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

Optimal. Leaf size=251 \[ \frac {2 b (a+b x) (A b-a B)}{\sqrt {a^2+2 a b x+b^2 x^2} \sqrt {d+e x} (b d-a e)^3}+\frac {2 (a+b x) (A b-a B)}{3 \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^{3/2} (b d-a e)^2}-\frac {2 (a+b x) (B d-A e)}{5 e \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^{5/2} (b d-a e)}-\frac {2 b^{3/2} (a+b x) (A b-a B) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^{7/2}} \]

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Rubi [A]  time = 0.16, antiderivative size = 251, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {770, 78, 51, 63, 208} \begin {gather*} \frac {2 b (a+b x) (A b-a B)}{\sqrt {a^2+2 a b x+b^2 x^2} \sqrt {d+e x} (b d-a e)^3}+\frac {2 (a+b x) (A b-a B)}{3 \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^{3/2} (b d-a e)^2}-\frac {2 (a+b x) (B d-A e)}{5 e \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^{5/2} (b d-a e)}-\frac {2 b^{3/2} (a+b x) (A b-a B) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 51

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

Rule 770

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dis
t[(a + b*x + c*x^2)^FracPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(f + g*x)*(b/2 + c
*x)^(2*p), x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && EqQ[b^2 - 4*a*c, 0]

Rubi steps

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

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Mathematica [C]  time = 0.05, size = 102, normalized size = 0.41 \begin {gather*} \frac {2 (a+b x) \left (5 e (d+e x) (A b-a B) \, _2F_1\left (-\frac {3}{2},1;-\frac {1}{2};\frac {b (d+e x)}{b d-a e}\right )-3 (b d-a e) (B d-A e)\right )}{15 e \sqrt {(a+b x)^2} (d+e x)^{5/2} (b d-a e)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(a + b*x)*(-3*(b*d - a*e)*(B*d - A*e) + 5*(A*b - a*B)*e*(d + e*x)*Hypergeometric2F1[-3/2, 1, -1/2, (b*(d +
e*x))/(b*d - a*e)]))/(15*e*(b*d - a*e)^2*Sqrt[(a + b*x)^2]*(d + e*x)^(5/2))

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IntegrateAlgebraic [A]  time = 60.56, size = 264, normalized size = 1.05 \begin {gather*} \frac {(-a e-b e x) \left (\frac {2 \left (3 a^2 A e^3+5 a^2 B e^2 (d+e x)-3 a^2 B d e^2-5 a A b e^2 (d+e x)-6 a A b d e^2+6 a b B d^2 e-5 a b B d e (d+e x)-15 a b B e (d+e x)^2+3 A b^2 d^2 e+5 A b^2 d e (d+e x)+15 A b^2 e (d+e x)^2-3 b^2 B d^3\right )}{15 e (d+e x)^{5/2} (a e-b d)^3}-\frac {2 \left (A b^{5/2}-a b^{3/2} B\right ) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x} \sqrt {a e-b d}}{b d-a e}\right )}{(a e-b d)^{7/2}}\right )}{e \sqrt {\frac {(a e+b e x)^2}{e^2}}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.49, size = 902, normalized size = 3.59 \begin {gather*} \left [\frac {15 \, {\left ({\left (B a b - A b^{2}\right )} e^{4} x^{3} + 3 \, {\left (B a b - A b^{2}\right )} d e^{3} x^{2} + 3 \, {\left (B a b - A b^{2}\right )} d^{2} e^{2} x + {\left (B a b - A b^{2}\right )} d^{3} e\right )} \sqrt {\frac {b}{b d - a e}} \log \left (\frac {b e x + 2 \, b d - a e + 2 \, {\left (b d - a e\right )} \sqrt {e x + d} \sqrt {\frac {b}{b d - a e}}}{b x + a}\right ) - 2 \, {\left (3 \, B b^{2} d^{3} - 3 \, A a^{2} e^{3} + 15 \, {\left (B a b - A b^{2}\right )} e^{3} x^{2} + {\left (14 \, B a b - 23 \, A b^{2}\right )} d^{2} e - {\left (2 \, B a^{2} - 11 \, A a b\right )} d e^{2} + 5 \, {\left (7 \, {\left (B a b - A b^{2}\right )} d e^{2} - {\left (B a^{2} - A a b\right )} e^{3}\right )} x\right )} \sqrt {e x + d}}{15 \, {\left (b^{3} d^{6} e - 3 \, a b^{2} d^{5} e^{2} + 3 \, a^{2} b d^{4} e^{3} - a^{3} d^{3} e^{4} + {\left (b^{3} d^{3} e^{4} - 3 \, a b^{2} d^{2} e^{5} + 3 \, a^{2} b d e^{6} - a^{3} e^{7}\right )} x^{3} + 3 \, {\left (b^{3} d^{4} e^{3} - 3 \, a b^{2} d^{3} e^{4} + 3 \, a^{2} b d^{2} e^{5} - a^{3} d e^{6}\right )} x^{2} + 3 \, {\left (b^{3} d^{5} e^{2} - 3 \, a b^{2} d^{4} e^{3} + 3 \, a^{2} b d^{3} e^{4} - a^{3} d^{2} e^{5}\right )} x\right )}}, \frac {2 \, {\left (15 \, {\left ({\left (B a b - A b^{2}\right )} e^{4} x^{3} + 3 \, {\left (B a b - A b^{2}\right )} d e^{3} x^{2} + 3 \, {\left (B a b - A b^{2}\right )} d^{2} e^{2} x + {\left (B a b - A b^{2}\right )} d^{3} e\right )} \sqrt {-\frac {b}{b d - a e}} \arctan \left (-\frac {{\left (b d - a e\right )} \sqrt {e x + d} \sqrt {-\frac {b}{b d - a e}}}{b e x + b d}\right ) - {\left (3 \, B b^{2} d^{3} - 3 \, A a^{2} e^{3} + 15 \, {\left (B a b - A b^{2}\right )} e^{3} x^{2} + {\left (14 \, B a b - 23 \, A b^{2}\right )} d^{2} e - {\left (2 \, B a^{2} - 11 \, A a b\right )} d e^{2} + 5 \, {\left (7 \, {\left (B a b - A b^{2}\right )} d e^{2} - {\left (B a^{2} - A a b\right )} e^{3}\right )} x\right )} \sqrt {e x + d}\right )}}{15 \, {\left (b^{3} d^{6} e - 3 \, a b^{2} d^{5} e^{2} + 3 \, a^{2} b d^{4} e^{3} - a^{3} d^{3} e^{4} + {\left (b^{3} d^{3} e^{4} - 3 \, a b^{2} d^{2} e^{5} + 3 \, a^{2} b d e^{6} - a^{3} e^{7}\right )} x^{3} + 3 \, {\left (b^{3} d^{4} e^{3} - 3 \, a b^{2} d^{3} e^{4} + 3 \, a^{2} b d^{2} e^{5} - a^{3} d e^{6}\right )} x^{2} + 3 \, {\left (b^{3} d^{5} e^{2} - 3 \, a b^{2} d^{4} e^{3} + 3 \, a^{2} b d^{3} e^{4} - a^{3} 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)/(e*x+d)^(7/2)/((b*x+a)^2)^(1/2),x, algorithm="fricas")

[Out]

[1/15*(15*((B*a*b - A*b^2)*e^4*x^3 + 3*(B*a*b - A*b^2)*d*e^3*x^2 + 3*(B*a*b - A*b^2)*d^2*e^2*x + (B*a*b - A*b^
2)*d^3*e)*sqrt(b/(b*d - a*e))*log((b*e*x + 2*b*d - a*e + 2*(b*d - a*e)*sqrt(e*x + d)*sqrt(b/(b*d - a*e)))/(b*x
+ a)) - 2*(3*B*b^2*d^3 - 3*A*a^2*e^3 + 15*(B*a*b - A*b^2)*e^3*x^2 + (14*B*a*b - 23*A*b^2)*d^2*e - (2*B*a^2 -
11*A*a*b)*d*e^2 + 5*(7*(B*a*b - A*b^2)*d*e^2 - (B*a^2 - A*a*b)*e^3)*x)*sqrt(e*x + d))/(b^3*d^6*e - 3*a*b^2*d^5
*e^2 + 3*a^2*b*d^4*e^3 - a^3*d^3*e^4 + (b^3*d^3*e^4 - 3*a*b^2*d^2*e^5 + 3*a^2*b*d*e^6 - a^3*e^7)*x^3 + 3*(b^3*
d^4*e^3 - 3*a*b^2*d^3*e^4 + 3*a^2*b*d^2*e^5 - a^3*d*e^6)*x^2 + 3*(b^3*d^5*e^2 - 3*a*b^2*d^4*e^3 + 3*a^2*b*d^3*
e^4 - a^3*d^2*e^5)*x), 2/15*(15*((B*a*b - A*b^2)*e^4*x^3 + 3*(B*a*b - A*b^2)*d*e^3*x^2 + 3*(B*a*b - A*b^2)*d^2
*e^2*x + (B*a*b - A*b^2)*d^3*e)*sqrt(-b/(b*d - a*e))*arctan(-(b*d - a*e)*sqrt(e*x + d)*sqrt(-b/(b*d - a*e))/(b
*e*x + b*d)) - (3*B*b^2*d^3 - 3*A*a^2*e^3 + 15*(B*a*b - A*b^2)*e^3*x^2 + (14*B*a*b - 23*A*b^2)*d^2*e - (2*B*a^
2 - 11*A*a*b)*d*e^2 + 5*(7*(B*a*b - A*b^2)*d*e^2 - (B*a^2 - A*a*b)*e^3)*x)*sqrt(e*x + d))/(b^3*d^6*e - 3*a*b^2
*d^5*e^2 + 3*a^2*b*d^4*e^3 - a^3*d^3*e^4 + (b^3*d^3*e^4 - 3*a*b^2*d^2*e^5 + 3*a^2*b*d*e^6 - a^3*e^7)*x^3 + 3*(
b^3*d^4*e^3 - 3*a*b^2*d^3*e^4 + 3*a^2*b*d^2*e^5 - a^3*d*e^6)*x^2 + 3*(b^3*d^5*e^2 - 3*a*b^2*d^4*e^3 + 3*a^2*b*
d^3*e^4 - a^3*d^2*e^5)*x)]

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giac [A]  time = 0.27, size = 368, normalized size = 1.47 \begin {gather*} -\frac {2 \, {\left (B a b^{2} \mathrm {sgn}\left (b x + a\right ) - A b^{3} \mathrm {sgn}\left (b x + a\right )\right )} \arctan \left (\frac {\sqrt {x e + d} b}{\sqrt {-b^{2} d + a b e}}\right )}{{\left (b^{3} d^{3} - 3 \, a b^{2} d^{2} e + 3 \, a^{2} b d e^{2} - a^{3} e^{3}\right )} \sqrt {-b^{2} d + a b e}} - \frac {2 \, {\left (3 \, B b^{2} d^{3} \mathrm {sgn}\left (b x + a\right ) + 15 \, {\left (x e + d\right )}^{2} B a b e \mathrm {sgn}\left (b x + a\right ) - 15 \, {\left (x e + d\right )}^{2} A b^{2} e \mathrm {sgn}\left (b x + a\right ) + 5 \, {\left (x e + d\right )} B a b d e \mathrm {sgn}\left (b x + a\right ) - 5 \, {\left (x e + d\right )} A b^{2} d e \mathrm {sgn}\left (b x + a\right ) - 6 \, B a b d^{2} e \mathrm {sgn}\left (b x + a\right ) - 3 \, A b^{2} d^{2} e \mathrm {sgn}\left (b x + a\right ) - 5 \, {\left (x e + d\right )} B a^{2} e^{2} \mathrm {sgn}\left (b x + a\right ) + 5 \, {\left (x e + d\right )} A a b e^{2} \mathrm {sgn}\left (b x + a\right ) + 3 \, B a^{2} d e^{2} \mathrm {sgn}\left (b x + a\right ) + 6 \, A a b d e^{2} \mathrm {sgn}\left (b x + a\right ) - 3 \, A a^{2} e^{3} \mathrm {sgn}\left (b x + a\right )\right )}}{15 \, {\left (b^{3} d^{3} e - 3 \, a b^{2} d^{2} e^{2} + 3 \, a^{2} b d e^{3} - a^{3} e^{4}\right )} {\left (x e + d\right )}^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/(e*x+d)^(7/2)/((b*x+a)^2)^(1/2),x, algorithm="giac")

[Out]

-2*(B*a*b^2*sgn(b*x + a) - A*b^3*sgn(b*x + a))*arctan(sqrt(x*e + d)*b/sqrt(-b^2*d + a*b*e))/((b^3*d^3 - 3*a*b^
2*d^2*e + 3*a^2*b*d*e^2 - a^3*e^3)*sqrt(-b^2*d + a*b*e)) - 2/15*(3*B*b^2*d^3*sgn(b*x + a) + 15*(x*e + d)^2*B*a
*b*e*sgn(b*x + a) - 15*(x*e + d)^2*A*b^2*e*sgn(b*x + a) + 5*(x*e + d)*B*a*b*d*e*sgn(b*x + a) - 5*(x*e + d)*A*b
^2*d*e*sgn(b*x + a) - 6*B*a*b*d^2*e*sgn(b*x + a) - 3*A*b^2*d^2*e*sgn(b*x + a) - 5*(x*e + d)*B*a^2*e^2*sgn(b*x
+ a) + 5*(x*e + d)*A*a*b*e^2*sgn(b*x + a) + 3*B*a^2*d*e^2*sgn(b*x + a) + 6*A*a*b*d*e^2*sgn(b*x + a) - 3*A*a^2*
e^3*sgn(b*x + a))/((b^3*d^3*e - 3*a*b^2*d^2*e^2 + 3*a^2*b*d*e^3 - a^3*e^4)*(x*e + d)^(5/2))

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maple [B]  time = 0.08, size = 386, normalized size = 1.54 \begin {gather*} -\frac {2 \left (b x +a \right ) \left (15 \sqrt {\left (a e -b d \right ) b}\, A \,b^{2} e^{3} x^{2}-15 \sqrt {\left (a e -b d \right ) b}\, B a b \,e^{3} x^{2}-5 \sqrt {\left (a e -b d \right ) b}\, A a b \,e^{3} x +35 \sqrt {\left (a e -b d \right ) b}\, A \,b^{2} d \,e^{2} x +5 \sqrt {\left (a e -b d \right ) b}\, B \,a^{2} e^{3} x -35 \sqrt {\left (a e -b d \right ) b}\, B a b d \,e^{2} x +3 \sqrt {\left (a e -b d \right ) b}\, A \,a^{2} e^{3}-11 \sqrt {\left (a e -b d \right ) b}\, A a b d \,e^{2}+15 \left (e x +d \right )^{\frac {5}{2}} A \,b^{3} e \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )+23 \sqrt {\left (a e -b d \right ) b}\, A \,b^{2} d^{2} e +2 \sqrt {\left (a e -b d \right ) b}\, B \,a^{2} d \,e^{2}-15 \left (e x +d \right )^{\frac {5}{2}} B a \,b^{2} e \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )-14 \sqrt {\left (a e -b d \right ) b}\, B a b \,d^{2} e -3 \sqrt {\left (a e -b d \right ) b}\, B \,b^{2} d^{3}\right )}{15 \sqrt {\left (b x +a \right )^{2}}\, \left (a e -b d \right )^{3} \sqrt {\left (a e -b d \right ) b}\, \left (e x +d \right )^{\frac {5}{2}} e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)/(e*x+d)^(7/2)/((b*x+a)^2)^(1/2),x)

[Out]

-2/15*(b*x+a)*(15*A*(e*x+d)^(5/2)*arctan((e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2)*b)*b^3*e-15*B*(e*x+d)^(5/2)*arctan(
(e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2)*b)*a*b^2*e+15*A*((a*e-b*d)*b)^(1/2)*x^2*b^2*e^3-15*B*((a*e-b*d)*b)^(1/2)*x^2
*a*b*e^3-5*A*((a*e-b*d)*b)^(1/2)*x*a*b*e^3+35*A*((a*e-b*d)*b)^(1/2)*x*b^2*d*e^2+5*B*((a*e-b*d)*b)^(1/2)*x*a^2*
e^3-35*B*((a*e-b*d)*b)^(1/2)*x*a*b*d*e^2+3*A*((a*e-b*d)*b)^(1/2)*a^2*e^3-11*A*((a*e-b*d)*b)^(1/2)*a*b*d*e^2+23
*A*((a*e-b*d)*b)^(1/2)*b^2*d^2*e+2*B*((a*e-b*d)*b)^(1/2)*a^2*d*e^2-14*B*((a*e-b*d)*b)^(1/2)*a*b*d^2*e-3*B*((a*
e-b*d)*b)^(1/2)*b^2*d^3)/((b*x+a)^2)^(1/2)/e/(a*e-b*d)^3/((a*e-b*d)*b)^(1/2)/(e*x+d)^(5/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {B x + A}{\sqrt {{\left (b x + a\right )}^{2}} {\left (e x + d\right )}^{\frac {7}{2}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/(e*x+d)^(7/2)/((b*x+a)^2)^(1/2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {A+B\,x}{\sqrt {{\left (a+b\,x\right )}^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)^2)^(1/2)*(d + e*x)^(7/2)),x)

[Out]

int((A + B*x)/(((a + b*x)^2)^(1/2)*(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)/(e*x+d)**(7/2)/((b*x+a)**2)**(1/2),x)

[Out]

Timed out

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