3.17.36 \(\int \frac {(a+b x)^2 (e+f x)^{3/2}}{c+d x} \, dx\)

Optimal. Leaf size=172 \[ -\frac {2 (b c-a d)^2 (d e-c f)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {e+f x}}{\sqrt {d e-c f}}\right )}{d^{9/2}}+\frac {2 \sqrt {e+f x} (b c-a d)^2 (d e-c f)}{d^4}+\frac {2 (e+f x)^{3/2} (b c-a d)^2}{3 d^3}-\frac {2 b (e+f x)^{5/2} (-2 a d f+b c f+b d e)}{5 d^2 f^2}+\frac {2 b^2 (e+f x)^{7/2}}{7 d f^2} \]

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Rubi [A]  time = 0.16, antiderivative size = 172, normalized size of antiderivative = 1.00, number of steps used = 6, 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)^{5/2} (-2 a d f+b c f+b d e)}{5 d^2 f^2}+\frac {2 (e+f x)^{3/2} (b c-a d)^2}{3 d^3}+\frac {2 \sqrt {e+f x} (b c-a d)^2 (d e-c f)}{d^4}-\frac {2 (b c-a d)^2 (d e-c f)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {e+f x}}{\sqrt {d e-c f}}\right )}{d^{9/2}}+\frac {2 b^2 (e+f x)^{7/2}}{7 d f^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((a + b*x)^2*(e + f*x)^(3/2))/(c + d*x),x]

[Out]

(2*(b*c - a*d)^2*(d*e - c*f)*Sqrt[e + f*x])/d^4 + (2*(b*c - a*d)^2*(e + f*x)^(3/2))/(3*d^3) - (2*b*(b*d*e + b*
c*f - 2*a*d*f)*(e + f*x)^(5/2))/(5*d^2*f^2) + (2*b^2*(e + f*x)^(7/2))/(7*d*f^2) - (2*(b*c - a*d)^2*(d*e - c*f)
^(3/2)*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 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)^2 (e+f x)^{3/2}}{c+d x} \, dx &=\int \left (-\frac {b (b d e+b c f-2 a d f) (e+f x)^{3/2}}{d^2 f}+\frac {(-b c+a d)^2 (e+f x)^{3/2}}{d^2 (c+d x)}+\frac {b^2 (e+f x)^{5/2}}{d f}\right ) \, dx\\ &=-\frac {2 b (b d e+b c f-2 a d f) (e+f x)^{5/2}}{5 d^2 f^2}+\frac {2 b^2 (e+f x)^{7/2}}{7 d f^2}+\frac {(b c-a d)^2 \int \frac {(e+f x)^{3/2}}{c+d x} \, dx}{d^2}\\ &=\frac {2 (b c-a d)^2 (e+f x)^{3/2}}{3 d^3}-\frac {2 b (b d e+b c f-2 a d f) (e+f x)^{5/2}}{5 d^2 f^2}+\frac {2 b^2 (e+f x)^{7/2}}{7 d f^2}+\frac {\left ((b c-a d)^2 (d e-c f)\right ) \int \frac {\sqrt {e+f x}}{c+d x} \, dx}{d^3}\\ &=\frac {2 (b c-a d)^2 (d e-c f) \sqrt {e+f x}}{d^4}+\frac {2 (b c-a d)^2 (e+f x)^{3/2}}{3 d^3}-\frac {2 b (b d e+b c f-2 a d f) (e+f x)^{5/2}}{5 d^2 f^2}+\frac {2 b^2 (e+f x)^{7/2}}{7 d f^2}+\frac {\left ((b c-a d)^2 (d e-c f)^2\right ) \int \frac {1}{(c+d x) \sqrt {e+f x}} \, dx}{d^4}\\ &=\frac {2 (b c-a d)^2 (d e-c f) \sqrt {e+f x}}{d^4}+\frac {2 (b c-a d)^2 (e+f x)^{3/2}}{3 d^3}-\frac {2 b (b d e+b c f-2 a d f) (e+f x)^{5/2}}{5 d^2 f^2}+\frac {2 b^2 (e+f x)^{7/2}}{7 d f^2}+\frac {\left (2 (b c-a d)^2 (d e-c f)^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 )}{d^4 f}\\ &=\frac {2 (b c-a d)^2 (d e-c f) \sqrt {e+f x}}{d^4}+\frac {2 (b c-a d)^2 (e+f x)^{3/2}}{3 d^3}-\frac {2 b (b d e+b c f-2 a d f) (e+f x)^{5/2}}{5 d^2 f^2}+\frac {2 b^2 (e+f x)^{7/2}}{7 d f^2}-\frac {2 (b c-a d)^2 (d e-c f)^{3/2} \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 = 0.23, size = 160, normalized size = 0.93 \begin {gather*} \frac {2 \left (105 (b c-a d)^2 (d e-c f) \left (\frac {\sqrt {e+f x}}{d}-\frac {\sqrt {d e-c f} \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {e+f x}}{\sqrt {d e-c f}}\right )}{d^{3/2}}\right )-\frac {21 b d (e+f x)^{5/2} (-2 a d f+b c f+b d e)}{f^2}+35 (e+f x)^{3/2} (b c-a d)^2+\frac {15 b^2 d^2 (e+f x)^{7/2}}{f^2}\right )}{105 d^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x)^2*(e + f*x)^(3/2))/(c + d*x),x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.22, size = 336, normalized size = 1.95 \begin {gather*} -\frac {2 \left (105 a^2 c d^2 f^3 \sqrt {e+f x}-35 a^2 d^3 f^2 (e+f x)^{3/2}-105 a^2 d^3 e f^2 \sqrt {e+f x}-210 a b c^2 d f^3 \sqrt {e+f x}+70 a b c d^2 f^2 (e+f x)^{3/2}+210 a b c d^2 e f^2 \sqrt {e+f x}-42 a b d^3 f (e+f x)^{5/2}+105 b^2 c^3 f^3 \sqrt {e+f x}-35 b^2 c^2 d f^2 (e+f x)^{3/2}-105 b^2 c^2 d e f^2 \sqrt {e+f x}+21 b^2 c d^2 f (e+f x)^{5/2}-15 b^2 d^3 (e+f x)^{7/2}+21 b^2 d^3 e (e+f x)^{5/2}\right )}{105 d^4 f^2}-\frac {2 (a d-b c)^2 (c f-d e)^{3/2} \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)^2*(e + f*x)^(3/2))/(c + d*x),x]

[Out]

(-2*(-105*b^2*c^2*d*e*f^2*Sqrt[e + f*x] + 210*a*b*c*d^2*e*f^2*Sqrt[e + f*x] - 105*a^2*d^3*e*f^2*Sqrt[e + f*x]
+ 105*b^2*c^3*f^3*Sqrt[e + f*x] - 210*a*b*c^2*d*f^3*Sqrt[e + f*x] + 105*a^2*c*d^2*f^3*Sqrt[e + f*x] - 35*b^2*c
^2*d*f^2*(e + f*x)^(3/2) + 70*a*b*c*d^2*f^2*(e + f*x)^(3/2) - 35*a^2*d^3*f^2*(e + f*x)^(3/2) + 21*b^2*d^3*e*(e
+ f*x)^(5/2) + 21*b^2*c*d^2*f*(e + f*x)^(5/2) - 42*a*b*d^3*f*(e + f*x)^(5/2) - 15*b^2*d^3*(e + f*x)^(7/2)))/(
105*d^4*f^2) - (2*(-(b*c) + a*d)^2*(-(d*e) + c*f)^(3/2)*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 = 1.42, size = 694, normalized size = 4.03 \begin {gather*} \left [-\frac {105 \, {\left ({\left (b^{2} c^{2} d - 2 \, a b c d^{2} + a^{2} d^{3}\right )} e f^{2} - {\left (b^{2} c^{3} - 2 \, a b c^{2} d + a^{2} c d^{2}\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^{2} d^{3} f^{3} x^{3} - 6 \, b^{2} d^{3} e^{3} - 21 \, {\left (b^{2} c d^{2} - 2 \, a b d^{3}\right )} e^{2} f + 140 \, {\left (b^{2} c^{2} d - 2 \, a b c d^{2} + a^{2} d^{3}\right )} e f^{2} - 105 \, {\left (b^{2} c^{3} - 2 \, a b c^{2} d + a^{2} c d^{2}\right )} f^{3} + 3 \, {\left (8 \, b^{2} d^{3} e f^{2} - 7 \, {\left (b^{2} c d^{2} - 2 \, a b d^{3}\right )} f^{3}\right )} x^{2} + {\left (3 \, b^{2} d^{3} e^{2} f - 42 \, {\left (b^{2} c d^{2} - 2 \, a b d^{3}\right )} e f^{2} + 35 \, {\left (b^{2} c^{2} d - 2 \, a b c d^{2} + a^{2} d^{3}\right )} f^{3}\right )} x\right )} \sqrt {f x + e}}{105 \, d^{4} f^{2}}, -\frac {2 \, {\left (105 \, {\left ({\left (b^{2} c^{2} d - 2 \, a b c d^{2} + a^{2} d^{3}\right )} e f^{2} - {\left (b^{2} c^{3} - 2 \, a b c^{2} d + a^{2} c d^{2}\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^{2} d^{3} f^{3} x^{3} - 6 \, b^{2} d^{3} e^{3} - 21 \, {\left (b^{2} c d^{2} - 2 \, a b d^{3}\right )} e^{2} f + 140 \, {\left (b^{2} c^{2} d - 2 \, a b c d^{2} + a^{2} d^{3}\right )} e f^{2} - 105 \, {\left (b^{2} c^{3} - 2 \, a b c^{2} d + a^{2} c d^{2}\right )} f^{3} + 3 \, {\left (8 \, b^{2} d^{3} e f^{2} - 7 \, {\left (b^{2} c d^{2} - 2 \, a b d^{3}\right )} f^{3}\right )} x^{2} + {\left (3 \, b^{2} d^{3} e^{2} f - 42 \, {\left (b^{2} c d^{2} - 2 \, a b d^{3}\right )} e f^{2} + 35 \, {\left (b^{2} c^{2} d - 2 \, a b c d^{2} + a^{2} d^{3}\right )} f^{3}\right )} x\right )} \sqrt {f x + e}\right )}}{105 \, d^{4} f^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/105*(105*((b^2*c^2*d - 2*a*b*c*d^2 + a^2*d^3)*e*f^2 - (b^2*c^3 - 2*a*b*c^2*d + a^2*c*d^2)*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^2*d^3*f^3*x^3
- 6*b^2*d^3*e^3 - 21*(b^2*c*d^2 - 2*a*b*d^3)*e^2*f + 140*(b^2*c^2*d - 2*a*b*c*d^2 + a^2*d^3)*e*f^2 - 105*(b^2*
c^3 - 2*a*b*c^2*d + a^2*c*d^2)*f^3 + 3*(8*b^2*d^3*e*f^2 - 7*(b^2*c*d^2 - 2*a*b*d^3)*f^3)*x^2 + (3*b^2*d^3*e^2*
f - 42*(b^2*c*d^2 - 2*a*b*d^3)*e*f^2 + 35*(b^2*c^2*d - 2*a*b*c*d^2 + a^2*d^3)*f^3)*x)*sqrt(f*x + e))/(d^4*f^2)
, -2/105*(105*((b^2*c^2*d - 2*a*b*c*d^2 + a^2*d^3)*e*f^2 - (b^2*c^3 - 2*a*b*c^2*d + a^2*c*d^2)*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^2*d^3*f^3*x^3 - 6*b^2*d^3*e^3 - 2
1*(b^2*c*d^2 - 2*a*b*d^3)*e^2*f + 140*(b^2*c^2*d - 2*a*b*c*d^2 + a^2*d^3)*e*f^2 - 105*(b^2*c^3 - 2*a*b*c^2*d +
a^2*c*d^2)*f^3 + 3*(8*b^2*d^3*e*f^2 - 7*(b^2*c*d^2 - 2*a*b*d^3)*f^3)*x^2 + (3*b^2*d^3*e^2*f - 42*(b^2*c*d^2 -
2*a*b*d^3)*e*f^2 + 35*(b^2*c^2*d - 2*a*b*c*d^2 + a^2*d^3)*f^3)*x)*sqrt(f*x + e))/(d^4*f^2)]

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giac [B]  time = 1.32, size = 424, normalized size = 2.47 \begin {gather*} \frac {2 \, {\left (b^{2} c^{4} f^{2} - 2 \, a b c^{3} d f^{2} + a^{2} c^{2} d^{2} f^{2} - 2 \, b^{2} c^{3} d f e + 4 \, a b c^{2} d^{2} f e - 2 \, a^{2} c d^{3} f e + b^{2} c^{2} d^{2} e^{2} - 2 \, a b c d^{3} e^{2} + a^{2} d^{4} e^{2}\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^{2} d^{6} f^{12} - 21 \, {\left (f x + e\right )}^{\frac {5}{2}} b^{2} c d^{5} f^{13} + 42 \, {\left (f x + e\right )}^{\frac {5}{2}} a b d^{6} f^{13} + 35 \, {\left (f x + e\right )}^{\frac {3}{2}} b^{2} c^{2} d^{4} f^{14} - 70 \, {\left (f x + e\right )}^{\frac {3}{2}} a b c d^{5} f^{14} + 35 \, {\left (f x + e\right )}^{\frac {3}{2}} a^{2} d^{6} f^{14} - 105 \, \sqrt {f x + e} b^{2} c^{3} d^{3} f^{15} + 210 \, \sqrt {f x + e} a b c^{2} d^{4} f^{15} - 105 \, \sqrt {f x + e} a^{2} c d^{5} f^{15} - 21 \, {\left (f x + e\right )}^{\frac {5}{2}} b^{2} d^{6} f^{12} e + 105 \, \sqrt {f x + e} b^{2} c^{2} d^{4} f^{14} e - 210 \, \sqrt {f x + e} a b c d^{5} f^{14} e + 105 \, \sqrt {f x + e} a^{2} d^{6} f^{14} e\right )}}{105 \, d^{7} f^{14}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(b^2*c^4*f^2 - 2*a*b*c^3*d*f^2 + a^2*c^2*d^2*f^2 - 2*b^2*c^3*d*f*e + 4*a*b*c^2*d^2*f*e - 2*a^2*c*d^3*f*e + b
^2*c^2*d^2*e^2 - 2*a*b*c*d^3*e^2 + a^2*d^4*e^2)*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^2*d^6*f^12 - 21*(f*x + e)^(5/2)*b^2*c*d^5*f^13 + 42*(f*x + e)^(5/2)*a*b*
d^6*f^13 + 35*(f*x + e)^(3/2)*b^2*c^2*d^4*f^14 - 70*(f*x + e)^(3/2)*a*b*c*d^5*f^14 + 35*(f*x + e)^(3/2)*a^2*d^
6*f^14 - 105*sqrt(f*x + e)*b^2*c^3*d^3*f^15 + 210*sqrt(f*x + e)*a*b*c^2*d^4*f^15 - 105*sqrt(f*x + e)*a^2*c*d^5
*f^15 - 21*(f*x + e)^(5/2)*b^2*d^6*f^12*e + 105*sqrt(f*x + e)*b^2*c^2*d^4*f^14*e - 210*sqrt(f*x + e)*a*b*c*d^5
*f^14*e + 105*sqrt(f*x + e)*a^2*d^6*f^14*e)/(d^7*f^14)

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maple [B]  time = 0.02, size = 644, normalized size = 3.74 \begin {gather*} \frac {2 a^{2} c^{2} 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 {4 a^{2} c e 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^{2} e^{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}}-\frac {4 a b \,c^{3} 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 {8 a b \,c^{2} e 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 {4 a b c \,e^{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}+\frac {2 b^{2} c^{4} 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^{4}}-\frac {4 b^{2} c^{3} e 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 {2 b^{2} c^{2} e^{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 {2 \sqrt {f x +e}\, a^{2} c f}{d^{2}}+\frac {2 \sqrt {f x +e}\, a^{2} e}{d}+\frac {4 \sqrt {f x +e}\, a b \,c^{2} f}{d^{3}}-\frac {4 \sqrt {f x +e}\, a b c e}{d^{2}}-\frac {2 \sqrt {f x +e}\, b^{2} c^{3} f}{d^{4}}+\frac {2 \sqrt {f x +e}\, b^{2} c^{2} e}{d^{3}}+\frac {2 \left (f x +e \right )^{\frac {3}{2}} a^{2}}{3 d}-\frac {4 \left (f x +e \right )^{\frac {3}{2}} a b c}{3 d^{2}}+\frac {2 \left (f x +e \right )^{\frac {3}{2}} b^{2} c^{2}}{3 d^{3}}+\frac {4 \left (f x +e \right )^{\frac {5}{2}} a b}{5 d f}-\frac {2 \left (f x +e \right )^{\frac {5}{2}} b^{2} c}{5 d^{2} f}-\frac {2 \left (f x +e \right )^{\frac {5}{2}} b^{2} e}{5 d \,f^{2}}+\frac {2 \left (f x +e \right )^{\frac {7}{2}} b^{2}}{7 d \,f^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/7*b^2*(f*x+e)^(7/2)/d/f^2+4/5/f/d*(f*x+e)^(5/2)*a*b-2/5/f/d^2*(f*x+e)^(5/2)*b^2*c-2/5/f^2/d*(f*x+e)^(5/2)*b^
2*e+2/3/d*(f*x+e)^(3/2)*a^2-4/3/d^2*(f*x+e)^(3/2)*a*b*c+2/3/d^3*(f*x+e)^(3/2)*b^2*c^2-2*f/d^2*a^2*c*(f*x+e)^(1
/2)+2/d*a^2*e*(f*x+e)^(1/2)+4*f/d^3*a*b*c^2*(f*x+e)^(1/2)-4/d^2*a*b*c*e*(f*x+e)^(1/2)-2*f/d^4*b^2*c^3*(f*x+e)^
(1/2)+2/d^3*b^2*c^2*e*(f*x+e)^(1/2)+2*f^2/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*c^2-4*f/d/((c*f-d*e)*d)^(1/2)*arctan((f*x+e)^(1/2)/((c*f-d*e)*d)^(1/2)*d)*a^2*c*e+2/((c*f-d*e)*d)^(1/2)*ar
ctan((f*x+e)^(1/2)/((c*f-d*e)*d)^(1/2)*d)*a^2*e^2-4*f^2/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*c^3+8*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*b*c^2*e-4/d/((
c*f-d*e)*d)^(1/2)*arctan((f*x+e)^(1/2)/((c*f-d*e)*d)^(1/2)*d)*a*b*c*e^2+2*f^2/d^4/((c*f-d*e)*d)^(1/2)*arctan((
f*x+e)^(1/2)/((c*f-d*e)*d)^(1/2)*d)*b^2*c^4-4*f/d^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^3*e+2/d^2/((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*e^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*(f*x+e)^(3/2)/(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 = 1.26, size = 438, normalized size = 2.55 \begin {gather*} {\left (e+f\,x\right )}^{3/2}\,\left (\frac {2\,{\left (a\,f-b\,e\right )}^2}{3\,d\,f^2}+\frac {\left (\frac {4\,b^2\,e-4\,a\,b\,f}{d\,f^2}+\frac {2\,b^2\,\left (c\,f^3-d\,e\,f^2\right )}{d^2\,f^4}\right )\,\left (c\,f^3-d\,e\,f^2\right )}{3\,d\,f^2}\right )-{\left (e+f\,x\right )}^{5/2}\,\left (\frac {4\,b^2\,e-4\,a\,b\,f}{5\,d\,f^2}+\frac {2\,b^2\,\left (c\,f^3-d\,e\,f^2\right )}{5\,d^2\,f^4}\right )+\frac {2\,b^2\,{\left (e+f\,x\right )}^{7/2}}{7\,d\,f^2}+\frac {2\,\mathrm {atan}\left (\frac {\sqrt {d}\,\sqrt {e+f\,x}\,{\left (a\,d-b\,c\right )}^2\,{\left (c\,f-d\,e\right )}^{3/2}}{a^2\,c^2\,d^2\,f^2-2\,a^2\,c\,d^3\,e\,f+a^2\,d^4\,e^2-2\,a\,b\,c^3\,d\,f^2+4\,a\,b\,c^2\,d^2\,e\,f-2\,a\,b\,c\,d^3\,e^2+b^2\,c^4\,f^2-2\,b^2\,c^3\,d\,e\,f+b^2\,c^2\,d^2\,e^2}\right )\,{\left (a\,d-b\,c\right )}^2\,{\left (c\,f-d\,e\right )}^{3/2}}{d^{9/2}}-\frac {\sqrt {e+f\,x}\,\left (\frac {2\,{\left (a\,f-b\,e\right )}^2}{d\,f^2}+\frac {\left (\frac {4\,b^2\,e-4\,a\,b\,f}{d\,f^2}+\frac {2\,b^2\,\left (c\,f^3-d\,e\,f^2\right )}{d^2\,f^4}\right )\,\left (c\,f^3-d\,e\,f^2\right )}{d\,f^2}\right )\,\left (c\,f^3-d\,e\,f^2\right )}{d\,f^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 70.19, size = 236, normalized size = 1.37 \begin {gather*} \frac {2 b^{2} \left (e + f x\right )^{\frac {7}{2}}}{7 d f^{2}} + \frac {\left (e + f x\right )^{\frac {5}{2}} \left (4 a b d f - 2 b^{2} c f - 2 b^{2} d e\right )}{5 d^{2} f^{2}} + \frac {\left (e + f x\right )^{\frac {3}{2}} \left (2 a^{2} d^{2} - 4 a b c d + 2 b^{2} c^{2}\right )}{3 d^{3}} + \frac {\sqrt {e + f x} \left (- 2 a^{2} c d^{2} f + 2 a^{2} d^{3} e + 4 a b c^{2} d f - 4 a b c d^{2} e - 2 b^{2} c^{3} f + 2 b^{2} c^{2} d e\right )}{d^{4}} + \frac {2 \left (a d - b c\right )^{2} \left (c f - d e\right )^{2} \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)**2*(f*x+e)**(3/2)/(d*x+c),x)

[Out]

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

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