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]

-2/5*x/b/sec(b*x+a)^(5/2)+4/25*sin(b*x+a)/b^2/sec(b*x+a)^(3/2)+12/25*(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, 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 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 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 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 \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*}

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Mathematica [B]  time = 8.10, size = 212, normalized size = 2.65 \[ \frac {\cos ^2\left (\frac {1}{2} (a+b x)\right ) \sqrt {\sec (a+b x)} \left (\left (5 (a+b x)-12 \tan \left (\frac {1}{2} (a+b x)\right )-5 a\right ) \left (\tan ^2\left (\frac {1}{2} (a+b x)\right )-1\right )-12 \sqrt {\cos (a+b x) \sec ^4\left (\frac {1}{2} (a+b x)\right )} F\left (\left .\sin ^{-1}\left (\tan \left (\frac {1}{2} (a+b x)\right )\right )\right |-1\right )+12 \sqrt {\cos (a+b x) \sec ^4\left (\frac {1}{2} (a+b x)\right )} E\left (\left .\sin ^{-1}\left (\tan \left (\frac {1}{2} (a+b x)\right )\right )\right |-1\right )\right )}{25 b^2}+\frac {\sqrt {\sec (a+b x)} \left (\frac {\sin (a+b x)}{25 b}+\frac {\sin (3 (a+b x))}{25 b}-\frac {1}{10} x \cos (a+b x)-\frac {1}{10} x \cos (3 (a+b x))\right )}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[Sec[a + b*x]]*(-1/10*(x*Cos[a + b*x]) - (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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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*sin(b*x+a)/sec(b*x+a)^(3/2),x, algorithm="fricas")

[Out]

Exception raised: TypeError >>  Error detected within library code:   integrate: implementation incomplete (ha
s polynomial part)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x \sin \left (b x + a\right )}{\sec \left (b x + a\right )^{\frac {3}{2}}}\,{d x} \]

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)

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maple [F]  time = 0.05, size = 0, normalized size = 0.00 \[ \int \frac {x \sin \left (b x +a \right )}{\sec \left (b x +a \right )^{\frac {3}{2}}}\, dx \]

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 = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x \sin \left (b x + a\right )}{\sec \left (b x + a\right )^{\frac {3}{2}}}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x\,\sin \left (a+b\,x\right )}{{\left (\frac {1}{\cos \left (a+b\,x\right )}\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x*sin(a + b*x))/(1/cos(a + b*x))^(3/2),x)

[Out]

int((x*sin(a + b*x))/(1/cos(a + b*x))^(3/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x \sin {\left (a + b x \right )}}{\sec ^{\frac {3}{2}}{\left (a + b x \right )}}\, dx \]

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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