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

Optimal. Leaf size=80 \[ -\frac {4 \sin (a+b x) \sqrt {\sec (a+b x)}}{3 b^2}+\frac {4 \sqrt {\cos (a+b x)} \sqrt {\sec (a+b x)} E\left (\left .\frac {1}{2} (a+b x)\right |2\right )}{3 b^2}+\frac {2 x \sec ^{\frac {3}{2}}(a+b x)}{3 b} \]

[Out]

2/3*x*sec(b*x+a)^(3/2)/b-4/3*sin(b*x+a)*sec(b*x+a)^(1/2)/b^2+4/3*(cos(1/2*b*x+1/2*a)^2)^(1/2)/cos(1/2*b*x+1/2*
a)*EllipticE(sin(1/2*b*x+1/2*a),2^(1/2))*cos(b*x+a)^(1/2)*sec(b*x+a)^(1/2)/b^2

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Rubi [A]  time = 0.05, antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {4212, 3768, 3771, 2639} \[ -\frac {4 \sin (a+b x) \sqrt {\sec (a+b x)}}{3 b^2}+\frac {4 \sqrt {\cos (a+b x)} \sqrt {\sec (a+b x)} E\left (\left .\frac {1}{2} (a+b x)\right |2\right )}{3 b^2}+\frac {2 x \sec ^{\frac {3}{2}}(a+b x)}{3 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*Sqrt[Cos[a + b*x]]*EllipticE[(a + b*x)/2, 2]*Sqrt[Sec[a + b*x]])/(3*b^2) + (2*x*Sec[a + b*x]^(3/2))/(3*b) -
(4*Sqrt[Sec[a + b*x]]*Sin[a + b*x])/(3*b^2)

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]

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 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]

Rubi steps

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

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Mathematica [A]  time = 0.18, size = 54, normalized size = 0.68 \[ \frac {2 \sec ^{\frac {3}{2}}(a+b x) \left (-\sin (2 (a+b x))+2 \cos ^{\frac {3}{2}}(a+b x) E\left (\left .\frac {1}{2} (a+b x)\right |2\right )+b x\right )}{3 b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sec[a + b*x]^(3/2)*(b*x + 2*Cos[a + b*x]^(3/2)*EllipticE[(a + b*x)/2, 2] - Sin[2*(a + b*x)]))/(3*b^2)

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fricas [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >>  Error detected within library code:   integrate: implementation incomplete (co
nstant residues)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x \sec \left (b x + a\right )^{\frac {5}{2}} \sin \left (b x + a\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.05, size = 0, normalized size = 0.00 \[ \int x \left (\sec ^{\frac {5}{2}}\left (b x +a \right )\right ) \sin \left (b x +a \right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x \sec \left (b x + a\right )^{\frac {5}{2}} \sin \left (b x + a\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x\,\sin \left (a+b\,x\right )\,{\left (\frac {1}{\cos \left (a+b\,x\right )}\right )}^{5/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*sin(a + b*x)*(1/cos(a + b*x))^(5/2),x)

[Out]

int(x*sin(a + b*x)*(1/cos(a + b*x))^(5/2), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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