3.79 \(\int \csc ^6(e+f x) \sqrt{a+b \sec ^2(e+f x)} \, dx\)

Optimal. Leaf size=149 \[ \frac{\sqrt{b} \tanh ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b \tan ^2(e+f x)+b}}\right )}{f}-\frac{\cot ^5(e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}{5 f (a+b)}-\frac{2 (5 a+4 b) \cot ^3(e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}{15 f (a+b)^2}-\frac{\cot (e+f x) \sqrt{a+b \tan ^2(e+f x)+b}}{f} \]

[Out]

(Sqrt[b]*ArcTanh[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b + b*Tan[e + f*x]^2]])/f - (Cot[e + f*x]*Sqrt[a + b + b*Tan[
e + f*x]^2])/f - (2*(5*a + 4*b)*Cot[e + f*x]^3*(a + b + b*Tan[e + f*x]^2)^(3/2))/(15*(a + b)^2*f) - (Cot[e + f
*x]^5*(a + b + b*Tan[e + f*x]^2)^(3/2))/(5*(a + b)*f)

________________________________________________________________________________________

Rubi [A]  time = 0.14012, antiderivative size = 149, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.24, Rules used = {4132, 462, 451, 277, 217, 206} \[ \frac{\sqrt{b} \tanh ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b \tan ^2(e+f x)+b}}\right )}{f}-\frac{\cot ^5(e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}{5 f (a+b)}-\frac{2 (5 a+4 b) \cot ^3(e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}{15 f (a+b)^2}-\frac{\cot (e+f x) \sqrt{a+b \tan ^2(e+f x)+b}}{f} \]

Antiderivative was successfully verified.

[In]

Int[Csc[e + f*x]^6*Sqrt[a + b*Sec[e + f*x]^2],x]

[Out]

(Sqrt[b]*ArcTanh[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b + b*Tan[e + f*x]^2]])/f - (Cot[e + f*x]*Sqrt[a + b + b*Tan[
e + f*x]^2])/f - (2*(5*a + 4*b)*Cot[e + f*x]^3*(a + b + b*Tan[e + f*x]^2)^(3/2))/(15*(a + b)^2*f) - (Cot[e + f
*x]^5*(a + b + b*Tan[e + f*x]^2)^(3/2))/(5*(a + b)*f)

Rule 4132

Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*sin[(e_.) + (f_.)*(x_)]^(m_), x_Symbol] :> With[{ff = Fr
eeFactors[Tan[e + f*x], x]}, Dist[ff^(m + 1)/f, Subst[Int[(x^m*ExpandToSum[a + b*(1 + ff^2*x^2)^(n/2), x]^p)/(
1 + ff^2*x^2)^(m/2 + 1), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[m/2] && Integer
Q[n/2]

Rule 462

Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^2, x_Symbol] :> Simp[(c^2*(e*x)^(
m + 1)*(a + b*x^n)^(p + 1))/(a*e*(m + 1)), x] - Dist[1/(a*e^n*(m + 1)), Int[(e*x)^(m + n)*(a + b*x^n)^p*Simp[b
*c^2*n*(p + 1) + c*(b*c - 2*a*d)*(m + 1) - a*(m + 1)*d^2*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, p}, x] && Ne
Q[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[m, -1] && GtQ[n, 0]

Rule 451

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

Rule 277

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

Rubi steps

\begin{align*} \int \csc ^6(e+f x) \sqrt{a+b \sec ^2(e+f x)} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (1+x^2\right )^2 \sqrt{a+b+b x^2}}{x^6} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac{\cot ^5(e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}{5 (a+b) f}+\frac{\operatorname{Subst}\left (\int \frac{\sqrt{a+b+b x^2} \left (2 (5 a+4 b)+5 (a+b) x^2\right )}{x^4} \, dx,x,\tan (e+f x)\right )}{5 (a+b) f}\\ &=-\frac{2 (5 a+4 b) \cot ^3(e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}{15 (a+b)^2 f}-\frac{\cot ^5(e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}{5 (a+b) f}+\frac{\operatorname{Subst}\left (\int \frac{\sqrt{a+b+b x^2}}{x^2} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac{\cot (e+f x) \sqrt{a+b+b \tan ^2(e+f x)}}{f}-\frac{2 (5 a+4 b) \cot ^3(e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}{15 (a+b)^2 f}-\frac{\cot ^5(e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}{5 (a+b) f}+\frac{b \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b+b x^2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac{\cot (e+f x) \sqrt{a+b+b \tan ^2(e+f x)}}{f}-\frac{2 (5 a+4 b) \cot ^3(e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}{15 (a+b)^2 f}-\frac{\cot ^5(e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}{5 (a+b) f}+\frac{b \operatorname{Subst}\left (\int \frac{1}{1-b x^2} \, dx,x,\frac{\tan (e+f x)}{\sqrt{a+b+b \tan ^2(e+f x)}}\right )}{f}\\ &=\frac{\sqrt{b} \tanh ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b+b \tan ^2(e+f x)}}\right )}{f}-\frac{\cot (e+f x) \sqrt{a+b+b \tan ^2(e+f x)}}{f}-\frac{2 (5 a+4 b) \cot ^3(e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}{15 (a+b)^2 f}-\frac{\cot ^5(e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}{5 (a+b) f}\\ \end{align*}

Mathematica [C]  time = 9.55979, size = 941, normalized size = 6.32 \[ \frac{\sqrt{2} \csc ^5(e+f x) \sec (e+f x) \sqrt{b \sec ^2(e+f x)+a} \left (1-\frac{a \sin ^2(e+f x)}{a+b}\right ) \left (\frac{24 a^2 b^2 \text{Hypergeometric2F1}\left (2,2,\frac{3}{2},-\frac{b \tan ^2(e+f x)}{a+b}\right ) \tan ^2(e+f x) \sin ^6(e+f x)}{(a+b)^3}+\frac{8 a^2 b^2 \text{HypergeometricPFQ}\left (\{2,2,2\},\left \{1,\frac{3}{2}\right \},-\frac{b \tan ^2(e+f x)}{a+b}\right ) \tan ^2(e+f x) \sin ^6(e+f x)}{(a+b)^3}+\frac{24 a^2 b \text{Hypergeometric2F1}\left (2,2,\frac{3}{2},-\frac{b \tan ^2(e+f x)}{a+b}\right ) \sin ^6(e+f x)}{(a+b)^2}+\frac{8 a^2 b \text{HypergeometricPFQ}\left (\{2,2,2\},\left \{1,\frac{3}{2}\right \},-\frac{b \tan ^2(e+f x)}{a+b}\right ) \sin ^6(e+f x)}{(a+b)^2}+\frac{8 a^2 b \sin ^{-1}\left (\sqrt{-\frac{b \tan ^2(e+f x)}{a+b}}\right ) \sin ^6(e+f x)}{(a+b)^2 \sqrt{-\frac{b \sec ^2(e+f x) \left (-a \sin ^2(e+f x)+a+b\right ) \tan ^2(e+f x)}{(a+b)^2}}}-\frac{8 a^2 \cos ^2(e+f x) \sin ^4(e+f x)}{a+b}-\frac{8 a b^2 \text{Hypergeometric2F1}\left (2,2,\frac{3}{2},-\frac{b \tan ^2(e+f x)}{a+b}\right ) \tan ^2(e+f x) \sin ^4(e+f x)}{(a+b)^2}-\frac{16 a b^2 \text{HypergeometricPFQ}\left (\{2,2,2\},\left \{1,\frac{3}{2}\right \},-\frac{b \tan ^2(e+f x)}{a+b}\right ) \tan ^2(e+f x) \sin ^4(e+f x)}{(a+b)^2}-\frac{8 a b \text{Hypergeometric2F1}\left (2,2,\frac{3}{2},-\frac{b \tan ^2(e+f x)}{a+b}\right ) \sin ^4(e+f x)}{a+b}-\frac{16 a b \text{HypergeometricPFQ}\left (\{2,2,2\},\left \{1,\frac{3}{2}\right \},-\frac{b \tan ^2(e+f x)}{a+b}\right ) \sin ^4(e+f x)}{a+b}-4 a \cos ^2(e+f x) \sin ^2(e+f x)-\frac{16 b^2 \text{Hypergeometric2F1}\left (2,2,\frac{3}{2},-\frac{b \tan ^2(e+f x)}{a+b}\right ) \tan ^2(e+f x) \sin ^2(e+f x)}{a+b}+\frac{8 b^2 \text{HypergeometricPFQ}\left (\{2,2,2\},\left \{1,\frac{3}{2}\right \},-\frac{b \tan ^2(e+f x)}{a+b}\right ) \tan ^2(e+f x) \sin ^2(e+f x)}{a+b}-16 b \text{Hypergeometric2F1}\left (2,2,\frac{3}{2},-\frac{b \tan ^2(e+f x)}{a+b}\right ) \sin ^2(e+f x)+8 b \text{HypergeometricPFQ}\left (\{2,2,2\},\left \{1,\frac{3}{2}\right \},-\frac{b \tan ^2(e+f x)}{a+b}\right ) \sin ^2(e+f x)-\frac{4 a \sin ^{-1}\left (\sqrt{-\frac{b \tan ^2(e+f x)}{a+b}}\right ) \cos ^2(e+f x) \sqrt{-\frac{b \tan ^2(e+f x)}{a+b}} \sin ^2(e+f x)}{\sqrt{\frac{b \sec ^2(e+f x)+a}{a+b}}}+\frac{3 b \sin ^{-1}\left (\sqrt{-\frac{b \tan ^2(e+f x)}{a+b}}\right ) \sin ^2(e+f x)}{\sqrt{-\frac{b \sec ^2(e+f x) \left (-a \sin ^2(e+f x)+a+b\right ) \tan ^2(e+f x)}{(a+b)^2}}}-3 (a+b) \cos ^2(e+f x)\right )}{15 f \sqrt{\cos (2 e+2 f x) a+a+2 b} \sqrt{-a \sin ^2(e+f x)+a+b}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Csc[e + f*x]^6*Sqrt[a + b*Sec[e + f*x]^2],x]

[Out]

(Sqrt[2]*Csc[e + f*x]^5*Sec[e + f*x]*Sqrt[a + b*Sec[e + f*x]^2]*(1 - (a*Sin[e + f*x]^2)/(a + b))*(-3*(a + b)*C
os[e + f*x]^2 - 4*a*Cos[e + f*x]^2*Sin[e + f*x]^2 - 16*b*Hypergeometric2F1[2, 2, 3/2, -((b*Tan[e + f*x]^2)/(a
+ b))]*Sin[e + f*x]^2 + 8*b*HypergeometricPFQ[{2, 2, 2}, {1, 3/2}, -((b*Tan[e + f*x]^2)/(a + b))]*Sin[e + f*x]
^2 - (8*a^2*Cos[e + f*x]^2*Sin[e + f*x]^4)/(a + b) - (8*a*b*Hypergeometric2F1[2, 2, 3/2, -((b*Tan[e + f*x]^2)/
(a + b))]*Sin[e + f*x]^4)/(a + b) - (16*a*b*HypergeometricPFQ[{2, 2, 2}, {1, 3/2}, -((b*Tan[e + f*x]^2)/(a + b
))]*Sin[e + f*x]^4)/(a + b) + (24*a^2*b*Hypergeometric2F1[2, 2, 3/2, -((b*Tan[e + f*x]^2)/(a + b))]*Sin[e + f*
x]^6)/(a + b)^2 + (8*a^2*b*HypergeometricPFQ[{2, 2, 2}, {1, 3/2}, -((b*Tan[e + f*x]^2)/(a + b))]*Sin[e + f*x]^
6)/(a + b)^2 - (16*b^2*Hypergeometric2F1[2, 2, 3/2, -((b*Tan[e + f*x]^2)/(a + b))]*Sin[e + f*x]^2*Tan[e + f*x]
^2)/(a + b) + (8*b^2*HypergeometricPFQ[{2, 2, 2}, {1, 3/2}, -((b*Tan[e + f*x]^2)/(a + b))]*Sin[e + f*x]^2*Tan[
e + f*x]^2)/(a + b) - (8*a*b^2*Hypergeometric2F1[2, 2, 3/2, -((b*Tan[e + f*x]^2)/(a + b))]*Sin[e + f*x]^4*Tan[
e + f*x]^2)/(a + b)^2 - (16*a*b^2*HypergeometricPFQ[{2, 2, 2}, {1, 3/2}, -((b*Tan[e + f*x]^2)/(a + b))]*Sin[e
+ f*x]^4*Tan[e + f*x]^2)/(a + b)^2 + (24*a^2*b^2*Hypergeometric2F1[2, 2, 3/2, -((b*Tan[e + f*x]^2)/(a + b))]*S
in[e + f*x]^6*Tan[e + f*x]^2)/(a + b)^3 + (8*a^2*b^2*HypergeometricPFQ[{2, 2, 2}, {1, 3/2}, -((b*Tan[e + f*x]^
2)/(a + b))]*Sin[e + f*x]^6*Tan[e + f*x]^2)/(a + b)^3 - (4*a*ArcSin[Sqrt[-((b*Tan[e + f*x]^2)/(a + b))]]*Cos[e
+ f*x]^2*Sin[e + f*x]^2*Sqrt[-((b*Tan[e + f*x]^2)/(a + b))])/Sqrt[(a + b*Sec[e + f*x]^2)/(a + b)] + (3*b*ArcS
in[Sqrt[-((b*Tan[e + f*x]^2)/(a + b))]]*Sin[e + f*x]^2)/Sqrt[-((b*Sec[e + f*x]^2*(a + b - a*Sin[e + f*x]^2)*Ta
n[e + f*x]^2)/(a + b)^2)] + (8*a^2*b*ArcSin[Sqrt[-((b*Tan[e + f*x]^2)/(a + b))]]*Sin[e + f*x]^6)/((a + b)^2*Sq
rt[-((b*Sec[e + f*x]^2*(a + b - a*Sin[e + f*x]^2)*Tan[e + f*x]^2)/(a + b)^2)])))/(15*f*Sqrt[a + 2*b + a*Cos[2*
e + 2*f*x]]*Sqrt[a + b - a*Sin[e + f*x]^2])

________________________________________________________________________________________

Maple [C]  time = 0.58, size = 8587, normalized size = 57.6 \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)^6*(a+b*sec(f*x+e)^2)^(1/2),x)

[Out]

result too large to display

________________________________________________________________________________________

Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^6*(a+b*sec(f*x+e)^2)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [B]  time = 4.45319, size = 1639, normalized size = 11. \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)^6*(a+b*sec(f*x+e)^2)^(1/2),x, algorithm="fricas")

[Out]

[1/60*(15*((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^2 - 6*a*b + b^2)*cos(f*x + e)^4 + 8*(a*b - b^2)*cos(f*x + e)^2 + 4*((a - b)*cos(f*x + e)^3 + 2*b*c
os(f*x + e))*sqrt(b)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*sin(f*x + e) + 8*b^2)/cos(f*x + e)^4)*sin(f*x
+ e) - 4*((8*a^2 + 25*a*b + 15*b^2)*cos(f*x + e)^5 - (20*a^2 + 59*a*b + 35*b^2)*cos(f*x + e)^3 + (15*a^2 + 40
*a*b + 23*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)*sin(f*x + e)), 1/30*(15*((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(-1/2*((a - b)*c
os(f*x + e)^3 + 2*b*cos(f*x + e))*sqrt(-b)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)/((a*b*cos(f*x + e)^2 +
b^2)*sin(f*x + e)))*sin(f*x + e) - 2*((8*a^2 + 25*a*b + 15*b^2)*cos(f*x + e)^5 - (20*a^2 + 59*a*b + 35*b^2)*co
s(f*x + e)^3 + (15*a^2 + 40*a*b + 23*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)*sin(f*x + e))]

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

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 )^{6}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^6*(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)^6, x)