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

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

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Rubi [A]  time = 0.22, 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 = {87, 63, 208} \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 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 87

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 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)^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 ) \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)^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 [C]  time = 0.13, size = 103, normalized size = 0.60 \begin {gather*} \frac {2 b (d e-c f) (6 a d f+b (-3 c f+2 d e+5 d f x))-6 f^2 (b c-a d)^2 \, _2F_1\left (-\frac {5}{2},1;-\frac {3}{2};\frac {d (e+f x)}{d e-c f}\right )}{15 d^2 f^2 (e+f x)^{5/2} (c f-d e)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*b*(d*e - c*f)*(6*a*d*f + b*(2*d*e - 3*c*f + 5*d*f*x)) - 6*(b*c - a*d)^2*f^2*Hypergeometric2F1[-5/2, 1, -3/2
, (d*(e + f*x))/(d*e - c*f)])/(15*d^2*f^2*(-(d*e) + c*f)*(e + f*x)^(5/2))

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IntegrateAlgebraic [B]  time = 0.31, size = 364, normalized size = 2.10 \begin {gather*} \frac {2 \sqrt {d} (a d-b c)^2 \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^2 c^2 f^4-5 a^2 c d f^3 (e+f x)-6 a^2 c d e f^3+3 a^2 d^2 e^2 f^2+5 a^2 d^2 e f^2 (e+f x)+15 a^2 d^2 f^2 (e+f x)^2+10 a b c^2 f^3 (e+f x)-6 a b c^2 e f^3+12 a b c d e^2 f^2-10 a b c d e f^2 (e+f x)-30 a b c d f^2 (e+f x)^2-6 a b d^2 e^3 f+3 b^2 c^2 e^2 f^2-10 b^2 c^2 e f^2 (e+f x)+15 b^2 c^2 f^2 (e+f x)^2-6 b^2 c d e^3 f+15 b^2 c d e^2 f (e+f x)+3 b^2 d^2 e^4-5 b^2 d^2 e^3 (e+f x)\right )}{15 f^2 (e+f x)^{5/2} (c f-d e)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(3*b^2*d^2*e^4 - 6*b^2*c*d*e^3*f - 6*a*b*d^2*e^3*f + 3*b^2*c^2*e^2*f^2 + 12*a*b*c*d*e^2*f^2 + 3*a^2*d^2*e^
2*f^2 - 6*a*b*c^2*e*f^3 - 6*a^2*c*d*e*f^3 + 3*a^2*c^2*f^4 - 5*b^2*d^2*e^3*(e + f*x) + 15*b^2*c*d*e^2*f*(e + f*
x) - 10*b^2*c^2*e*f^2*(e + f*x) - 10*a*b*c*d*e*f^2*(e + f*x) + 5*a^2*d^2*e*f^2*(e + f*x) + 10*a*b*c^2*f^3*(e +
f*x) - 5*a^2*c*d*f^3*(e + f*x) + 15*b^2*c^2*f^2*(e + f*x)^2 - 30*a*b*c*d*f^2*(e + f*x)^2 + 15*a^2*d^2*f^2*(e
+ f*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[-(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 = 1.70, size = 1173, normalized size = 6.78 \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 )} e f^{4} x^{2} + 3 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} e^{2} f^{3} x + {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} e^{3} f^{2}\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 (2 \, b^{2} d^{2} e^{4} - 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} - 3 \, {\left (3 \, b^{2} c d - 2 \, a b d^{2}\right )} e^{3} f - {\left (8 \, b^{2} c^{2} - 28 \, a b c d + 23 \, a^{2} d^{2}\right )} e^{2} f^{2} - {\left (4 \, a b c^{2} - 11 \, a^{2} c d\right )} e f^{3} + 5 \, {\left (b^{2} d^{2} e^{3} f - 3 \, b^{2} c d e^{2} f^{2} - {\left (4 \, b^{2} c^{2} - 14 \, a b c d + 7 \, a^{2} d^{2}\right )} e f^{3} - {\left (2 \, a b c^{2} - a^{2} c d\right )} f^{4}\right )} x\right )} \sqrt {f x + e}}{15 \, {\left (d^{3} e^{6} f^{2} - 3 \, c d^{2} e^{5} f^{3} + 3 \, c^{2} d e^{4} f^{4} - c^{3} e^{3} f^{5} + {\left (d^{3} e^{3} f^{5} - 3 \, c d^{2} e^{2} f^{6} + 3 \, c^{2} d e f^{7} - c^{3} f^{8}\right )} x^{3} + 3 \, {\left (d^{3} e^{4} f^{4} - 3 \, c d^{2} e^{3} f^{5} + 3 \, c^{2} d e^{2} f^{6} - c^{3} e f^{7}\right )} x^{2} + 3 \, {\left (d^{3} e^{5} f^{3} - 3 \, c d^{2} e^{4} f^{4} + 3 \, c^{2} d e^{3} f^{5} - c^{3} e^{2} f^{6}\right )} x\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 )} e f^{4} x^{2} + 3 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} e^{2} f^{3} x + {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} e^{3} f^{2}\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 (2 \, b^{2} d^{2} e^{4} - 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} - 3 \, {\left (3 \, b^{2} c d - 2 \, a b d^{2}\right )} e^{3} f - {\left (8 \, b^{2} c^{2} - 28 \, a b c d + 23 \, a^{2} d^{2}\right )} e^{2} f^{2} - {\left (4 \, a b c^{2} - 11 \, a^{2} c d\right )} e f^{3} + 5 \, {\left (b^{2} d^{2} e^{3} f - 3 \, b^{2} c d e^{2} f^{2} - {\left (4 \, b^{2} c^{2} - 14 \, a b c d + 7 \, a^{2} d^{2}\right )} e f^{3} - {\left (2 \, a b c^{2} - a^{2} c d\right )} f^{4}\right )} x\right )} \sqrt {f x + e}\right )}}{15 \, {\left (d^{3} e^{6} f^{2} - 3 \, c d^{2} e^{5} f^{3} + 3 \, c^{2} d e^{4} f^{4} - c^{3} e^{3} f^{5} + {\left (d^{3} e^{3} f^{5} - 3 \, c d^{2} e^{2} f^{6} + 3 \, c^{2} d e f^{7} - c^{3} f^{8}\right )} x^{3} + 3 \, {\left (d^{3} e^{4} f^{4} - 3 \, c d^{2} e^{3} f^{5} + 3 \, c^{2} d e^{2} f^{6} - c^{3} e f^{7}\right )} x^{2} + 3 \, {\left (d^{3} e^{5} f^{3} - 3 \, c d^{2} e^{4} f^{4} + 3 \, c^{2} d e^{3} f^{5} - c^{3} e^{2} f^{6}\right )} x\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)*e*f^4*x^2 + 3*(b^2*c^2
- 2*a*b*c*d + a^2*d^2)*e^2*f^3*x + (b^2*c^2 - 2*a*b*c*d + a^2*d^2)*e^3*f^2)*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*(2*b^2*d^2*e^4 - 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 - 3*(3*b^2*c*d - 2*a*b*d^2)*e^3*f - (8*b^2*c^2 - 28*a*b*c*d + 23*a
^2*d^2)*e^2*f^2 - (4*a*b*c^2 - 11*a^2*c*d)*e*f^3 + 5*(b^2*d^2*e^3*f - 3*b^2*c*d*e^2*f^2 - (4*b^2*c^2 - 14*a*b*
c*d + 7*a^2*d^2)*e*f^3 - (2*a*b*c^2 - a^2*c*d)*f^4)*x)*sqrt(f*x + e))/(d^3*e^6*f^2 - 3*c*d^2*e^5*f^3 + 3*c^2*d
*e^4*f^4 - c^3*e^3*f^5 + (d^3*e^3*f^5 - 3*c*d^2*e^2*f^6 + 3*c^2*d*e*f^7 - c^3*f^8)*x^3 + 3*(d^3*e^4*f^4 - 3*c*
d^2*e^3*f^5 + 3*c^2*d*e^2*f^6 - c^3*e*f^7)*x^2 + 3*(d^3*e^5*f^3 - 3*c*d^2*e^4*f^4 + 3*c^2*d*e^3*f^5 - c^3*e^2*
f^6)*x), -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)*e*f^4*x^2 + 3*
(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*e^2*f^3*x + (b^2*c^2 - 2*a*b*c*d + a^2*d^2)*e^3*f^2)*sqrt(-d/(d*e - c*f))*arct
an(-(d*e - c*f)*sqrt(f*x + e)*sqrt(-d/(d*e - c*f))/(d*f*x + d*e)) + (2*b^2*d^2*e^4 - 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 - 3*(3*b^2*c*d - 2*a*b*d^2)*e^3*f - (8*b^2*c^2 - 28*a*b*c*d + 23*a^2*d^2)*e^
2*f^2 - (4*a*b*c^2 - 11*a^2*c*d)*e*f^3 + 5*(b^2*d^2*e^3*f - 3*b^2*c*d*e^2*f^2 - (4*b^2*c^2 - 14*a*b*c*d + 7*a^
2*d^2)*e*f^3 - (2*a*b*c^2 - a^2*c*d)*f^4)*x)*sqrt(f*x + e))/(d^3*e^6*f^2 - 3*c*d^2*e^5*f^3 + 3*c^2*d*e^4*f^4 -
c^3*e^3*f^5 + (d^3*e^3*f^5 - 3*c*d^2*e^2*f^6 + 3*c^2*d*e*f^7 - c^3*f^8)*x^3 + 3*(d^3*e^4*f^4 - 3*c*d^2*e^3*f^
5 + 3*c^2*d*e^2*f^6 - c^3*e*f^7)*x^2 + 3*(d^3*e^5*f^3 - 3*c*d^2*e^4*f^4 + 3*c^2*d*e^3*f^5 - c^3*e^2*f^6)*x)]

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giac [B]  time = 1.44, 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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maple [B]  time = 0.02, size = 408, normalized size = 2.36 \begin {gather*} -\frac {2 a^{2} 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 {4 a 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 b^{2} c^{2} d \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^{2} d^{2}}{\left (c f -d e \right )^{3} \sqrt {f x +e}}+\frac {4 a b c d}{\left (c f -d e \right )^{3} \sqrt {f x +e}}-\frac {2 b^{2} c^{2}}{\left (c f -d e \right )^{3} \sqrt {f x +e}}+\frac {2 a^{2} d}{3 \left (c f -d e \right )^{2} \left (f x +e \right )^{\frac {3}{2}}}-\frac {4 a b c}{3 \left (c f -d e \right )^{2} \left (f x +e \right )^{\frac {3}{2}}}+\frac {4 b^{2} c e}{3 \left (c f -d e \right )^{2} \left (f x +e \right )^{\frac {3}{2}} f}-\frac {2 b^{2} d \,e^{2}}{3 \left (c f -d e \right )^{2} \left (f x +e \right )^{\frac {3}{2}} f^{2}}-\frac {2 a^{2}}{5 \left (c f -d e \right ) \left (f x +e \right )^{\frac {5}{2}}}+\frac {4 a b e}{5 \left (c f -d e \right ) \left (f x +e \right )^{\frac {5}{2}} f}-\frac {2 b^{2} e^{2}}{5 \left (c f -d e \right ) \left (f x +e \right )^{\frac {5}{2}} f^{2}} \end {gather*}

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)

[Out]

-2/5/(c*f-d*e)/(f*x+e)^(5/2)*a^2+4/5/f/(c*f-d*e)/(f*x+e)^(5/2)*a*b*e-2/5/f^2/(c*f-d*e)/(f*x+e)^(5/2)*b^2*e^2+2
/3/(c*f-d*e)^2/(f*x+e)^(3/2)*a^2*d-4/3/(c*f-d*e)^2/(f*x+e)^(3/2)*a*b*c+4/3/f/(c*f-d*e)^2/(f*x+e)^(3/2)*b^2*c*e
-2/3/f^2/(c*f-d*e)^2/(f*x+e)^(3/2)*b^2*d*e^2-2/(c*f-d*e)^3/(f*x+e)^(1/2)*a^2*d^2+4/(c*f-d*e)^3/(f*x+e)^(1/2)*a
*b*c*d-2/(c*f-d*e)^3/(f*x+e)^(1/2)*b^2*c^2-2*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+4*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)*a*b*c-2*d
/(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^2*c^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-d*e>0)', see `assume?` for
more details)Is c*f-d*e positive or negative?

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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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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \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]

Timed out

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