\(\int (d+e x) (a+c x^2)^{5/2} \, dx\) [548]

Optimal result

Integrand size = 17, antiderivative size = 107 \[ \int (d+e x) \left (a+c x^2\right )^{5/2} \, dx=\frac {5}{16} a^2 d x \sqrt {a+c x^2}+\frac {5}{24} a d x \left (a+c x^2\right )^{3/2}+\frac {1}{6} d x \left (a+c x^2\right )^{5/2}+\frac {e \left (a+c x^2\right )^{7/2}}{7 c}+\frac {5 a^3 d \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{16 \sqrt {c}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {655, 201, 223, 212} \[ \int (d+e x) \left (a+c x^2\right )^{5/2} \, dx=\frac {5 a^3 d \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{16 \sqrt {c}}+\frac {5}{16} a^2 d x \sqrt {a+c x^2}+\frac {1}{6} d x \left (a+c x^2\right )^{5/2}+\frac {5}{24} a d x \left (a+c x^2\right )^{3/2}+\frac {e \left (a+c x^2\right )^{7/2}}{7 c} \]

[In]

Int[(d + e*x)*(a + c*x^2)^(5/2),x]

[Out]

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

Mathematica [A] (verified)

Time = 0.47 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.00 \[ \int (d+e x) \left (a+c x^2\right )^{5/2} \, dx=\frac {\sqrt {a+c x^2} \left (48 a^3 e+8 c^3 x^5 (7 d+6 e x)+3 a^2 c x (77 d+48 e x)+2 a c^2 x^3 (91 d+72 e x)\right )-105 a^3 \sqrt {c} d \log \left (-\sqrt {c} x+\sqrt {a+c x^2}\right )}{336 c} \]

[In]

Integrate[(d + e*x)*(a + c*x^2)^(5/2),x]

[Out]

(Sqrt[a + c*x^2]*(48*a^3*e + 8*c^3*x^5*(7*d + 6*e*x) + 3*a^2*c*x*(77*d + 48*e*x) + 2*a*c^2*x^3*(91*d + 72*e*x)
) - 105*a^3*Sqrt[c]*d*Log[-(Sqrt[c]*x) + Sqrt[a + c*x^2]])/(336*c)

Maple [A] (verified)

Time = 1.89 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.80

method	result	size
default	\(d \left (\frac {x \left (c \,x^{2}+a \right )^{\frac {5}{2}}}{6}+\frac {5 a \left (\frac {x \left (c \,x^{2}+a \right )^{\frac {3}{2}}}{4}+\frac {3 a \left (\frac {x \sqrt {c \,x^{2}+a}}{2}+\frac {a \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+a}\right )}{2 \sqrt {c}}\right )}{4}\right )}{6}\right )+\frac {e \left (c \,x^{2}+a \right )^{\frac {7}{2}}}{7 c}\)	\(86\)
risch	\(\frac {\left (48 e \,c^{3} x^{6}+56 c^{3} d \,x^{5}+144 c^{2} a e \,x^{4}+182 a \,c^{2} d \,x^{3}+144 x^{2} a^{2} c e +231 a^{2} c d x +48 e \,a^{3}\right ) \sqrt {c \,x^{2}+a}}{336 c}+\frac {5 a^{3} d \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+a}\right )}{16 \sqrt {c}}\)	\(104\)

[In]

int((e*x+d)*(c*x^2+a)^(5/2),x,method=_RETURNVERBOSE)

[Out]

d*(1/6*x*(c*x^2+a)^(5/2)+5/6*a*(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)))))+1/7*e*(c*x^2+a)^(7/2)/c

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 224, normalized size of antiderivative = 2.09 \[ \int (d+e x) \left (a+c x^2\right )^{5/2} \, dx=\left [\frac {105 \, a^{3} \sqrt {c} d \log \left (-2 \, c x^{2} - 2 \, \sqrt {c x^{2} + a} \sqrt {c} x - a\right ) + 2 \, {\left (48 \, c^{3} e x^{6} + 56 \, c^{3} d x^{5} + 144 \, a c^{2} e x^{4} + 182 \, a c^{2} d x^{3} + 144 \, a^{2} c e x^{2} + 231 \, a^{2} c d x + 48 \, a^{3} e\right )} \sqrt {c x^{2} + a}}{672 \, c}, -\frac {105 \, a^{3} \sqrt {-c} d \arctan \left (\frac {\sqrt {-c} x}{\sqrt {c x^{2} + a}}\right ) - {\left (48 \, c^{3} e x^{6} + 56 \, c^{3} d x^{5} + 144 \, a c^{2} e x^{4} + 182 \, a c^{2} d x^{3} + 144 \, a^{2} c e x^{2} + 231 \, a^{2} c d x + 48 \, a^{3} e\right )} \sqrt {c x^{2} + a}}{336 \, c}\right ] \]

[In]

integrate((e*x+d)*(c*x^2+a)^(5/2),x, algorithm="fricas")

[Out]

[1/672*(105*a^3*sqrt(c)*d*log(-2*c*x^2 - 2*sqrt(c*x^2 + a)*sqrt(c)*x - a) + 2*(48*c^3*e*x^6 + 56*c^3*d*x^5 + 1
44*a*c^2*e*x^4 + 182*a*c^2*d*x^3 + 144*a^2*c*e*x^2 + 231*a^2*c*d*x + 48*a^3*e)*sqrt(c*x^2 + a))/c, -1/336*(105
*a^3*sqrt(-c)*d*arctan(sqrt(-c)*x/sqrt(c*x^2 + a)) - (48*c^3*e*x^6 + 56*c^3*d*x^5 + 144*a*c^2*e*x^4 + 182*a*c^
2*d*x^3 + 144*a^2*c*e*x^2 + 231*a^2*c*d*x + 48*a^3*e)*sqrt(c*x^2 + a))/c]

Sympy [A] (verification not implemented)

Time = 0.48 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.40 \[ \int (d+e x) \left (a+c x^2\right )^{5/2} \, dx=\begin {cases} \frac {5 a^{3} d \left (\begin {cases} \frac {\log {\left (2 \sqrt {c} \sqrt {a + c x^{2}} + 2 c x \right )}}{\sqrt {c}} & \text {for}\: a \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {c x^{2}}} & \text {otherwise} \end {cases}\right )}{16} + \sqrt {a + c x^{2}} \left (\frac {a^{3} e}{7 c} + \frac {11 a^{2} d x}{16} + \frac {3 a^{2} e x^{2}}{7} + \frac {13 a c d x^{3}}{24} + \frac {3 a c e x^{4}}{7} + \frac {c^{2} d x^{5}}{6} + \frac {c^{2} e x^{6}}{7}\right ) & \text {for}\: c \neq 0 \\a^{\frac {5}{2}} \left (d x + \frac {e x^{2}}{2}\right ) & \text {otherwise} \end {cases} \]

[In]

integrate((e*x+d)*(c*x**2+a)**(5/2),x)

[Out]

Piecewise((5*a**3*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))/16 + sqrt(a + c*x**2)*(a**3*e/(7*c) + 11*a**2*d*x/16 + 3*a**2*e*x**2/7 + 13*a*c*d*x**3/24 + 3*a*c*
e*x**4/7 + c**2*d*x**5/6 + c**2*e*x**6/7), Ne(c, 0)), (a**(5/2)*(d*x + e*x**2/2), True))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.72 \[ \int (d+e x) \left (a+c x^2\right )^{5/2} \, dx=\frac {1}{6} \, {\left (c x^{2} + a\right )}^{\frac {5}{2}} d x + \frac {5}{24} \, {\left (c x^{2} + a\right )}^{\frac {3}{2}} a d x + \frac {5}{16} \, \sqrt {c x^{2} + a} a^{2} d x + \frac {5 \, a^{3} d \operatorname {arsinh}\left (\frac {c x}{\sqrt {a c}}\right )}{16 \, \sqrt {c}} + \frac {{\left (c x^{2} + a\right )}^{\frac {7}{2}} e}{7 \, c} \]

[In]

integrate((e*x+d)*(c*x^2+a)^(5/2),x, algorithm="maxima")

[Out]

1/6*(c*x^2 + a)^(5/2)*d*x + 5/24*(c*x^2 + a)^(3/2)*a*d*x + 5/16*sqrt(c*x^2 + a)*a^2*d*x + 5/16*a^3*d*arcsinh(c
*x/sqrt(a*c))/sqrt(c) + 1/7*(c*x^2 + a)^(7/2)*e/c

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.94 \[ \int (d+e x) \left (a+c x^2\right )^{5/2} \, dx=-\frac {5 \, a^{3} d \log \left ({\left | -\sqrt {c} x + \sqrt {c x^{2} + a} \right |}\right )}{16 \, \sqrt {c}} + \frac {1}{336} \, {\left (\frac {48 \, a^{3} e}{c} + {\left (231 \, a^{2} d + 2 \, {\left (72 \, a^{2} e + {\left (91 \, a c d + 4 \, {\left (18 \, a c e + {\left (6 \, c^{2} e x + 7 \, c^{2} d\right )} x\right )} x\right )} x\right )} x\right )} x\right )} \sqrt {c x^{2} + a} \]

[In]

integrate((e*x+d)*(c*x^2+a)^(5/2),x, algorithm="giac")

[Out]

-5/16*a^3*d*log(abs(-sqrt(c)*x + sqrt(c*x^2 + a)))/sqrt(c) + 1/336*(48*a^3*e/c + (231*a^2*d + 2*(72*a^2*e + (9
1*a*c*d + 4*(18*a*c*e + (6*c^2*e*x + 7*c^2*d)*x)*x)*x)*x)*x)*sqrt(c*x^2 + a)

Mupad [B] (verification not implemented)

Time = 9.47 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.50 \[ \int (d+e x) \left (a+c x^2\right )^{5/2} \, dx=\frac {e\,{\left (c\,x^2+a\right )}^{7/2}}{7\,c}+\frac {d\,x\,{\left (c\,x^2+a\right )}^{5/2}\,{{}}_2{\mathrm {F}}_1\left (-\frac {5}{2},\frac {1}{2};\ \frac {3}{2};\ -\frac {c\,x^2}{a}\right )}{{\left (\frac {c\,x^2}{a}+1\right )}^{5/2}} \]

[In]

int((a + c*x^2)^(5/2)*(d + e*x),x)

[Out]

(e*(a + c*x^2)^(7/2))/(7*c) + (d*x*(a + c*x^2)^(5/2)*hypergeom([-5/2, 1/2], 3/2, -(c*x^2)/a))/((c*x^2)/a + 1)^
(5/2)