3.21.77 \(\int \frac {(a+b x)^{3/2} (A+B x)}{(d+e x)^{5/2}} \, dx\)

Optimal. Leaf size=202 \[ -\frac {\sqrt {b} (-3 a B e-2 A b e+5 b B d) \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {a+b x}}{\sqrt {b} \sqrt {d+e x}}\right )}{e^{7/2}}+\frac {b \sqrt {a+b x} \sqrt {d+e x} (-3 a B e-2 A b e+5 b B d)}{e^3 (b d-a e)}-\frac {2 (a+b x)^{3/2} (-3 a B e-2 A b e+5 b B d)}{3 e^2 \sqrt {d+e x} (b d-a e)}-\frac {2 (a+b x)^{5/2} (B d-A e)}{3 e (d+e x)^{3/2} (b d-a e)} \]

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Rubi [A]  time = 0.16, antiderivative size = 202, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {78, 47, 50, 63, 217, 206} \begin {gather*} -\frac {2 (a+b x)^{3/2} (-3 a B e-2 A b e+5 b B d)}{3 e^2 \sqrt {d+e x} (b d-a e)}+\frac {b \sqrt {a+b x} \sqrt {d+e x} (-3 a B e-2 A b e+5 b B d)}{e^3 (b d-a e)}-\frac {\sqrt {b} (-3 a B e-2 A b e+5 b B d) \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {a+b x}}{\sqrt {b} \sqrt {d+e x}}\right )}{e^{7/2}}-\frac {2 (a+b x)^{5/2} (B d-A e)}{3 e (d+e x)^{3/2} (b d-a e)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((a + b*x)^(3/2)*(A + B*x))/(d + e*x)^(5/2),x]

[Out]

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

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 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rubi steps

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

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Mathematica [C]  time = 0.18, size = 113, normalized size = 0.56 \begin {gather*} \frac {2 (a+b x)^{5/2} \left (-\frac {\left (\frac {b (d+e x)}{b d-a e}\right )^{3/2} (-3 a B e-2 A b e+5 b B d) \, _2F_1\left (\frac {3}{2},\frac {5}{2};\frac {7}{2};\frac {e (a+b x)}{a e-b d}\right )}{b}-5 A e+5 B d\right )}{15 e (d+e x)^{3/2} (a e-b d)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x)^(3/2)*(A + B*x))/(d + e*x)^(5/2),x]

[Out]

(2*(a + b*x)^(5/2)*(5*B*d - 5*A*e - ((5*b*B*d - 2*A*b*e - 3*a*B*e)*((b*(d + e*x))/(b*d - a*e))^(3/2)*Hypergeom
etric2F1[3/2, 5/2, 7/2, (e*(a + b*x))/(-(b*d) + a*e)])/b))/(15*e*(-(b*d) + a*e)*(d + e*x)^(3/2))

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IntegrateAlgebraic [A]  time = 0.30, size = 243, normalized size = 1.20 \begin {gather*} \frac {\left (3 a \sqrt {b} B e+2 A b^{3/2} e-5 b^{3/2} B d\right ) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {e} \sqrt {a+b x}}\right )}{e^{7/2}}+\frac {(a+b x)^{3/2} \left (-\frac {6 A b^2 e (d+e x)^2}{(a+b x)^2}+\frac {4 A b e^2 (d+e x)}{a+b x}+\frac {15 b^2 B d (d+e x)^2}{(a+b x)^2}+\frac {6 a B e^2 (d+e x)}{a+b x}-\frac {9 a b B e (d+e x)^2}{(a+b x)^2}-\frac {10 b B d e (d+e x)}{a+b x}+2 A e^3-2 B d e^2\right )}{3 e^3 (d+e x)^{3/2} \left (\frac {b (d+e x)}{a+b x}-e\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((a + b*x)^(3/2)*(A + B*x))/(d + e*x)^(5/2),x]

[Out]

((a + b*x)^(3/2)*(-2*B*d*e^2 + 2*A*e^3 - (10*b*B*d*e*(d + e*x))/(a + b*x) + (4*A*b*e^2*(d + e*x))/(a + b*x) +
(6*a*B*e^2*(d + e*x))/(a + b*x) + (15*b^2*B*d*(d + e*x)^2)/(a + b*x)^2 - (6*A*b^2*e*(d + e*x)^2)/(a + b*x)^2 -
(9*a*b*B*e*(d + e*x)^2)/(a + b*x)^2))/(3*e^3*(d + e*x)^(3/2)*(-e + (b*(d + e*x))/(a + b*x))) + ((-5*b^(3/2)*B
*d + 2*A*b^(3/2)*e + 3*a*Sqrt[b]*B*e)*ArcTanh[(Sqrt[b]*Sqrt[d + e*x])/(Sqrt[e]*Sqrt[a + b*x])])/e^(7/2)

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fricas [A]  time = 5.39, size = 537, normalized size = 2.66 \begin {gather*} \left [-\frac {3 \, {\left (5 \, B b d^{3} - {\left (3 \, B a + 2 \, A b\right )} d^{2} e + {\left (5 \, B b d e^{2} - {\left (3 \, B a + 2 \, A b\right )} e^{3}\right )} x^{2} + 2 \, {\left (5 \, B b d^{2} e - {\left (3 \, B a + 2 \, A b\right )} d e^{2}\right )} x\right )} \sqrt {\frac {b}{e}} \log \left (8 \, b^{2} e^{2} x^{2} + b^{2} d^{2} + 6 \, a b d e + a^{2} e^{2} + 4 \, {\left (2 \, b e^{2} x + b d e + a e^{2}\right )} \sqrt {b x + a} \sqrt {e x + d} \sqrt {\frac {b}{e}} + 8 \, {\left (b^{2} d e + a b e^{2}\right )} x\right ) - 4 \, {\left (3 \, B b e^{2} x^{2} + 15 \, B b d^{2} - 2 \, A a e^{2} - 2 \, {\left (2 \, B a + 3 \, A b\right )} d e + 2 \, {\left (10 \, B b d e - {\left (3 \, B a + 4 \, A b\right )} e^{2}\right )} x\right )} \sqrt {b x + a} \sqrt {e x + d}}{12 \, {\left (e^{5} x^{2} + 2 \, d e^{4} x + d^{2} e^{3}\right )}}, \frac {3 \, {\left (5 \, B b d^{3} - {\left (3 \, B a + 2 \, A b\right )} d^{2} e + {\left (5 \, B b d e^{2} - {\left (3 \, B a + 2 \, A b\right )} e^{3}\right )} x^{2} + 2 \, {\left (5 \, B b d^{2} e - {\left (3 \, B a + 2 \, A b\right )} d e^{2}\right )} x\right )} \sqrt {-\frac {b}{e}} \arctan \left (\frac {{\left (2 \, b e x + b d + a e\right )} \sqrt {b x + a} \sqrt {e x + d} \sqrt {-\frac {b}{e}}}{2 \, {\left (b^{2} e x^{2} + a b d + {\left (b^{2} d + a b e\right )} x\right )}}\right ) + 2 \, {\left (3 \, B b e^{2} x^{2} + 15 \, B b d^{2} - 2 \, A a e^{2} - 2 \, {\left (2 \, B a + 3 \, A b\right )} d e + 2 \, {\left (10 \, B b d e - {\left (3 \, B a + 4 \, A b\right )} e^{2}\right )} x\right )} \sqrt {b x + a} \sqrt {e x + d}}{6 \, {\left (e^{5} x^{2} + 2 \, d e^{4} x + d^{2} e^{3}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(3/2)*(B*x+A)/(e*x+d)^(5/2),x, algorithm="fricas")

[Out]

[-1/12*(3*(5*B*b*d^3 - (3*B*a + 2*A*b)*d^2*e + (5*B*b*d*e^2 - (3*B*a + 2*A*b)*e^3)*x^2 + 2*(5*B*b*d^2*e - (3*B
*a + 2*A*b)*d*e^2)*x)*sqrt(b/e)*log(8*b^2*e^2*x^2 + b^2*d^2 + 6*a*b*d*e + a^2*e^2 + 4*(2*b*e^2*x + b*d*e + a*e
^2)*sqrt(b*x + a)*sqrt(e*x + d)*sqrt(b/e) + 8*(b^2*d*e + a*b*e^2)*x) - 4*(3*B*b*e^2*x^2 + 15*B*b*d^2 - 2*A*a*e
^2 - 2*(2*B*a + 3*A*b)*d*e + 2*(10*B*b*d*e - (3*B*a + 4*A*b)*e^2)*x)*sqrt(b*x + a)*sqrt(e*x + d))/(e^5*x^2 + 2
*d*e^4*x + d^2*e^3), 1/6*(3*(5*B*b*d^3 - (3*B*a + 2*A*b)*d^2*e + (5*B*b*d*e^2 - (3*B*a + 2*A*b)*e^3)*x^2 + 2*(
5*B*b*d^2*e - (3*B*a + 2*A*b)*d*e^2)*x)*sqrt(-b/e)*arctan(1/2*(2*b*e*x + b*d + a*e)*sqrt(b*x + a)*sqrt(e*x + d
)*sqrt(-b/e)/(b^2*e*x^2 + a*b*d + (b^2*d + a*b*e)*x)) + 2*(3*B*b*e^2*x^2 + 15*B*b*d^2 - 2*A*a*e^2 - 2*(2*B*a +
3*A*b)*d*e + 2*(10*B*b*d*e - (3*B*a + 4*A*b)*e^2)*x)*sqrt(b*x + a)*sqrt(e*x + d))/(e^5*x^2 + 2*d*e^4*x + d^2*
e^3)]

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giac [B]  time = 2.09, size = 352, normalized size = 1.74 \begin {gather*} \frac {{\left (5 \, B b d {\left | b \right |} - 3 \, B a {\left | b \right |} e - 2 \, A b {\left | b \right |} e\right )} e^{\left (-\frac {7}{2}\right )} \log \left ({\left | -\sqrt {b x + a} \sqrt {b} e^{\frac {1}{2}} + \sqrt {b^{2} d + {\left (b x + a\right )} b e - a b e} \right |}\right )}{\sqrt {b}} + \frac {{\left ({\left (b x + a\right )} {\left (\frac {3 \, {\left (B b^{5} d {\left | b \right |} e^{4} - B a b^{4} {\left | b \right |} e^{5}\right )} {\left (b x + a\right )}}{b^{4} d e^{5} - a b^{3} e^{6}} + \frac {4 \, {\left (5 \, B b^{6} d^{2} {\left | b \right |} e^{3} - 8 \, B a b^{5} d {\left | b \right |} e^{4} - 2 \, A b^{6} d {\left | b \right |} e^{4} + 3 \, B a^{2} b^{4} {\left | b \right |} e^{5} + 2 \, A a b^{5} {\left | b \right |} e^{5}\right )}}{b^{4} d e^{5} - a b^{3} e^{6}}\right )} + \frac {3 \, {\left (5 \, B b^{7} d^{3} {\left | b \right |} e^{2} - 13 \, B a b^{6} d^{2} {\left | b \right |} e^{3} - 2 \, A b^{7} d^{2} {\left | b \right |} e^{3} + 11 \, B a^{2} b^{5} d {\left | b \right |} e^{4} + 4 \, A a b^{6} d {\left | b \right |} e^{4} - 3 \, B a^{3} b^{4} {\left | b \right |} e^{5} - 2 \, A a^{2} b^{5} {\left | b \right |} e^{5}\right )}}{b^{4} d e^{5} - a b^{3} e^{6}}\right )} \sqrt {b x + a}}{3 \, {\left (b^{2} d + {\left (b x + a\right )} b e - a b e\right )}^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(3/2)*(B*x+A)/(e*x+d)^(5/2),x, algorithm="giac")

[Out]

(5*B*b*d*abs(b) - 3*B*a*abs(b)*e - 2*A*b*abs(b)*e)*e^(-7/2)*log(abs(-sqrt(b*x + a)*sqrt(b)*e^(1/2) + sqrt(b^2*
d + (b*x + a)*b*e - a*b*e)))/sqrt(b) + 1/3*((b*x + a)*(3*(B*b^5*d*abs(b)*e^4 - B*a*b^4*abs(b)*e^5)*(b*x + a)/(
b^4*d*e^5 - a*b^3*e^6) + 4*(5*B*b^6*d^2*abs(b)*e^3 - 8*B*a*b^5*d*abs(b)*e^4 - 2*A*b^6*d*abs(b)*e^4 + 3*B*a^2*b
^4*abs(b)*e^5 + 2*A*a*b^5*abs(b)*e^5)/(b^4*d*e^5 - a*b^3*e^6)) + 3*(5*B*b^7*d^3*abs(b)*e^2 - 13*B*a*b^6*d^2*ab
s(b)*e^3 - 2*A*b^7*d^2*abs(b)*e^3 + 11*B*a^2*b^5*d*abs(b)*e^4 + 4*A*a*b^6*d*abs(b)*e^4 - 3*B*a^3*b^4*abs(b)*e^
5 - 2*A*a^2*b^5*abs(b)*e^5)/(b^4*d*e^5 - a*b^3*e^6))*sqrt(b*x + a)/(b^2*d + (b*x + a)*b*e - a*b*e)^(3/2)

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maple [B]  time = 0.02, size = 698, normalized size = 3.46 \begin {gather*} \frac {\sqrt {b x +a}\, \left (6 A \,b^{2} e^{3} x^{2} \ln \left (\frac {2 b e x +a e +b d +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}}{2 \sqrt {b e}}\right )+9 B a b \,e^{3} x^{2} \ln \left (\frac {2 b e x +a e +b d +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}}{2 \sqrt {b e}}\right )-15 B \,b^{2} d \,e^{2} x^{2} \ln \left (\frac {2 b e x +a e +b d +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}}{2 \sqrt {b e}}\right )+12 A \,b^{2} d \,e^{2} x \ln \left (\frac {2 b e x +a e +b d +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}}{2 \sqrt {b e}}\right )+18 B a b d \,e^{2} x \ln \left (\frac {2 b e x +a e +b d +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}}{2 \sqrt {b e}}\right )-30 B \,b^{2} d^{2} e x \ln \left (\frac {2 b e x +a e +b d +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}}{2 \sqrt {b e}}\right )+6 A \,b^{2} d^{2} e \ln \left (\frac {2 b e x +a e +b d +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}}{2 \sqrt {b e}}\right )+9 B a b \,d^{2} e \ln \left (\frac {2 b e x +a e +b d +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}}{2 \sqrt {b e}}\right )-15 B \,b^{2} d^{3} \ln \left (\frac {2 b e x +a e +b d +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}}{2 \sqrt {b e}}\right )+6 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}\, B b \,e^{2} x^{2}-16 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}\, A b \,e^{2} x -12 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}\, B a \,e^{2} x +40 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}\, B b d e x -4 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}\, A a \,e^{2}-12 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}\, A b d e -8 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}\, B a d e +30 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}\, B b \,d^{2}\right )}{6 \sqrt {b e}\, \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \left (e x +d \right )^{\frac {3}{2}} e^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^(3/2)*(B*x+A)/(e*x+d)^(5/2),x)

[Out]

1/6*(b*x+a)^(1/2)*(6*A*ln(1/2*(2*b*e*x+a*e+b*d+2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2))/(b*e)^(1/2))*x^2*b^2*e^3
+9*B*a*b*e^3*x^2*ln(1/2*(2*b*e*x+a*e+b*d+2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2))/(b*e)^(1/2))-15*B*b^2*d*e^2*x^
2*ln(1/2*(2*b*e*x+a*e+b*d+2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2))/(b*e)^(1/2))+12*A*b^2*d*e^2*x*ln(1/2*(2*b*e*x
+a*e+b*d+2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2))/(b*e)^(1/2))+18*B*a*b*d*e^2*x*ln(1/2*(2*b*e*x+a*e+b*d+2*((b*x+
a)*(e*x+d))^(1/2)*(b*e)^(1/2))/(b*e)^(1/2))-30*B*b^2*d^2*e*x*ln(1/2*(2*b*e*x+a*e+b*d+2*((b*x+a)*(e*x+d))^(1/2)
*(b*e)^(1/2))/(b*e)^(1/2))+6*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)*B*b*e^2*x^2+6*A*b^2*d^2*e*ln(1/2*(2*b*e*x+a*e
+b*d+2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2))/(b*e)^(1/2))-16*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)*A*b*e^2*x+9*B*
a*b*d^2*e*ln(1/2*(2*b*e*x+a*e+b*d+2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2))/(b*e)^(1/2))-15*B*b^2*d^3*ln(1/2*(2*b
*e*x+a*e+b*d+2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2))/(b*e)^(1/2))-12*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)*B*a*e^
2*x+40*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)*B*b*d*e*x-4*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)*A*a*e^2-12*((b*x+a)
*(e*x+d))^(1/2)*(b*e)^(1/2)*A*b*d*e-8*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)*B*a*d*e+30*((b*x+a)*(e*x+d))^(1/2)*(
b*e)^(1/2)*B*b*d^2)/(b*e)^(1/2)/((b*x+a)*(e*x+d))^(1/2)/e^3/(e*x+d)^(3/2)

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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)^(3/2)*(B*x+A)/(e*x+d)^(5/2),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(a*e-b*d>0)', see `assume?` for
more details)Is a*e-b*d zero or nonzero?

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {\left (A+B\,x\right )\,{\left (a+b\,x\right )}^{3/2}}{{\left (d+e\,x\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*x)*(a + b*x)^(3/2))/(d + e*x)^(5/2),x)

[Out]

int(((A + B*x)*(a + b*x)^(3/2))/(d + e*x)^(5/2), x)

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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)**(3/2)*(B*x+A)/(e*x+d)**(5/2),x)

[Out]

Timed out

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