3.1.18 \(\int (A+B x) (a+b x^2)^{5/2} \, dx\)

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

________________________________________________________________________________________

Rubi [A]  time = 0.04, 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 A x \sqrt {a+b x^2}+\frac {5 a^3 A \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{16 \sqrt {b}}+\frac {1}{6} A x \left (a+b x^2\right )^{5/2}+\frac {5}{24} a A x \left (a+b x^2\right )^{3/2}+\frac {B \left (a+b x^2\right )^{7/2}}{7 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(A + B*x)*(a + b*x^2)^(5/2),x]

[Out]

(5*a^2*A*x*Sqrt[a + b*x^2])/16 + (5*a*A*x*(a + b*x^2)^(3/2))/24 + (A*x*(a + b*x^2)^(5/2))/6 + (B*(a + b*x^2)^(
7/2))/(7*b) + (5*a^3*A*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/(16*Sqrt[b])

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

________________________________________________________________________________________

Mathematica [A]  time = 0.09, size = 108, normalized size = 1.01 \begin {gather*} \frac {105 a^3 A \sqrt {b} \log \left (\sqrt {b} \sqrt {a+b x^2}+b x\right )+\sqrt {a+b x^2} \left (48 a^3 B+3 a^2 b x (77 A+48 B x)+2 a b^2 x^3 (91 A+72 B x)+8 b^3 x^5 (7 A+6 B x)\right )}{336 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x)*(a + b*x^2)^(5/2),x]

[Out]

(Sqrt[a + b*x^2]*(48*a^3*B + 8*b^3*x^5*(7*A + 6*B*x) + 3*a^2*b*x*(77*A + 48*B*x) + 2*a*b^2*x^3*(91*A + 72*B*x)
) + 105*a^3*A*Sqrt[b]*Log[b*x + Sqrt[b]*Sqrt[a + b*x^2]])/(336*b)

________________________________________________________________________________________

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

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(A + B*x)*(a + b*x^2)^(5/2),x]

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.56, size = 224, normalized size = 2.09 \begin {gather*} \left [\frac {105 \, A a^{3} \sqrt {b} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) + 2 \, {\left (48 \, B b^{3} x^{6} + 56 \, A b^{3} x^{5} + 144 \, B a b^{2} x^{4} + 182 \, A a b^{2} x^{3} + 144 \, B a^{2} b x^{2} + 231 \, A a^{2} b x + 48 \, B a^{3}\right )} \sqrt {b x^{2} + a}}{672 \, b}, -\frac {105 \, A a^{3} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) - {\left (48 \, B b^{3} x^{6} + 56 \, A b^{3} x^{5} + 144 \, B a b^{2} x^{4} + 182 \, A a b^{2} x^{3} + 144 \, B a^{2} b x^{2} + 231 \, A a^{2} b x + 48 \, B a^{3}\right )} \sqrt {b x^{2} + a}}{336 \, b}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(b*x^2+a)^(5/2),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

giac [A]  time = 0.61, size = 101, normalized size = 0.94 \begin {gather*} -\frac {5 \, A a^{3} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{16 \, \sqrt {b}} + \frac {1}{336} \, {\left (\frac {48 \, B a^{3}}{b} + {\left (231 \, A a^{2} + 2 \, {\left (72 \, B a^{2} + {\left (91 \, A a b + 4 \, {\left (18 \, B a b + {\left (6 \, B b^{2} x + 7 \, A b^{2}\right )} x\right )} x\right )} x\right )} x\right )} x\right )} \sqrt {b x^{2} + a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(b*x^2+a)^(5/2),x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.00, size = 85, normalized size = 0.79 \begin {gather*} \frac {5 A \,a^{3} \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{16 \sqrt {b}}+\frac {5 \sqrt {b \,x^{2}+a}\, A \,a^{2} x}{16}+\frac {5 \left (b \,x^{2}+a \right )^{\frac {3}{2}} A a x}{24}+\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}} A x}{6}+\frac {\left (b \,x^{2}+a \right )^{\frac {7}{2}} B}{7 b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(b*x^2+a)^(5/2),x)

[Out]

1/7*B*(b*x^2+a)^(7/2)/b+1/6*A*x*(b*x^2+a)^(5/2)+5/24*a*A*x*(b*x^2+a)^(3/2)+5/16*a^2*A*x*(b*x^2+a)^(1/2)+5/16*A
*a^3/b^(1/2)*ln(b^(1/2)*x+(b*x^2+a)^(1/2))

________________________________________________________________________________________

maxima [A]  time = 1.40, size = 77, normalized size = 0.72 \begin {gather*} \frac {1}{6} \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} A x + \frac {5}{24} \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} A a x + \frac {5}{16} \, \sqrt {b x^{2} + a} A a^{2} x + \frac {5 \, A a^{3} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{16 \, \sqrt {b}} + \frac {{\left (b x^{2} + a\right )}^{\frac {7}{2}} B}{7 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(b*x^2+a)^(5/2),x, algorithm="maxima")

[Out]

1/6*(b*x^2 + a)^(5/2)*A*x + 5/24*(b*x^2 + a)^(3/2)*A*a*x + 5/16*sqrt(b*x^2 + a)*A*a^2*x + 5/16*A*a^3*arcsinh(b
*x/sqrt(a*b))/sqrt(b) + 1/7*(b*x^2 + a)^(7/2)*B/b

________________________________________________________________________________________

mupad [B]  time = 1.16, size = 54, normalized size = 0.50 \begin {gather*} \frac {B\,{\left (b\,x^2+a\right )}^{7/2}}{7\,b}+\frac {A\,x\,{\left (b\,x^2+a\right )}^{5/2}\,{{}}_2{\mathrm {F}}_1\left (-\frac {5}{2},\frac {1}{2};\ \frac {3}{2};\ -\frac {b\,x^2}{a}\right )}{{\left (\frac {b\,x^2}{a}+1\right )}^{5/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^2)^(5/2)*(A + B*x),x)

[Out]

(B*(a + b*x^2)^(7/2))/(7*b) + (A*x*(a + b*x^2)^(5/2)*hypergeom([-5/2, 1/2], 3/2, -(b*x^2)/a))/((b*x^2)/a + 1)^
(5/2)

________________________________________________________________________________________

sympy [A]  time = 26.05, size = 348, normalized size = 3.25 \begin {gather*} \frac {A a^{\frac {5}{2}} x \sqrt {1 + \frac {b x^{2}}{a}}}{2} + \frac {3 A a^{\frac {5}{2}} x}{16 \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {35 A a^{\frac {3}{2}} b x^{3}}{48 \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {17 A \sqrt {a} b^{2} x^{5}}{24 \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {5 A a^{3} \operatorname {asinh}{\left (\frac {\sqrt {b} x}{\sqrt {a}} \right )}}{16 \sqrt {b}} + \frac {A b^{3} x^{7}}{6 \sqrt {a} \sqrt {1 + \frac {b x^{2}}{a}}} + B a^{2} \left (\begin {cases} \frac {\sqrt {a} x^{2}}{2} & \text {for}\: b = 0 \\\frac {\left (a + b x^{2}\right )^{\frac {3}{2}}}{3 b} & \text {otherwise} \end {cases}\right ) + 2 B a b \left (\begin {cases} - \frac {2 a^{2} \sqrt {a + b x^{2}}}{15 b^{2}} + \frac {a x^{2} \sqrt {a + b x^{2}}}{15 b} + \frac {x^{4} \sqrt {a + b x^{2}}}{5} & \text {for}\: b \neq 0 \\\frac {\sqrt {a} x^{4}}{4} & \text {otherwise} \end {cases}\right ) + B b^{2} \left (\begin {cases} \frac {8 a^{3} \sqrt {a + b x^{2}}}{105 b^{3}} - \frac {4 a^{2} x^{2} \sqrt {a + b x^{2}}}{105 b^{2}} + \frac {a x^{4} \sqrt {a + b x^{2}}}{35 b} + \frac {x^{6} \sqrt {a + b x^{2}}}{7} & \text {for}\: b \neq 0 \\\frac {\sqrt {a} x^{6}}{6} & \text {otherwise} \end {cases}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(b*x**2+a)**(5/2),x)

[Out]

A*a**(5/2)*x*sqrt(1 + b*x**2/a)/2 + 3*A*a**(5/2)*x/(16*sqrt(1 + b*x**2/a)) + 35*A*a**(3/2)*b*x**3/(48*sqrt(1 +
b*x**2/a)) + 17*A*sqrt(a)*b**2*x**5/(24*sqrt(1 + b*x**2/a)) + 5*A*a**3*asinh(sqrt(b)*x/sqrt(a))/(16*sqrt(b))
+ A*b**3*x**7/(6*sqrt(a)*sqrt(1 + b*x**2/a)) + B*a**2*Piecewise((sqrt(a)*x**2/2, Eq(b, 0)), ((a + b*x**2)**(3/
2)/(3*b), True)) + 2*B*a*b*Piecewise((-2*a**2*sqrt(a + b*x**2)/(15*b**2) + a*x**2*sqrt(a + b*x**2)/(15*b) + x*
*4*sqrt(a + b*x**2)/5, Ne(b, 0)), (sqrt(a)*x**4/4, True)) + B*b**2*Piecewise((8*a**3*sqrt(a + b*x**2)/(105*b**
3) - 4*a**2*x**2*sqrt(a + b*x**2)/(105*b**2) + a*x**4*sqrt(a + b*x**2)/(35*b) + x**6*sqrt(a + b*x**2)/7, Ne(b,
0)), (sqrt(a)*x**6/6, True))

________________________________________________________________________________________