3.343 \(\int \frac{x \sin (a+b x)}{\sec ^{\frac{5}{2}}(a+b x)} \, dx\)

Optimal. Leaf size=103 \[ \frac{20 \sqrt{\cos (a+b x)} \sqrt{\sec (a+b x)} \text{EllipticF}\left (\frac{1}{2} (a+b x),2\right )}{147 b^2}+\frac{4 \sin (a+b x)}{49 b^2 \sec ^{\frac{5}{2}}(a+b x)}+\frac{20 \sin (a+b x)}{147 b^2 \sqrt{\sec (a+b x)}}-\frac{2 x}{7 b \sec ^{\frac{7}{2}}(a+b x)} \]

[Out]

(-2*x)/(7*b*Sec[a + b*x]^(7/2)) + (20*Sqrt[Cos[a + b*x]]*EllipticF[(a + b*x)/2, 2]*Sqrt[Sec[a + b*x]])/(147*b^
2) + (4*Sin[a + b*x])/(49*b^2*Sec[a + b*x]^(5/2)) + (20*Sin[a + b*x])/(147*b^2*Sqrt[Sec[a + b*x]])

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Rubi [A]  time = 0.0600873, antiderivative size = 103, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {4212, 3769, 3771, 2641} \[ \frac{4 \sin (a+b x)}{49 b^2 \sec ^{\frac{5}{2}}(a+b x)}+\frac{20 \sin (a+b x)}{147 b^2 \sqrt{\sec (a+b x)}}+\frac{20 \sqrt{\cos (a+b x)} \sqrt{\sec (a+b x)} F\left (\left .\frac{1}{2} (a+b x)\right |2\right )}{147 b^2}-\frac{2 x}{7 b \sec ^{\frac{7}{2}}(a+b x)} \]

Antiderivative was successfully verified.

[In]

Int[(x*Sin[a + b*x])/Sec[a + b*x]^(5/2),x]

[Out]

(-2*x)/(7*b*Sec[a + b*x]^(7/2)) + (20*Sqrt[Cos[a + b*x]]*EllipticF[(a + b*x)/2, 2]*Sqrt[Sec[a + b*x]])/(147*b^
2) + (4*Sin[a + b*x])/(49*b^2*Sec[a + b*x]^(5/2)) + (20*Sin[a + b*x])/(147*b^2*Sqrt[Sec[a + b*x]])

Rule 4212

Int[(x_)^(m_.)*Sec[(a_.) + (b_.)*(x_)^(n_.)]^(p_)*Sin[(a_.) + (b_.)*(x_)^(n_.)], x_Symbol] :> Simp[(x^(m - n +
1)*Sec[a + b*x^n]^(p - 1))/(b*n*(p - 1)), x] - Dist[(m - n + 1)/(b*n*(p - 1)), Int[x^(m - n)*Sec[a + b*x^n]^(
p - 1), x], x] /; FreeQ[{a, b, p}, x] && IntegerQ[n] && GeQ[m - n, 0] && NeQ[p, 1]

Rule 3769

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(Cos[c + d*x]*(b*Csc[c + d*x])^(n + 1))/(b*d*n), x
] + Dist[(n + 1)/(b^2*n), Int[(b*Csc[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1] && Integer
Q[2*n]

Rule 3771

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[(b*Csc[c + d*x])^n*Sin[c + d*x]^n, Int[1/Sin[c + d
*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]

Rule 2641

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ
[{c, d}, x]

Rubi steps

\begin{align*} \int \frac{x \sin (a+b x)}{\sec ^{\frac{5}{2}}(a+b x)} \, dx &=-\frac{2 x}{7 b \sec ^{\frac{7}{2}}(a+b x)}+\frac{2 \int \frac{1}{\sec ^{\frac{7}{2}}(a+b x)} \, dx}{7 b}\\ &=-\frac{2 x}{7 b \sec ^{\frac{7}{2}}(a+b x)}+\frac{4 \sin (a+b x)}{49 b^2 \sec ^{\frac{5}{2}}(a+b x)}+\frac{10 \int \frac{1}{\sec ^{\frac{3}{2}}(a+b x)} \, dx}{49 b}\\ &=-\frac{2 x}{7 b \sec ^{\frac{7}{2}}(a+b x)}+\frac{4 \sin (a+b x)}{49 b^2 \sec ^{\frac{5}{2}}(a+b x)}+\frac{20 \sin (a+b x)}{147 b^2 \sqrt{\sec (a+b x)}}+\frac{10 \int \sqrt{\sec (a+b x)} \, dx}{147 b}\\ &=-\frac{2 x}{7 b \sec ^{\frac{7}{2}}(a+b x)}+\frac{4 \sin (a+b x)}{49 b^2 \sec ^{\frac{5}{2}}(a+b x)}+\frac{20 \sin (a+b x)}{147 b^2 \sqrt{\sec (a+b x)}}+\frac{\left (10 \sqrt{\cos (a+b x)} \sqrt{\sec (a+b x)}\right ) \int \frac{1}{\sqrt{\cos (a+b x)}} \, dx}{147 b}\\ &=-\frac{2 x}{7 b \sec ^{\frac{7}{2}}(a+b x)}+\frac{20 \sqrt{\cos (a+b x)} F\left (\left .\frac{1}{2} (a+b x)\right |2\right ) \sqrt{\sec (a+b x)}}{147 b^2}+\frac{4 \sin (a+b x)}{49 b^2 \sec ^{\frac{5}{2}}(a+b x)}+\frac{20 \sin (a+b x)}{147 b^2 \sqrt{\sec (a+b x)}}\\ \end{align*}

Mathematica [A]  time = 0.339256, size = 89, normalized size = 0.86 \[ \frac{\sqrt{\sec (a+b x)} \left (80 \sqrt{\cos (a+b x)} \text{EllipticF}\left (\frac{1}{2} (a+b x),2\right )+52 \sin (2 (a+b x))+6 \sin (4 (a+b x))-84 b x \cos (2 (a+b x))-21 b x \cos (4 (a+b x))-63 b x\right )}{588 b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(x*Sin[a + b*x])/Sec[a + b*x]^(5/2),x]

[Out]

(Sqrt[Sec[a + b*x]]*(-63*b*x - 84*b*x*Cos[2*(a + b*x)] - 21*b*x*Cos[4*(a + b*x)] + 80*Sqrt[Cos[a + b*x]]*Ellip
ticF[(a + b*x)/2, 2] + 52*Sin[2*(a + b*x)] + 6*Sin[4*(a + b*x)]))/(588*b^2)

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Maple [F]  time = 0.099, size = 0, normalized size = 0. \begin{align*} \int{x\sin \left ( bx+a \right ) \left ( \sec \left ( bx+a \right ) \right ) ^{-{\frac{5}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*sin(b*x+a)/sec(b*x+a)^(5/2),x)

[Out]

int(x*sin(b*x+a)/sec(b*x+a)^(5/2),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x \sin \left (b x + a\right )}{\sec \left (b x + a\right )^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x*sin(b*x + a)/sec(b*x + a)^(5/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: UnboundLocalError

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*sin(b*x+a)/sec(b*x+a)**(5/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x \sin \left (b x + a\right )}{\sec \left (b x + a\right )^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x*sin(b*x + a)/sec(b*x + a)^(5/2), x)