3.21.78 \(\int \frac {(a+b x)^{3/2} (A+B x)}{(d+e x)^{7/2}} \, dx\)

Optimal. Leaf size=138 \[ -\frac {2 (a+b x)^{5/2} (B d-A e)}{5 e (d+e x)^{5/2} (b d-a e)}+\frac {2 b^{3/2} B \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {a+b x}}{\sqrt {b} \sqrt {d+e x}}\right )}{e^{7/2}}-\frac {2 b B \sqrt {a+b x}}{e^3 \sqrt {d+e x}}-\frac {2 B (a+b x)^{3/2}}{3 e^2 (d+e x)^{3/2}} \]

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Rubi [A]  time = 0.08, antiderivative size = 138, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {78, 47, 63, 217, 206} \begin {gather*} -\frac {2 (a+b x)^{5/2} (B d-A e)}{5 e (d+e x)^{5/2} (b d-a e)}+\frac {2 b^{3/2} B \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {a+b x}}{\sqrt {b} \sqrt {d+e x}}\right )}{e^{7/2}}-\frac {2 B (a+b x)^{3/2}}{3 e^2 (d+e x)^{3/2}}-\frac {2 b B \sqrt {a+b x}}{e^3 \sqrt {d+e x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((a + b*x)^(3/2)*(A + B*x))/(d + e*x)^(7/2),x]

[Out]

(-2*(B*d - A*e)*(a + b*x)^(5/2))/(5*e*(b*d - a*e)*(d + e*x)^(5/2)) - (2*B*(a + b*x)^(3/2))/(3*e^2*(d + e*x)^(3
/2)) - (2*b*B*Sqrt[a + b*x])/(e^3*Sqrt[d + e*x]) + (2*b^(3/2)*B*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 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)^{7/2}} \, dx &=-\frac {2 (B d-A e) (a+b x)^{5/2}}{5 e (b d-a e) (d+e x)^{5/2}}+\frac {B \int \frac {(a+b x)^{3/2}}{(d+e x)^{5/2}} \, dx}{e}\\ &=-\frac {2 (B d-A e) (a+b x)^{5/2}}{5 e (b d-a e) (d+e x)^{5/2}}-\frac {2 B (a+b x)^{3/2}}{3 e^2 (d+e x)^{3/2}}+\frac {(b B) \int \frac {\sqrt {a+b x}}{(d+e x)^{3/2}} \, dx}{e^2}\\ &=-\frac {2 (B d-A e) (a+b x)^{5/2}}{5 e (b d-a e) (d+e x)^{5/2}}-\frac {2 B (a+b x)^{3/2}}{3 e^2 (d+e x)^{3/2}}-\frac {2 b B \sqrt {a+b x}}{e^3 \sqrt {d+e x}}+\frac {\left (b^2 B\right ) \int \frac {1}{\sqrt {a+b x} \sqrt {d+e x}} \, dx}{e^3}\\ &=-\frac {2 (B d-A e) (a+b x)^{5/2}}{5 e (b d-a e) (d+e x)^{5/2}}-\frac {2 B (a+b x)^{3/2}}{3 e^2 (d+e x)^{3/2}}-\frac {2 b B \sqrt {a+b x}}{e^3 \sqrt {d+e x}}+\frac {(2 b B) \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}}{5 e (b d-a e) (d+e x)^{5/2}}-\frac {2 B (a+b x)^{3/2}}{3 e^2 (d+e x)^{3/2}}-\frac {2 b B \sqrt {a+b x}}{e^3 \sqrt {d+e x}}+\frac {(2 b B) \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}}{5 e (b d-a e) (d+e x)^{5/2}}-\frac {2 B (a+b x)^{3/2}}{3 e^2 (d+e x)^{3/2}}-\frac {2 b B \sqrt {a+b x}}{e^3 \sqrt {d+e x}}+\frac {2 b^{3/2} B \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 [A]  time = 1.49, size = 173, normalized size = 1.25 \begin {gather*} \frac {2 \left (e^3 (a+b x)^3 (A e-B d)+\frac {5}{3} B e (a+b x) (d+e x) (a e-b d) (a e+3 b d+4 b e x)+\frac {5 B \sqrt {e} \sqrt {a+b x} (b d-a e)^{7/2} \left (\frac {b (d+e x)}{b d-a e}\right )^{5/2} \sinh ^{-1}\left (\frac {\sqrt {e} \sqrt {a+b x}}{\sqrt {b d-a e}}\right )}{b}\right )}{5 e^4 \sqrt {a+b x} (d+e x)^{5/2} (b d-a e)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x)^(3/2)*(A + B*x))/(d + e*x)^(7/2),x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.24, size = 171, normalized size = 1.24 \begin {gather*} \frac {2 b^{3/2} B \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {e} \sqrt {a+b x}}\right )}{e^{7/2}}-\frac {2 (a+b x)^{5/2} \left (-\frac {15 b^2 B d (d+e x)^2}{(a+b x)^2}+\frac {5 a B e^2 (d+e x)}{a+b x}+\frac {15 a b B e (d+e x)^2}{(a+b x)^2}-\frac {5 b B d e (d+e x)}{a+b x}+3 A e^3-3 B d e^2\right )}{15 e^3 (d+e x)^{5/2} (a e-b d)} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((a + b*x)^(3/2)*(A + B*x))/(d + e*x)^(7/2),x]

[Out]

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

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fricas [B]  time = 14.79, size = 767, normalized size = 5.56 \begin {gather*} \left [\frac {15 \, {\left (B b^{2} d^{4} - B a b d^{3} e + {\left (B b^{2} d e^{3} - B a b e^{4}\right )} x^{3} + 3 \, {\left (B b^{2} d^{2} e^{2} - B a b d e^{3}\right )} x^{2} + 3 \, {\left (B b^{2} d^{3} e - B a b d^{2} 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 (15 \, B b^{2} d^{3} - 10 \, B a b d^{2} e - 2 \, B a^{2} d e^{2} - 3 \, A a^{2} e^{3} + {\left (23 \, B b^{2} d e^{2} - {\left (20 \, B a b + 3 \, A b^{2}\right )} e^{3}\right )} x^{2} + {\left (35 \, B b^{2} d^{2} e - 24 \, B a b d e^{2} - {\left (5 \, B a^{2} + 6 \, A a b\right )} e^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {e x + d}}{30 \, {\left (b d^{4} e^{3} - a d^{3} e^{4} + {\left (b d e^{6} - a e^{7}\right )} x^{3} + 3 \, {\left (b d^{2} e^{5} - a d e^{6}\right )} x^{2} + 3 \, {\left (b d^{3} e^{4} - a d^{2} e^{5}\right )} x\right )}}, -\frac {15 \, {\left (B b^{2} d^{4} - B a b d^{3} e + {\left (B b^{2} d e^{3} - B a b e^{4}\right )} x^{3} + 3 \, {\left (B b^{2} d^{2} e^{2} - B a b d e^{3}\right )} x^{2} + 3 \, {\left (B b^{2} d^{3} e - B a b d^{2} 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 (15 \, B b^{2} d^{3} - 10 \, B a b d^{2} e - 2 \, B a^{2} d e^{2} - 3 \, A a^{2} e^{3} + {\left (23 \, B b^{2} d e^{2} - {\left (20 \, B a b + 3 \, A b^{2}\right )} e^{3}\right )} x^{2} + {\left (35 \, B b^{2} d^{2} e - 24 \, B a b d e^{2} - {\left (5 \, B a^{2} + 6 \, A a b\right )} e^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {e x + d}}{15 \, {\left (b d^{4} e^{3} - a d^{3} e^{4} + {\left (b d e^{6} - a e^{7}\right )} x^{3} + 3 \, {\left (b d^{2} e^{5} - a d e^{6}\right )} x^{2} + 3 \, {\left (b d^{3} e^{4} - a d^{2} e^{5}\right )} x\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)^(7/2),x, algorithm="fricas")

[Out]

[1/30*(15*(B*b^2*d^4 - B*a*b*d^3*e + (B*b^2*d*e^3 - B*a*b*e^4)*x^3 + 3*(B*b^2*d^2*e^2 - B*a*b*d*e^3)*x^2 + 3*(
B*b^2*d^3*e - B*a*b*d^2*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*(15*B*b^2*d^3 - 10*B*a*b*d^
2*e - 2*B*a^2*d*e^2 - 3*A*a^2*e^3 + (23*B*b^2*d*e^2 - (20*B*a*b + 3*A*b^2)*e^3)*x^2 + (35*B*b^2*d^2*e - 24*B*a
*b*d*e^2 - (5*B*a^2 + 6*A*a*b)*e^3)*x)*sqrt(b*x + a)*sqrt(e*x + d))/(b*d^4*e^3 - a*d^3*e^4 + (b*d*e^6 - a*e^7)
*x^3 + 3*(b*d^2*e^5 - a*d*e^6)*x^2 + 3*(b*d^3*e^4 - a*d^2*e^5)*x), -1/15*(15*(B*b^2*d^4 - B*a*b*d^3*e + (B*b^2
*d*e^3 - B*a*b*e^4)*x^3 + 3*(B*b^2*d^2*e^2 - B*a*b*d*e^3)*x^2 + 3*(B*b^2*d^3*e - B*a*b*d^2*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*(15*B*b^2*d^3 - 10*B*a*b*d^2*e - 2*B*a^2*d*e^2 - 3*A*a^2*e^3 + (23*B*b^2*d*e^2 - (20*B*a*b + 3*A*b^2)*e
^3)*x^2 + (35*B*b^2*d^2*e - 24*B*a*b*d*e^2 - (5*B*a^2 + 6*A*a*b)*e^3)*x)*sqrt(b*x + a)*sqrt(e*x + d))/(b*d^4*e
^3 - a*d^3*e^4 + (b*d*e^6 - a*e^7)*x^3 + 3*(b*d^2*e^5 - a*d*e^6)*x^2 + 3*(b*d^3*e^4 - a*d^2*e^5)*x)]

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giac [B]  time = 2.55, size = 375, normalized size = 2.72 \begin {gather*} -2 \, B \sqrt {b} {\left | b \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 ) - \frac {2 \, {\left ({\left (b x + a\right )} {\left (\frac {{\left (23 \, B b^{7} d^{2} {\left | b \right |} e^{4} - 43 \, B a b^{6} d {\left | b \right |} e^{5} - 3 \, A b^{7} d {\left | b \right |} e^{5} + 20 \, B a^{2} b^{5} {\left | b \right |} e^{6} + 3 \, A a b^{6} {\left | b \right |} e^{6}\right )} {\left (b x + a\right )}}{b^{4} d^{2} e^{5} - 2 \, a b^{3} d e^{6} + a^{2} b^{2} e^{7}} + \frac {35 \, {\left (B b^{8} d^{3} {\left | b \right |} e^{3} - 3 \, B a b^{7} d^{2} {\left | b \right |} e^{4} + 3 \, B a^{2} b^{6} d {\left | b \right |} e^{5} - B a^{3} b^{5} {\left | b \right |} e^{6}\right )}}{b^{4} d^{2} e^{5} - 2 \, a b^{3} d e^{6} + a^{2} b^{2} e^{7}}\right )} + \frac {15 \, {\left (B b^{9} d^{4} {\left | b \right |} e^{2} - 4 \, B a b^{8} d^{3} {\left | b \right |} e^{3} + 6 \, B a^{2} b^{7} d^{2} {\left | b \right |} e^{4} - 4 \, B a^{3} b^{6} d {\left | b \right |} e^{5} + B a^{4} b^{5} {\left | b \right |} e^{6}\right )}}{b^{4} d^{2} e^{5} - 2 \, a b^{3} d e^{6} + a^{2} b^{2} e^{7}}\right )} \sqrt {b x + a}}{15 \, {\left (b^{2} d + {\left (b x + a\right )} b e - a b e\right )}^{\frac {5}{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)^(7/2),x, algorithm="giac")

[Out]

-2*B*sqrt(b)*abs(b)*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))) -
2/15*((b*x + a)*((23*B*b^7*d^2*abs(b)*e^4 - 43*B*a*b^6*d*abs(b)*e^5 - 3*A*b^7*d*abs(b)*e^5 + 20*B*a^2*b^5*abs(
b)*e^6 + 3*A*a*b^6*abs(b)*e^6)*(b*x + a)/(b^4*d^2*e^5 - 2*a*b^3*d*e^6 + a^2*b^2*e^7) + 35*(B*b^8*d^3*abs(b)*e^
3 - 3*B*a*b^7*d^2*abs(b)*e^4 + 3*B*a^2*b^6*d*abs(b)*e^5 - B*a^3*b^5*abs(b)*e^6)/(b^4*d^2*e^5 - 2*a*b^3*d*e^6 +
a^2*b^2*e^7)) + 15*(B*b^9*d^4*abs(b)*e^2 - 4*B*a*b^8*d^3*abs(b)*e^3 + 6*B*a^2*b^7*d^2*abs(b)*e^4 - 4*B*a^3*b^
6*d*abs(b)*e^5 + B*a^4*b^5*abs(b)*e^6)/(b^4*d^2*e^5 - 2*a*b^3*d*e^6 + a^2*b^2*e^7))*sqrt(b*x + a)/(b^2*d + (b*
x + a)*b*e - a*b*e)^(5/2)

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maple [B]  time = 0.03, size = 780, normalized size = 5.65 \begin {gather*} -\frac {\sqrt {b x +a}\, \left (-15 B a \,b^{2} e^{4} x^{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 )+15 B \,b^{3} d \,e^{3} x^{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 )-45 B a \,b^{2} d \,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 )+45 B \,b^{3} d^{2} 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 )-45 B a \,b^{2} d^{2} 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 )+45 B \,b^{3} d^{3} 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 )-15 B a \,b^{2} d^{3} 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^{3} d^{4} \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}\, A \,b^{2} e^{3} x^{2}+40 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}\, B a b \,e^{3} x^{2}-46 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}\, B \,b^{2} d \,e^{2} x^{2}+12 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}\, A a b \,e^{3} x +10 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}\, B \,a^{2} e^{3} x +48 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}\, B a b d \,e^{2} x -70 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}\, B \,b^{2} d^{2} e x +6 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}\, A \,a^{2} e^{3}+4 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}\, B \,a^{2} d \,e^{2}+20 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}\, B a b \,d^{2} e -30 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}\, B \,b^{2} d^{3}\right )}{15 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \left (a e -b d \right ) \sqrt {b e}\, \left (e x +d \right )^{\frac {5}{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)^(7/2),x)

[Out]

-1/15*(b*x+a)^(1/2)*(-15*B*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^3*a*b
^2*e^4+15*B*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^3*b^3*d*e^3-45*B*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*a*b^2*d*e^3+45*B*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^3*d^2*e^2+6*A*x^2*b^2*e^3*((b*x+a)*(e*x+d))^
(1/2)*(b*e)^(1/2)-45*B*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*a*b^2*d^2
*e^2+45*B*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*b^3*d^3*e+40*B*x^2*a*b
*e^3*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)-46*B*x^2*b^2*d*e^2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+12*A*x*a*b*e^3
*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)-15*B*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))*a*b^2*d^3*e+15*B*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))*b^3*d^4+1
0*B*x*a^2*e^3*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+48*B*x*a*b*d*e^2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)-70*B*x*
b^2*d^2*e*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+6*A*a^2*e^3*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+4*B*a^2*d*e^2*((
b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+20*B*a*b*d^2*e*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)-30*B*b^2*d^3*((b*x+a)*(e*
x+d))^(1/2)*(b*e)^(1/2))/((b*x+a)*(e*x+d))^(1/2)/(a*e-b*d)/(b*e)^(1/2)/(e*x+d)^(5/2)/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)^(3/2)*(B*x+A)/(e*x+d)^(7/2),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 = -0.01 \begin {gather*} \int \frac {\left (A+B\,x\right )\,{\left (a+b\,x\right )}^{3/2}}{{\left (d+e\,x\right )}^{7/2}} \,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)^(7/2),x)

[Out]

int(((A + B*x)*(a + b*x)^(3/2))/(d + e*x)^(7/2), 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)**(7/2),x)

[Out]

Timed out

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