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

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

[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))

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Rubi [A]  time = 0.0291233, antiderivative size = 111, normalized size of antiderivative = 1., 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 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]

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 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]

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*}

Mathematica [C]  time = 0.0071599, 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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Maple [F]  time = 0.033, size = 0, normalized size = 0. \begin{align*} \int \left ( b{x}^{2}+a \right ) ^{{\frac{7}{4}}}\, dx \end{align*}

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., size = 0, normalized size = 0. \begin{align*} \int{\left (b x^{2} + a\right )}^{\frac{7}{4}}\,{d x} \end{align*}

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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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (b x^{2} + a\right )}^{\frac{7}{4}}, x\right ) \end{align*}

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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Sympy [C]  time = 2.20992, size = 26, normalized size = 0.23 \begin{align*} 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 )} \end{align*}

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError