - 2 \, a b c d^{2} + a^{2} 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^{2} c^{2} f^{2} - 30 \, {\left (f x + e\right )}^{2} a b c d f^{2} + 15 \, {\left (f x + e\right )}^{2} a^{2} d^{2} f^{2} + 10 \, {\left (f x + e\right )} a b c^{2} f^{3} - 5 \, {\left (f x + e\right )} a^{2} c d f^{3} + 3 \, a^{2} c^{2} f^{4} - 10 \, {\left (f x + e\right )} b^{2} c^{2} f^{2} e - 10 \, {\left (f x + e\right )} a b c d f^{2} e + 5 \, {\left (f x + e\right )} a^{2} d^{2} f^{2} e - 6 \, a b c^{2} f^{3} e - 6 \, a^{2} c d f^{3} e + 15 \, {\left (f x + e\right )} b^{2} c d f e^{2} + 3 \, b^{2} c^{2} f^{2} e^{2} + 12 \, a b c d f^{2} e^{2} + 3 \, a^{2} d^{2} f^{2} e^{2} - 5 \, {\left (f x + e\right )} b^{2} d^{2} e^{3} - 6 \, b^{2} c d f e^{3} - 6 \, a b d^{2} f e^{3} + 3 \, b^{2} d^{2} e^{4}\right )}}{15 \, {\left (c^{3} f^{5} - 3 \, c^{2} d f^{4} e + 3 \, c d^{2} f^{3} e^{2} - d^{3} f^{2} 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)^2/(d*x+c)/(f*x+e)^(7/2),x, algorithm="giac")

[Out]

-2*(b^2*c^2*d - 2*a*b*c*d^2 + a^2*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^2*c^2*f^2 - 30*(f*x + e)^2*a*b*c*d*f^2
+ 15*(f*x + e)^2*a^2*d^2*f^2 + 10*(f*x + e)*a*b*c^2*f^3 - 5*(f*x + e)*a^2*c*d*f^3 + 3*a^2*c^2*f^4 - 10*(f*x +
e)*b^2*c^2*f^2*e - 10*(f*x + e)*a*b*c*d*f^2*e + 5*(f*x + e)*a^2*d^2*f^2*e - 6*a*b*c^2*f^3*e - 6*a^2*c*d*f^3*e
+ 15*(f*x + e)*b^2*c*d*f*e^2 + 3*b^2*c^2*f^2*e^2 + 12*a*b*c*d*f^2*e^2 + 3*a^2*d^2*f^2*e^2 - 5*(f*x + e)*b^2*d
^2*e^3 - 6*b^2*c*d*f*e^3 - 6*a*b*d^2*f*e^3 + 3*b^2*d^2*e^4)/((c^3*f^5 - 3*c^2*d*f^4*e + 3*c*d^2*f^3*e^2 - d^3*
f^2*e^3)*(f*x + e)^(5/2))

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Mupad [B]
time = 1.41, size = 264, normalized size = 1.53 \begin {gather*} -\frac {\frac {2\,\left (a^2\,f^2-2\,a\,b\,e\,f+b^2\,e^2\right )}{5\,\left (c\,f-d\,e\right )}+\frac {2\,{\left (e+f\,x\right )}^2\,\left (a^2\,d^2\,f^2-2\,a\,b\,c\,d\,f^2+b^2\,c^2\,f^2\right )}{{\left (c\,f-d\,e\right )}^3}-\frac {2\,\left (e+f\,x\right )\,\left (d\,a^2\,f^2-2\,c\,a\,b\,f^2-d\,b^2\,e^2+2\,c\,b^2\,e\,f\right )}{3\,{\left (c\,f-d\,e\right )}^2}}{f^2\,{\left (e+f\,x\right )}^{5/2}}-\frac {2\,\sqrt {d}\,\mathrm {atan}\left (\frac {\sqrt {d}\,\sqrt {e+f\,x}\,{\left (a\,d-b\,c\right )}^2\,\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}\,\left (a^2\,d^2-2\,a\,b\,c\,d+b^2\,c^2\right )}\right )\,{\left (a\,d-b\,c\right )}^2}{{\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)^2/((e + f*x)^(7/2)*(c + d*x)),x)

[Out]

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

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