3.341 \(\int \frac{x \sin (a+b x)}{\sqrt{\sec (a+b x)}} \, dx\)

Optimal. Leaf size=80 \[ \frac{4 \sqrt{\cos (a+b x)} \sqrt{\sec (a+b x)} \text{EllipticF}\left (\frac{1}{2} (a+b x),2\right )}{9 b^2}+\frac{4 \sin (a+b x)}{9 b^2 \sqrt{\sec (a+b x)}}-\frac{2 x}{3 b \sec ^{\frac{3}{2}}(a+b x)} \]

[Out]

(-2*x)/(3*b*Sec[a + b*x]^(3/2)) + (4*Sqrt[Cos[a + b*x]]*EllipticF[(a + b*x)/2, 2]*Sqrt[Sec[a + b*x]])/(9*b^2)
+ (4*Sin[a + b*x])/(9*b^2*Sqrt[Sec[a + b*x]])

________________________________________________________________________________________

Rubi [A]  time = 0.0456498, 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, 2641} \[ \frac{4 \sin (a+b x)}{9 b^2 \sqrt{\sec (a+b x)}}+\frac{4 \sqrt{\cos (a+b x)} \sqrt{\sec (a+b x)} F\left (\left .\frac{1}{2} (a+b x)\right |2\right )}{9 b^2}-\frac{2 x}{3 b \sec ^{\frac{3}{2}}(a+b x)} \]

Antiderivative was successfully verified.

[In]

Int[(x*Sin[a + b*x])/Sqrt[Sec[a + b*x]],x]

[Out]

(-2*x)/(3*b*Sec[a + b*x]^(3/2)) + (4*Sqrt[Cos[a + b*x]]*EllipticF[(a + b*x)/2, 2]*Sqrt[Sec[a + b*x]])/(9*b^2)
+ (4*Sin[a + b*x])/(9*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)}{\sqrt{\sec (a+b x)}} \, dx &=-\frac{2 x}{3 b \sec ^{\frac{3}{2}}(a+b x)}+\frac{2 \int \frac{1}{\sec ^{\frac{3}{2}}(a+b x)} \, dx}{3 b}\\ &=-\frac{2 x}{3 b \sec ^{\frac{3}{2}}(a+b x)}+\frac{4 \sin (a+b x)}{9 b^2 \sqrt{\sec (a+b x)}}+\frac{2 \int \sqrt{\sec (a+b x)} \, dx}{9 b}\\ &=-\frac{2 x}{3 b \sec ^{\frac{3}{2}}(a+b x)}+\frac{4 \sin (a+b x)}{9 b^2 \sqrt{\sec (a+b x)}}+\frac{\left (2 \sqrt{\cos (a+b x)} \sqrt{\sec (a+b x)}\right ) \int \frac{1}{\sqrt{\cos (a+b x)}} \, dx}{9 b}\\ &=-\frac{2 x}{3 b \sec ^{\frac{3}{2}}(a+b x)}+\frac{4 \sqrt{\cos (a+b x)} F\left (\left .\frac{1}{2} (a+b x)\right |2\right ) \sqrt{\sec (a+b x)}}{9 b^2}+\frac{4 \sin (a+b x)}{9 b^2 \sqrt{\sec (a+b x)}}\\ \end{align*}

Mathematica [A]  time = 0.247597, size = 63, normalized size = 0.79 \[ \frac{\sqrt{\sec (a+b x)} \left (4 \sqrt{\cos (a+b x)} \text{EllipticF}\left (\frac{1}{2} (a+b x),2\right )+2 \sin (2 (a+b x))-6 b x \cos ^2(a+b x)\right )}{9 b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(x*Sin[a + b*x])/Sqrt[Sec[a + b*x]],x]

[Out]

(Sqrt[Sec[a + b*x]]*(-6*b*x*Cos[a + b*x]^2 + 4*Sqrt[Cos[a + b*x]]*EllipticF[(a + b*x)/2, 2] + 2*Sin[2*(a + b*x
)]))/(9*b^2)

________________________________________________________________________________________

Maple [F]  time = 0.097, size = 0, normalized size = 0. \begin{align*} \int{x\sin \left ( bx+a \right ){\frac{1}{\sqrt{\sec \left ( bx+a \right ) }}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x \sin \left (b x + a\right )}{\sqrt{\sec \left (b x + a\right )}}\,{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)^(1/2),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

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)^(1/2),x, algorithm="fricas")

[Out]

Exception raised: UnboundLocalError

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x \sin{\left (a + b x \right )}}{\sqrt{\sec{\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)**(1/2),x)

[Out]

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

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x \sin \left (b x + a\right )}{\sqrt{\sec \left (b x + a\right )}}\,{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)^(1/2),x, algorithm="giac")

[Out]

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