\(\int (d+e x) (a+c x^2)^{3/2} \, dx\) [537]

Optimal result

Integrand size = 17, antiderivative size = 87 \[ \int (d+e x) \left (a+c x^2\right )^{3/2} \, dx=\frac {3}{8} a d x \sqrt {a+c x^2}+\frac {1}{4} d x \left (a+c x^2\right )^{3/2}+\frac {e \left (a+c x^2\right )^{5/2}}{5 c}+\frac {3 a^2 d \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{8 \sqrt {c}} \]

[Out]

1/4*d*x*(c*x^2+a)^(3/2)+1/5*e*(c*x^2+a)^(5/2)/c+3/8*a^2*d*arctanh(x*c^(1/2)/(c*x^2+a)^(1/2))/c^(1/2)+3/8*a*d*x
*(c*x^2+a)^(1/2)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {655, 201, 223, 212} \[ \int (d+e x) \left (a+c x^2\right )^{3/2} \, dx=\frac {3 a^2 d \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{8 \sqrt {c}}+\frac {1}{4} d x \left (a+c x^2\right )^{3/2}+\frac {3}{8} a d x \sqrt {a+c x^2}+\frac {e \left (a+c x^2\right )^{5/2}}{5 c} \]

[In]

Int[(d + e*x)*(a + c*x^2)^(3/2),x]

[Out]

(3*a*d*x*Sqrt[a + c*x^2])/8 + (d*x*(a + c*x^2)^(3/2))/4 + (e*(a + c*x^2)^(5/2))/(5*c) + (3*a^2*d*ArcTanh[(Sqrt
[c]*x)/Sqrt[a + c*x^2]])/(8*Sqrt[c])

Rule 201

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 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 223

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 655

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]

Rubi steps \begin{align*} \text {integral}& = \frac {e \left (a+c x^2\right )^{5/2}}{5 c}+d \int \left (a+c x^2\right )^{3/2} \, dx \\ & = \frac {1}{4} d x \left (a+c x^2\right )^{3/2}+\frac {e \left (a+c x^2\right )^{5/2}}{5 c}+\frac {1}{4} (3 a d) \int \sqrt {a+c x^2} \, dx \\ & = \frac {3}{8} a d x \sqrt {a+c x^2}+\frac {1}{4} d x \left (a+c x^2\right )^{3/2}+\frac {e \left (a+c x^2\right )^{5/2}}{5 c}+\frac {1}{8} \left (3 a^2 d\right ) \int \frac {1}{\sqrt {a+c x^2}} \, dx \\ & = \frac {3}{8} a d x \sqrt {a+c x^2}+\frac {1}{4} d x \left (a+c x^2\right )^{3/2}+\frac {e \left (a+c x^2\right )^{5/2}}{5 c}+\frac {1}{8} \left (3 a^2 d\right ) \text {Subst}\left (\int \frac {1}{1-c x^2} \, dx,x,\frac {x}{\sqrt {a+c x^2}}\right ) \\ & = \frac {3}{8} a d x \sqrt {a+c x^2}+\frac {1}{4} d x \left (a+c x^2\right )^{3/2}+\frac {e \left (a+c x^2\right )^{5/2}}{5 c}+\frac {3 a^2 d \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{8 \sqrt {c}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.36 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.00 \[ \int (d+e x) \left (a+c x^2\right )^{3/2} \, dx=\frac {\sqrt {a+c x^2} \left (8 a^2 e+2 c^2 x^3 (5 d+4 e x)+a c x (25 d+16 e x)\right )-15 a^2 \sqrt {c} d \log \left (-\sqrt {c} x+\sqrt {a+c x^2}\right )}{40 c} \]

[In]

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

[Out]

(Sqrt[a + c*x^2]*(8*a^2*e + 2*c^2*x^3*(5*d + 4*e*x) + a*c*x*(25*d + 16*e*x)) - 15*a^2*Sqrt[c]*d*Log[-(Sqrt[c]*
x) + Sqrt[a + c*x^2]])/(40*c)

Maple [A] (verified)

Time = 2.39 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.80

method	result	size
default	\(d \left (\frac {x \left (c \,x^{2}+a \right )^{\frac {3}{2}}}{4}+\frac {3 a \left (\frac {x \sqrt {c \,x^{2}+a}}{2}+\frac {a \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+a}\right )}{2 \sqrt {c}}\right )}{4}\right )+\frac {e \left (c \,x^{2}+a \right )^{\frac {5}{2}}}{5 c}\)	\(70\)
risch	\(\frac {\left (8 c^{2} e \,x^{4}+10 c^{2} d \,x^{3}+16 x^{2} a c e +25 a d x c +8 a^{2} e \right ) \sqrt {c \,x^{2}+a}}{40 c}+\frac {3 a^{2} d \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+a}\right )}{8 \sqrt {c}}\)	\(80\)

[In]

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

[Out]

d*(1/4*x*(c*x^2+a)^(3/2)+3/4*a*(1/2*x*(c*x^2+a)^(1/2)+1/2*a/c^(1/2)*ln(c^(1/2)*x+(c*x^2+a)^(1/2))))+1/5*e*(c*x
^2+a)^(5/2)/c

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 176, normalized size of antiderivative = 2.02 \[ \int (d+e x) \left (a+c x^2\right )^{3/2} \, dx=\left [\frac {15 \, a^{2} \sqrt {c} d \log \left (-2 \, c x^{2} - 2 \, \sqrt {c x^{2} + a} \sqrt {c} x - a\right ) + 2 \, {\left (8 \, c^{2} e x^{4} + 10 \, c^{2} d x^{3} + 16 \, a c e x^{2} + 25 \, a c d x + 8 \, a^{2} e\right )} \sqrt {c x^{2} + a}}{80 \, c}, -\frac {15 \, a^{2} \sqrt {-c} d \arctan \left (\frac {\sqrt {-c} x}{\sqrt {c x^{2} + a}}\right ) - {\left (8 \, c^{2} e x^{4} + 10 \, c^{2} d x^{3} + 16 \, a c e x^{2} + 25 \, a c d x + 8 \, a^{2} e\right )} \sqrt {c x^{2} + a}}{40 \, c}\right ] \]

[In]

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

[Out]

[1/80*(15*a^2*sqrt(c)*d*log(-2*c*x^2 - 2*sqrt(c*x^2 + a)*sqrt(c)*x - a) + 2*(8*c^2*e*x^4 + 10*c^2*d*x^3 + 16*a
*c*e*x^2 + 25*a*c*d*x + 8*a^2*e)*sqrt(c*x^2 + a))/c, -1/40*(15*a^2*sqrt(-c)*d*arctan(sqrt(-c)*x/sqrt(c*x^2 + a
)) - (8*c^2*e*x^4 + 10*c^2*d*x^3 + 16*a*c*e*x^2 + 25*a*c*d*x + 8*a^2*e)*sqrt(c*x^2 + a))/c]

Sympy [A] (verification not implemented)

Time = 0.40 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.37 \[ \int (d+e x) \left (a+c x^2\right )^{3/2} \, dx=\begin {cases} \frac {3 a^{2} d \left (\begin {cases} \frac {\log {\left (2 \sqrt {c} \sqrt {a + c x^{2}} + 2 c x \right )}}{\sqrt {c}} & \text {for}\: a \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {c x^{2}}} & \text {otherwise} \end {cases}\right )}{8} + \sqrt {a + c x^{2}} \left (\frac {a^{2} e}{5 c} + \frac {5 a d x}{8} + \frac {2 a e x^{2}}{5} + \frac {c d x^{3}}{4} + \frac {c e x^{4}}{5}\right ) & \text {for}\: c \neq 0 \\a^{\frac {3}{2}} \left (d x + \frac {e x^{2}}{2}\right ) & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((3*a**2*d*Piecewise((log(2*sqrt(c)*sqrt(a + c*x**2) + 2*c*x)/sqrt(c), Ne(a, 0)), (x*log(x)/sqrt(c*x*
*2), True))/8 + sqrt(a + c*x**2)*(a**2*e/(5*c) + 5*a*d*x/8 + 2*a*e*x**2/5 + c*d*x**3/4 + c*e*x**4/5), Ne(c, 0)
), (a**(3/2)*(d*x + e*x**2/2), True))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.70 \[ \int (d+e x) \left (a+c x^2\right )^{3/2} \, dx=\frac {1}{4} \, {\left (c x^{2} + a\right )}^{\frac {3}{2}} d x + \frac {3}{8} \, \sqrt {c x^{2} + a} a d x + \frac {3 \, a^{2} d \operatorname {arsinh}\left (\frac {c x}{\sqrt {a c}}\right )}{8 \, \sqrt {c}} + \frac {{\left (c x^{2} + a\right )}^{\frac {5}{2}} e}{5 \, c} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.87 \[ \int (d+e x) \left (a+c x^2\right )^{3/2} \, dx=-\frac {3 \, a^{2} d \log \left ({\left | -\sqrt {c} x + \sqrt {c x^{2} + a} \right |}\right )}{8 \, \sqrt {c}} + \frac {1}{40} \, \sqrt {c x^{2} + a} {\left (\frac {8 \, a^{2} e}{c} + {\left (25 \, a d + 2 \, {\left (8 \, a e + {\left (4 \, c e x + 5 \, c d\right )} x\right )} x\right )} x\right )} \]

[In]

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

[Out]

-3/8*a^2*d*log(abs(-sqrt(c)*x + sqrt(c*x^2 + a)))/sqrt(c) + 1/40*sqrt(c*x^2 + a)*(8*a^2*e/c + (25*a*d + 2*(8*a
*e + (4*c*e*x + 5*c*d)*x)*x)*x)

Mupad [B] (verification not implemented)

Time = 9.48 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.62 \[ \int (d+e x) \left (a+c x^2\right )^{3/2} \, dx=\frac {e\,{\left (c\,x^2+a\right )}^{5/2}}{5\,c}+\frac {d\,x\,{\left (c\,x^2+a\right )}^{3/2}\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{2},\frac {1}{2};\ \frac {3}{2};\ -\frac {c\,x^2}{a}\right )}{{\left (\frac {c\,x^2}{a}+1\right )}^{3/2}} \]

[In]

int((a + c*x^2)^(3/2)*(d + e*x),x)

[Out]

(e*(a + c*x^2)^(5/2))/(5*c) + (d*x*(a + c*x^2)^(3/2)*hypergeom([-3/2, 1/2], 3/2, -(c*x^2)/a))/((c*x^2)/a + 1)^
(3/2)