3.1836 \(\int (A+B x) (d+e x)^{5/2} \sqrt{a^2+2 a b x+b^2 x^2} \, dx\)

Optimal. Leaf size=164 \[ -\frac{2 \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{9/2} (-a B e-A b e+2 b B d)}{9 e^3 (a+b x)}+\frac{2 \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{7/2} (b d-a e) (B d-A e)}{7 e^3 (a+b x)}+\frac{2 b B \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{11/2}}{11 e^3 (a+b x)} \]

[Out]

(2*(b*d - a*e)*(B*d - A*e)*(d + e*x)^(7/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(7*e^3*(a + b*x)) - (2*(2*b*B*d - A*
b*e - a*B*e)*(d + e*x)^(9/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(9*e^3*(a + b*x)) + (2*b*B*(d + e*x)^(11/2)*Sqrt[a
^2 + 2*a*b*x + b^2*x^2])/(11*e^3*(a + b*x))

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Rubi [A]  time = 0.0821467, antiderivative size = 164, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.057, Rules used = {770, 77} \[ -\frac{2 \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{9/2} (-a B e-A b e+2 b B d)}{9 e^3 (a+b x)}+\frac{2 \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{7/2} (b d-a e) (B d-A e)}{7 e^3 (a+b x)}+\frac{2 b B \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{11/2}}{11 e^3 (a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(b*d - a*e)*(B*d - A*e)*(d + e*x)^(7/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(7*e^3*(a + b*x)) - (2*(2*b*B*d - A*
b*e - a*B*e)*(d + e*x)^(9/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(9*e^3*(a + b*x)) + (2*b*B*(d + e*x)^(11/2)*Sqrt[a
^2 + 2*a*b*x + b^2*x^2])/(11*e^3*(a + b*x))

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]

Rule 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rubi steps

\begin{align*} \int (A+B x) (d+e x)^{5/2} \sqrt{a^2+2 a b x+b^2 x^2} \, dx &=\frac{\sqrt{a^2+2 a b x+b^2 x^2} \int \left (a b+b^2 x\right ) (A+B x) (d+e x)^{5/2} \, dx}{a b+b^2 x}\\ &=\frac{\sqrt{a^2+2 a b x+b^2 x^2} \int \left (-\frac{b (b d-a e) (-B d+A e) (d+e x)^{5/2}}{e^2}+\frac{b (-2 b B d+A b e+a B e) (d+e x)^{7/2}}{e^2}+\frac{b^2 B (d+e x)^{9/2}}{e^2}\right ) \, dx}{a b+b^2 x}\\ &=\frac{2 (b d-a e) (B d-A e) (d+e x)^{7/2} \sqrt{a^2+2 a b x+b^2 x^2}}{7 e^3 (a+b x)}-\frac{2 (2 b B d-A b e-a B e) (d+e x)^{9/2} \sqrt{a^2+2 a b x+b^2 x^2}}{9 e^3 (a+b x)}+\frac{2 b B (d+e x)^{11/2} \sqrt{a^2+2 a b x+b^2 x^2}}{11 e^3 (a+b x)}\\ \end{align*}

Mathematica [A]  time = 0.079309, size = 88, normalized size = 0.54 \[ \frac{2 \sqrt{(a+b x)^2} (d+e x)^{7/2} \left (11 a e (9 A e-2 B d+7 B e x)+11 A b e (7 e x-2 d)+b B \left (8 d^2-28 d e x+63 e^2 x^2\right )\right )}{693 e^3 (a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[(a + b*x)^2]*(d + e*x)^(7/2)*(11*A*b*e*(-2*d + 7*e*x) + 11*a*e*(-2*B*d + 9*A*e + 7*B*e*x) + b*B*(8*d^2
- 28*d*e*x + 63*e^2*x^2)))/(693*e^3*(a + b*x))

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Maple [A]  time = 0.004, size = 89, normalized size = 0.5 \begin{align*}{\frac{126\,B{x}^{2}b{e}^{2}+154\,Axb{e}^{2}+154\,aB{e}^{2}x-56\,Bxbde+198\,aA{e}^{2}-44\,Abde-44\,aBde+16\,Bb{d}^{2}}{693\,{e}^{3} \left ( bx+a \right ) } \left ( ex+d \right ) ^{{\frac{7}{2}}}\sqrt{ \left ( bx+a \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(e*x+d)^(5/2)*((b*x+a)^2)^(1/2),x)

[Out]

2/693*(e*x+d)^(7/2)*(63*B*b*e^2*x^2+77*A*b*e^2*x+77*B*a*e^2*x-28*B*b*d*e*x+99*A*a*e^2-22*A*b*d*e-22*B*a*d*e+8*
B*b*d^2)*((b*x+a)^2)^(1/2)/e^3/(b*x+a)

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Maxima [A]  time = 1.03834, size = 289, normalized size = 1.76 \begin{align*} \frac{2 \,{\left (7 \, b e^{4} x^{4} - 2 \, b d^{4} + 9 \, a d^{3} e +{\left (19 \, b d e^{3} + 9 \, a e^{4}\right )} x^{3} + 3 \,{\left (5 \, b d^{2} e^{2} + 9 \, a d e^{3}\right )} x^{2} +{\left (b d^{3} e + 27 \, a d^{2} e^{2}\right )} x\right )} \sqrt{e x + d} A}{63 \, e^{2}} + \frac{2 \,{\left (63 \, b e^{5} x^{5} + 8 \, b d^{5} - 22 \, a d^{4} e + 7 \,{\left (23 \, b d e^{4} + 11 \, a e^{5}\right )} x^{4} +{\left (113 \, b d^{2} e^{3} + 209 \, a d e^{4}\right )} x^{3} + 3 \,{\left (b d^{3} e^{2} + 55 \, a d^{2} e^{3}\right )} x^{2} -{\left (4 \, b d^{4} e - 11 \, a d^{3} e^{2}\right )} x\right )} \sqrt{e x + d} B}{693 \, e^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)^(5/2)*((b*x+a)^2)^(1/2),x, algorithm="maxima")

[Out]

2/63*(7*b*e^4*x^4 - 2*b*d^4 + 9*a*d^3*e + (19*b*d*e^3 + 9*a*e^4)*x^3 + 3*(5*b*d^2*e^2 + 9*a*d*e^3)*x^2 + (b*d^
3*e + 27*a*d^2*e^2)*x)*sqrt(e*x + d)*A/e^2 + 2/693*(63*b*e^5*x^5 + 8*b*d^5 - 22*a*d^4*e + 7*(23*b*d*e^4 + 11*a
*e^5)*x^4 + (113*b*d^2*e^3 + 209*a*d*e^4)*x^3 + 3*(b*d^3*e^2 + 55*a*d^2*e^3)*x^2 - (4*b*d^4*e - 11*a*d^3*e^2)*
x)*sqrt(e*x + d)*B/e^3

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Fricas [A]  time = 1.29877, size = 435, normalized size = 2.65 \begin{align*} \frac{2 \,{\left (63 \, B b e^{5} x^{5} + 8 \, B b d^{5} + 99 \, A a d^{3} e^{2} - 22 \,{\left (B a + A b\right )} d^{4} e + 7 \,{\left (23 \, B b d e^{4} + 11 \,{\left (B a + A b\right )} e^{5}\right )} x^{4} +{\left (113 \, B b d^{2} e^{3} + 99 \, A a e^{5} + 209 \,{\left (B a + A b\right )} d e^{4}\right )} x^{3} + 3 \,{\left (B b d^{3} e^{2} + 99 \, A a d e^{4} + 55 \,{\left (B a + A b\right )} d^{2} e^{3}\right )} x^{2} -{\left (4 \, B b d^{4} e - 297 \, A a d^{2} e^{3} - 11 \,{\left (B a + A b\right )} d^{3} e^{2}\right )} x\right )} \sqrt{e x + d}}{693 \, e^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)^(5/2)*((b*x+a)^2)^(1/2),x, algorithm="fricas")

[Out]

2/693*(63*B*b*e^5*x^5 + 8*B*b*d^5 + 99*A*a*d^3*e^2 - 22*(B*a + A*b)*d^4*e + 7*(23*B*b*d*e^4 + 11*(B*a + A*b)*e
^5)*x^4 + (113*B*b*d^2*e^3 + 99*A*a*e^5 + 209*(B*a + A*b)*d*e^4)*x^3 + 3*(B*b*d^3*e^2 + 99*A*a*d*e^4 + 55*(B*a
+ A*b)*d^2*e^3)*x^2 - (4*B*b*d^4*e - 297*A*a*d^2*e^3 - 11*(B*a + A*b)*d^3*e^2)*x)*sqrt(e*x + d)/e^3

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)**(5/2)*((b*x+a)**2)**(1/2),x)

[Out]

Timed out

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Giac [B]  time = 1.21454, size = 779, normalized size = 4.75 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)^(5/2)*((b*x+a)^2)^(1/2),x, algorithm="giac")

[Out]

2/3465*(231*(3*(x*e + d)^(5/2) - 5*(x*e + d)^(3/2)*d)*B*a*d^2*e^(-1)*sgn(b*x + a) + 231*(3*(x*e + d)^(5/2) - 5
*(x*e + d)^(3/2)*d)*A*b*d^2*e^(-1)*sgn(b*x + a) + 33*(15*(x*e + d)^(7/2) - 42*(x*e + d)^(5/2)*d + 35*(x*e + d)
^(3/2)*d^2)*B*b*d^2*e^(-2)*sgn(b*x + a) + 1155*(x*e + d)^(3/2)*A*a*d^2*sgn(b*x + a) + 66*(15*(x*e + d)^(7/2) -
42*(x*e + d)^(5/2)*d + 35*(x*e + d)^(3/2)*d^2)*B*a*d*e^(-1)*sgn(b*x + a) + 66*(15*(x*e + d)^(7/2) - 42*(x*e +
d)^(5/2)*d + 35*(x*e + d)^(3/2)*d^2)*A*b*d*e^(-1)*sgn(b*x + a) + 22*(35*(x*e + d)^(9/2) - 135*(x*e + d)^(7/2)
*d + 189*(x*e + d)^(5/2)*d^2 - 105*(x*e + d)^(3/2)*d^3)*B*b*d*e^(-2)*sgn(b*x + a) + 462*(3*(x*e + d)^(5/2) - 5
*(x*e + d)^(3/2)*d)*A*a*d*sgn(b*x + a) + 11*(35*(x*e + d)^(9/2) - 135*(x*e + d)^(7/2)*d + 189*(x*e + d)^(5/2)*
d^2 - 105*(x*e + d)^(3/2)*d^3)*B*a*e^(-1)*sgn(b*x + a) + 11*(35*(x*e + d)^(9/2) - 135*(x*e + d)^(7/2)*d + 189*
(x*e + d)^(5/2)*d^2 - 105*(x*e + d)^(3/2)*d^3)*A*b*e^(-1)*sgn(b*x + a) + (315*(x*e + d)^(11/2) - 1540*(x*e + d
)^(9/2)*d + 2970*(x*e + d)^(7/2)*d^2 - 2772*(x*e + d)^(5/2)*d^3 + 1155*(x*e + d)^(3/2)*d^4)*B*b*e^(-2)*sgn(b*x
+ a) + 33*(15*(x*e + d)^(7/2) - 42*(x*e + d)^(5/2)*d + 35*(x*e + d)^(3/2)*d^2)*A*a*sgn(b*x + a))*e^(-1)