3.72 \(\int \csc ^5(e+f x) \sqrt{a+b \sec ^2(e+f x)} \, dx\)

Optimal. Leaf size=183 \[ -\frac{\left (3 a^2+12 a b+8 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a+b} \sec (e+f x)}{\sqrt{a+b \sec ^2(e+f x)}}\right )}{8 f (a+b)^{3/2}}+\frac{\sqrt{b} \tanh ^{-1}\left (\frac{\sqrt{b} \sec (e+f x)}{\sqrt{a+b \sec ^2(e+f x)}}\right )}{f}-\frac{\cot (e+f x) \csc ^3(e+f x) \sqrt{a+b \sec ^2(e+f x)}}{4 f}-\frac{(3 a+4 b) \cot (e+f x) \csc (e+f x) \sqrt{a+b \sec ^2(e+f x)}}{8 f (a+b)} \]

[Out]

(Sqrt[b]*ArcTanh[(Sqrt[b]*Sec[e + f*x])/Sqrt[a + b*Sec[e + f*x]^2]])/f - ((3*a^2 + 12*a*b + 8*b^2)*ArcTanh[(Sq
rt[a + b]*Sec[e + f*x])/Sqrt[a + b*Sec[e + f*x]^2]])/(8*(a + b)^(3/2)*f) - ((3*a + 4*b)*Cot[e + f*x]*Csc[e + f
*x]*Sqrt[a + b*Sec[e + f*x]^2])/(8*(a + b)*f) - (Cot[e + f*x]*Csc[e + f*x]^3*Sqrt[a + b*Sec[e + f*x]^2])/(4*f)

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Rubi [A]  time = 0.22016, antiderivative size = 183, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.32, Rules used = {4134, 467, 578, 523, 217, 206, 377, 207} \[ -\frac{\left (3 a^2+12 a b+8 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a+b} \sec (e+f x)}{\sqrt{a+b \sec ^2(e+f x)}}\right )}{8 f (a+b)^{3/2}}+\frac{\sqrt{b} \tanh ^{-1}\left (\frac{\sqrt{b} \sec (e+f x)}{\sqrt{a+b \sec ^2(e+f x)}}\right )}{f}-\frac{\cot (e+f x) \csc ^3(e+f x) \sqrt{a+b \sec ^2(e+f x)}}{4 f}-\frac{(3 a+4 b) \cot (e+f x) \csc (e+f x) \sqrt{a+b \sec ^2(e+f x)}}{8 f (a+b)} \]

Antiderivative was successfully verified.

[In]

Int[Csc[e + f*x]^5*Sqrt[a + b*Sec[e + f*x]^2],x]

[Out]

(Sqrt[b]*ArcTanh[(Sqrt[b]*Sec[e + f*x])/Sqrt[a + b*Sec[e + f*x]^2]])/f - ((3*a^2 + 12*a*b + 8*b^2)*ArcTanh[(Sq
rt[a + b]*Sec[e + f*x])/Sqrt[a + b*Sec[e + f*x]^2]])/(8*(a + b)^(3/2)*f) - ((3*a + 4*b)*Cot[e + f*x]*Csc[e + f
*x]*Sqrt[a + b*Sec[e + f*x]^2])/(8*(a + b)*f) - (Cot[e + f*x]*Csc[e + f*x]^3*Sqrt[a + b*Sec[e + f*x]^2])/(4*f)

Rule 4134

Int[((a_) + (b_.)*((c_.)*sec[(e_.) + (f_.)*(x_)])^(n_))^(p_.)*sin[(e_.) + (f_.)*(x_)]^(m_.), x_Symbol] :> With
[{ff = FreeFactors[Cos[e + f*x], x]}, Dist[1/(f*ff^m), Subst[Int[((-1 + ff^2*x^2)^((m - 1)/2)*(a + b*(c*ff*x)^
n)^p)/x^(m + 1), x], x, Sec[e + f*x]/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && IntegerQ[(m - 1)/2] && (Gt
Q[m, 0] || EqQ[n, 2] || EqQ[n, 4])

Rule 467

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

Rule 578

Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_)*((e_) + (f_.)*(x_)^(n_)), x
_Symbol] :> Simp[(g^(n - 1)*(b*e - a*f)*(g*x)^(m - n + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(b*n*(b*c -
a*d)*(p + 1)), x] - Dist[g^n/(b*n*(b*c - a*d)*(p + 1)), Int[(g*x)^(m - n)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*S
imp[c*(b*e - a*f)*(m - n + 1) + (d*(b*e - a*f)*(m + n*q + 1) - b*n*(c*f - d*e)*(p + 1))*x^n, x], x], x] /; Fre
eQ[{a, b, c, d, e, f, g, q}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m - n + 1, 0]

Rule 523

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*Sqrt[(c_) + (d_.)*(x_)^(n_)]), x_Symbol] :> Dist[f/b, I
nt[1/Sqrt[c + d*x^n], x], x] + Dist[(b*e - a*f)/b, Int[1/((a + b*x^n)*Sqrt[c + d*x^n]), x], x] /; FreeQ[{a, b,
c, d, e, f, n}, x]

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 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 377

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]

Rule 207

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

Rubi steps

\begin{align*} \int \csc ^5(e+f x) \sqrt{a+b \sec ^2(e+f x)} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^4 \sqrt{a+b x^2}}{\left (-1+x^2\right )^3} \, dx,x,\sec (e+f x)\right )}{f}\\ &=-\frac{\cot (e+f x) \csc ^3(e+f x) \sqrt{a+b \sec ^2(e+f x)}}{4 f}+\frac{\operatorname{Subst}\left (\int \frac{x^2 \left (3 a+4 b x^2\right )}{\left (-1+x^2\right )^2 \sqrt{a+b x^2}} \, dx,x,\sec (e+f x)\right )}{4 f}\\ &=-\frac{(3 a+4 b) \cot (e+f x) \csc (e+f x) \sqrt{a+b \sec ^2(e+f x)}}{8 (a+b) f}-\frac{\cot (e+f x) \csc ^3(e+f x) \sqrt{a+b \sec ^2(e+f x)}}{4 f}+\frac{\operatorname{Subst}\left (\int \frac{a (3 a+4 b)+8 b (a+b) x^2}{\left (-1+x^2\right ) \sqrt{a+b x^2}} \, dx,x,\sec (e+f x)\right )}{8 (a+b) f}\\ &=-\frac{(3 a+4 b) \cot (e+f x) \csc (e+f x) \sqrt{a+b \sec ^2(e+f x)}}{8 (a+b) f}-\frac{\cot (e+f x) \csc ^3(e+f x) \sqrt{a+b \sec ^2(e+f x)}}{4 f}+\frac{b \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x^2}} \, dx,x,\sec (e+f x)\right )}{f}+\frac{\left (3 a^2+12 a b+8 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{\left (-1+x^2\right ) \sqrt{a+b x^2}} \, dx,x,\sec (e+f x)\right )}{8 (a+b) f}\\ &=-\frac{(3 a+4 b) \cot (e+f x) \csc (e+f x) \sqrt{a+b \sec ^2(e+f x)}}{8 (a+b) f}-\frac{\cot (e+f x) \csc ^3(e+f x) \sqrt{a+b \sec ^2(e+f x)}}{4 f}+\frac{b \operatorname{Subst}\left (\int \frac{1}{1-b x^2} \, dx,x,\frac{\sec (e+f x)}{\sqrt{a+b \sec ^2(e+f x)}}\right )}{f}+\frac{\left (3 a^2+12 a b+8 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{-1-(-a-b) x^2} \, dx,x,\frac{\sec (e+f x)}{\sqrt{a+b \sec ^2(e+f x)}}\right )}{8 (a+b) f}\\ &=\frac{\sqrt{b} \tanh ^{-1}\left (\frac{\sqrt{b} \sec (e+f x)}{\sqrt{a+b \sec ^2(e+f x)}}\right )}{f}-\frac{\left (3 a^2+12 a b+8 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a+b} \sec (e+f x)}{\sqrt{a+b \sec ^2(e+f x)}}\right )}{8 (a+b)^{3/2} f}-\frac{(3 a+4 b) \cot (e+f x) \csc (e+f x) \sqrt{a+b \sec ^2(e+f x)}}{8 (a+b) f}-\frac{\cot (e+f x) \csc ^3(e+f x) \sqrt{a+b \sec ^2(e+f x)}}{4 f}\\ \end{align*}

Mathematica [A]  time = 1.40722, size = 198, normalized size = 1.08 \[ \frac{\cos (e+f x) \sqrt{a+b \sec ^2(e+f x)} \left (-\left (3 a^2+12 a b+8 b^2\right ) \sqrt{a+b} \tanh ^{-1}\left (\frac{\sqrt{-a \sin ^2(e+f x)+a+b}}{\sqrt{a+b}}\right )+8 \sqrt{b} (a+b)^2 \tanh ^{-1}\left (\frac{\sqrt{-a \sin ^2(e+f x)+a+b}}{\sqrt{b}}\right )-(a+b) \csc ^2(e+f x) \sqrt{-a \sin ^2(e+f x)+a+b} \left (2 (a+b) \csc ^2(e+f x)+3 a+4 b\right )\right )}{4 \sqrt{2} f (a+b)^2 \sqrt{a \cos (2 (e+f x))+a+2 b}} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[e + f*x]^5*Sqrt[a + b*Sec[e + f*x]^2],x]

[Out]

(Cos[e + f*x]*Sqrt[a + b*Sec[e + f*x]^2]*(8*Sqrt[b]*(a + b)^2*ArcTanh[Sqrt[a + b - a*Sin[e + f*x]^2]/Sqrt[b]]
- Sqrt[a + b]*(3*a^2 + 12*a*b + 8*b^2)*ArcTanh[Sqrt[a + b - a*Sin[e + f*x]^2]/Sqrt[a + b]] - (a + b)*Csc[e + f
*x]^2*(3*a + 4*b + 2*(a + b)*Csc[e + f*x]^2)*Sqrt[a + b - a*Sin[e + f*x]^2]))/(4*Sqrt[2]*(a + b)^2*f*Sqrt[a +
2*b + a*Cos[2*(e + f*x)]])

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Maple [B]  time = 0.472, size = 9758, normalized size = 53.3 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{b \sec \left (f x + e\right )^{2} + a} \csc \left (f x + e\right )^{5}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(b*sec(f*x + e)^2 + a)*csc(f*x + e)^5, x)

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Fricas [B]  time = 2.16584, size = 3656, normalized size = 19.98 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/16*(((3*a^2 + 12*a*b + 8*b^2)*cos(f*x + e)^4 - 2*(3*a^2 + 12*a*b + 8*b^2)*cos(f*x + e)^2 + 3*a^2 + 12*a*b +
8*b^2)*sqrt(a + b)*log(2*(a*cos(f*x + e)^2 - 2*sqrt(a + b)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*
x + e) + a + 2*b)/(cos(f*x + e)^2 - 1)) + 8*((a^2 + 2*a*b + b^2)*cos(f*x + e)^4 - 2*(a^2 + 2*a*b + b^2)*cos(f*
x + e)^2 + a^2 + 2*a*b + b^2)*sqrt(b)*log((a*cos(f*x + e)^2 + 2*sqrt(b)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x +
e)^2)*cos(f*x + e) + 2*b)/cos(f*x + e)^2) + 2*((3*a^2 + 7*a*b + 4*b^2)*cos(f*x + e)^3 - (5*a^2 + 11*a*b + 6*b^
2)*cos(f*x + e))*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/((a^2 + 2*a*b + b^2)*f*cos(f*x + e)^4 - 2*(a^2 +
2*a*b + b^2)*f*cos(f*x + e)^2 + (a^2 + 2*a*b + b^2)*f), 1/8*(((3*a^2 + 12*a*b + 8*b^2)*cos(f*x + e)^4 - 2*(3*
a^2 + 12*a*b + 8*b^2)*cos(f*x + e)^2 + 3*a^2 + 12*a*b + 8*b^2)*sqrt(-a - b)*arctan(sqrt(-a - b)*sqrt((a*cos(f*
x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e)/(a + b)) + 4*((a^2 + 2*a*b + b^2)*cos(f*x + e)^4 - 2*(a^2 + 2*a*b +
b^2)*cos(f*x + e)^2 + a^2 + 2*a*b + b^2)*sqrt(b)*log((a*cos(f*x + e)^2 + 2*sqrt(b)*sqrt((a*cos(f*x + e)^2 + b
)/cos(f*x + e)^2)*cos(f*x + e) + 2*b)/cos(f*x + e)^2) + ((3*a^2 + 7*a*b + 4*b^2)*cos(f*x + e)^3 - (5*a^2 + 11*
a*b + 6*b^2)*cos(f*x + e))*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/((a^2 + 2*a*b + b^2)*f*cos(f*x + e)^4
- 2*(a^2 + 2*a*b + b^2)*f*cos(f*x + e)^2 + (a^2 + 2*a*b + b^2)*f), -1/16*(16*((a^2 + 2*a*b + b^2)*cos(f*x + e)
^4 - 2*(a^2 + 2*a*b + b^2)*cos(f*x + e)^2 + a^2 + 2*a*b + b^2)*sqrt(-b)*arctan(sqrt(-b)*sqrt((a*cos(f*x + e)^2
+ b)/cos(f*x + e)^2)*cos(f*x + e)/b) - ((3*a^2 + 12*a*b + 8*b^2)*cos(f*x + e)^4 - 2*(3*a^2 + 12*a*b + 8*b^2)*
cos(f*x + e)^2 + 3*a^2 + 12*a*b + 8*b^2)*sqrt(a + b)*log(2*(a*cos(f*x + e)^2 - 2*sqrt(a + b)*sqrt((a*cos(f*x +
e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e) + a + 2*b)/(cos(f*x + e)^2 - 1)) - 2*((3*a^2 + 7*a*b + 4*b^2)*cos(f*x
+ e)^3 - (5*a^2 + 11*a*b + 6*b^2)*cos(f*x + e))*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/((a^2 + 2*a*b + b
^2)*f*cos(f*x + e)^4 - 2*(a^2 + 2*a*b + b^2)*f*cos(f*x + e)^2 + (a^2 + 2*a*b + b^2)*f), 1/8*(((3*a^2 + 12*a*b
+ 8*b^2)*cos(f*x + e)^4 - 2*(3*a^2 + 12*a*b + 8*b^2)*cos(f*x + e)^2 + 3*a^2 + 12*a*b + 8*b^2)*sqrt(-a - b)*arc
tan(sqrt(-a - b)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e)/(a + b)) - 8*((a^2 + 2*a*b + b^2)*co
s(f*x + e)^4 - 2*(a^2 + 2*a*b + b^2)*cos(f*x + e)^2 + a^2 + 2*a*b + b^2)*sqrt(-b)*arctan(sqrt(-b)*sqrt((a*cos(
f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e)/b) + ((3*a^2 + 7*a*b + 4*b^2)*cos(f*x + e)^3 - (5*a^2 + 11*a*b +
6*b^2)*cos(f*x + e))*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/((a^2 + 2*a*b + b^2)*f*cos(f*x + e)^4 - 2*(a
^2 + 2*a*b + b^2)*f*cos(f*x + e)^2 + (a^2 + 2*a*b + b^2)*f)]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{b \sec \left (f x + e\right )^{2} + a} \csc \left (f x + e\right )^{5}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(b*sec(f*x + e)^2 + a)*csc(f*x + e)^5, x)