3.813 \(\int (a+b x^2)^{7/4} \, dx\)

Optimal. Leaf size=111 \[ -\frac {14 a^{5/2} \sqrt [4]{\frac {b x^2}{a}+1} E\left (\left .\frac {1}{2} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{15 \sqrt {b} \sqrt [4]{a+b x^2}}+\frac {14 a^2 x}{15 \sqrt [4]{a+b x^2}}+\frac {14}{45} a x \left (a+b x^2\right )^{3/4}+\frac {2}{9} x \left (a+b x^2\right )^{7/4} \]

[Out]

14/15*a^2*x/(b*x^2+a)^(1/4)+14/45*a*x*(b*x^2+a)^(3/4)+2/9*x*(b*x^2+a)^(7/4)-14/15*a^(5/2)*(1+b*x^2/a)^(1/4)*(c
os(1/2*arctan(x*b^(1/2)/a^(1/2)))^2)^(1/2)/cos(1/2*arctan(x*b^(1/2)/a^(1/2)))*EllipticE(sin(1/2*arctan(x*b^(1/
2)/a^(1/2))),2^(1/2))/(b*x^2+a)^(1/4)/b^(1/2)

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Rubi [A]  time = 0.03, antiderivative size = 111, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {195, 229, 227, 196} \[ \frac {14 a^2 x}{15 \sqrt [4]{a+b x^2}}-\frac {14 a^{5/2} \sqrt [4]{\frac {b x^2}{a}+1} E\left (\left .\frac {1}{2} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{15 \sqrt {b} \sqrt [4]{a+b x^2}}+\frac {14}{45} a x \left (a+b x^2\right )^{3/4}+\frac {2}{9} x \left (a+b x^2\right )^{7/4} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^(7/4),x]

[Out]

(14*a^2*x)/(15*(a + b*x^2)^(1/4)) + (14*a*x*(a + b*x^2)^(3/4))/45 + (2*x*(a + b*x^2)^(7/4))/9 - (14*a^(5/2)*(1
+ (b*x^2)/a)^(1/4)*EllipticE[ArcTan[(Sqrt[b]*x)/Sqrt[a]]/2, 2])/(15*Sqrt[b]*(a + b*x^2)^(1/4))

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 196

Int[((a_) + (b_.)*(x_)^2)^(-5/4), x_Symbol] :> Simp[(2*EllipticE[(1*ArcTan[Rt[b/a, 2]*x])/2, 2])/(a^(5/4)*Rt[b
/a, 2]), x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b/a]

Rule 227

Int[((a_) + (b_.)*(x_)^2)^(-1/4), x_Symbol] :> Simp[(2*x)/(a + b*x^2)^(1/4), x] - Dist[a, Int[1/(a + b*x^2)^(5
/4), x], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b/a]

Rule 229

Int[((a_) + (b_.)*(x_)^2)^(-1/4), x_Symbol] :> Dist[(1 + (b*x^2)/a)^(1/4)/(a + b*x^2)^(1/4), Int[1/(1 + (b*x^2
)/a)^(1/4), x], x] /; FreeQ[{a, b}, x] && PosQ[a]

Rubi steps

\begin {align*} \int \left (a+b x^2\right )^{7/4} \, dx &=\frac {2}{9} x \left (a+b x^2\right )^{7/4}+\frac {1}{9} (7 a) \int \left (a+b x^2\right )^{3/4} \, dx\\ &=\frac {14}{45} a x \left (a+b x^2\right )^{3/4}+\frac {2}{9} x \left (a+b x^2\right )^{7/4}+\frac {1}{15} \left (7 a^2\right ) \int \frac {1}{\sqrt [4]{a+b x^2}} \, dx\\ &=\frac {14}{45} a x \left (a+b x^2\right )^{3/4}+\frac {2}{9} x \left (a+b x^2\right )^{7/4}+\frac {\left (7 a^2 \sqrt [4]{1+\frac {b x^2}{a}}\right ) \int \frac {1}{\sqrt [4]{1+\frac {b x^2}{a}}} \, dx}{15 \sqrt [4]{a+b x^2}}\\ &=\frac {14 a^2 x}{15 \sqrt [4]{a+b x^2}}+\frac {14}{45} a x \left (a+b x^2\right )^{3/4}+\frac {2}{9} x \left (a+b x^2\right )^{7/4}-\frac {\left (7 a^2 \sqrt [4]{1+\frac {b x^2}{a}}\right ) \int \frac {1}{\left (1+\frac {b x^2}{a}\right )^{5/4}} \, dx}{15 \sqrt [4]{a+b x^2}}\\ &=\frac {14 a^2 x}{15 \sqrt [4]{a+b x^2}}+\frac {14}{45} a x \left (a+b x^2\right )^{3/4}+\frac {2}{9} x \left (a+b x^2\right )^{7/4}-\frac {14 a^{5/2} \sqrt [4]{1+\frac {b x^2}{a}} E\left (\left .\frac {1}{2} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{15 \sqrt {b} \sqrt [4]{a+b x^2}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 47, normalized size = 0.42 \[ \frac {a x \left (a+b x^2\right )^{3/4} \, _2F_1\left (-\frac {7}{4},\frac {1}{2};\frac {3}{2};-\frac {b x^2}{a}\right )}{\left (\frac {b x^2}{a}+1\right )^{3/4}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2)^(7/4),x]

[Out]

(a*x*(a + b*x^2)^(3/4)*Hypergeometric2F1[-7/4, 1/2, 3/2, -((b*x^2)/a)])/(1 + (b*x^2)/a)^(3/4)

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fricas [F]  time = 0.69, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (b x^{2} + a\right )}^{\frac {7}{4}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(7/4),x, algorithm="fricas")

[Out]

integral((b*x^2 + a)^(7/4), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b x^{2} + a\right )}^{\frac {7}{4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(7/4),x, algorithm="giac")

[Out]

integrate((b*x^2 + a)^(7/4), x)

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maple [F]  time = 0.30, size = 0, normalized size = 0.00 \[ \int \left (b \,x^{2}+a \right )^{\frac {7}{4}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)^(7/4),x)

[Out]

int((b*x^2+a)^(7/4),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b x^{2} + a\right )}^{\frac {7}{4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(7/4),x, algorithm="maxima")

[Out]

integrate((b*x^2 + a)^(7/4), x)

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mupad [B]  time = 4.82, size = 37, normalized size = 0.33 \[ \frac {x\,{\left (b\,x^2+a\right )}^{7/4}\,{{}}_2{\mathrm {F}}_1\left (-\frac {7}{4},\frac {1}{2};\ \frac {3}{2};\ -\frac {b\,x^2}{a}\right )}{{\left (\frac {b\,x^2}{a}+1\right )}^{7/4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^2)^(7/4),x)

[Out]

(x*(a + b*x^2)^(7/4)*hypergeom([-7/4, 1/2], 3/2, -(b*x^2)/a))/((b*x^2)/a + 1)^(7/4)

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sympy [C]  time = 1.54, size = 26, normalized size = 0.23 \[ a^{\frac {7}{4}} x {{}_{2}F_{1}\left (\begin {matrix} - \frac {7}{4}, \frac {1}{2} \\ \frac {3}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)**(7/4),x)

[Out]

a**(7/4)*x*hyper((-7/4, 1/2), (3/2,), b*x**2*exp_polar(I*pi)/a)

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