\(\int x \sec ^{\frac {7}{2}}(a+b x) \sin (a+b x) \, dx\) [337]

Optimal result

Integrand size = 18, antiderivative size = 80 \[ \int x \sec ^{\frac {7}{2}}(a+b x) \sin (a+b x) \, dx=-\frac {4 \sqrt {\cos (a+b x)} \operatorname {EllipticF}\left (\frac {1}{2} (a+b x),2\right ) \sqrt {\sec (a+b x)}}{15 b^2}+\frac {2 x \sec ^{\frac {5}{2}}(a+b x)}{5 b}-\frac {4 \sec ^{\frac {3}{2}}(a+b x) \sin (a+b x)}{15 b^2} \]

[Out]

2/5*x*sec(b*x+a)^(5/2)/b-4/15*sec(b*x+a)^(3/2)*sin(b*x+a)/b^2-4/15*(cos(1/2*a+1/2*b*x)^2)^(1/2)/cos(1/2*a+1/2*
b*x)*EllipticF(sin(1/2*a+1/2*b*x),2^(1/2))*cos(b*x+a)^(1/2)*sec(b*x+a)^(1/2)/b^2

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {4297, 3853, 3856, 2720} \[ \int x \sec ^{\frac {7}{2}}(a+b x) \sin (a+b x) \, dx=-\frac {4 \sin (a+b x) \sec ^{\frac {3}{2}}(a+b x)}{15 b^2}-\frac {4 \sqrt {\cos (a+b x)} \sqrt {\sec (a+b x)} \operatorname {EllipticF}\left (\frac {1}{2} (a+b x),2\right )}{15 b^2}+\frac {2 x \sec ^{\frac {5}{2}}(a+b x)}{5 b} \]

[In]

Int[x*Sec[a + b*x]^(7/2)*Sin[a + b*x],x]

[Out]

(-4*Sqrt[Cos[a + b*x]]*EllipticF[(a + b*x)/2, 2]*Sqrt[Sec[a + b*x]])/(15*b^2) + (2*x*Sec[a + b*x]^(5/2))/(5*b)
- (4*Sec[a + b*x]^(3/2)*Sin[a + b*x])/(15*b^2)

Rule 2720

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ
[{c, d}, x]

Rule 3853

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 3856

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 4297

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*} \text {integral}& = \frac {2 x \sec ^{\frac {5}{2}}(a+b x)}{5 b}-\frac {2 \int \sec ^{\frac {5}{2}}(a+b x) \, dx}{5 b} \\ & = \frac {2 x \sec ^{\frac {5}{2}}(a+b x)}{5 b}-\frac {4 \sec ^{\frac {3}{2}}(a+b x) \sin (a+b x)}{15 b^2}-\frac {2 \int \sqrt {\sec (a+b x)} \, dx}{15 b} \\ & = \frac {2 x \sec ^{\frac {5}{2}}(a+b x)}{5 b}-\frac {4 \sec ^{\frac {3}{2}}(a+b x) \sin (a+b x)}{15 b^2}-\frac {\left (2 \sqrt {\cos (a+b x)} \sqrt {\sec (a+b x)}\right ) \int \frac {1}{\sqrt {\cos (a+b x)}} \, dx}{15 b} \\ & = -\frac {4 \sqrt {\cos (a+b x)} \operatorname {EllipticF}\left (\frac {1}{2} (a+b x),2\right ) \sqrt {\sec (a+b x)}}{15 b^2}+\frac {2 x \sec ^{\frac {5}{2}}(a+b x)}{5 b}-\frac {4 \sec ^{\frac {3}{2}}(a+b x) \sin (a+b x)}{15 b^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.30 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.76 \[ \int x \sec ^{\frac {7}{2}}(a+b x) \sin (a+b x) \, dx=\frac {2 \sqrt {\sec (a+b x)} \left (-2 \sqrt {\cos (a+b x)} \operatorname {EllipticF}\left (\frac {1}{2} (a+b x),2\right )+3 b x \sec ^2(a+b x)-2 \tan (a+b x)\right )}{15 b^2} \]

[In]

Integrate[x*Sec[a + b*x]^(7/2)*Sin[a + b*x],x]

[Out]

(2*Sqrt[Sec[a + b*x]]*(-2*Sqrt[Cos[a + b*x]]*EllipticF[(a + b*x)/2, 2] + 3*b*x*Sec[a + b*x]^2 - 2*Tan[a + b*x]
))/(15*b^2)

Maple [F]

\[\int x \sec \left (x b +a \right )^{\frac {7}{2}} \sin \left (x b +a \right )d x\]

[In]

int(x*sec(b*x+a)^(7/2)*sin(b*x+a),x)

[Out]

int(x*sec(b*x+a)^(7/2)*sin(b*x+a),x)

Fricas [F(-2)]

Exception generated. \[ \int x \sec ^{\frac {7}{2}}(a+b x) \sin (a+b x) \, dx=\text {Exception raised: TypeError} \]

[In]

integrate(x*sec(b*x+a)^(7/2)*sin(b*x+a),x, algorithm="fricas")

[Out]

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

Sympy [F(-1)]

Timed out. \[ \int x \sec ^{\frac {7}{2}}(a+b x) \sin (a+b x) \, dx=\text {Timed out} \]

[In]

integrate(x*sec(b*x+a)**(7/2)*sin(b*x+a),x)

[Out]

Timed out

Maxima [F]

\[ \int x \sec ^{\frac {7}{2}}(a+b x) \sin (a+b x) \, dx=\int { x \sec \left (b x + a\right )^{\frac {7}{2}} \sin \left (b x + a\right ) \,d x } \]

[In]

integrate(x*sec(b*x+a)^(7/2)*sin(b*x+a),x, algorithm="maxima")

[Out]

integrate(x*sec(b*x + a)^(7/2)*sin(b*x + a), x)

Giac [F]

\[ \int x \sec ^{\frac {7}{2}}(a+b x) \sin (a+b x) \, dx=\int { x \sec \left (b x + a\right )^{\frac {7}{2}} \sin \left (b x + a\right ) \,d x } \]

[In]

integrate(x*sec(b*x+a)^(7/2)*sin(b*x+a),x, algorithm="giac")

[Out]

integrate(x*sec(b*x + a)^(7/2)*sin(b*x + a), x)

Mupad [F(-1)]

Timed out. \[ \int x \sec ^{\frac {7}{2}}(a+b x) \sin (a+b x) \, dx=\int x\,\sin \left (a+b\,x\right )\,{\left (\frac {1}{\cos \left (a+b\,x\right )}\right )}^{7/2} \,d x \]

[In]

int(x*sin(a + b*x)*(1/cos(a + b*x))^(7/2),x)

[Out]

int(x*sin(a + b*x)*(1/cos(a + b*x))^(7/2), x)