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

Optimal. Leaf size=164 \[ \frac {2 (b c-a d) (d e-c f)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {e+f x}}{\sqrt {d e-c f}}\right )}{d^{9/2}}-\frac {2 \sqrt {e+f x} (b c-a d) (d e-c f)^2}{d^4}-\frac {2 (e+f x)^{3/2} (b c-a d) (d e-c f)}{3 d^3}-\frac {2 (e+f x)^{5/2} (b c-a d)}{5 d^2}+\frac {2 b (e+f x)^{7/2}}{7 d f} \]

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Rubi [A]  time = 0.19, antiderivative size = 164, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {80, 50, 63, 208} \begin {gather*} -\frac {2 (e+f x)^{5/2} (b c-a d)}{5 d^2}-\frac {2 (e+f x)^{3/2} (b c-a d) (d e-c f)}{3 d^3}-\frac {2 \sqrt {e+f x} (b c-a d) (d e-c f)^2}{d^4}+\frac {2 (b c-a d) (d e-c f)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {e+f x}}{\sqrt {d e-c f}}\right )}{d^{9/2}}+\frac {2 b (e+f x)^{7/2}}{7 d f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(b*c - a*d)*(d*e - c*f)^2*Sqrt[e + f*x])/d^4 - (2*(b*c - a*d)*(d*e - c*f)*(e + f*x)^(3/2))/(3*d^3) - (2*(b
*c - a*d)*(e + f*x)^(5/2))/(5*d^2) + (2*b*(e + f*x)^(7/2))/(7*d*f) + (2*(b*c - a*d)*(d*e - c*f)^(5/2)*ArcTanh[
(Sqrt[d]*Sqrt[e + f*x])/Sqrt[d*e - c*f]])/d^(9/2)

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && 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 80

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(c + d*x)
^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 2)), x] + Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(
d*f*(n + p + 2)), Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2,
0]

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) (e+f x)^{5/2}}{c+d x} \, dx &=\frac {2 b (e+f x)^{7/2}}{7 d f}+\frac {\left (2 \left (-\frac {7}{2} b c f+\frac {7 a d f}{2}\right )\right ) \int \frac {(e+f x)^{5/2}}{c+d x} \, dx}{7 d f}\\ &=-\frac {2 (b c-a d) (e+f x)^{5/2}}{5 d^2}+\frac {2 b (e+f x)^{7/2}}{7 d f}-\frac {((b c-a d) (d e-c f)) \int \frac {(e+f x)^{3/2}}{c+d x} \, dx}{d^2}\\ &=-\frac {2 (b c-a d) (d e-c f) (e+f x)^{3/2}}{3 d^3}-\frac {2 (b c-a d) (e+f x)^{5/2}}{5 d^2}+\frac {2 b (e+f x)^{7/2}}{7 d f}-\frac {\left ((b c-a d) (d e-c f)^2\right ) \int \frac {\sqrt {e+f x}}{c+d x} \, dx}{d^3}\\ &=-\frac {2 (b c-a d) (d e-c f)^2 \sqrt {e+f x}}{d^4}-\frac {2 (b c-a d) (d e-c f) (e+f x)^{3/2}}{3 d^3}-\frac {2 (b c-a d) (e+f x)^{5/2}}{5 d^2}+\frac {2 b (e+f x)^{7/2}}{7 d f}-\frac {\left ((b c-a d) (d e-c f)^3\right ) \int \frac {1}{(c+d x) \sqrt {e+f x}} \, dx}{d^4}\\ &=-\frac {2 (b c-a d) (d e-c f)^2 \sqrt {e+f x}}{d^4}-\frac {2 (b c-a d) (d e-c f) (e+f x)^{3/2}}{3 d^3}-\frac {2 (b c-a d) (e+f x)^{5/2}}{5 d^2}+\frac {2 b (e+f x)^{7/2}}{7 d f}-\frac {\left (2 (b c-a d) (d e-c f)^3\right ) \operatorname {Subst}\left (\int \frac {1}{c-\frac {d e}{f}+\frac {d x^2}{f}} \, dx,x,\sqrt {e+f x}\right )}{d^4 f}\\ &=-\frac {2 (b c-a d) (d e-c f)^2 \sqrt {e+f x}}{d^4}-\frac {2 (b c-a d) (d e-c f) (e+f x)^{3/2}}{3 d^3}-\frac {2 (b c-a d) (e+f x)^{5/2}}{5 d^2}+\frac {2 b (e+f x)^{7/2}}{7 d f}+\frac {2 (b c-a d) (d e-c f)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {e+f x}}{\sqrt {d e-c f}}\right )}{d^{9/2}}\\ \end {align*}

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Mathematica [A]  time = 0.37, size = 136, normalized size = 0.83 \begin {gather*} \frac {2 b (e+f x)^{7/2}}{7 d f}-\frac {2 (b c-a d) \left (5 (d e-c f) \left (\sqrt {d} \sqrt {e+f x} (-3 c f+4 d e+d f x)-3 (d e-c f)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {e+f x}}{\sqrt {d e-c f}}\right )\right )+3 d^{5/2} (e+f x)^{5/2}\right )}{15 d^{9/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*b*(e + f*x)^(7/2))/(7*d*f) - (2*(b*c - a*d)*(3*d^(5/2)*(e + f*x)^(5/2) + 5*(d*e - c*f)*(Sqrt[d]*Sqrt[e + f*
x]*(4*d*e - 3*c*f + d*f*x) - 3*(d*e - c*f)^(3/2)*ArcTanh[(Sqrt[d]*Sqrt[e + f*x])/Sqrt[d*e - c*f]])))/(15*d^(9/
2))

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IntegrateAlgebraic [A]  time = 0.23, size = 314, normalized size = 1.91 \begin {gather*} \frac {2 (a d-b c) (c f-d e)^{5/2} \tan ^{-1}\left (\frac {\sqrt {d} \sqrt {e+f x} \sqrt {c f-d e}}{d e-c f}\right )}{d^{9/2}}-\frac {2 \left (-105 a c^2 d f^3 \sqrt {e+f x}+35 a c d^2 f^2 (e+f x)^{3/2}+210 a c d^2 e f^2 \sqrt {e+f x}-105 a d^3 e^2 f \sqrt {e+f x}-21 a d^3 f (e+f x)^{5/2}-35 a d^3 e f (e+f x)^{3/2}+105 b c^3 f^3 \sqrt {e+f x}-35 b c^2 d f^2 (e+f x)^{3/2}-210 b c^2 d e f^2 \sqrt {e+f x}+105 b c d^2 e^2 f \sqrt {e+f x}+21 b c d^2 f (e+f x)^{5/2}+35 b c d^2 e f (e+f x)^{3/2}-15 b d^3 (e+f x)^{7/2}\right )}{105 d^4 f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(105*b*c*d^2*e^2*f*Sqrt[e + f*x] - 105*a*d^3*e^2*f*Sqrt[e + f*x] - 210*b*c^2*d*e*f^2*Sqrt[e + f*x] + 210*a
*c*d^2*e*f^2*Sqrt[e + f*x] + 105*b*c^3*f^3*Sqrt[e + f*x] - 105*a*c^2*d*f^3*Sqrt[e + f*x] + 35*b*c*d^2*e*f*(e +
f*x)^(3/2) - 35*a*d^3*e*f*(e + f*x)^(3/2) - 35*b*c^2*d*f^2*(e + f*x)^(3/2) + 35*a*c*d^2*f^2*(e + f*x)^(3/2) +
21*b*c*d^2*f*(e + f*x)^(5/2) - 21*a*d^3*f*(e + f*x)^(5/2) - 15*b*d^3*(e + f*x)^(7/2)))/(105*d^4*f) + (2*(-(b*
c) + a*d)*(-(d*e) + c*f)^(5/2)*ArcTan[(Sqrt[d]*Sqrt[-(d*e) + c*f]*Sqrt[e + f*x])/(d*e - c*f)])/d^(9/2)

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fricas [B]  time = 1.31, size = 591, normalized size = 3.60 \begin {gather*} \left [\frac {105 \, {\left ({\left (b c d^{2} - a d^{3}\right )} e^{2} f - 2 \, {\left (b c^{2} d - a c d^{2}\right )} e f^{2} + {\left (b c^{3} - a c^{2} d\right )} f^{3}\right )} \sqrt {\frac {d e - c f}{d}} \log \left (\frac {d f x + 2 \, d e - c f + 2 \, \sqrt {f x + e} d \sqrt {\frac {d e - c f}{d}}}{d x + c}\right ) + 2 \, {\left (15 \, b d^{3} f^{3} x^{3} + 15 \, b d^{3} e^{3} - 161 \, {\left (b c d^{2} - a d^{3}\right )} e^{2} f + 245 \, {\left (b c^{2} d - a c d^{2}\right )} e f^{2} - 105 \, {\left (b c^{3} - a c^{2} d\right )} f^{3} + 3 \, {\left (15 \, b d^{3} e f^{2} - 7 \, {\left (b c d^{2} - a d^{3}\right )} f^{3}\right )} x^{2} + {\left (45 \, b d^{3} e^{2} f - 77 \, {\left (b c d^{2} - a d^{3}\right )} e f^{2} + 35 \, {\left (b c^{2} d - a c d^{2}\right )} f^{3}\right )} x\right )} \sqrt {f x + e}}{105 \, d^{4} f}, \frac {2 \, {\left (105 \, {\left ({\left (b c d^{2} - a d^{3}\right )} e^{2} f - 2 \, {\left (b c^{2} d - a c d^{2}\right )} e f^{2} + {\left (b c^{3} - a c^{2} d\right )} f^{3}\right )} \sqrt {-\frac {d e - c f}{d}} \arctan \left (-\frac {\sqrt {f x + e} d \sqrt {-\frac {d e - c f}{d}}}{d e - c f}\right ) + {\left (15 \, b d^{3} f^{3} x^{3} + 15 \, b d^{3} e^{3} - 161 \, {\left (b c d^{2} - a d^{3}\right )} e^{2} f + 245 \, {\left (b c^{2} d - a c d^{2}\right )} e f^{2} - 105 \, {\left (b c^{3} - a c^{2} d\right )} f^{3} + 3 \, {\left (15 \, b d^{3} e f^{2} - 7 \, {\left (b c d^{2} - a d^{3}\right )} f^{3}\right )} x^{2} + {\left (45 \, b d^{3} e^{2} f - 77 \, {\left (b c d^{2} - a d^{3}\right )} e f^{2} + 35 \, {\left (b c^{2} d - a c d^{2}\right )} f^{3}\right )} x\right )} \sqrt {f x + e}\right )}}{105 \, d^{4} f}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/105*(105*((b*c*d^2 - a*d^3)*e^2*f - 2*(b*c^2*d - a*c*d^2)*e*f^2 + (b*c^3 - a*c^2*d)*f^3)*sqrt((d*e - c*f)/d
)*log((d*f*x + 2*d*e - c*f + 2*sqrt(f*x + e)*d*sqrt((d*e - c*f)/d))/(d*x + c)) + 2*(15*b*d^3*f^3*x^3 + 15*b*d^
3*e^3 - 161*(b*c*d^2 - a*d^3)*e^2*f + 245*(b*c^2*d - a*c*d^2)*e*f^2 - 105*(b*c^3 - a*c^2*d)*f^3 + 3*(15*b*d^3*
e*f^2 - 7*(b*c*d^2 - a*d^3)*f^3)*x^2 + (45*b*d^3*e^2*f - 77*(b*c*d^2 - a*d^3)*e*f^2 + 35*(b*c^2*d - a*c*d^2)*f
^3)*x)*sqrt(f*x + e))/(d^4*f), 2/105*(105*((b*c*d^2 - a*d^3)*e^2*f - 2*(b*c^2*d - a*c*d^2)*e*f^2 + (b*c^3 - a*
c^2*d)*f^3)*sqrt(-(d*e - c*f)/d)*arctan(-sqrt(f*x + e)*d*sqrt(-(d*e - c*f)/d)/(d*e - c*f)) + (15*b*d^3*f^3*x^3
+ 15*b*d^3*e^3 - 161*(b*c*d^2 - a*d^3)*e^2*f + 245*(b*c^2*d - a*c*d^2)*e*f^2 - 105*(b*c^3 - a*c^2*d)*f^3 + 3*
(15*b*d^3*e*f^2 - 7*(b*c*d^2 - a*d^3)*f^3)*x^2 + (45*b*d^3*e^2*f - 77*(b*c*d^2 - a*d^3)*e*f^2 + 35*(b*c^2*d -
a*c*d^2)*f^3)*x)*sqrt(f*x + e))/(d^4*f)]

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giac [B]  time = 1.30, size = 387, normalized size = 2.36 \begin {gather*} \frac {2 \, {\left (b c^{4} f^{3} - a c^{3} d f^{3} - 3 \, b c^{3} d f^{2} e + 3 \, a c^{2} d^{2} f^{2} e + 3 \, b c^{2} d^{2} f e^{2} - 3 \, a c d^{3} f e^{2} - b c d^{3} e^{3} + a d^{4} e^{3}\right )} \arctan \left (\frac {\sqrt {f x + e} d}{\sqrt {c d f - d^{2} e}}\right )}{\sqrt {c d f - d^{2} e} d^{4}} + \frac {2 \, {\left (15 \, {\left (f x + e\right )}^{\frac {7}{2}} b d^{6} f^{6} - 21 \, {\left (f x + e\right )}^{\frac {5}{2}} b c d^{5} f^{7} + 21 \, {\left (f x + e\right )}^{\frac {5}{2}} a d^{6} f^{7} + 35 \, {\left (f x + e\right )}^{\frac {3}{2}} b c^{2} d^{4} f^{8} - 35 \, {\left (f x + e\right )}^{\frac {3}{2}} a c d^{5} f^{8} - 105 \, \sqrt {f x + e} b c^{3} d^{3} f^{9} + 105 \, \sqrt {f x + e} a c^{2} d^{4} f^{9} - 35 \, {\left (f x + e\right )}^{\frac {3}{2}} b c d^{5} f^{7} e + 35 \, {\left (f x + e\right )}^{\frac {3}{2}} a d^{6} f^{7} e + 210 \, \sqrt {f x + e} b c^{2} d^{4} f^{8} e - 210 \, \sqrt {f x + e} a c d^{5} f^{8} e - 105 \, \sqrt {f x + e} b c d^{5} f^{7} e^{2} + 105 \, \sqrt {f x + e} a d^{6} f^{7} e^{2}\right )}}{105 \, d^{7} f^{7}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(b*c^4*f^3 - a*c^3*d*f^3 - 3*b*c^3*d*f^2*e + 3*a*c^2*d^2*f^2*e + 3*b*c^2*d^2*f*e^2 - 3*a*c*d^3*f*e^2 - b*c*d
^3*e^3 + a*d^4*e^3)*arctan(sqrt(f*x + e)*d/sqrt(c*d*f - d^2*e))/(sqrt(c*d*f - d^2*e)*d^4) + 2/105*(15*(f*x + e
)^(7/2)*b*d^6*f^6 - 21*(f*x + e)^(5/2)*b*c*d^5*f^7 + 21*(f*x + e)^(5/2)*a*d^6*f^7 + 35*(f*x + e)^(3/2)*b*c^2*d
^4*f^8 - 35*(f*x + e)^(3/2)*a*c*d^5*f^8 - 105*sqrt(f*x + e)*b*c^3*d^3*f^9 + 105*sqrt(f*x + e)*a*c^2*d^4*f^9 -
35*(f*x + e)^(3/2)*b*c*d^5*f^7*e + 35*(f*x + e)^(3/2)*a*d^6*f^7*e + 210*sqrt(f*x + e)*b*c^2*d^4*f^8*e - 210*sq
rt(f*x + e)*a*c*d^5*f^8*e - 105*sqrt(f*x + e)*b*c*d^5*f^7*e^2 + 105*sqrt(f*x + e)*a*d^6*f^7*e^2)/(d^7*f^7)

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maple [B]  time = 0.02, size = 573, normalized size = 3.49 \begin {gather*} -\frac {2 a \,c^{3} f^{3} \arctan \left (\frac {\sqrt {f x +e}\, d}{\sqrt {\left (c f -d e \right ) d}}\right )}{\sqrt {\left (c f -d e \right ) d}\, d^{3}}+\frac {6 a \,c^{2} e \,f^{2} \arctan \left (\frac {\sqrt {f x +e}\, d}{\sqrt {\left (c f -d e \right ) d}}\right )}{\sqrt {\left (c f -d e \right ) d}\, d^{2}}-\frac {6 a c \,e^{2} f \arctan \left (\frac {\sqrt {f x +e}\, d}{\sqrt {\left (c f -d e \right ) d}}\right )}{\sqrt {\left (c f -d e \right ) d}\, d}+\frac {2 a \,e^{3} \arctan \left (\frac {\sqrt {f x +e}\, d}{\sqrt {\left (c f -d e \right ) d}}\right )}{\sqrt {\left (c f -d e \right ) d}}+\frac {2 b \,c^{4} f^{3} \arctan \left (\frac {\sqrt {f x +e}\, d}{\sqrt {\left (c f -d e \right ) d}}\right )}{\sqrt {\left (c f -d e \right ) d}\, d^{4}}-\frac {6 b \,c^{3} e \,f^{2} \arctan \left (\frac {\sqrt {f x +e}\, d}{\sqrt {\left (c f -d e \right ) d}}\right )}{\sqrt {\left (c f -d e \right ) d}\, d^{3}}+\frac {6 b \,c^{2} e^{2} f \arctan \left (\frac {\sqrt {f x +e}\, d}{\sqrt {\left (c f -d e \right ) d}}\right )}{\sqrt {\left (c f -d e \right ) d}\, d^{2}}-\frac {2 b c \,e^{3} \arctan \left (\frac {\sqrt {f x +e}\, d}{\sqrt {\left (c f -d e \right ) d}}\right )}{\sqrt {\left (c f -d e \right ) d}\, d}+\frac {2 \sqrt {f x +e}\, a \,c^{2} f^{2}}{d^{3}}-\frac {4 \sqrt {f x +e}\, a c e f}{d^{2}}+\frac {2 \sqrt {f x +e}\, a \,e^{2}}{d}-\frac {2 \sqrt {f x +e}\, b \,c^{3} f^{2}}{d^{4}}+\frac {4 \sqrt {f x +e}\, b \,c^{2} e f}{d^{3}}-\frac {2 \sqrt {f x +e}\, b c \,e^{2}}{d^{2}}-\frac {2 \left (f x +e \right )^{\frac {3}{2}} a c f}{3 d^{2}}+\frac {2 \left (f x +e \right )^{\frac {3}{2}} a e}{3 d}+\frac {2 \left (f x +e \right )^{\frac {3}{2}} b \,c^{2} f}{3 d^{3}}-\frac {2 \left (f x +e \right )^{\frac {3}{2}} b c e}{3 d^{2}}+\frac {2 \left (f x +e \right )^{\frac {5}{2}} a}{5 d}-\frac {2 \left (f x +e \right )^{\frac {5}{2}} b c}{5 d^{2}}+\frac {2 \left (f x +e \right )^{\frac {7}{2}} b}{7 d f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/7*b*(f*x+e)^(7/2)/d/f+2/5/d*(f*x+e)^(5/2)*a-2/5/d^2*(f*x+e)^(5/2)*b*c-2/3*f/d^2*(f*x+e)^(3/2)*a*c+2/3/d*(f*x
+e)^(3/2)*a*e+2/3*f/d^3*(f*x+e)^(3/2)*b*c^2-2/3/d^2*(f*x+e)^(3/2)*b*c*e+2*f^2/d^3*a*c^2*(f*x+e)^(1/2)-4*f/d^2*
a*c*e*(f*x+e)^(1/2)+2/d*a*e^2*(f*x+e)^(1/2)-2*f^2/d^4*b*c^3*(f*x+e)^(1/2)+4*f/d^3*b*c^2*e*(f*x+e)^(1/2)-2/d^2*
b*c*e^2*(f*x+e)^(1/2)-2*f^3/d^3/((c*f-d*e)*d)^(1/2)*arctan((f*x+e)^(1/2)*d/((c*f-d*e)*d)^(1/2))*a*c^3+6*f^2/d^
2/((c*f-d*e)*d)^(1/2)*arctan((f*x+e)^(1/2)*d/((c*f-d*e)*d)^(1/2))*a*c^2*e-6*f/d/((c*f-d*e)*d)^(1/2)*arctan((f*
x+e)^(1/2)*d/((c*f-d*e)*d)^(1/2))*a*c*e^2+2/((c*f-d*e)*d)^(1/2)*arctan((f*x+e)^(1/2)*d/((c*f-d*e)*d)^(1/2))*a*
e^3+2*f^3/d^4/((c*f-d*e)*d)^(1/2)*arctan((f*x+e)^(1/2)*d/((c*f-d*e)*d)^(1/2))*b*c^4-6*f^2/d^3/((c*f-d*e)*d)^(1
/2)*arctan((f*x+e)^(1/2)*d/((c*f-d*e)*d)^(1/2))*b*c^3*e+6*f/d^2/((c*f-d*e)*d)^(1/2)*arctan((f*x+e)^(1/2)*d/((c
*f-d*e)*d)^(1/2))*b*c^2*e^2-2/d/((c*f-d*e)*d)^(1/2)*arctan((f*x+e)^(1/2)*d/((c*f-d*e)*d)^(1/2))*b*c*e^3

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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)*(f*x+e)^(5/2)/(d*x+c),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 = 0.13, size = 330, normalized size = 2.01 \begin {gather*} {\left (e+f\,x\right )}^{5/2}\,\left (\frac {2\,a\,f-2\,b\,e}{5\,d\,f}-\frac {2\,b\,\left (c\,f^2-d\,e\,f\right )}{5\,d^2\,f^2}\right )+\frac {2\,\mathrm {atan}\left (\frac {\sqrt {d}\,\sqrt {e+f\,x}\,\left (a\,d-b\,c\right )\,{\left (c\,f-d\,e\right )}^{5/2}}{b\,c^4\,f^3-3\,b\,c^3\,d\,e\,f^2-a\,c^3\,d\,f^3+3\,b\,c^2\,d^2\,e^2\,f+3\,a\,c^2\,d^2\,e\,f^2-b\,c\,d^3\,e^3-3\,a\,c\,d^3\,e^2\,f+a\,d^4\,e^3}\right )\,\left (a\,d-b\,c\right )\,{\left (c\,f-d\,e\right )}^{5/2}}{d^{9/2}}+\frac {2\,b\,{\left (e+f\,x\right )}^{7/2}}{7\,d\,f}-\frac {{\left (e+f\,x\right )}^{3/2}\,\left (c\,f^2-d\,e\,f\right )\,\left (\frac {2\,a\,f-2\,b\,e}{d\,f}-\frac {2\,b\,\left (c\,f^2-d\,e\,f\right )}{d^2\,f^2}\right )}{3\,d\,f}+\frac {\sqrt {e+f\,x}\,{\left (c\,f^2-d\,e\,f\right )}^2\,\left (\frac {2\,a\,f-2\,b\,e}{d\,f}-\frac {2\,b\,\left (c\,f^2-d\,e\,f\right )}{d^2\,f^2}\right )}{d^2\,f^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(e + f*x)^(5/2)*((2*a*f - 2*b*e)/(5*d*f) - (2*b*(c*f^2 - d*e*f))/(5*d^2*f^2)) + (2*atan((d^(1/2)*(e + f*x)^(1/
2)*(a*d - b*c)*(c*f - d*e)^(5/2))/(a*d^4*e^3 + b*c^4*f^3 - a*c^3*d*f^3 - b*c*d^3*e^3 - 3*a*c*d^3*e^2*f - 3*b*c
^3*d*e*f^2 + 3*a*c^2*d^2*e*f^2 + 3*b*c^2*d^2*e^2*f))*(a*d - b*c)*(c*f - d*e)^(5/2))/d^(9/2) + (2*b*(e + f*x)^(
7/2))/(7*d*f) - ((e + f*x)^(3/2)*(c*f^2 - d*e*f)*((2*a*f - 2*b*e)/(d*f) - (2*b*(c*f^2 - d*e*f))/(d^2*f^2)))/(3
*d*f) + ((e + f*x)^(1/2)*(c*f^2 - d*e*f)^2*((2*a*f - 2*b*e)/(d*f) - (2*b*(c*f^2 - d*e*f))/(d^2*f^2)))/(d^2*f^2
)

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sympy [A]  time = 69.78, size = 221, normalized size = 1.35 \begin {gather*} \frac {2 b \left (e + f x\right )^{\frac {7}{2}}}{7 d f} + \frac {\left (e + f x\right )^{\frac {5}{2}} \left (2 a d - 2 b c\right )}{5 d^{2}} + \frac {\left (e + f x\right )^{\frac {3}{2}} \left (- 2 a c d f + 2 a d^{2} e + 2 b c^{2} f - 2 b c d e\right )}{3 d^{3}} + \frac {\sqrt {e + f x} \left (2 a c^{2} d f^{2} - 4 a c d^{2} e f + 2 a d^{3} e^{2} - 2 b c^{3} f^{2} + 4 b c^{2} d e f - 2 b c d^{2} e^{2}\right )}{d^{4}} - \frac {2 \left (a d - b c\right ) \left (c f - d e\right )^{3} \operatorname {atan}{\left (\frac {\sqrt {e + f x}}{\sqrt {\frac {c f - d e}{d}}} \right )}}{d^{5} \sqrt {\frac {c f - d e}{d}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*b*(e + f*x)**(7/2)/(7*d*f) + (e + f*x)**(5/2)*(2*a*d - 2*b*c)/(5*d**2) + (e + f*x)**(3/2)*(-2*a*c*d*f + 2*a*
d**2*e + 2*b*c**2*f - 2*b*c*d*e)/(3*d**3) + sqrt(e + f*x)*(2*a*c**2*d*f**2 - 4*a*c*d**2*e*f + 2*a*d**3*e**2 -
2*b*c**3*f**2 + 4*b*c**2*d*e*f - 2*b*c*d**2*e**2)/d**4 - 2*(a*d - b*c)*(c*f - d*e)**3*atan(sqrt(e + f*x)/sqrt(
(c*f - d*e)/d))/(d**5*sqrt((c*f - d*e)/d))

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