3.377 \(\int \csc ^5(e+f x) \sqrt {b \sec (e+f x)} \, dx\)

Optimal. Leaf size=123 \[ -\frac {\cot ^4(e+f x) (b \sec (e+f x))^{7/2}}{4 b^3 f}-\frac {7 \cot ^2(e+f x) (b \sec (e+f x))^{3/2}}{16 b f}+\frac {21 \sqrt {b} \tan ^{-1}\left (\frac {\sqrt {b \sec (e+f x)}}{\sqrt {b}}\right )}{32 f}-\frac {21 \sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b \sec (e+f x)}}{\sqrt {b}}\right )}{32 f} \]

[Out]

-7/16*cot(f*x+e)^2*(b*sec(f*x+e))^(3/2)/b/f-1/4*cot(f*x+e)^4*(b*sec(f*x+e))^(7/2)/b^3/f+21/32*arctan((b*sec(f*
x+e))^(1/2)/b^(1/2))*b^(1/2)/f-21/32*arctanh((b*sec(f*x+e))^(1/2)/b^(1/2))*b^(1/2)/f

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Rubi [A]  time = 0.09, antiderivative size = 123, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {2622, 288, 329, 298, 203, 206} \[ -\frac {\cot ^4(e+f x) (b \sec (e+f x))^{7/2}}{4 b^3 f}-\frac {7 \cot ^2(e+f x) (b \sec (e+f x))^{3/2}}{16 b f}+\frac {21 \sqrt {b} \tan ^{-1}\left (\frac {\sqrt {b \sec (e+f x)}}{\sqrt {b}}\right )}{32 f}-\frac {21 \sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b \sec (e+f x)}}{\sqrt {b}}\right )}{32 f} \]

Antiderivative was successfully verified.

[In]

Int[Csc[e + f*x]^5*Sqrt[b*Sec[e + f*x]],x]

[Out]

(21*Sqrt[b]*ArcTan[Sqrt[b*Sec[e + f*x]]/Sqrt[b]])/(32*f) - (21*Sqrt[b]*ArcTanh[Sqrt[b*Sec[e + f*x]]/Sqrt[b]])/
(32*f) - (7*Cot[e + f*x]^2*(b*Sec[e + f*x])^(3/2))/(16*b*f) - (Cot[e + f*x]^4*(b*Sec[e + f*x])^(7/2))/(4*b^3*f
)

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 288

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^
n)^(p + 1))/(b*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rule 298

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]], s = Denominator[Rt[-(a/b),
2]]}, Dist[s/(2*b), Int[1/(r + s*x^2), x], x] - Dist[s/(2*b), Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] &
&  !GtQ[a/b, 0]

Rule 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 2622

Int[csc[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Dist[1/(f*a^n), Subst[Int
[x^(m + n - 1)/(-1 + x^2/a^2)^((n + 1)/2), x], x, a*Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n
+ 1)/2] &&  !(IntegerQ[(m + 1)/2] && LtQ[0, m, n])

Rubi steps

\begin {align*} \int \csc ^5(e+f x) \sqrt {b \sec (e+f x)} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {x^{9/2}}{\left (-1+\frac {x^2}{b^2}\right )^3} \, dx,x,b \sec (e+f x)\right )}{b^5 f}\\ &=-\frac {\cot ^4(e+f x) (b \sec (e+f x))^{7/2}}{4 b^3 f}+\frac {7 \operatorname {Subst}\left (\int \frac {x^{5/2}}{\left (-1+\frac {x^2}{b^2}\right )^2} \, dx,x,b \sec (e+f x)\right )}{8 b^3 f}\\ &=-\frac {7 \cot ^2(e+f x) (b \sec (e+f x))^{3/2}}{16 b f}-\frac {\cot ^4(e+f x) (b \sec (e+f x))^{7/2}}{4 b^3 f}+\frac {21 \operatorname {Subst}\left (\int \frac {\sqrt {x}}{-1+\frac {x^2}{b^2}} \, dx,x,b \sec (e+f x)\right )}{32 b f}\\ &=-\frac {7 \cot ^2(e+f x) (b \sec (e+f x))^{3/2}}{16 b f}-\frac {\cot ^4(e+f x) (b \sec (e+f x))^{7/2}}{4 b^3 f}+\frac {21 \operatorname {Subst}\left (\int \frac {x^2}{-1+\frac {x^4}{b^2}} \, dx,x,\sqrt {b \sec (e+f x)}\right )}{16 b f}\\ &=-\frac {7 \cot ^2(e+f x) (b \sec (e+f x))^{3/2}}{16 b f}-\frac {\cot ^4(e+f x) (b \sec (e+f x))^{7/2}}{4 b^3 f}-\frac {(21 b) \operatorname {Subst}\left (\int \frac {1}{b-x^2} \, dx,x,\sqrt {b \sec (e+f x)}\right )}{32 f}+\frac {(21 b) \operatorname {Subst}\left (\int \frac {1}{b+x^2} \, dx,x,\sqrt {b \sec (e+f x)}\right )}{32 f}\\ &=\frac {21 \sqrt {b} \tan ^{-1}\left (\frac {\sqrt {b \sec (e+f x)}}{\sqrt {b}}\right )}{32 f}-\frac {21 \sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b \sec (e+f x)}}{\sqrt {b}}\right )}{32 f}-\frac {7 \cot ^2(e+f x) (b \sec (e+f x))^{3/2}}{16 b f}-\frac {\cot ^4(e+f x) (b \sec (e+f x))^{7/2}}{4 b^3 f}\\ \end {align*}

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Mathematica [A]  time = 1.01, size = 107, normalized size = 0.87 \[ \frac {b \left (-16 \csc ^4(e+f x)-28 \csc ^2(e+f x)+21 \sqrt {\sec (e+f x)} \left (\log \left (1-\sqrt {\sec (e+f x)}\right )-\log \left (\sqrt {\sec (e+f x)}+1\right )\right )+42 \sqrt {\sec (e+f x)} \tan ^{-1}\left (\sqrt {\sec (e+f x)}\right )\right )}{64 f \sqrt {b \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[e + f*x]^5*Sqrt[b*Sec[e + f*x]],x]

[Out]

(b*(-28*Csc[e + f*x]^2 - 16*Csc[e + f*x]^4 + 42*ArcTan[Sqrt[Sec[e + f*x]]]*Sqrt[Sec[e + f*x]] + 21*(Log[1 - Sq
rt[Sec[e + f*x]]] - Log[1 + Sqrt[Sec[e + f*x]]])*Sqrt[Sec[e + f*x]]))/(64*f*Sqrt[b*Sec[e + f*x]])

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fricas [B]  time = 0.93, size = 438, normalized size = 3.56 \[ \left [\frac {42 \, {\left (\cos \left (f x + e\right )^{4} - 2 \, \cos \left (f x + e\right )^{2} + 1\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} \sqrt {\frac {b}{\cos \left (f x + e\right )}} {\left (\cos \left (f x + e\right ) + 1\right )}}{2 \, b}\right ) + 21 \, {\left (\cos \left (f x + e\right )^{4} - 2 \, \cos \left (f x + e\right )^{2} + 1\right )} \sqrt {-b} \log \left (\frac {b \cos \left (f x + e\right )^{2} - 4 \, {\left (\cos \left (f x + e\right )^{2} - \cos \left (f x + e\right )\right )} \sqrt {-b} \sqrt {\frac {b}{\cos \left (f x + e\right )}} - 6 \, b \cos \left (f x + e\right ) + b}{\cos \left (f x + e\right )^{2} + 2 \, \cos \left (f x + e\right ) + 1}\right ) + 8 \, {\left (7 \, \cos \left (f x + e\right )^{3} - 11 \, \cos \left (f x + e\right )\right )} \sqrt {\frac {b}{\cos \left (f x + e\right )}}}{128 \, {\left (f \cos \left (f x + e\right )^{4} - 2 \, f \cos \left (f x + e\right )^{2} + f\right )}}, -\frac {42 \, {\left (\cos \left (f x + e\right )^{4} - 2 \, \cos \left (f x + e\right )^{2} + 1\right )} \sqrt {b} \arctan \left (\frac {\sqrt {\frac {b}{\cos \left (f x + e\right )}} {\left (\cos \left (f x + e\right ) - 1\right )}}{2 \, \sqrt {b}}\right ) - 21 \, {\left (\cos \left (f x + e\right )^{4} - 2 \, \cos \left (f x + e\right )^{2} + 1\right )} \sqrt {b} \log \left (\frac {b \cos \left (f x + e\right )^{2} - 4 \, {\left (\cos \left (f x + e\right )^{2} + \cos \left (f x + e\right )\right )} \sqrt {b} \sqrt {\frac {b}{\cos \left (f x + e\right )}} + 6 \, b \cos \left (f x + e\right ) + b}{\cos \left (f x + e\right )^{2} - 2 \, \cos \left (f x + e\right ) + 1}\right ) - 8 \, {\left (7 \, \cos \left (f x + e\right )^{3} - 11 \, \cos \left (f x + e\right )\right )} \sqrt {\frac {b}{\cos \left (f x + e\right )}}}{128 \, {\left (f \cos \left (f x + e\right )^{4} - 2 \, f \cos \left (f x + e\right )^{2} + f\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^5*(b*sec(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

[1/128*(42*(cos(f*x + e)^4 - 2*cos(f*x + e)^2 + 1)*sqrt(-b)*arctan(1/2*sqrt(-b)*sqrt(b/cos(f*x + e))*(cos(f*x
+ e) + 1)/b) + 21*(cos(f*x + e)^4 - 2*cos(f*x + e)^2 + 1)*sqrt(-b)*log((b*cos(f*x + e)^2 - 4*(cos(f*x + e)^2 -
cos(f*x + e))*sqrt(-b)*sqrt(b/cos(f*x + e)) - 6*b*cos(f*x + e) + b)/(cos(f*x + e)^2 + 2*cos(f*x + e) + 1)) +
8*(7*cos(f*x + e)^3 - 11*cos(f*x + e))*sqrt(b/cos(f*x + e)))/(f*cos(f*x + e)^4 - 2*f*cos(f*x + e)^2 + f), -1/1
28*(42*(cos(f*x + e)^4 - 2*cos(f*x + e)^2 + 1)*sqrt(b)*arctan(1/2*sqrt(b/cos(f*x + e))*(cos(f*x + e) - 1)/sqrt
(b)) - 21*(cos(f*x + e)^4 - 2*cos(f*x + e)^2 + 1)*sqrt(b)*log((b*cos(f*x + e)^2 - 4*(cos(f*x + e)^2 + cos(f*x
+ e))*sqrt(b)*sqrt(b/cos(f*x + e)) + 6*b*cos(f*x + e) + b)/(cos(f*x + e)^2 - 2*cos(f*x + e) + 1)) - 8*(7*cos(f
*x + e)^3 - 11*cos(f*x + e))*sqrt(b/cos(f*x + e)))/(f*cos(f*x + e)^4 - 2*f*cos(f*x + e)^2 + f)]

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: NotImplementedError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^5*(b*sec(f*x+e))^(1/2),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: Unable to check sign: (4*pi/x/2)>(-4*pi/
x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)2*(2*sqrt(-b*tan(1
/2*(f*x+exp(1)))^4+b)*(-5/128-1/256*tan(1/2*(f*x+exp(1)))^2)+2*(21/256*sqrt(-b)*ln(abs(-sqrt(-b)*tan(1/2*(f*x+
exp(1)))^2+sqrt(-b*tan(1/2*(f*x+exp(1)))^4+b)))-1/128*(-b^2*(-sqrt(-b)*tan(1/2*(f*x+exp(1)))^2+sqrt(-b*tan(1/2
*(f*x+exp(1)))^4+b))-b*(-sqrt(-b)*tan(1/2*(f*x+exp(1)))^2+sqrt(-b*tan(1/2*(f*x+exp(1)))^4+b))^3-10*b*sqrt(-b)*
(-sqrt(-b)*tan(1/2*(f*x+exp(1)))^2+sqrt(-b*tan(1/2*(f*x+exp(1)))^4+b))^2+10*b^2*sqrt(-b))/((-sqrt(-b)*tan(1/2*
(f*x+exp(1)))^2+sqrt(-b*tan(1/2*(f*x+exp(1)))^4+b))^2-b)^2+21/128*b*atan((-sqrt(-b)*tan(1/2*(f*x+exp(1)))^2+sq
rt(-b*tan(1/2*(f*x+exp(1)))^4+b))/sqrt(-b))/sqrt(-b)))*sign(cos(f*x+exp(1)))/f

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maple [B]  time = 0.23, size = 1089, normalized size = 8.85 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(f*x+e)^5*(b*sec(f*x+e))^(1/2),x)

[Out]

-1/64/f*(72*cos(f*x+e)^3*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(3/2)+56*cos(f*x+e)^2*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(
3/2)-11*cos(f*x+e)^3*ln(-(2*cos(f*x+e)^2*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)-cos(f*x+e)^2+2*cos(f*x+e)-2*(-co
s(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)-1)/sin(f*x+e)^2)+32*cos(f*x+e)^3*ln(-2*(2*cos(f*x+e)^2*(-cos(f*x+e)/(cos(f*x+
e)+1)^2)^(1/2)-cos(f*x+e)^2+2*cos(f*x+e)-2*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)-1)/sin(f*x+e)^2)-21*cos(f*x+e)
^3*arctan(1/2/(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2))-104*cos(f*x+e)*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(3/2)+44*cos
(f*x+e)^2*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)+11*cos(f*x+e)^2*ln(-(2*cos(f*x+e)^2*(-cos(f*x+e)/(cos(f*x+e)+1)
^2)^(1/2)-cos(f*x+e)^2+2*cos(f*x+e)-2*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)-1)/sin(f*x+e)^2)-32*cos(f*x+e)^2*ln
(-2*(2*cos(f*x+e)^2*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)-cos(f*x+e)^2+2*cos(f*x+e)-2*(-cos(f*x+e)/(cos(f*x+e)+
1)^2)^(1/2)-1)/sin(f*x+e)^2)+21*cos(f*x+e)^2*arctan(1/2/(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2))-88*(-cos(f*x+e)/
(cos(f*x+e)+1)^2)^(3/2)-88*cos(f*x+e)*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)+11*cos(f*x+e)*ln(-(2*cos(f*x+e)^2*(
-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)-cos(f*x+e)^2+2*cos(f*x+e)-2*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)-1)/sin(f*
x+e)^2)-32*cos(f*x+e)*ln(-2*(2*cos(f*x+e)^2*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)-cos(f*x+e)^2+2*cos(f*x+e)-2*(
-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)-1)/sin(f*x+e)^2)+21*cos(f*x+e)*arctan(1/2/(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(
1/2))+44*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)-11*ln(-(2*cos(f*x+e)^2*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)-cos(
f*x+e)^2+2*cos(f*x+e)-2*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)-1)/sin(f*x+e)^2)+32*ln(-2*(2*cos(f*x+e)^2*(-cos(f
*x+e)/(cos(f*x+e)+1)^2)^(1/2)-cos(f*x+e)^2+2*cos(f*x+e)-2*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)-1)/sin(f*x+e)^2
)-21*arctan(1/2/(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)))*cos(f*x+e)*(b/cos(f*x+e))^(1/2)/(-cos(f*x+e)/(cos(f*x+e
)+1)^2)^(1/2)/sin(f*x+e)^4

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maxima [A]  time = 0.42, size = 138, normalized size = 1.12 \[ \frac {b {\left (\frac {42 \, \arctan \left (\frac {\sqrt {\frac {b}{\cos \left (f x + e\right )}}}{\sqrt {b}}\right )}{\sqrt {b}} + \frac {21 \, \log \left (-\frac {\sqrt {b} - \sqrt {\frac {b}{\cos \left (f x + e\right )}}}{\sqrt {b} + \sqrt {\frac {b}{\cos \left (f x + e\right )}}}\right )}{\sqrt {b}} + \frac {4 \, {\left (7 \, b^{2} \left (\frac {b}{\cos \left (f x + e\right )}\right )^{\frac {3}{2}} - 11 \, \left (\frac {b}{\cos \left (f x + e\right )}\right )^{\frac {7}{2}}\right )}}{b^{4} - \frac {2 \, b^{4}}{\cos \left (f x + e\right )^{2}} + \frac {b^{4}}{\cos \left (f x + e\right )^{4}}}\right )}}{64 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^5*(b*sec(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

1/64*b*(42*arctan(sqrt(b/cos(f*x + e))/sqrt(b))/sqrt(b) + 21*log(-(sqrt(b) - sqrt(b/cos(f*x + e)))/(sqrt(b) +
sqrt(b/cos(f*x + e))))/sqrt(b) + 4*(7*b^2*(b/cos(f*x + e))^(3/2) - 11*(b/cos(f*x + e))^(7/2))/(b^4 - 2*b^4/cos
(f*x + e)^2 + b^4/cos(f*x + e)^4))/f

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\sqrt {\frac {b}{\cos \left (e+f\,x\right )}}}{{\sin \left (e+f\,x\right )}^5} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b/cos(e + f*x))^(1/2)/sin(e + f*x)^5,x)

[Out]

int((b/cos(e + f*x))^(1/2)/sin(e + f*x)^5, x)

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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(csc(f*x+e)**5*(b*sec(f*x+e))**(1/2),x)

[Out]

Timed out

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