3.15.37 \(\int (d+e x)^m (a^2+2 a b x+b^2 x^2)^{3/2} \, dx\)

Optimal. Leaf size=219 \[ -\frac {\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^3 (d+e x)^{m+1}}{e^4 (m+1) (a+b x)}+\frac {3 b \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^2 (d+e x)^{m+2}}{e^4 (m+2) (a+b x)}-\frac {3 b^2 \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e) (d+e x)^{m+3}}{e^4 (m+3) (a+b x)}+\frac {b^3 \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^{m+4}}{e^4 (m+4) (a+b x)} \]

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Rubi [A]  time = 0.09, antiderivative size = 219, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {646, 43} \begin {gather*} -\frac {\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^3 (d+e x)^{m+1}}{e^4 (m+1) (a+b x)}+\frac {3 b \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^2 (d+e x)^{m+2}}{e^4 (m+2) (a+b x)}-\frac {3 b^2 \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e) (d+e x)^{m+3}}{e^4 (m+3) (a+b x)}+\frac {b^3 \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^{m+4}}{e^4 (m+4) (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(((b*d - a*e)^3*(d + e*x)^(1 + m)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(e^4*(1 + m)*(a + b*x))) + (3*b*(b*d - a*e)^
2*(d + e*x)^(2 + m)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(e^4*(2 + m)*(a + b*x)) - (3*b^2*(b*d - a*e)*(d + e*x)^(3 +
m)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(e^4*(3 + m)*(a + b*x)) + (b^3*(d + e*x)^(4 + m)*Sqrt[a^2 + 2*a*b*x + b^2*x
^2])/(e^4*(4 + m)*(a + b*x))

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 646

Int[((d_.) + (e_.)*(x_))^(m_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(a + b*x + c*x^2)^Fra
cPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(b/2 + c*x)^(2*p), x], x] /; FreeQ[{a, b,
c, d, e, m, p}, x] && EqQ[b^2 - 4*a*c, 0] &&  !IntegerQ[p] && NeQ[2*c*d - b*e, 0]

Rubi steps

\begin {align*} \int (d+e x)^m \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx &=\frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \left (a b+b^2 x\right )^3 (d+e x)^m \, dx}{b^2 \left (a b+b^2 x\right )}\\ &=\frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \left (-\frac {b^3 (b d-a e)^3 (d+e x)^m}{e^3}+\frac {3 b^4 (b d-a e)^2 (d+e x)^{1+m}}{e^3}-\frac {3 b^5 (b d-a e) (d+e x)^{2+m}}{e^3}+\frac {b^6 (d+e x)^{3+m}}{e^3}\right ) \, dx}{b^2 \left (a b+b^2 x\right )}\\ &=-\frac {(b d-a e)^3 (d+e x)^{1+m} \sqrt {a^2+2 a b x+b^2 x^2}}{e^4 (1+m) (a+b x)}+\frac {3 b (b d-a e)^2 (d+e x)^{2+m} \sqrt {a^2+2 a b x+b^2 x^2}}{e^4 (2+m) (a+b x)}-\frac {3 b^2 (b d-a e) (d+e x)^{3+m} \sqrt {a^2+2 a b x+b^2 x^2}}{e^4 (3+m) (a+b x)}+\frac {b^3 (d+e x)^{4+m} \sqrt {a^2+2 a b x+b^2 x^2}}{e^4 (4+m) (a+b x)}\\ \end {align*}

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Mathematica [A]  time = 0.12, size = 113, normalized size = 0.52 \begin {gather*} \frac {\left ((a+b x)^2\right )^{3/2} (d+e x)^{m+1} \left (-\frac {3 b^2 (d+e x)^2 (b d-a e)}{m+3}+\frac {3 b (d+e x) (b d-a e)^2}{m+2}-\frac {(b d-a e)^3}{m+1}+\frac {b^3 (d+e x)^3}{m+4}\right )}{e^4 (a+b x)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(((a + b*x)^2)^(3/2)*(d + e*x)^(1 + m)*(-((b*d - a*e)^3/(1 + m)) + (3*b*(b*d - a*e)^2*(d + e*x))/(2 + m) - (3*
b^2*(b*d - a*e)*(d + e*x)^2)/(3 + m) + (b^3*(d + e*x)^3)/(4 + m)))/(e^4*(a + b*x)^3)

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IntegrateAlgebraic [F]  time = 1.19, size = 0, normalized size = 0.00 \begin {gather*} \int (d+e x)^m \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [B]  time = 0.42, size = 496, normalized size = 2.26 \begin {gather*} \frac {{\left (a^{3} d e^{3} m^{3} - 6 \, b^{3} d^{4} + 24 \, a b^{2} d^{3} e - 36 \, a^{2} b d^{2} e^{2} + 24 \, a^{3} d e^{3} + {\left (b^{3} e^{4} m^{3} + 6 \, b^{3} e^{4} m^{2} + 11 \, b^{3} e^{4} m + 6 \, b^{3} e^{4}\right )} x^{4} + {\left (24 \, a b^{2} e^{4} + {\left (b^{3} d e^{3} + 3 \, a b^{2} e^{4}\right )} m^{3} + 3 \, {\left (b^{3} d e^{3} + 7 \, a b^{2} e^{4}\right )} m^{2} + 2 \, {\left (b^{3} d e^{3} + 21 \, a b^{2} e^{4}\right )} m\right )} x^{3} - 3 \, {\left (a^{2} b d^{2} e^{2} - 3 \, a^{3} d e^{3}\right )} m^{2} + 3 \, {\left (12 \, a^{2} b e^{4} + {\left (a b^{2} d e^{3} + a^{2} b e^{4}\right )} m^{3} - {\left (b^{3} d^{2} e^{2} - 5 \, a b^{2} d e^{3} - 8 \, a^{2} b e^{4}\right )} m^{2} - {\left (b^{3} d^{2} e^{2} - 4 \, a b^{2} d e^{3} - 19 \, a^{2} b e^{4}\right )} m\right )} x^{2} + {\left (6 \, a b^{2} d^{3} e - 21 \, a^{2} b d^{2} e^{2} + 26 \, a^{3} d e^{3}\right )} m + {\left (24 \, a^{3} e^{4} + {\left (3 \, a^{2} b d e^{3} + a^{3} e^{4}\right )} m^{3} - 3 \, {\left (2 \, a b^{2} d^{2} e^{2} - 7 \, a^{2} b d e^{3} - 3 \, a^{3} e^{4}\right )} m^{2} + 2 \, {\left (3 \, b^{3} d^{3} e - 12 \, a b^{2} d^{2} e^{2} + 18 \, a^{2} b d e^{3} + 13 \, a^{3} e^{4}\right )} m\right )} x\right )} {\left (e x + d\right )}^{m}}{e^{4} m^{4} + 10 \, e^{4} m^{3} + 35 \, e^{4} m^{2} + 50 \, e^{4} m + 24 \, e^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a^3*d*e^3*m^3 - 6*b^3*d^4 + 24*a*b^2*d^3*e - 36*a^2*b*d^2*e^2 + 24*a^3*d*e^3 + (b^3*e^4*m^3 + 6*b^3*e^4*m^2 +
11*b^3*e^4*m + 6*b^3*e^4)*x^4 + (24*a*b^2*e^4 + (b^3*d*e^3 + 3*a*b^2*e^4)*m^3 + 3*(b^3*d*e^3 + 7*a*b^2*e^4)*m
^2 + 2*(b^3*d*e^3 + 21*a*b^2*e^4)*m)*x^3 - 3*(a^2*b*d^2*e^2 - 3*a^3*d*e^3)*m^2 + 3*(12*a^2*b*e^4 + (a*b^2*d*e^
3 + a^2*b*e^4)*m^3 - (b^3*d^2*e^2 - 5*a*b^2*d*e^3 - 8*a^2*b*e^4)*m^2 - (b^3*d^2*e^2 - 4*a*b^2*d*e^3 - 19*a^2*b
*e^4)*m)*x^2 + (6*a*b^2*d^3*e - 21*a^2*b*d^2*e^2 + 26*a^3*d*e^3)*m + (24*a^3*e^4 + (3*a^2*b*d*e^3 + a^3*e^4)*m
^3 - 3*(2*a*b^2*d^2*e^2 - 7*a^2*b*d*e^3 - 3*a^3*e^4)*m^2 + 2*(3*b^3*d^3*e - 12*a*b^2*d^2*e^2 + 18*a^2*b*d*e^3
+ 13*a^3*e^4)*m)*x)*(e*x + d)^m/(e^4*m^4 + 10*e^4*m^3 + 35*e^4*m^2 + 50*e^4*m + 24*e^4)

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giac [B]  time = 0.25, size = 1075, normalized size = 4.91

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((x*e + d)^m*b^3*m^3*x^4*e^4*sgn(b*x + a) + (x*e + d)^m*b^3*d*m^3*x^3*e^3*sgn(b*x + a) + 3*(x*e + d)^m*a*b^2*m
^3*x^3*e^4*sgn(b*x + a) + 6*(x*e + d)^m*b^3*m^2*x^4*e^4*sgn(b*x + a) + 3*(x*e + d)^m*a*b^2*d*m^3*x^2*e^3*sgn(b
*x + a) + 3*(x*e + d)^m*b^3*d*m^2*x^3*e^3*sgn(b*x + a) - 3*(x*e + d)^m*b^3*d^2*m^2*x^2*e^2*sgn(b*x + a) + 3*(x
*e + d)^m*a^2*b*m^3*x^2*e^4*sgn(b*x + a) + 21*(x*e + d)^m*a*b^2*m^2*x^3*e^4*sgn(b*x + a) + 11*(x*e + d)^m*b^3*
m*x^4*e^4*sgn(b*x + a) + 3*(x*e + d)^m*a^2*b*d*m^3*x*e^3*sgn(b*x + a) + 15*(x*e + d)^m*a*b^2*d*m^2*x^2*e^3*sgn
(b*x + a) + 2*(x*e + d)^m*b^3*d*m*x^3*e^3*sgn(b*x + a) - 6*(x*e + d)^m*a*b^2*d^2*m^2*x*e^2*sgn(b*x + a) - 3*(x
*e + d)^m*b^3*d^2*m*x^2*e^2*sgn(b*x + a) + 6*(x*e + d)^m*b^3*d^3*m*x*e*sgn(b*x + a) + (x*e + d)^m*a^3*m^3*x*e^
4*sgn(b*x + a) + 24*(x*e + d)^m*a^2*b*m^2*x^2*e^4*sgn(b*x + a) + 42*(x*e + d)^m*a*b^2*m*x^3*e^4*sgn(b*x + a) +
6*(x*e + d)^m*b^3*x^4*e^4*sgn(b*x + a) + (x*e + d)^m*a^3*d*m^3*e^3*sgn(b*x + a) + 21*(x*e + d)^m*a^2*b*d*m^2*
x*e^3*sgn(b*x + a) + 12*(x*e + d)^m*a*b^2*d*m*x^2*e^3*sgn(b*x + a) - 3*(x*e + d)^m*a^2*b*d^2*m^2*e^2*sgn(b*x +
a) - 24*(x*e + d)^m*a*b^2*d^2*m*x*e^2*sgn(b*x + a) + 6*(x*e + d)^m*a*b^2*d^3*m*e*sgn(b*x + a) - 6*(x*e + d)^m
*b^3*d^4*sgn(b*x + a) + 9*(x*e + d)^m*a^3*m^2*x*e^4*sgn(b*x + a) + 57*(x*e + d)^m*a^2*b*m*x^2*e^4*sgn(b*x + a)
+ 24*(x*e + d)^m*a*b^2*x^3*e^4*sgn(b*x + a) + 9*(x*e + d)^m*a^3*d*m^2*e^3*sgn(b*x + a) + 36*(x*e + d)^m*a^2*b
*d*m*x*e^3*sgn(b*x + a) - 21*(x*e + d)^m*a^2*b*d^2*m*e^2*sgn(b*x + a) + 24*(x*e + d)^m*a*b^2*d^3*e*sgn(b*x + a
) + 26*(x*e + d)^m*a^3*m*x*e^4*sgn(b*x + a) + 36*(x*e + d)^m*a^2*b*x^2*e^4*sgn(b*x + a) + 26*(x*e + d)^m*a^3*d
*m*e^3*sgn(b*x + a) - 36*(x*e + d)^m*a^2*b*d^2*e^2*sgn(b*x + a) + 24*(x*e + d)^m*a^3*x*e^4*sgn(b*x + a) + 24*(
x*e + d)^m*a^3*d*e^3*sgn(b*x + a))/(m^4*e^4 + 10*m^3*e^4 + 35*m^2*e^4 + 50*m*e^4 + 24*e^4)

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maple [B]  time = 0.06, size = 402, normalized size = 1.84 \begin {gather*} \frac {\left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}} \left (b^{3} e^{3} m^{3} x^{3}+3 a \,b^{2} e^{3} m^{3} x^{2}+6 b^{3} e^{3} m^{2} x^{3}+3 a^{2} b \,e^{3} m^{3} x +21 a \,b^{2} e^{3} m^{2} x^{2}-3 b^{3} d \,e^{2} m^{2} x^{2}+11 b^{3} e^{3} m \,x^{3}+a^{3} e^{3} m^{3}+24 a^{2} b \,e^{3} m^{2} x -6 a \,b^{2} d \,e^{2} m^{2} x +42 a \,b^{2} e^{3} m \,x^{2}-9 b^{3} d \,e^{2} m \,x^{2}+6 b^{3} e^{3} x^{3}+9 a^{3} e^{3} m^{2}-3 a^{2} b d \,e^{2} m^{2}+57 a^{2} b \,e^{3} m x -30 a \,b^{2} d \,e^{2} m x +24 a \,b^{2} e^{3} x^{2}+6 b^{3} d^{2} e m x -6 b^{3} d \,e^{2} x^{2}+26 a^{3} e^{3} m -21 a^{2} b d \,e^{2} m +36 a^{2} b \,e^{3} x +6 a \,b^{2} d^{2} e m -24 a \,b^{2} d \,e^{2} x +6 b^{3} d^{2} e x +24 a^{3} e^{3}-36 a^{2} b d \,e^{2}+24 a \,b^{2} d^{2} e -6 b^{3} d^{3}\right ) \left (e x +d \right )^{m +1}}{\left (b x +a \right )^{3} \left (m^{4}+10 m^{3}+35 m^{2}+50 m +24\right ) e^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((b*x+a)^2)^(3/2)*(e*x+d)^(m+1)*(b^3*e^3*m^3*x^3+3*a*b^2*e^3*m^3*x^2+6*b^3*e^3*m^2*x^3+3*a^2*b*e^3*m^3*x+21*a*
b^2*e^3*m^2*x^2-3*b^3*d*e^2*m^2*x^2+11*b^3*e^3*m*x^3+a^3*e^3*m^3+24*a^2*b*e^3*m^2*x-6*a*b^2*d*e^2*m^2*x+42*a*b
^2*e^3*m*x^2-9*b^3*d*e^2*m*x^2+6*b^3*e^3*x^3+9*a^3*e^3*m^2-3*a^2*b*d*e^2*m^2+57*a^2*b*e^3*m*x-30*a*b^2*d*e^2*m
*x+24*a*b^2*e^3*x^2+6*b^3*d^2*e*m*x-6*b^3*d*e^2*x^2+26*a^3*e^3*m-21*a^2*b*d*e^2*m+36*a^2*b*e^3*x+6*a*b^2*d^2*e
*m-24*a*b^2*d*e^2*x+6*b^3*d^2*e*x+24*a^3*e^3-36*a^2*b*d*e^2+24*a*b^2*d^2*e-6*b^3*d^3)/(b*x+a)^3/e^4/(m^4+10*m^
3+35*m^2+50*m+24)

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maxima [A]  time = 1.23, size = 305, normalized size = 1.39 \begin {gather*} \frac {{\left ({\left (m^{3} + 6 \, m^{2} + 11 \, m + 6\right )} b^{3} e^{4} x^{4} - 3 \, {\left (m^{2} + 7 \, m + 12\right )} a^{2} b d^{2} e^{2} + {\left (m^{3} + 9 \, m^{2} + 26 \, m + 24\right )} a^{3} d e^{3} + 6 \, a b^{2} d^{3} e {\left (m + 4\right )} - 6 \, b^{3} d^{4} + {\left ({\left (m^{3} + 3 \, m^{2} + 2 \, m\right )} b^{3} d e^{3} + 3 \, {\left (m^{3} + 7 \, m^{2} + 14 \, m + 8\right )} a b^{2} e^{4}\right )} x^{3} - 3 \, {\left ({\left (m^{2} + m\right )} b^{3} d^{2} e^{2} - {\left (m^{3} + 5 \, m^{2} + 4 \, m\right )} a b^{2} d e^{3} - {\left (m^{3} + 8 \, m^{2} + 19 \, m + 12\right )} a^{2} b e^{4}\right )} x^{2} - {\left (6 \, {\left (m^{2} + 4 \, m\right )} a b^{2} d^{2} e^{2} - 3 \, {\left (m^{3} + 7 \, m^{2} + 12 \, m\right )} a^{2} b d e^{3} - {\left (m^{3} + 9 \, m^{2} + 26 \, m + 24\right )} a^{3} e^{4} - 6 \, b^{3} d^{3} e m\right )} x\right )} {\left (e x + d\right )}^{m}}{{\left (m^{4} + 10 \, m^{3} + 35 \, m^{2} + 50 \, m + 24\right )} e^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^m*(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="maxima")

[Out]

((m^3 + 6*m^2 + 11*m + 6)*b^3*e^4*x^4 - 3*(m^2 + 7*m + 12)*a^2*b*d^2*e^2 + (m^3 + 9*m^2 + 26*m + 24)*a^3*d*e^3
+ 6*a*b^2*d^3*e*(m + 4) - 6*b^3*d^4 + ((m^3 + 3*m^2 + 2*m)*b^3*d*e^3 + 3*(m^3 + 7*m^2 + 14*m + 8)*a*b^2*e^4)*
x^3 - 3*((m^2 + m)*b^3*d^2*e^2 - (m^3 + 5*m^2 + 4*m)*a*b^2*d*e^3 - (m^3 + 8*m^2 + 19*m + 12)*a^2*b*e^4)*x^2 -
(6*(m^2 + 4*m)*a*b^2*d^2*e^2 - 3*(m^3 + 7*m^2 + 12*m)*a^2*b*d*e^3 - (m^3 + 9*m^2 + 26*m + 24)*a^3*e^4 - 6*b^3*
d^3*e*m)*x)*(e*x + d)^m/((m^4 + 10*m^3 + 35*m^2 + 50*m + 24)*e^4)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (d + e x\right )^{m} \left (\left (a + b x\right )^{2}\right )^{\frac {3}{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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