3.336 \(\int x \sec ^{\frac{9}{2}}(a+b x) \sin (a+b x) \, dx\)

Optimal. Leaf size=103 \[ -\frac{4 \sin (a+b x) \sec ^{\frac{5}{2}}(a+b x)}{35 b^2}-\frac{12 \sin (a+b x) \sqrt{\sec (a+b x)}}{35 b^2}+\frac{12 \sqrt{\cos (a+b x)} \sqrt{\sec (a+b x)} E\left (\left .\frac{1}{2} (a+b x)\right |2\right )}{35 b^2}+\frac{2 x \sec ^{\frac{7}{2}}(a+b x)}{7 b} \]

[Out]

(12*Sqrt[Cos[a + b*x]]*EllipticE[(a + b*x)/2, 2]*Sqrt[Sec[a + b*x]])/(35*b^2) + (2*x*Sec[a + b*x]^(7/2))/(7*b)
- (12*Sqrt[Sec[a + b*x]]*Sin[a + b*x])/(35*b^2) - (4*Sec[a + b*x]^(5/2)*Sin[a + b*x])/(35*b^2)

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Rubi [A]  time = 0.0620826, 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, 3768, 3771, 2639} \[ -\frac{4 \sin (a+b x) \sec ^{\frac{5}{2}}(a+b x)}{35 b^2}-\frac{12 \sin (a+b x) \sqrt{\sec (a+b x)}}{35 b^2}+\frac{12 \sqrt{\cos (a+b x)} \sqrt{\sec (a+b x)} E\left (\left .\frac{1}{2} (a+b x)\right |2\right )}{35 b^2}+\frac{2 x \sec ^{\frac{7}{2}}(a+b x)}{7 b} \]

Antiderivative was successfully verified.

[In]

Int[x*Sec[a + b*x]^(9/2)*Sin[a + b*x],x]

[Out]

(12*Sqrt[Cos[a + b*x]]*EllipticE[(a + b*x)/2, 2]*Sqrt[Sec[a + b*x]])/(35*b^2) + (2*x*Sec[a + b*x]^(7/2))/(7*b)
- (12*Sqrt[Sec[a + b*x]]*Sin[a + b*x])/(35*b^2) - (4*Sec[a + b*x]^(5/2)*Sin[a + b*x])/(35*b^2)

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 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[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 2639

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

Rubi steps

\begin{align*} \int x \sec ^{\frac{9}{2}}(a+b x) \sin (a+b x) \, dx &=\frac{2 x \sec ^{\frac{7}{2}}(a+b x)}{7 b}-\frac{2 \int \sec ^{\frac{7}{2}}(a+b x) \, dx}{7 b}\\ &=\frac{2 x \sec ^{\frac{7}{2}}(a+b x)}{7 b}-\frac{4 \sec ^{\frac{5}{2}}(a+b x) \sin (a+b x)}{35 b^2}-\frac{6 \int \sec ^{\frac{3}{2}}(a+b x) \, dx}{35 b}\\ &=\frac{2 x \sec ^{\frac{7}{2}}(a+b x)}{7 b}-\frac{12 \sqrt{\sec (a+b x)} \sin (a+b x)}{35 b^2}-\frac{4 \sec ^{\frac{5}{2}}(a+b x) \sin (a+b x)}{35 b^2}+\frac{6 \int \frac{1}{\sqrt{\sec (a+b x)}} \, dx}{35 b}\\ &=\frac{2 x \sec ^{\frac{7}{2}}(a+b x)}{7 b}-\frac{12 \sqrt{\sec (a+b x)} \sin (a+b x)}{35 b^2}-\frac{4 \sec ^{\frac{5}{2}}(a+b x) \sin (a+b x)}{35 b^2}+\frac{\left (6 \sqrt{\cos (a+b x)} \sqrt{\sec (a+b x)}\right ) \int \sqrt{\cos (a+b x)} \, dx}{35 b}\\ &=\frac{12 \sqrt{\cos (a+b x)} E\left (\left .\frac{1}{2} (a+b x)\right |2\right ) \sqrt{\sec (a+b x)}}{35 b^2}+\frac{2 x \sec ^{\frac{7}{2}}(a+b x)}{7 b}-\frac{12 \sqrt{\sec (a+b x)} \sin (a+b x)}{35 b^2}-\frac{4 \sec ^{\frac{5}{2}}(a+b x) \sin (a+b x)}{35 b^2}\\ \end{align*}

Mathematica [A]  time = 0.286994, size = 65, normalized size = 0.63 \[ \frac{\sec ^{\frac{7}{2}}(a+b x) \left (-10 \sin (2 (a+b x))-3 \sin (4 (a+b x))+24 \cos ^{\frac{7}{2}}(a+b x) E\left (\left .\frac{1}{2} (a+b x)\right |2\right )+20 b x\right )}{70 b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x*Sec[a + b*x]^(9/2)*Sin[a + b*x],x]

[Out]

(Sec[a + b*x]^(7/2)*(20*b*x + 24*Cos[a + b*x]^(7/2)*EllipticE[(a + b*x)/2, 2] - 10*Sin[2*(a + b*x)] - 3*Sin[4*
(a + b*x)]))/(70*b^2)

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Maple [F]  time = 0.101, size = 0, normalized size = 0. \begin{align*} \int x \left ( \sec \left ( bx+a \right ) \right ) ^{{\frac{9}{2}}}\sin \left ( bx+a \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*sec(b*x+a)^(9/2)*sin(b*x+a),x)

[Out]

int(x*sec(b*x+a)^(9/2)*sin(b*x+a),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x \sec \left (b x + a\right )^{\frac{9}{2}} \sin \left (b x + a\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*sec(b*x+a)^(9/2)*sin(b*x+a),x, algorithm="maxima")

[Out]

integrate(x*sec(b*x + a)^(9/2)*sin(b*x + a), 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*sec(b*x+a)^(9/2)*sin(b*x+a),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*sec(b*x+a)**(9/2)*sin(b*x+a),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x \sec \left (b x + a\right )^{\frac{9}{2}} \sin \left (b x + a\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*sec(b*x+a)^(9/2)*sin(b*x+a),x, algorithm="giac")

[Out]

integrate(x*sec(b*x + a)^(9/2)*sin(b*x + a), x)