3.17.45 \(\int \frac {(a+b x)^3 \sqrt {e+f x}}{c+d x} \, dx\)

Optimal. Leaf size=210 \[ \frac {2 b (e+f x)^{3/2} \left (3 a^2 d^2 f^2-3 a b d f (c f+d e)+b^2 \left (c^2 f^2+c d e f+d^2 e^2\right )\right )}{3 d^3 f^3}-\frac {2 b^2 (e+f x)^{5/2} (-3 a d f+b c f+2 b d e)}{5 d^2 f^3}+\frac {2 (b c-a d)^3 \sqrt {d e-c f} \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {e+f x}}{\sqrt {d e-c f}}\right )}{d^{9/2}}-\frac {2 \sqrt {e+f x} (b c-a d)^3}{d^4}+\frac {2 b^3 (e+f x)^{7/2}}{7 d f^3} \]

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Rubi [A]  time = 0.18, antiderivative size = 210, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {88, 50, 63, 208} \begin {gather*} \frac {2 b (e+f x)^{3/2} \left (3 a^2 d^2 f^2-3 a b d f (c f+d e)+b^2 \left (c^2 f^2+c d e f+d^2 e^2\right )\right )}{3 d^3 f^3}-\frac {2 b^2 (e+f x)^{5/2} (-3 a d f+b c f+2 b d e)}{5 d^2 f^3}-\frac {2 \sqrt {e+f x} (b c-a d)^3}{d^4}+\frac {2 (b c-a d)^3 \sqrt {d e-c f} \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {e+f x}}{\sqrt {d e-c f}}\right )}{d^{9/2}}+\frac {2 b^3 (e+f x)^{7/2}}{7 d f^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((a + b*x)^3*Sqrt[e + f*x])/(c + d*x),x]

[Out]

(-2*(b*c - a*d)^3*Sqrt[e + f*x])/d^4 + (2*b*(3*a^2*d^2*f^2 - 3*a*b*d*f*(d*e + c*f) + b^2*(d^2*e^2 + c*d*e*f +
c^2*f^2))*(e + f*x)^(3/2))/(3*d^3*f^3) - (2*b^2*(2*b*d*e + b*c*f - 3*a*d*f)*(e + f*x)^(5/2))/(5*d^2*f^3) + (2*
b^3*(e + f*x)^(7/2))/(7*d*f^3) + (2*(b*c - a*d)^3*Sqrt[d*e - c*f]*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 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

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)^3 \sqrt {e+f x}}{c+d x} \, dx &=\int \left (\frac {b \left (3 a^2 d^2 f^2-3 a b d f (d e+c f)+b^2 \left (d^2 e^2+c d e f+c^2 f^2\right )\right ) \sqrt {e+f x}}{d^3 f^2}+\frac {(-b c+a d)^3 \sqrt {e+f x}}{d^3 (c+d x)}-\frac {b^2 (2 b d e+b c f-3 a d f) (e+f x)^{3/2}}{d^2 f^2}+\frac {b^3 (e+f x)^{5/2}}{d f^2}\right ) \, dx\\ &=\frac {2 b \left (3 a^2 d^2 f^2-3 a b d f (d e+c f)+b^2 \left (d^2 e^2+c d e f+c^2 f^2\right )\right ) (e+f x)^{3/2}}{3 d^3 f^3}-\frac {2 b^2 (2 b d e+b c f-3 a d f) (e+f x)^{5/2}}{5 d^2 f^3}+\frac {2 b^3 (e+f x)^{7/2}}{7 d f^3}-\frac {(b c-a d)^3 \int \frac {\sqrt {e+f x}}{c+d x} \, dx}{d^3}\\ &=-\frac {2 (b c-a d)^3 \sqrt {e+f x}}{d^4}+\frac {2 b \left (3 a^2 d^2 f^2-3 a b d f (d e+c f)+b^2 \left (d^2 e^2+c d e f+c^2 f^2\right )\right ) (e+f x)^{3/2}}{3 d^3 f^3}-\frac {2 b^2 (2 b d e+b c f-3 a d f) (e+f x)^{5/2}}{5 d^2 f^3}+\frac {2 b^3 (e+f x)^{7/2}}{7 d f^3}-\frac {\left ((b c-a d)^3 (d e-c f)\right ) \int \frac {1}{(c+d x) \sqrt {e+f x}} \, dx}{d^4}\\ &=-\frac {2 (b c-a d)^3 \sqrt {e+f x}}{d^4}+\frac {2 b \left (3 a^2 d^2 f^2-3 a b d f (d e+c f)+b^2 \left (d^2 e^2+c d e f+c^2 f^2\right )\right ) (e+f x)^{3/2}}{3 d^3 f^3}-\frac {2 b^2 (2 b d e+b c f-3 a d f) (e+f x)^{5/2}}{5 d^2 f^3}+\frac {2 b^3 (e+f x)^{7/2}}{7 d f^3}-\frac {\left (2 (b c-a d)^3 (d e-c f)\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)^3 \sqrt {e+f x}}{d^4}+\frac {2 b \left (3 a^2 d^2 f^2-3 a b d f (d e+c f)+b^2 \left (d^2 e^2+c d e f+c^2 f^2\right )\right ) (e+f x)^{3/2}}{3 d^3 f^3}-\frac {2 b^2 (2 b d e+b c f-3 a d f) (e+f x)^{5/2}}{5 d^2 f^3}+\frac {2 b^3 (e+f x)^{7/2}}{7 d f^3}+\frac {2 (b c-a d)^3 \sqrt {d e-c f} \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 = 0.28, size = 210, normalized size = 1.00 \begin {gather*} \frac {2 b (e+f x)^{3/2} \left (3 a^2 d^2 f^2-3 a b d f (c f+d e)+b^2 \left (c^2 f^2+c d e f+d^2 e^2\right )\right )}{3 d^3 f^3}-\frac {2 b^2 (e+f x)^{5/2} (-3 a d f+b c f+2 b d e)}{5 d^2 f^3}+\frac {2 (b c-a d)^3 \sqrt {d e-c f} \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)^3}{d^4}+\frac {2 b^3 (e+f x)^{7/2}}{7 d f^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x)^3*Sqrt[e + f*x])/(c + d*x),x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.34, size = 301, normalized size = 1.43 \begin {gather*} \frac {2 \sqrt {e+f x} \left (105 a^3 d^3 f^3-315 a^2 b c d^2 f^3+105 a^2 b d^3 f^2 (e+f x)+315 a b^2 c^2 d f^3-105 a b^2 c d^2 f^2 (e+f x)+63 a b^2 d^3 f (e+f x)^2-105 a b^2 d^3 e f (e+f x)-105 b^3 c^3 f^3+35 b^3 c^2 d f^2 (e+f x)-21 b^3 c d^2 f (e+f x)^2+35 b^3 c d^2 e f (e+f x)+35 b^3 d^3 e^2 (e+f x)+15 b^3 d^3 (e+f x)^3-42 b^3 d^3 e (e+f x)^2\right )}{105 d^4 f^3}+\frac {2 (a d-b c)^3 \sqrt {c f-d e} \tan ^{-1}\left (\frac {\sqrt {d} \sqrt {e+f x} \sqrt {c f-d e}}{d e-c f}\right )}{d^{9/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((a + b*x)^3*Sqrt[e + f*x])/(c + d*x),x]

[Out]

(2*Sqrt[e + f*x]*(-105*b^3*c^3*f^3 + 315*a*b^2*c^2*d*f^3 - 315*a^2*b*c*d^2*f^3 + 105*a^3*d^3*f^3 + 35*b^3*d^3*
e^2*(e + f*x) + 35*b^3*c*d^2*e*f*(e + f*x) - 105*a*b^2*d^3*e*f*(e + f*x) + 35*b^3*c^2*d*f^2*(e + f*x) - 105*a*
b^2*c*d^2*f^2*(e + f*x) + 105*a^2*b*d^3*f^2*(e + f*x) - 42*b^3*d^3*e*(e + f*x)^2 - 21*b^3*c*d^2*f*(e + f*x)^2
+ 63*a*b^2*d^3*f*(e + f*x)^2 + 15*b^3*d^3*(e + f*x)^3))/(105*d^4*f^3) + (2*(-(b*c) + a*d)^3*Sqrt[-(d*e) + c*f]
*ArcTan[(Sqrt[d]*Sqrt[-(d*e) + c*f]*Sqrt[e + f*x])/(d*e - c*f)])/d^(9/2)

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fricas [A]  time = 1.64, size = 705, normalized size = 3.36 \begin {gather*} \left [-\frac {105 \, {\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} f^{3} \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^{3} d^{3} f^{3} x^{3} + 8 \, b^{3} d^{3} e^{3} + 14 \, {\left (b^{3} c d^{2} - 3 \, a b^{2} d^{3}\right )} e^{2} f + 35 \, {\left (b^{3} c^{2} d - 3 \, a b^{2} c d^{2} + 3 \, a^{2} b d^{3}\right )} e f^{2} - 105 \, {\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} f^{3} + 3 \, {\left (b^{3} d^{3} e f^{2} - 7 \, {\left (b^{3} c d^{2} - 3 \, a b^{2} d^{3}\right )} f^{3}\right )} x^{2} - {\left (4 \, b^{3} d^{3} e^{2} f + 7 \, {\left (b^{3} c d^{2} - 3 \, a b^{2} d^{3}\right )} e f^{2} - 35 \, {\left (b^{3} c^{2} d - 3 \, a b^{2} c d^{2} + 3 \, a^{2} b d^{3}\right )} f^{3}\right )} x\right )} \sqrt {f x + e}}{105 \, d^{4} f^{3}}, \frac {2 \, {\left (105 \, {\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} f^{3} \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^{3} d^{3} f^{3} x^{3} + 8 \, b^{3} d^{3} e^{3} + 14 \, {\left (b^{3} c d^{2} - 3 \, a b^{2} d^{3}\right )} e^{2} f + 35 \, {\left (b^{3} c^{2} d - 3 \, a b^{2} c d^{2} + 3 \, a^{2} b d^{3}\right )} e f^{2} - 105 \, {\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} f^{3} + 3 \, {\left (b^{3} d^{3} e f^{2} - 7 \, {\left (b^{3} c d^{2} - 3 \, a b^{2} d^{3}\right )} f^{3}\right )} x^{2} - {\left (4 \, b^{3} d^{3} e^{2} f + 7 \, {\left (b^{3} c d^{2} - 3 \, a b^{2} d^{3}\right )} e f^{2} - 35 \, {\left (b^{3} c^{2} d - 3 \, a b^{2} c d^{2} + 3 \, a^{2} b d^{3}\right )} f^{3}\right )} x\right )} \sqrt {f x + e}\right )}}{105 \, d^{4} f^{3}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^3*(f*x+e)^(1/2)/(d*x+c),x, algorithm="fricas")

[Out]

[-1/105*(105*(b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*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^3*d^3*f^3*x^3 + 8*b^3*d^3*e^3 + 14*(b^3*c*d^
2 - 3*a*b^2*d^3)*e^2*f + 35*(b^3*c^2*d - 3*a*b^2*c*d^2 + 3*a^2*b*d^3)*e*f^2 - 105*(b^3*c^3 - 3*a*b^2*c^2*d + 3
*a^2*b*c*d^2 - a^3*d^3)*f^3 + 3*(b^3*d^3*e*f^2 - 7*(b^3*c*d^2 - 3*a*b^2*d^3)*f^3)*x^2 - (4*b^3*d^3*e^2*f + 7*(
b^3*c*d^2 - 3*a*b^2*d^3)*e*f^2 - 35*(b^3*c^2*d - 3*a*b^2*c*d^2 + 3*a^2*b*d^3)*f^3)*x)*sqrt(f*x + e))/(d^4*f^3)
, 2/105*(105*(b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*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^3*d^3*f^3*x^3 + 8*b^3*d^3*e^3 + 14*(b^3*c*d^2 - 3*a*b^2*d^3)*e^2
*f + 35*(b^3*c^2*d - 3*a*b^2*c*d^2 + 3*a^2*b*d^3)*e*f^2 - 105*(b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d
^3)*f^3 + 3*(b^3*d^3*e*f^2 - 7*(b^3*c*d^2 - 3*a*b^2*d^3)*f^3)*x^2 - (4*b^3*d^3*e^2*f + 7*(b^3*c*d^2 - 3*a*b^2*
d^3)*e*f^2 - 35*(b^3*c^2*d - 3*a*b^2*c*d^2 + 3*a^2*b*d^3)*f^3)*x)*sqrt(f*x + e))/(d^4*f^3)]

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giac [B]  time = 1.34, size = 436, normalized size = 2.08 \begin {gather*} \frac {2 \, {\left (b^{3} c^{4} f - 3 \, a b^{2} c^{3} d f + 3 \, a^{2} b c^{2} d^{2} f - a^{3} c d^{3} f - b^{3} c^{3} d e + 3 \, a b^{2} c^{2} d^{2} e - 3 \, a^{2} b c d^{3} e + a^{3} d^{4} e\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^{3} d^{6} f^{18} - 21 \, {\left (f x + e\right )}^{\frac {5}{2}} b^{3} c d^{5} f^{19} + 63 \, {\left (f x + e\right )}^{\frac {5}{2}} a b^{2} d^{6} f^{19} + 35 \, {\left (f x + e\right )}^{\frac {3}{2}} b^{3} c^{2} d^{4} f^{20} - 105 \, {\left (f x + e\right )}^{\frac {3}{2}} a b^{2} c d^{5} f^{20} + 105 \, {\left (f x + e\right )}^{\frac {3}{2}} a^{2} b d^{6} f^{20} - 105 \, \sqrt {f x + e} b^{3} c^{3} d^{3} f^{21} + 315 \, \sqrt {f x + e} a b^{2} c^{2} d^{4} f^{21} - 315 \, \sqrt {f x + e} a^{2} b c d^{5} f^{21} + 105 \, \sqrt {f x + e} a^{3} d^{6} f^{21} - 42 \, {\left (f x + e\right )}^{\frac {5}{2}} b^{3} d^{6} f^{18} e + 35 \, {\left (f x + e\right )}^{\frac {3}{2}} b^{3} c d^{5} f^{19} e - 105 \, {\left (f x + e\right )}^{\frac {3}{2}} a b^{2} d^{6} f^{19} e + 35 \, {\left (f x + e\right )}^{\frac {3}{2}} b^{3} d^{6} f^{18} e^{2}\right )}}{105 \, d^{7} f^{21}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^3*(f*x+e)^(1/2)/(d*x+c),x, algorithm="giac")

[Out]

2*(b^3*c^4*f - 3*a*b^2*c^3*d*f + 3*a^2*b*c^2*d^2*f - a^3*c*d^3*f - b^3*c^3*d*e + 3*a*b^2*c^2*d^2*e - 3*a^2*b*c
*d^3*e + a^3*d^4*e)*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^3*d^6*f^18 - 21*(f*x + e)^(5/2)*b^3*c*d^5*f^19 + 63*(f*x + e)^(5/2)*a*b^2*d^6*f^19 + 35*(f*x + e)^(3
/2)*b^3*c^2*d^4*f^20 - 105*(f*x + e)^(3/2)*a*b^2*c*d^5*f^20 + 105*(f*x + e)^(3/2)*a^2*b*d^6*f^20 - 105*sqrt(f*
x + e)*b^3*c^3*d^3*f^21 + 315*sqrt(f*x + e)*a*b^2*c^2*d^4*f^21 - 315*sqrt(f*x + e)*a^2*b*c*d^5*f^21 + 105*sqrt
(f*x + e)*a^3*d^6*f^21 - 42*(f*x + e)^(5/2)*b^3*d^6*f^18*e + 35*(f*x + e)^(3/2)*b^3*c*d^5*f^19*e - 105*(f*x +
e)^(3/2)*a*b^2*d^6*f^19*e + 35*(f*x + e)^(3/2)*b^3*d^6*f^18*e^2)/(d^7*f^21)

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maple [B]  time = 0.02, size = 629, normalized size = 3.00 \begin {gather*} -\frac {2 a^{3} c 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^{3} e \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 {6 a^{2} b \,c^{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 {6 a^{2} b c e \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 {6 a \,b^{2} c^{3} 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^{3}}+\frac {6 a \,b^{2} c^{2} e \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^{3} c^{4} 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^{4}}-\frac {2 b^{3} c^{3} e \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 {2 \sqrt {f x +e}\, a^{3}}{d}-\frac {6 \sqrt {f x +e}\, a^{2} b c}{d^{2}}+\frac {6 \sqrt {f x +e}\, a \,b^{2} c^{2}}{d^{3}}-\frac {2 \sqrt {f x +e}\, b^{3} c^{3}}{d^{4}}+\frac {2 \left (f x +e \right )^{\frac {3}{2}} a^{2} b}{d f}-\frac {2 \left (f x +e \right )^{\frac {3}{2}} a \,b^{2} c}{d^{2} f}-\frac {2 \left (f x +e \right )^{\frac {3}{2}} a \,b^{2} e}{d \,f^{2}}+\frac {2 \left (f x +e \right )^{\frac {3}{2}} b^{3} c^{2}}{3 d^{3} f}+\frac {2 \left (f x +e \right )^{\frac {3}{2}} b^{3} c e}{3 d^{2} f^{2}}+\frac {2 \left (f x +e \right )^{\frac {3}{2}} b^{3} e^{2}}{3 d \,f^{3}}+\frac {6 \left (f x +e \right )^{\frac {5}{2}} a \,b^{2}}{5 d \,f^{2}}-\frac {2 \left (f x +e \right )^{\frac {5}{2}} b^{3} c}{5 d^{2} f^{2}}-\frac {4 \left (f x +e \right )^{\frac {5}{2}} b^{3} e}{5 d \,f^{3}}+\frac {2 \left (f x +e \right )^{\frac {7}{2}} b^{3}}{7 d \,f^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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)^3*(f*x+e)^(1/2)/(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 = 0.14, size = 451, normalized size = 2.15 \begin {gather*} {\left (e+f\,x\right )}^{3/2}\,\left (\frac {\left (\frac {6\,b^3\,e-6\,a\,b^2\,f}{d\,f^3}+\frac {2\,b^3\,\left (c\,f^4-d\,e\,f^3\right )}{d^2\,f^6}\right )\,\left (c\,f^4-d\,e\,f^3\right )}{3\,d\,f^3}+\frac {2\,b\,{\left (a\,f-b\,e\right )}^2}{d\,f^3}\right )-{\left (e+f\,x\right )}^{5/2}\,\left (\frac {6\,b^3\,e-6\,a\,b^2\,f}{5\,d\,f^3}+\frac {2\,b^3\,\left (c\,f^4-d\,e\,f^3\right )}{5\,d^2\,f^6}\right )+\sqrt {e+f\,x}\,\left (\frac {2\,{\left (a\,f-b\,e\right )}^3}{d\,f^3}-\frac {\left (\frac {\left (\frac {6\,b^3\,e-6\,a\,b^2\,f}{d\,f^3}+\frac {2\,b^3\,\left (c\,f^4-d\,e\,f^3\right )}{d^2\,f^6}\right )\,\left (c\,f^4-d\,e\,f^3\right )}{d\,f^3}+\frac {6\,b\,{\left (a\,f-b\,e\right )}^2}{d\,f^3}\right )\,\left (c\,f^4-d\,e\,f^3\right )}{d\,f^3}\right )+\frac {2\,b^3\,{\left (e+f\,x\right )}^{7/2}}{7\,d\,f^3}+\frac {\mathrm {atan}\left (\frac {\sqrt {d}\,\sqrt {e+f\,x}\,{\left (a\,d-b\,c\right )}^3\,\sqrt {d\,e-c\,f}\,1{}\mathrm {i}}{-f\,a^3\,c\,d^3+e\,a^3\,d^4+3\,f\,a^2\,b\,c^2\,d^2-3\,e\,a^2\,b\,c\,d^3-3\,f\,a\,b^2\,c^3\,d+3\,e\,a\,b^2\,c^2\,d^2+f\,b^3\,c^4-e\,b^3\,c^3\,d}\right )\,{\left (a\,d-b\,c\right )}^3\,\sqrt {d\,e-c\,f}\,2{}\mathrm {i}}{d^{9/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(e + f*x)^(3/2)*((((6*b^3*e - 6*a*b^2*f)/(d*f^3) + (2*b^3*(c*f^4 - d*e*f^3))/(d^2*f^6))*(c*f^4 - d*e*f^3))/(3*
d*f^3) + (2*b*(a*f - b*e)^2)/(d*f^3)) - (e + f*x)^(5/2)*((6*b^3*e - 6*a*b^2*f)/(5*d*f^3) + (2*b^3*(c*f^4 - d*e
*f^3))/(5*d^2*f^6)) + (e + f*x)^(1/2)*((2*(a*f - b*e)^3)/(d*f^3) - (((((6*b^3*e - 6*a*b^2*f)/(d*f^3) + (2*b^3*
(c*f^4 - d*e*f^3))/(d^2*f^6))*(c*f^4 - d*e*f^3))/(d*f^3) + (6*b*(a*f - b*e)^2)/(d*f^3))*(c*f^4 - d*e*f^3))/(d*
f^3)) + (2*b^3*(e + f*x)^(7/2))/(7*d*f^3) + (atan((d^(1/2)*(e + f*x)^(1/2)*(a*d - b*c)^3*(d*e - c*f)^(1/2)*1i)
/(a^3*d^4*e + b^3*c^4*f - a^3*c*d^3*f - b^3*c^3*d*e - 3*a^2*b*c*d^3*e - 3*a*b^2*c^3*d*f + 3*a*b^2*c^2*d^2*e +
3*a^2*b*c^2*d^2*f))*(a*d - b*c)^3*(d*e - c*f)^(1/2)*2i)/d^(9/2)

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sympy [A]  time = 16.41, size = 269, normalized size = 1.28 \begin {gather*} \frac {2 \left (\frac {b^{3} \left (e + f x\right )^{\frac {7}{2}}}{7 d f^{2}} + \frac {\left (e + f x\right )^{\frac {5}{2}} \left (3 a b^{2} d f - b^{3} c f - 2 b^{3} d e\right )}{5 d^{2} f^{2}} + \frac {\left (e + f x\right )^{\frac {3}{2}} \left (3 a^{2} b d^{2} f^{2} - 3 a b^{2} c d f^{2} - 3 a b^{2} d^{2} e f + b^{3} c^{2} f^{2} + b^{3} c d e f + b^{3} d^{2} e^{2}\right )}{3 d^{3} f^{2}} + \frac {\sqrt {e + f x} \left (a^{3} d^{3} f - 3 a^{2} b c d^{2} f + 3 a b^{2} c^{2} d f - b^{3} c^{3} f\right )}{d^{4}} - \frac {f \left (a d - b c\right )^{3} \left (c f - d e\right ) \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}}}\right )}{f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)**3*(f*x+e)**(1/2)/(d*x+c),x)

[Out]

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

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