3.342 \(\int \frac{x \sin (a+b x)}{\sec ^{\frac{3}{2}}(a+b x)} \, dx\)

Optimal. Leaf size=80 \[ \frac{4 \sin (a+b x)}{25 b^2 \sec ^{\frac{3}{2}}(a+b x)}+\frac{12 \sqrt{\cos (a+b x)} \sqrt{\sec (a+b x)} E\left (\left .\frac{1}{2} (a+b x)\right |2\right )}{25 b^2}-\frac{2 x}{5 b \sec ^{\frac{5}{2}}(a+b x)} \]

[Out]

(-2*x)/(5*b*Sec[a + b*x]^(5/2)) + (12*Sqrt[Cos[a + b*x]]*EllipticE[(a + b*x)/2, 2]*Sqrt[Sec[a + b*x]])/(25*b^2
) + (4*Sin[a + b*x])/(25*b^2*Sec[a + b*x]^(3/2))

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Rubi [A]  time = 0.0465426, antiderivative size = 80, normalized size of antiderivative = 1., 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, 3769, 3771, 2639} \[ \frac{4 \sin (a+b x)}{25 b^2 \sec ^{\frac{3}{2}}(a+b x)}+\frac{12 \sqrt{\cos (a+b x)} \sqrt{\sec (a+b x)} E\left (\left .\frac{1}{2} (a+b x)\right |2\right )}{25 b^2}-\frac{2 x}{5 b \sec ^{\frac{5}{2}}(a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*x)/(5*b*Sec[a + b*x]^(5/2)) + (12*Sqrt[Cos[a + b*x]]*EllipticE[(a + b*x)/2, 2]*Sqrt[Sec[a + b*x]])/(25*b^2
) + (4*Sin[a + b*x])/(25*b^2*Sec[a + b*x]^(3/2))

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 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 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 \frac{x \sin (a+b x)}{\sec ^{\frac{3}{2}}(a+b x)} \, dx &=-\frac{2 x}{5 b \sec ^{\frac{5}{2}}(a+b x)}+\frac{2 \int \frac{1}{\sec ^{\frac{5}{2}}(a+b x)} \, dx}{5 b}\\ &=-\frac{2 x}{5 b \sec ^{\frac{5}{2}}(a+b x)}+\frac{4 \sin (a+b x)}{25 b^2 \sec ^{\frac{3}{2}}(a+b x)}+\frac{6 \int \frac{1}{\sqrt{\sec (a+b x)}} \, dx}{25 b}\\ &=-\frac{2 x}{5 b \sec ^{\frac{5}{2}}(a+b x)}+\frac{4 \sin (a+b x)}{25 b^2 \sec ^{\frac{3}{2}}(a+b x)}+\frac{\left (6 \sqrt{\cos (a+b x)} \sqrt{\sec (a+b x)}\right ) \int \sqrt{\cos (a+b x)} \, dx}{25 b}\\ &=-\frac{2 x}{5 b \sec ^{\frac{5}{2}}(a+b x)}+\frac{12 \sqrt{\cos (a+b x)} E\left (\left .\frac{1}{2} (a+b x)\right |2\right ) \sqrt{\sec (a+b x)}}{25 b^2}+\frac{4 \sin (a+b x)}{25 b^2 \sec ^{\frac{3}{2}}(a+b x)}\\ \end{align*}

Mathematica [B]  time = 8.09831, size = 212, normalized size = 2.65 \[ \frac{\cos ^2\left (\frac{1}{2} (a+b x)\right ) \sqrt{\sec (a+b x)} \left (-12 \sqrt{\cos (a+b x) \sec ^4\left (\frac{1}{2} (a+b x)\right )} \text{EllipticF}\left (\sin ^{-1}\left (\tan \left (\frac{1}{2} (a+b x)\right )\right ),-1\right )+\left (5 (a+b x)-12 \tan \left (\frac{1}{2} (a+b x)\right )-5 a\right ) \left (\tan ^2\left (\frac{1}{2} (a+b x)\right )-1\right )+12 \sqrt{\cos (a+b x) \sec ^4\left (\frac{1}{2} (a+b x)\right )} E\left (\left .\sin ^{-1}\left (\tan \left (\frac{1}{2} (a+b x)\right )\right )\right |-1\right )\right )}{25 b^2}+\frac{\sqrt{\sec (a+b x)} \left (\frac{\sin (a+b x)}{25 b}+\frac{\sin (3 (a+b x))}{25 b}-\frac{1}{10} x \cos (a+b x)-\frac{1}{10} x \cos (3 (a+b x))\right )}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[Sec[a + b*x]]*(-(x*Cos[a + b*x])/10 - (x*Cos[3*(a + b*x)])/10 + Sin[a + b*x]/(25*b) + Sin[3*(a + b*x)]/(
25*b)))/b + (Cos[(a + b*x)/2]^2*Sqrt[Sec[a + b*x]]*(12*EllipticE[ArcSin[Tan[(a + b*x)/2]], -1]*Sqrt[Cos[a + b*
x]*Sec[(a + b*x)/2]^4] - 12*EllipticF[ArcSin[Tan[(a + b*x)/2]], -1]*Sqrt[Cos[a + b*x]*Sec[(a + b*x)/2]^4] + (-
5*a + 5*(a + b*x) - 12*Tan[(a + b*x)/2])*(-1 + Tan[(a + b*x)/2]^2)))/(25*b^2)

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Maple [F]  time = 0.097, size = 0, normalized size = 0. \begin{align*} \int{x\sin \left ( bx+a \right ) \left ( \sec \left ( bx+a \right ) \right ) ^{-{\frac{3}{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)^(3/2),x)

[Out]

int(x*sin(b*x+a)/sec(b*x+a)^(3/2),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{3}{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)^(3/2),x, algorithm="maxima")

[Out]

integrate(x*sin(b*x + a)/sec(b*x + a)^(3/2), 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)^(3/2),x, algorithm="fricas")

[Out]

Exception raised: UnboundLocalError

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x \sin{\left (a + b x \right )}}{\sec ^{\frac{3}{2}}{\left (a + b x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*sin(b*x+a)/sec(b*x+a)**(3/2),x)

[Out]

Integral(x*sin(a + b*x)/sec(a + b*x)**(3/2), x)

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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{3}{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)^(3/2),x, algorithm="giac")

[Out]

integrate(x*sin(b*x + a)/sec(b*x + a)^(3/2), x)