3.17.21 \(\int \frac {(A+B x) (d+e x)^{3/2}}{(a+b x)^3} \, dx\)

Optimal. Leaf size=197 \[ -\frac {3 e (-5 a B e+A b e+4 b B d) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{4 b^{7/2} \sqrt {b d-a e}}+\frac {3 e \sqrt {d+e x} (-5 a B e+A b e+4 b B d)}{4 b^3 (b d-a e)}-\frac {(d+e x)^{3/2} (-5 a B e+A b e+4 b B d)}{4 b^2 (a+b x) (b d-a e)}-\frac {(d+e x)^{5/2} (A b-a B)}{2 b (a+b x)^2 (b d-a e)} \]

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Rubi [A]  time = 0.15, antiderivative size = 197, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {78, 47, 50, 63, 208} \begin {gather*} -\frac {(d+e x)^{3/2} (-5 a B e+A b e+4 b B d)}{4 b^2 (a+b x) (b d-a e)}+\frac {3 e \sqrt {d+e x} (-5 a B e+A b e+4 b B d)}{4 b^3 (b d-a e)}-\frac {3 e (-5 a B e+A b e+4 b B d) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{4 b^{7/2} \sqrt {b d-a e}}-\frac {(d+e x)^{5/2} (A b-a B)}{2 b (a+b x)^2 (b d-a e)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 47

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

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 78

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

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) (d+e x)^{3/2}}{(a+b x)^3} \, dx &=-\frac {(A b-a B) (d+e x)^{5/2}}{2 b (b d-a e) (a+b x)^2}+\frac {(4 b B d+A b e-5 a B e) \int \frac {(d+e x)^{3/2}}{(a+b x)^2} \, dx}{4 b (b d-a e)}\\ &=-\frac {(4 b B d+A b e-5 a B e) (d+e x)^{3/2}}{4 b^2 (b d-a e) (a+b x)}-\frac {(A b-a B) (d+e x)^{5/2}}{2 b (b d-a e) (a+b x)^2}+\frac {(3 e (4 b B d+A b e-5 a B e)) \int \frac {\sqrt {d+e x}}{a+b x} \, dx}{8 b^2 (b d-a e)}\\ &=\frac {3 e (4 b B d+A b e-5 a B e) \sqrt {d+e x}}{4 b^3 (b d-a e)}-\frac {(4 b B d+A b e-5 a B e) (d+e x)^{3/2}}{4 b^2 (b d-a e) (a+b x)}-\frac {(A b-a B) (d+e x)^{5/2}}{2 b (b d-a e) (a+b x)^2}+\frac {(3 e (4 b B d+A b e-5 a B e)) \int \frac {1}{(a+b x) \sqrt {d+e x}} \, dx}{8 b^3}\\ &=\frac {3 e (4 b B d+A b e-5 a B e) \sqrt {d+e x}}{4 b^3 (b d-a e)}-\frac {(4 b B d+A b e-5 a B e) (d+e x)^{3/2}}{4 b^2 (b d-a e) (a+b x)}-\frac {(A b-a B) (d+e x)^{5/2}}{2 b (b d-a e) (a+b x)^2}+\frac {(3 (4 b B d+A b e-5 a B e)) \operatorname {Subst}\left (\int \frac {1}{a-\frac {b d}{e}+\frac {b x^2}{e}} \, dx,x,\sqrt {d+e x}\right )}{4 b^3}\\ &=\frac {3 e (4 b B d+A b e-5 a B e) \sqrt {d+e x}}{4 b^3 (b d-a e)}-\frac {(4 b B d+A b e-5 a B e) (d+e x)^{3/2}}{4 b^2 (b d-a e) (a+b x)}-\frac {(A b-a B) (d+e x)^{5/2}}{2 b (b d-a e) (a+b x)^2}-\frac {3 e (4 b B d+A b e-5 a B e) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{4 b^{7/2} \sqrt {b d-a e}}\\ \end {align*}

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Mathematica [C]  time = 0.09, size = 96, normalized size = 0.49 \begin {gather*} \frac {(d+e x)^{5/2} \left (\frac {e (-5 a B e+A b e+4 b B d) \, _2F_1\left (2,\frac {5}{2};\frac {7}{2};\frac {b (d+e x)}{b d-a e}\right )}{(b d-a e)^2}+\frac {5 a B-5 A b}{(a+b x)^2}\right )}{10 b (b d-a e)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((d + e*x)^(5/2)*((-5*A*b + 5*a*B)/(a + b*x)^2 + (e*(4*b*B*d + A*b*e - 5*a*B*e)*Hypergeometric2F1[2, 5/2, 7/2,
(b*(d + e*x))/(b*d - a*e)])/(b*d - a*e)^2))/(10*b*(b*d - a*e))

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IntegrateAlgebraic [A]  time = 0.81, size = 204, normalized size = 1.04 \begin {gather*} -\frac {e \sqrt {d+e x} \left (-15 a^2 B e^2+3 a A b e^2-25 a b B e (d+e x)+27 a b B d e+5 A b^2 e (d+e x)-3 A b^2 d e-12 b^2 B d^2-8 b^2 B (d+e x)^2+20 b^2 B d (d+e x)\right )}{4 b^3 (a e+b (d+e x)-b d)^2}-\frac {3 \left (-5 a B e^2+A b e^2+4 b B d e\right ) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x} \sqrt {a e-b d}}{b d-a e}\right )}{4 b^{7/2} \sqrt {a e-b d}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((A + B*x)*(d + e*x)^(3/2))/(a + b*x)^3,x]

[Out]

-1/4*(e*Sqrt[d + e*x]*(-12*b^2*B*d^2 - 3*A*b^2*d*e + 27*a*b*B*d*e + 3*a*A*b*e^2 - 15*a^2*B*e^2 + 20*b^2*B*d*(d
+ e*x) + 5*A*b^2*e*(d + e*x) - 25*a*b*B*e*(d + e*x) - 8*b^2*B*(d + e*x)^2))/(b^3*(-(b*d) + a*e + b*(d + e*x))
^2) - (3*(4*b*B*d*e + A*b*e^2 - 5*a*B*e^2)*ArcTan[(Sqrt[b]*Sqrt[-(b*d) + a*e]*Sqrt[d + e*x])/(b*d - a*e)])/(4*
b^(7/2)*Sqrt[-(b*d) + a*e])

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fricas [A]  time = 0.91, size = 703, normalized size = 3.57 \begin {gather*} \left [\frac {3 \, {\left (4 \, B a^{2} b d e - {\left (5 \, B a^{3} - A a^{2} b\right )} e^{2} + {\left (4 \, B b^{3} d e - {\left (5 \, B a b^{2} - A b^{3}\right )} e^{2}\right )} x^{2} + 2 \, {\left (4 \, B a b^{2} d e - {\left (5 \, B a^{2} b - A a b^{2}\right )} e^{2}\right )} x\right )} \sqrt {b^{2} d - a b e} \log \left (\frac {b e x + 2 \, b d - a e - 2 \, \sqrt {b^{2} d - a b e} \sqrt {e x + d}}{b x + a}\right ) - 2 \, {\left (2 \, {\left (B a b^{3} + A b^{4}\right )} d^{2} - {\left (17 \, B a^{2} b^{2} - A a b^{3}\right )} d e + 3 \, {\left (5 \, B a^{3} b - A a^{2} b^{2}\right )} e^{2} - 8 \, {\left (B b^{4} d e - B a b^{3} e^{2}\right )} x^{2} + {\left (4 \, B b^{4} d^{2} - {\left (29 \, B a b^{3} - 5 \, A b^{4}\right )} d e + 5 \, {\left (5 \, B a^{2} b^{2} - A a b^{3}\right )} e^{2}\right )} x\right )} \sqrt {e x + d}}{8 \, {\left (a^{2} b^{5} d - a^{3} b^{4} e + {\left (b^{7} d - a b^{6} e\right )} x^{2} + 2 \, {\left (a b^{6} d - a^{2} b^{5} e\right )} x\right )}}, \frac {3 \, {\left (4 \, B a^{2} b d e - {\left (5 \, B a^{3} - A a^{2} b\right )} e^{2} + {\left (4 \, B b^{3} d e - {\left (5 \, B a b^{2} - A b^{3}\right )} e^{2}\right )} x^{2} + 2 \, {\left (4 \, B a b^{2} d e - {\left (5 \, B a^{2} b - A a b^{2}\right )} e^{2}\right )} x\right )} \sqrt {-b^{2} d + a b e} \arctan \left (\frac {\sqrt {-b^{2} d + a b e} \sqrt {e x + d}}{b e x + b d}\right ) - {\left (2 \, {\left (B a b^{3} + A b^{4}\right )} d^{2} - {\left (17 \, B a^{2} b^{2} - A a b^{3}\right )} d e + 3 \, {\left (5 \, B a^{3} b - A a^{2} b^{2}\right )} e^{2} - 8 \, {\left (B b^{4} d e - B a b^{3} e^{2}\right )} x^{2} + {\left (4 \, B b^{4} d^{2} - {\left (29 \, B a b^{3} - 5 \, A b^{4}\right )} d e + 5 \, {\left (5 \, B a^{2} b^{2} - A a b^{3}\right )} e^{2}\right )} x\right )} \sqrt {e x + d}}{4 \, {\left (a^{2} b^{5} d - a^{3} b^{4} e + {\left (b^{7} d - a b^{6} e\right )} x^{2} + 2 \, {\left (a b^{6} d - a^{2} b^{5} e\right )} x\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/8*(3*(4*B*a^2*b*d*e - (5*B*a^3 - A*a^2*b)*e^2 + (4*B*b^3*d*e - (5*B*a*b^2 - A*b^3)*e^2)*x^2 + 2*(4*B*a*b^2*
d*e - (5*B*a^2*b - A*a*b^2)*e^2)*x)*sqrt(b^2*d - a*b*e)*log((b*e*x + 2*b*d - a*e - 2*sqrt(b^2*d - a*b*e)*sqrt(
e*x + d))/(b*x + a)) - 2*(2*(B*a*b^3 + A*b^4)*d^2 - (17*B*a^2*b^2 - A*a*b^3)*d*e + 3*(5*B*a^3*b - A*a^2*b^2)*e
^2 - 8*(B*b^4*d*e - B*a*b^3*e^2)*x^2 + (4*B*b^4*d^2 - (29*B*a*b^3 - 5*A*b^4)*d*e + 5*(5*B*a^2*b^2 - A*a*b^3)*e
^2)*x)*sqrt(e*x + d))/(a^2*b^5*d - a^3*b^4*e + (b^7*d - a*b^6*e)*x^2 + 2*(a*b^6*d - a^2*b^5*e)*x), 1/4*(3*(4*B
*a^2*b*d*e - (5*B*a^3 - A*a^2*b)*e^2 + (4*B*b^3*d*e - (5*B*a*b^2 - A*b^3)*e^2)*x^2 + 2*(4*B*a*b^2*d*e - (5*B*a
^2*b - A*a*b^2)*e^2)*x)*sqrt(-b^2*d + a*b*e)*arctan(sqrt(-b^2*d + a*b*e)*sqrt(e*x + d)/(b*e*x + b*d)) - (2*(B*
a*b^3 + A*b^4)*d^2 - (17*B*a^2*b^2 - A*a*b^3)*d*e + 3*(5*B*a^3*b - A*a^2*b^2)*e^2 - 8*(B*b^4*d*e - B*a*b^3*e^2
)*x^2 + (4*B*b^4*d^2 - (29*B*a*b^3 - 5*A*b^4)*d*e + 5*(5*B*a^2*b^2 - A*a*b^3)*e^2)*x)*sqrt(e*x + d))/(a^2*b^5*
d - a^3*b^4*e + (b^7*d - a*b^6*e)*x^2 + 2*(a*b^6*d - a^2*b^5*e)*x)]

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giac [A]  time = 1.31, size = 236, normalized size = 1.20 \begin {gather*} \frac {2 \, \sqrt {x e + d} B e}{b^{3}} + \frac {3 \, {\left (4 \, B b d e - 5 \, B a e^{2} + A b e^{2}\right )} \arctan \left (\frac {\sqrt {x e + d} b}{\sqrt {-b^{2} d + a b e}}\right )}{4 \, \sqrt {-b^{2} d + a b e} b^{3}} - \frac {4 \, {\left (x e + d\right )}^{\frac {3}{2}} B b^{2} d e - 4 \, \sqrt {x e + d} B b^{2} d^{2} e - 9 \, {\left (x e + d\right )}^{\frac {3}{2}} B a b e^{2} + 5 \, {\left (x e + d\right )}^{\frac {3}{2}} A b^{2} e^{2} + 11 \, \sqrt {x e + d} B a b d e^{2} - 3 \, \sqrt {x e + d} A b^{2} d e^{2} - 7 \, \sqrt {x e + d} B a^{2} e^{3} + 3 \, \sqrt {x e + d} A a b e^{3}}{4 \, {\left ({\left (x e + d\right )} b - b d + a e\right )}^{2} b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*sqrt(x*e + d)*B*e/b^3 + 3/4*(4*B*b*d*e - 5*B*a*e^2 + A*b*e^2)*arctan(sqrt(x*e + d)*b/sqrt(-b^2*d + a*b*e))/(
sqrt(-b^2*d + a*b*e)*b^3) - 1/4*(4*(x*e + d)^(3/2)*B*b^2*d*e - 4*sqrt(x*e + d)*B*b^2*d^2*e - 9*(x*e + d)^(3/2)
*B*a*b*e^2 + 5*(x*e + d)^(3/2)*A*b^2*e^2 + 11*sqrt(x*e + d)*B*a*b*d*e^2 - 3*sqrt(x*e + d)*A*b^2*d*e^2 - 7*sqrt
(x*e + d)*B*a^2*e^3 + 3*sqrt(x*e + d)*A*a*b*e^3)/(((x*e + d)*b - b*d + a*e)^2*b^3)

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maple [B]  time = 0.02, size = 360, normalized size = 1.83 \begin {gather*} -\frac {3 \sqrt {e x +d}\, A a \,e^{3}}{4 \left (b x e +a e \right )^{2} b^{2}}+\frac {3 \sqrt {e x +d}\, A d \,e^{2}}{4 \left (b x e +a e \right )^{2} b}+\frac {7 \sqrt {e x +d}\, B \,a^{2} e^{3}}{4 \left (b x e +a e \right )^{2} b^{3}}-\frac {11 \sqrt {e x +d}\, B a d \,e^{2}}{4 \left (b x e +a e \right )^{2} b^{2}}+\frac {\sqrt {e x +d}\, B \,d^{2} e}{\left (b x e +a e \right )^{2} b}-\frac {5 \left (e x +d \right )^{\frac {3}{2}} A \,e^{2}}{4 \left (b x e +a e \right )^{2} b}+\frac {3 A \,e^{2} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )}{4 \sqrt {\left (a e -b d \right ) b}\, b^{2}}+\frac {9 \left (e x +d \right )^{\frac {3}{2}} B a \,e^{2}}{4 \left (b x e +a e \right )^{2} b^{2}}-\frac {15 B a \,e^{2} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )}{4 \sqrt {\left (a e -b d \right ) b}\, b^{3}}-\frac {\left (e x +d \right )^{\frac {3}{2}} B d e}{\left (b x e +a e \right )^{2} b}+\frac {3 B d e \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )}{\sqrt {\left (a e -b d \right ) b}\, b^{2}}+\frac {2 \sqrt {e x +d}\, B e}{b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(e*x+d)^(3/2)/(b*x+a)^3,x)

[Out]

2*e*B/b^3*(e*x+d)^(1/2)-5/4/b/(b*e*x+a*e)^2*(e*x+d)^(3/2)*A*e^2+9/4/b^2/(b*e*x+a*e)^2*(e*x+d)^(3/2)*B*a*e^2-e/
b/(b*e*x+a*e)^2*(e*x+d)^(3/2)*B*d-3/4/b^2/(b*e*x+a*e)^2*(e*x+d)^(1/2)*A*a*e^3+3/4/b/(b*e*x+a*e)^2*(e*x+d)^(1/2
)*A*d*e^2+7/4/b^3/(b*e*x+a*e)^2*(e*x+d)^(1/2)*B*a^2*e^3-11/4/b^2/(b*e*x+a*e)^2*(e*x+d)^(1/2)*B*a*d*e^2+e/b/(b*
e*x+a*e)^2*(e*x+d)^(1/2)*B*d^2+3/4/b^2/((a*e-b*d)*b)^(1/2)*arctan((e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2)*b)*A*e^2-1
5/4/b^3/((a*e-b*d)*b)^(1/2)*arctan((e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2)*b)*B*a*e^2+3*e/b^2/((a*e-b*d)*b)^(1/2)*ar
ctan((e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2)*b)*B*d

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

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mupad [B]  time = 1.30, size = 256, normalized size = 1.30 \begin {gather*} \frac {\sqrt {d+e\,x}\,\left (\frac {7\,B\,a^2\,e^3}{4}-\frac {11\,B\,a\,b\,d\,e^2}{4}-\frac {3\,A\,a\,b\,e^3}{4}+B\,b^2\,d^2\,e+\frac {3\,A\,b^2\,d\,e^2}{4}\right )-{\left (d+e\,x\right )}^{3/2}\,\left (\frac {5\,A\,b^2\,e^2}{4}+B\,d\,b^2\,e-\frac {9\,B\,a\,b\,e^2}{4}\right )}{b^5\,{\left (d+e\,x\right )}^2-\left (2\,b^5\,d-2\,a\,b^4\,e\right )\,\left (d+e\,x\right )+b^5\,d^2+a^2\,b^3\,e^2-2\,a\,b^4\,d\,e}+\frac {2\,B\,e\,\sqrt {d+e\,x}}{b^3}+\frac {3\,e\,\mathrm {atan}\left (\frac {\sqrt {b}\,e\,\sqrt {d+e\,x}\,\left (A\,b\,e-5\,B\,a\,e+4\,B\,b\,d\right )}{\sqrt {a\,e-b\,d}\,\left (A\,b\,e^2-5\,B\,a\,e^2+4\,B\,b\,d\,e\right )}\right )\,\left (A\,b\,e-5\,B\,a\,e+4\,B\,b\,d\right )}{4\,b^{7/2}\,\sqrt {a\,e-b\,d}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*x)*(d + e*x)^(3/2))/(a + b*x)^3,x)

[Out]

((d + e*x)^(1/2)*((7*B*a^2*e^3)/4 - (3*A*a*b*e^3)/4 + (3*A*b^2*d*e^2)/4 + B*b^2*d^2*e - (11*B*a*b*d*e^2)/4) -
(d + e*x)^(3/2)*((5*A*b^2*e^2)/4 - (9*B*a*b*e^2)/4 + B*b^2*d*e))/(b^5*(d + e*x)^2 - (2*b^5*d - 2*a*b^4*e)*(d +
e*x) + b^5*d^2 + a^2*b^3*e^2 - 2*a*b^4*d*e) + (2*B*e*(d + e*x)^(1/2))/b^3 + (3*e*atan((b^(1/2)*e*(d + e*x)^(1
/2)*(A*b*e - 5*B*a*e + 4*B*b*d))/((a*e - b*d)^(1/2)*(A*b*e^2 - 5*B*a*e^2 + 4*B*b*d*e)))*(A*b*e - 5*B*a*e + 4*B
*b*d))/(4*b^(7/2)*(a*e - b*d)^(1/2))

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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)*(e*x+d)**(3/2)/(b*x+a)**3,x)

[Out]

Timed out

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