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

Optimal. Leaf size=107 \[ \frac {5 a^3 d \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{16 \sqrt {c}}+\frac {5}{16} a^2 d x \sqrt {a+c x^2}+\frac {1}{6} d x \left (a+c x^2\right )^{5/2}+\frac {5}{24} a d x \left (a+c x^2\right )^{3/2}+\frac {e \left (a+c x^2\right )^{7/2}}{7 c} \]

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Rubi [A]  time = 0.03, antiderivative size = 107, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {641, 195, 217, 206} \begin {gather*} \frac {5}{16} a^2 d x \sqrt {a+c x^2}+\frac {5 a^3 d \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{16 \sqrt {c}}+\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} \end {gather*}

Antiderivative was successfully verified.

[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 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]

Rubi steps

\begin {align*} \int (d+e x) \left (a+c x^2\right )^{5/2} \, dx &=\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 ) \operatorname {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*}

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Mathematica [A]  time = 0.07, size = 108, normalized size = 1.01 \begin {gather*} \frac {105 a^3 \sqrt {c} d \log \left (\sqrt {c} \sqrt {a+c x^2}+c x\right )+\sqrt {a+c x^2} \left (48 a^3 e+3 a^2 c x (77 d+48 e x)+2 a c^2 x^3 (91 d+72 e x)+8 c^3 x^5 (7 d+6 e x)\right )}{336 c} \end {gather*}

Antiderivative was successfully verified.

[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[c*x + Sqrt[c]*Sqrt[a + c*x^2]])/(336*c)

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IntegrateAlgebraic [A]  time = 0.42, size = 116, normalized size = 1.08 \begin {gather*} \frac {\sqrt {a+c x^2} \left (48 a^3 e+231 a^2 c d x+144 a^2 c e x^2+182 a c^2 d x^3+144 a c^2 e x^4+56 c^3 d x^5+48 c^3 e x^6\right )}{336 c}-\frac {5 a^3 d \log \left (\sqrt {a+c x^2}-\sqrt {c} x\right )}{16 \sqrt {c}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + c*x^2]*(48*a^3*e + 231*a^2*c*d*x + 144*a^2*c*e*x^2 + 182*a*c^2*d*x^3 + 144*a*c^2*e*x^4 + 56*c^3*d*x^
5 + 48*c^3*e*x^6))/(336*c) - (5*a^3*d*Log[-(Sqrt[c]*x) + Sqrt[a + c*x^2]])/(16*Sqrt[c])

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fricas [A]  time = 0.43, size = 224, normalized size = 2.09 \begin {gather*} \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 ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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]

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giac [A]  time = 0.20, size = 105, normalized size = 0.98 \begin {gather*} -\frac {5 \, a^{3} d \log \left ({\left | -\sqrt {c} x + \sqrt {c x^{2} + a} \right |}\right )}{16 \, \sqrt {c}} + \frac {1}{336} \, \sqrt {c x^{2} + a} {\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} x e + 7 \, c^{2} d\right )} x\right )} x\right )} x\right )} x\right )} x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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*sqrt(c*x^2 + a)*(48*a^3*e/c + (231*a^2*d +
2*(72*a^2*e + (91*a*c*d + 4*(18*a*c*e + (6*c^2*x*e + 7*c^2*d)*x)*x)*x)*x)*x)

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maple [A]  time = 0.05, size = 85, normalized size = 0.79 \begin {gather*} \frac {5 a^{3} d \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+a}\right )}{16 \sqrt {c}}+\frac {5 \sqrt {c \,x^{2}+a}\, a^{2} d x}{16}+\frac {5 \left (c \,x^{2}+a \right )^{\frac {3}{2}} a d x}{24}+\frac {\left (c \,x^{2}+a \right )^{\frac {5}{2}} d x}{6}+\frac {\left (c \,x^{2}+a \right )^{\frac {7}{2}} e}{7 c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.34, size = 77, normalized size = 0.72 \begin {gather*} \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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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mupad [B]  time = 0.59, size = 54, normalized size = 0.50 \begin {gather*} \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}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [A]  time = 14.06, size = 348, normalized size = 3.25 \begin {gather*} \frac {a^{\frac {5}{2}} d x \sqrt {1 + \frac {c x^{2}}{a}}}{2} + \frac {3 a^{\frac {5}{2}} d x}{16 \sqrt {1 + \frac {c x^{2}}{a}}} + \frac {35 a^{\frac {3}{2}} c d x^{3}}{48 \sqrt {1 + \frac {c x^{2}}{a}}} + \frac {17 \sqrt {a} c^{2} d x^{5}}{24 \sqrt {1 + \frac {c x^{2}}{a}}} + \frac {5 a^{3} d \operatorname {asinh}{\left (\frac {\sqrt {c} x}{\sqrt {a}} \right )}}{16 \sqrt {c}} + a^{2} 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 a c 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 ) + c^{2} e \left (\begin {cases} \frac {8 a^{3} \sqrt {a + c x^{2}}}{105 c^{3}} - \frac {4 a^{2} x^{2} \sqrt {a + c x^{2}}}{105 c^{2}} + \frac {a x^{4} \sqrt {a + c x^{2}}}{35 c} + \frac {x^{6} \sqrt {a + c x^{2}}}{7} & \text {for}\: c \neq 0 \\\frac {\sqrt {a} x^{6}}{6} & \text {otherwise} \end {cases}\right ) + \frac {c^{3} d x^{7}}{6 \sqrt {a} \sqrt {1 + \frac {c x^{2}}{a}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**(5/2)*d*x*sqrt(1 + c*x**2/a)/2 + 3*a**(5/2)*d*x/(16*sqrt(1 + c*x**2/a)) + 35*a**(3/2)*c*d*x**3/(48*sqrt(1 +
c*x**2/a)) + 17*sqrt(a)*c**2*d*x**5/(24*sqrt(1 + c*x**2/a)) + 5*a**3*d*asinh(sqrt(c)*x/sqrt(a))/(16*sqrt(c))
+ a**2*e*Piecewise((sqrt(a)*x**2/2, Eq(c, 0)), ((a + c*x**2)**(3/2)/(3*c), True)) + 2*a*c*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*sqrt(a + c*x**2)/5, Ne(c, 0)), (sqrt(a)*x*
*4/4, True)) + c**2*e*Piecewise((8*a**3*sqrt(a + c*x**2)/(105*c**3) - 4*a**2*x**2*sqrt(a + c*x**2)/(105*c**2)
+ a*x**4*sqrt(a + c*x**2)/(35*c) + x**6*sqrt(a + c*x**2)/7, Ne(c, 0)), (sqrt(a)*x**6/6, True)) + c**3*d*x**7/(
6*sqrt(a)*sqrt(1 + c*x**2/a))

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