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

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

[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])

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Rubi [A]  time = 0.0369465, antiderivative size = 107, normalized size of antiderivative = 1., 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} \[ \frac{5}{16} a^2 A x \sqrt{a+b x^2}+\frac{5 a^3 A \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right )}{16 \sqrt{b}}+\frac{1}{6} A x \left (a+b x^2\right )^{5/2}+\frac{5}{24} a A x \left (a+b x^2\right )^{3/2}+\frac{B \left (a+b x^2\right )^{7/2}}{7 b} \]

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 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 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 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 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])

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 = 0.0805573, size = 108, normalized size = 1.01 \[ \frac{\sqrt{a+b x^2} \left (3 a^2 b x (77 A+48 B x)+48 a^3 B+2 a b^2 x^3 (91 A+72 B x)+8 b^3 x^5 (7 A+6 B x)\right )+105 a^3 A \sqrt{b} \log \left (\sqrt{b} \sqrt{a+b x^2}+b x\right )}{336 b} \]

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)

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Maple [A]  time = 0.005, size = 85, normalized size = 0.8 \begin{align*}{\frac{B}{7\,b} \left ( b{x}^{2}+a \right ) ^{{\frac{7}{2}}}}+{\frac{Ax}{6} \left ( b{x}^{2}+a \right ) ^{{\frac{5}{2}}}}+{\frac{5\,aAx}{24} \left ( b{x}^{2}+a \right ) ^{{\frac{3}{2}}}}+{\frac{5\,{a}^{2}Ax}{16}\sqrt{b{x}^{2}+a}}+{\frac{5\,A{a}^{3}}{16}\ln \left ( x\sqrt{b}+\sqrt{b{x}^{2}+a} \right ){\frac{1}{\sqrt{b}}}} \end{align*}

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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]

Exception raised: ValueError

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Fricas [A]  time = 1.58966, size = 549, normalized size = 5.13 \begin{align*} \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{align*}

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]

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Sympy [A]  time = 12.5701, size = 348, normalized size = 3.25 \begin{align*} \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{align*}

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))

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Giac [A]  time = 1.21431, size = 136, normalized size = 1.27 \begin{align*} -\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{align*}

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)