3.536 \(\int (d+e x)^2 (a+c x^2)^{3/2} \, dx\)

Optimal. Leaf size=154 \[ \frac {a^2 \left (6 c d^2-a e^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{16 c^{3/2}}+\frac {x \left (a+c x^2\right )^{3/2} \left (6 c d^2-a e^2\right )}{24 c}+\frac {a x \sqrt {a+c x^2} \left (6 c d^2-a e^2\right )}{16 c}+\frac {7 d e \left (a+c x^2\right )^{5/2}}{30 c}+\frac {e \left (a+c x^2\right )^{5/2} (d+e x)}{6 c} \]

[Out]

1/24*(-a*e^2+6*c*d^2)*x*(c*x^2+a)^(3/2)/c+7/30*d*e*(c*x^2+a)^(5/2)/c+1/6*e*(e*x+d)*(c*x^2+a)^(5/2)/c+1/16*a^2*
(-a*e^2+6*c*d^2)*arctanh(x*c^(1/2)/(c*x^2+a)^(1/2))/c^(3/2)+1/16*a*(-a*e^2+6*c*d^2)*x*(c*x^2+a)^(1/2)/c

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Rubi [A]  time = 0.07, antiderivative size = 154, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {743, 641, 195, 217, 206} \[ \frac {a^2 \left (6 c d^2-a e^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{16 c^{3/2}}+\frac {x \left (a+c x^2\right )^{3/2} \left (6 c d^2-a e^2\right )}{24 c}+\frac {a x \sqrt {a+c x^2} \left (6 c d^2-a e^2\right )}{16 c}+\frac {7 d e \left (a+c x^2\right )^{5/2}}{30 c}+\frac {e \left (a+c x^2\right )^{5/2} (d+e x)}{6 c} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^2*(a + c*x^2)^(3/2),x]

[Out]

(a*(6*c*d^2 - a*e^2)*x*Sqrt[a + c*x^2])/(16*c) + ((6*c*d^2 - a*e^2)*x*(a + c*x^2)^(3/2))/(24*c) + (7*d*e*(a +
c*x^2)^(5/2))/(30*c) + (e*(d + e*x)*(a + c*x^2)^(5/2))/(6*c) + (a^2*(6*c*d^2 - a*e^2)*ArcTanh[(Sqrt[c]*x)/Sqrt
[a + c*x^2]])/(16*c^(3/2))

Rule 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

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]

Rule 641

Int[((d_) + (e_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(e*(a + c*x^2)^(p + 1))/(2*c*(p + 1)),
x] + Dist[d, Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, p}, x] && NeQ[p, -1]

Rule 743

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^(m - 1)*(a + c*x^2)^(p
+ 1))/(c*(m + 2*p + 1)), x] + Dist[1/(c*(m + 2*p + 1)), Int[(d + e*x)^(m - 2)*Simp[c*d^2*(m + 2*p + 1) - a*e^
2*(m - 1) + 2*c*d*e*(m + p)*x, x]*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, m, p}, x] && NeQ[c*d^2 + a*e^2,
0] && If[RationalQ[m], GtQ[m, 1], SumSimplerQ[m, -2]] && NeQ[m + 2*p + 1, 0] && IntQuadraticQ[a, 0, c, d, e, m
, p, x]

Rubi steps

\begin {align*} \int (d+e x)^2 \left (a+c x^2\right )^{3/2} \, dx &=\frac {e (d+e x) \left (a+c x^2\right )^{5/2}}{6 c}+\frac {\int \left (6 c d^2-a e^2+7 c d e x\right ) \left (a+c x^2\right )^{3/2} \, dx}{6 c}\\ &=\frac {7 d e \left (a+c x^2\right )^{5/2}}{30 c}+\frac {e (d+e x) \left (a+c x^2\right )^{5/2}}{6 c}+\frac {\left (6 c d^2-a e^2\right ) \int \left (a+c x^2\right )^{3/2} \, dx}{6 c}\\ &=\frac {\left (6 c d^2-a e^2\right ) x \left (a+c x^2\right )^{3/2}}{24 c}+\frac {7 d e \left (a+c x^2\right )^{5/2}}{30 c}+\frac {e (d+e x) \left (a+c x^2\right )^{5/2}}{6 c}+\frac {\left (a \left (6 c d^2-a e^2\right )\right ) \int \sqrt {a+c x^2} \, dx}{8 c}\\ &=\frac {a \left (6 c d^2-a e^2\right ) x \sqrt {a+c x^2}}{16 c}+\frac {\left (6 c d^2-a e^2\right ) x \left (a+c x^2\right )^{3/2}}{24 c}+\frac {7 d e \left (a+c x^2\right )^{5/2}}{30 c}+\frac {e (d+e x) \left (a+c x^2\right )^{5/2}}{6 c}+\frac {\left (a^2 \left (6 c d^2-a e^2\right )\right ) \int \frac {1}{\sqrt {a+c x^2}} \, dx}{16 c}\\ &=\frac {a \left (6 c d^2-a e^2\right ) x \sqrt {a+c x^2}}{16 c}+\frac {\left (6 c d^2-a e^2\right ) x \left (a+c x^2\right )^{3/2}}{24 c}+\frac {7 d e \left (a+c x^2\right )^{5/2}}{30 c}+\frac {e (d+e x) \left (a+c x^2\right )^{5/2}}{6 c}+\frac {\left (a^2 \left (6 c d^2-a e^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{1-c x^2} \, dx,x,\frac {x}{\sqrt {a+c x^2}}\right )}{16 c}\\ &=\frac {a \left (6 c d^2-a e^2\right ) x \sqrt {a+c x^2}}{16 c}+\frac {\left (6 c d^2-a e^2\right ) x \left (a+c x^2\right )^{3/2}}{24 c}+\frac {7 d e \left (a+c x^2\right )^{5/2}}{30 c}+\frac {e (d+e x) \left (a+c x^2\right )^{5/2}}{6 c}+\frac {a^2 \left (6 c d^2-a e^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{16 c^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.09, size = 132, normalized size = 0.86 \[ \frac {\sqrt {c} \sqrt {a+c x^2} \left (3 a^2 e (32 d+5 e x)+2 a c x \left (75 d^2+96 d e x+35 e^2 x^2\right )+4 c^2 x^3 \left (15 d^2+24 d e x+10 e^2 x^2\right )\right )-15 a^2 \left (a e^2-6 c d^2\right ) \log \left (\sqrt {c} \sqrt {a+c x^2}+c x\right )}{240 c^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[c]*Sqrt[a + c*x^2]*(3*a^2*e*(32*d + 5*e*x) + 4*c^2*x^3*(15*d^2 + 24*d*e*x + 10*e^2*x^2) + 2*a*c*x*(75*d^
2 + 96*d*e*x + 35*e^2*x^2)) - 15*a^2*(-6*c*d^2 + a*e^2)*Log[c*x + Sqrt[c]*Sqrt[a + c*x^2]])/(240*c^(3/2))

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fricas [A]  time = 0.79, size = 296, normalized size = 1.92 \[ \left [-\frac {15 \, {\left (6 \, a^{2} c d^{2} - a^{3} e^{2}\right )} \sqrt {c} \log \left (-2 \, c x^{2} + 2 \, \sqrt {c x^{2} + a} \sqrt {c} x - a\right ) - 2 \, {\left (40 \, c^{3} e^{2} x^{5} + 96 \, c^{3} d e x^{4} + 192 \, a c^{2} d e x^{2} + 96 \, a^{2} c d e + 10 \, {\left (6 \, c^{3} d^{2} + 7 \, a c^{2} e^{2}\right )} x^{3} + 15 \, {\left (10 \, a c^{2} d^{2} + a^{2} c e^{2}\right )} x\right )} \sqrt {c x^{2} + a}}{480 \, c^{2}}, -\frac {15 \, {\left (6 \, a^{2} c d^{2} - a^{3} e^{2}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {-c} x}{\sqrt {c x^{2} + a}}\right ) - {\left (40 \, c^{3} e^{2} x^{5} + 96 \, c^{3} d e x^{4} + 192 \, a c^{2} d e x^{2} + 96 \, a^{2} c d e + 10 \, {\left (6 \, c^{3} d^{2} + 7 \, a c^{2} e^{2}\right )} x^{3} + 15 \, {\left (10 \, a c^{2} d^{2} + a^{2} c e^{2}\right )} x\right )} \sqrt {c x^{2} + a}}{240 \, c^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/480*(15*(6*a^2*c*d^2 - a^3*e^2)*sqrt(c)*log(-2*c*x^2 + 2*sqrt(c*x^2 + a)*sqrt(c)*x - a) - 2*(40*c^3*e^2*x^
5 + 96*c^3*d*e*x^4 + 192*a*c^2*d*e*x^2 + 96*a^2*c*d*e + 10*(6*c^3*d^2 + 7*a*c^2*e^2)*x^3 + 15*(10*a*c^2*d^2 +
a^2*c*e^2)*x)*sqrt(c*x^2 + a))/c^2, -1/240*(15*(6*a^2*c*d^2 - a^3*e^2)*sqrt(-c)*arctan(sqrt(-c)*x/sqrt(c*x^2 +
a)) - (40*c^3*e^2*x^5 + 96*c^3*d*e*x^4 + 192*a*c^2*d*e*x^2 + 96*a^2*c*d*e + 10*(6*c^3*d^2 + 7*a*c^2*e^2)*x^3
+ 15*(10*a*c^2*d^2 + a^2*c*e^2)*x)*sqrt(c*x^2 + a))/c^2]

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giac [A]  time = 0.23, size = 142, normalized size = 0.92 \[ \frac {1}{240} \, \sqrt {c x^{2} + a} {\left (\frac {96 \, a^{2} d e}{c} + {\left (2 \, {\left (96 \, a d e + {\left (4 \, {\left (5 \, c x e^{2} + 12 \, c d e\right )} x + \frac {5 \, {\left (6 \, c^{5} d^{2} + 7 \, a c^{4} e^{2}\right )}}{c^{4}}\right )} x\right )} x + \frac {15 \, {\left (10 \, a c^{4} d^{2} + a^{2} c^{3} e^{2}\right )}}{c^{4}}\right )} x\right )} - \frac {{\left (6 \, a^{2} c d^{2} - a^{3} e^{2}\right )} \log \left ({\left | -\sqrt {c} x + \sqrt {c x^{2} + a} \right |}\right )}{16 \, c^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/240*sqrt(c*x^2 + a)*(96*a^2*d*e/c + (2*(96*a*d*e + (4*(5*c*x*e^2 + 12*c*d*e)*x + 5*(6*c^5*d^2 + 7*a*c^4*e^2)
/c^4)*x)*x + 15*(10*a*c^4*d^2 + a^2*c^3*e^2)/c^4)*x) - 1/16*(6*a^2*c*d^2 - a^3*e^2)*log(abs(-sqrt(c)*x + sqrt(
c*x^2 + a)))/c^(3/2)

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maple [A]  time = 0.06, size = 161, normalized size = 1.05 \[ -\frac {a^{3} e^{2} \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+a}\right )}{16 c^{\frac {3}{2}}}+\frac {3 a^{2} d^{2} \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+a}\right )}{8 \sqrt {c}}-\frac {\sqrt {c \,x^{2}+a}\, a^{2} e^{2} x}{16 c}+\frac {3 \sqrt {c \,x^{2}+a}\, a \,d^{2} x}{8}-\frac {\left (c \,x^{2}+a \right )^{\frac {3}{2}} a \,e^{2} x}{24 c}+\frac {\left (c \,x^{2}+a \right )^{\frac {3}{2}} d^{2} x}{4}+\frac {\left (c \,x^{2}+a \right )^{\frac {5}{2}} e^{2} x}{6 c}+\frac {2 \left (c \,x^{2}+a \right )^{\frac {5}{2}} d e}{5 c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*e^2*x*(c*x^2+a)^(5/2)/c-1/24*e^2*a/c*x*(c*x^2+a)^(3/2)-1/16*e^2*a^2/c*x*(c*x^2+a)^(1/2)-1/16*e^2*a^3/c^(3/
2)*ln(c^(1/2)*x+(c*x^2+a)^(1/2))+2/5*d*e*(c*x^2+a)^(5/2)/c+1/4*d^2*x*(c*x^2+a)^(3/2)+3/8*d^2*a*x*(c*x^2+a)^(1/
2)+3/8*d^2*a^2/c^(1/2)*ln(c^(1/2)*x+(c*x^2+a)^(1/2))

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maxima [A]  time = 1.35, size = 146, normalized size = 0.95 \[ \frac {1}{4} \, {\left (c x^{2} + a\right )}^{\frac {3}{2}} d^{2} x + \frac {3}{8} \, \sqrt {c x^{2} + a} a d^{2} x + \frac {{\left (c x^{2} + a\right )}^{\frac {5}{2}} e^{2} x}{6 \, c} - \frac {{\left (c x^{2} + a\right )}^{\frac {3}{2}} a e^{2} x}{24 \, c} - \frac {\sqrt {c x^{2} + a} a^{2} e^{2} x}{16 \, c} + \frac {3 \, a^{2} d^{2} \operatorname {arsinh}\left (\frac {c x}{\sqrt {a c}}\right )}{8 \, \sqrt {c}} - \frac {a^{3} e^{2} \operatorname {arsinh}\left (\frac {c x}{\sqrt {a c}}\right )}{16 \, c^{\frac {3}{2}}} + \frac {2 \, {\left (c x^{2} + a\right )}^{\frac {5}{2}} d e}{5 \, c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(c*x^2 + a)^(3/2)*d^2*x + 3/8*sqrt(c*x^2 + a)*a*d^2*x + 1/6*(c*x^2 + a)^(5/2)*e^2*x/c - 1/24*(c*x^2 + a)^(
3/2)*a*e^2*x/c - 1/16*sqrt(c*x^2 + a)*a^2*e^2*x/c + 3/8*a^2*d^2*arcsinh(c*x/sqrt(a*c))/sqrt(c) - 1/16*a^3*e^2*
arcsinh(c*x/sqrt(a*c))/c^(3/2) + 2/5*(c*x^2 + a)^(5/2)*d*e/c

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (c\,x^2+a\right )}^{3/2}\,{\left (d+e\,x\right )}^2 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + c*x^2)^(3/2)*(d + e*x)^2,x)

[Out]

int((a + c*x^2)^(3/2)*(d + e*x)^2, x)

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sympy [A]  time = 15.78, size = 372, normalized size = 2.42 \[ \frac {a^{\frac {5}{2}} e^{2} x}{16 c \sqrt {1 + \frac {c x^{2}}{a}}} + \frac {a^{\frac {3}{2}} d^{2} x \sqrt {1 + \frac {c x^{2}}{a}}}{2} + \frac {a^{\frac {3}{2}} d^{2} x}{8 \sqrt {1 + \frac {c x^{2}}{a}}} + \frac {17 a^{\frac {3}{2}} e^{2} x^{3}}{48 \sqrt {1 + \frac {c x^{2}}{a}}} + \frac {3 \sqrt {a} c d^{2} x^{3}}{8 \sqrt {1 + \frac {c x^{2}}{a}}} + \frac {11 \sqrt {a} c e^{2} x^{5}}{24 \sqrt {1 + \frac {c x^{2}}{a}}} - \frac {a^{3} e^{2} \operatorname {asinh}{\left (\frac {\sqrt {c} x}{\sqrt {a}} \right )}}{16 c^{\frac {3}{2}}} + \frac {3 a^{2} d^{2} \operatorname {asinh}{\left (\frac {\sqrt {c} x}{\sqrt {a}} \right )}}{8 \sqrt {c}} + 2 a d e \left (\begin {cases} \frac {\sqrt {a} x^{2}}{2} & \text {for}\: c = 0 \\\frac {\left (a + c x^{2}\right )^{\frac {3}{2}}}{3 c} & \text {otherwise} \end {cases}\right ) + 2 c d e \left (\begin {cases} - \frac {2 a^{2} \sqrt {a + c x^{2}}}{15 c^{2}} + \frac {a x^{2} \sqrt {a + c x^{2}}}{15 c} + \frac {x^{4} \sqrt {a + c x^{2}}}{5} & \text {for}\: c \neq 0 \\\frac {\sqrt {a} x^{4}}{4} & \text {otherwise} \end {cases}\right ) + \frac {c^{2} d^{2} x^{5}}{4 \sqrt {a} \sqrt {1 + \frac {c x^{2}}{a}}} + \frac {c^{2} e^{2} x^{7}}{6 \sqrt {a} \sqrt {1 + \frac {c x^{2}}{a}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**(5/2)*e**2*x/(16*c*sqrt(1 + c*x**2/a)) + a**(3/2)*d**2*x*sqrt(1 + c*x**2/a)/2 + a**(3/2)*d**2*x/(8*sqrt(1 +
c*x**2/a)) + 17*a**(3/2)*e**2*x**3/(48*sqrt(1 + c*x**2/a)) + 3*sqrt(a)*c*d**2*x**3/(8*sqrt(1 + c*x**2/a)) + 1
1*sqrt(a)*c*e**2*x**5/(24*sqrt(1 + c*x**2/a)) - a**3*e**2*asinh(sqrt(c)*x/sqrt(a))/(16*c**(3/2)) + 3*a**2*d**2
*asinh(sqrt(c)*x/sqrt(a))/(8*sqrt(c)) + 2*a*d*e*Piecewise((sqrt(a)*x**2/2, Eq(c, 0)), ((a + c*x**2)**(3/2)/(3*
c), True)) + 2*c*d*e*Piecewise((-2*a**2*sqrt(a + c*x**2)/(15*c**2) + a*x**2*sqrt(a + c*x**2)/(15*c) + x**4*sqr
t(a + c*x**2)/5, Ne(c, 0)), (sqrt(a)*x**4/4, True)) + c**2*d**2*x**5/(4*sqrt(a)*sqrt(1 + c*x**2/a)) + c**2*e**
2*x**7/(6*sqrt(a)*sqrt(1 + c*x**2/a))

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