3.471 \(\int \frac {1}{\sqrt {b \sec (e+f x)} \sin ^{\frac {17}{2}}(e+f x)} \, dx\)

Optimal. Leaf size=121 \[ -\frac {256 b}{1155 f \sin ^{\frac {3}{2}}(e+f x) (b \sec (e+f x))^{3/2}}-\frac {64 b}{385 f \sin ^{\frac {7}{2}}(e+f x) (b \sec (e+f x))^{3/2}}-\frac {8 b}{55 f \sin ^{\frac {11}{2}}(e+f x) (b \sec (e+f x))^{3/2}}-\frac {2 b}{15 f \sin ^{\frac {15}{2}}(e+f x) (b \sec (e+f x))^{3/2}} \]

[Out]

-2/15*b/f/(b*sec(f*x+e))^(3/2)/sin(f*x+e)^(15/2)-8/55*b/f/(b*sec(f*x+e))^(3/2)/sin(f*x+e)^(11/2)-64/385*b/f/(b
*sec(f*x+e))^(3/2)/sin(f*x+e)^(7/2)-256/1155*b/f/(b*sec(f*x+e))^(3/2)/sin(f*x+e)^(3/2)

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Rubi [A]  time = 0.16, antiderivative size = 121, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {2584, 2578} \[ -\frac {256 b}{1155 f \sin ^{\frac {3}{2}}(e+f x) (b \sec (e+f x))^{3/2}}-\frac {64 b}{385 f \sin ^{\frac {7}{2}}(e+f x) (b \sec (e+f x))^{3/2}}-\frac {8 b}{55 f \sin ^{\frac {11}{2}}(e+f x) (b \sec (e+f x))^{3/2}}-\frac {2 b}{15 f \sin ^{\frac {15}{2}}(e+f x) (b \sec (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[b*Sec[e + f*x]]*Sin[e + f*x]^(17/2)),x]

[Out]

(-2*b)/(15*f*(b*Sec[e + f*x])^(3/2)*Sin[e + f*x]^(15/2)) - (8*b)/(55*f*(b*Sec[e + f*x])^(3/2)*Sin[e + f*x]^(11
/2)) - (64*b)/(385*f*(b*Sec[e + f*x])^(3/2)*Sin[e + f*x]^(7/2)) - (256*b)/(1155*f*(b*Sec[e + f*x])^(3/2)*Sin[e
+ f*x]^(3/2))

Rule 2578

Int[((b_.)*sec[(e_.) + (f_.)*(x_)])^(n_.)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Simp[(b*(a*Sin[e
+ f*x])^(m + 1)*(b*Sec[e + f*x])^(n - 1))/(a*f*(m + 1)), x] /; FreeQ[{a, b, e, f, m, n}, x] && EqQ[m - n + 2,
0] && NeQ[m, -1]

Rule 2584

Int[((b_.)*sec[(e_.) + (f_.)*(x_)])^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(b*(a*Sin[e +
f*x])^(m + 1)*(b*Sec[e + f*x])^(n - 1))/(a*f*(m + 1)), x] + Dist[(m - n + 2)/(a^2*(m + 1)), Int[(a*Sin[e + f*
x])^(m + 2)*(b*Sec[e + f*x])^n, x], x] /; FreeQ[{a, b, e, f, n}, x] && LtQ[m, -1] && IntegersQ[2*m, 2*n]

Rubi steps

\begin {align*} \int \frac {1}{\sqrt {b \sec (e+f x)} \sin ^{\frac {17}{2}}(e+f x)} \, dx &=-\frac {2 b}{15 f (b \sec (e+f x))^{3/2} \sin ^{\frac {15}{2}}(e+f x)}+\frac {4}{5} \int \frac {1}{\sqrt {b \sec (e+f x)} \sin ^{\frac {13}{2}}(e+f x)} \, dx\\ &=-\frac {2 b}{15 f (b \sec (e+f x))^{3/2} \sin ^{\frac {15}{2}}(e+f x)}-\frac {8 b}{55 f (b \sec (e+f x))^{3/2} \sin ^{\frac {11}{2}}(e+f x)}+\frac {32}{55} \int \frac {1}{\sqrt {b \sec (e+f x)} \sin ^{\frac {9}{2}}(e+f x)} \, dx\\ &=-\frac {2 b}{15 f (b \sec (e+f x))^{3/2} \sin ^{\frac {15}{2}}(e+f x)}-\frac {8 b}{55 f (b \sec (e+f x))^{3/2} \sin ^{\frac {11}{2}}(e+f x)}-\frac {64 b}{385 f (b \sec (e+f x))^{3/2} \sin ^{\frac {7}{2}}(e+f x)}+\frac {128}{385} \int \frac {1}{\sqrt {b \sec (e+f x)} \sin ^{\frac {5}{2}}(e+f x)} \, dx\\ &=-\frac {2 b}{15 f (b \sec (e+f x))^{3/2} \sin ^{\frac {15}{2}}(e+f x)}-\frac {8 b}{55 f (b \sec (e+f x))^{3/2} \sin ^{\frac {11}{2}}(e+f x)}-\frac {64 b}{385 f (b \sec (e+f x))^{3/2} \sin ^{\frac {7}{2}}(e+f x)}-\frac {256 b}{1155 f (b \sec (e+f x))^{3/2} \sin ^{\frac {3}{2}}(e+f x)}\\ \end {align*}

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Mathematica [A]  time = 0.26, size = 62, normalized size = 0.51 \[ \frac {2 b (150 \cos (2 (e+f x))-36 \cos (4 (e+f x))+4 \cos (6 (e+f x))-195)}{1155 f \sin ^{\frac {15}{2}}(e+f x) (b \sec (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(Sqrt[b*Sec[e + f*x]]*Sin[e + f*x]^(17/2)),x]

[Out]

(2*b*(-195 + 150*Cos[2*(e + f*x)] - 36*Cos[4*(e + f*x)] + 4*Cos[6*(e + f*x)]))/(1155*f*(b*Sec[e + f*x])^(3/2)*
Sin[e + f*x]^(15/2))

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fricas [A]  time = 0.55, size = 116, normalized size = 0.96 \[ \frac {2 \, {\left (128 \, \cos \left (f x + e\right )^{8} - 480 \, \cos \left (f x + e\right )^{6} + 660 \, \cos \left (f x + e\right )^{4} - 385 \, \cos \left (f x + e\right )^{2}\right )} \sqrt {\frac {b}{\cos \left (f x + e\right )}} \sqrt {\sin \left (f x + e\right )}}{1155 \, {\left (b f \cos \left (f x + e\right )^{8} - 4 \, b f \cos \left (f x + e\right )^{6} + 6 \, b f \cos \left (f x + e\right )^{4} - 4 \, b f \cos \left (f x + e\right )^{2} + b f\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/sin(f*x+e)^(17/2)/(b*sec(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

2/1155*(128*cos(f*x + e)^8 - 480*cos(f*x + e)^6 + 660*cos(f*x + e)^4 - 385*cos(f*x + e)^2)*sqrt(b/cos(f*x + e)
)*sqrt(sin(f*x + e))/(b*f*cos(f*x + e)^8 - 4*b*f*cos(f*x + e)^6 + 6*b*f*cos(f*x + e)^4 - 4*b*f*cos(f*x + e)^2
+ b*f)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {b \sec \left (f x + e\right )} \sin \left (f x + e\right )^{\frac {17}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/sin(f*x+e)^(17/2)/(b*sec(f*x+e))^(1/2),x, algorithm="giac")

[Out]

integrate(1/(sqrt(b*sec(f*x + e))*sin(f*x + e)^(17/2)), x)

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maple [A]  time = 0.21, size = 102, normalized size = 0.84 \[ \frac {512 \cos \left (f x +e \right ) \left (128 \left (\cos ^{6}\left (f x +e \right )\right )-480 \left (\cos ^{4}\left (f x +e \right )\right )+660 \left (\cos ^{2}\left (f x +e \right )\right )-385\right ) \left (-1+\cos \left (f x +e \right )\right )^{8}}{1155 f \sin \left (f x +e \right )^{\frac {15}{2}} \left (\sin ^{2}\left (f x +e \right )+\cos ^{2}\left (f x +e \right )-2 \cos \left (f x +e \right )+1\right )^{8} \sqrt {\frac {b}{\cos \left (f x +e \right )}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/sin(f*x+e)^(17/2)/(b*sec(f*x+e))^(1/2),x)

[Out]

512/1155/f*cos(f*x+e)*(128*cos(f*x+e)^6-480*cos(f*x+e)^4+660*cos(f*x+e)^2-385)*(-1+cos(f*x+e))^8/sin(f*x+e)^(1
5/2)/(sin(f*x+e)^2+cos(f*x+e)^2-2*cos(f*x+e)+1)^8/(b/cos(f*x+e))^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {b \sec \left (f x + e\right )} \sin \left (f x + e\right )^{\frac {17}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/sin(f*x+e)^(17/2)/(b*sec(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

integrate(1/(sqrt(b*sec(f*x + e))*sin(f*x + e)^(17/2)), x)

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mupad [B]  time = 6.41, size = 192, normalized size = 1.59 \[ \frac {{\mathrm {e}}^{-e\,8{}\mathrm {i}-f\,x\,8{}\mathrm {i}}\,\sqrt {\frac {b}{\frac {{\mathrm {e}}^{-e\,1{}\mathrm {i}-f\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}}{2}}}\,\left (\frac {{\mathrm {e}}^{e\,8{}\mathrm {i}+f\,x\,8{}\mathrm {i}}\,1024{}\mathrm {i}}{77\,b\,f}+\frac {{\mathrm {e}}^{e\,8{}\mathrm {i}+f\,x\,8{}\mathrm {i}}\,\cos \left (2\,e+2\,f\,x\right )\,384{}\mathrm {i}}{55\,b\,f}-\frac {{\mathrm {e}}^{e\,8{}\mathrm {i}+f\,x\,8{}\mathrm {i}}\,\cos \left (4\,e+4\,f\,x\right )\,5248{}\mathrm {i}}{1155\,b\,f}+\frac {{\mathrm {e}}^{e\,8{}\mathrm {i}+f\,x\,8{}\mathrm {i}}\,\cos \left (6\,e+6\,f\,x\right )\,256{}\mathrm {i}}{165\,b\,f}-\frac {{\mathrm {e}}^{e\,8{}\mathrm {i}+f\,x\,8{}\mathrm {i}}\,\cos \left (8\,e+8\,f\,x\right )\,256{}\mathrm {i}}{1155\,b\,f}\right )\,1{}\mathrm {i}}{128\,{\sin \left (e+f\,x\right )}^{15/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sin(e + f*x)^(17/2)*(b/cos(e + f*x))^(1/2)),x)

[Out]

(exp(- e*8i - f*x*8i)*(b/(exp(- e*1i - f*x*1i)/2 + exp(e*1i + f*x*1i)/2))^(1/2)*((exp(e*8i + f*x*8i)*1024i)/(7
7*b*f) + (exp(e*8i + f*x*8i)*cos(2*e + 2*f*x)*384i)/(55*b*f) - (exp(e*8i + f*x*8i)*cos(4*e + 4*f*x)*5248i)/(11
55*b*f) + (exp(e*8i + f*x*8i)*cos(6*e + 6*f*x)*256i)/(165*b*f) - (exp(e*8i + f*x*8i)*cos(8*e + 8*f*x)*256i)/(1
155*b*f))*1i)/(128*sin(e + f*x)^(15/2))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/sin(f*x+e)**(17/2)/(b*sec(f*x+e))**(1/2),x)

[Out]

Timed out

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