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

Optimal. Leaf size=53 \[ \frac{2 x \sqrt{\sec (a+b x)}}{b}-\frac{4 \sqrt{\cos (a+b x)} \sqrt{\sec (a+b x)} \text{EllipticF}\left (\frac{1}{2} (a+b x),2\right )}{b^2} \]

[Out]

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

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Rubi [A]  time = 0.0358781, antiderivative size = 53, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {4212, 3771, 2641} \[ \frac{2 x \sqrt{\sec (a+b x)}}{b}-\frac{4 \sqrt{\cos (a+b x)} \sqrt{\sec (a+b x)} F\left (\left .\frac{1}{2} (a+b x)\right |2\right )}{b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.140606, size = 42, normalized size = 0.79 \[ \frac{2 \sqrt{\sec (a+b x)} \left (b x-2 \sqrt{\cos (a+b x)} \text{EllipticF}\left (\frac{1}{2} (a+b x),2\right )\right )}{b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(b*x - 2*Sqrt[Cos[a + b*x]]*EllipticF[(a + b*x)/2, 2])*Sqrt[Sec[a + b*x]])/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{3}{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)^(3/2)*sin(b*x+a),x)

[Out]

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

[Out]

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

[Out]

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