\(\int \csc ^4(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx\) [78]

Optimal result

Integrand size = 25, antiderivative size = 105 \[ \int \csc ^4(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx=\frac {\sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{f}-\frac {\cot (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{f}-\frac {\cot ^3(e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}{3 (a+b) f} \]

[Out]

arctanh(b^(1/2)*tan(f*x+e)/(a+b+b*tan(f*x+e)^2)^(1/2))*b^(1/2)/f-cot(f*x+e)*(a+b+b*tan(f*x+e)^2)^(1/2)/f-1/3*c
ot(f*x+e)^3*(a+b+b*tan(f*x+e)^2)^(3/2)/(a+b)/f

Rubi [A] (verified)

Time = 0.13 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {4217, 462, 283, 223, 212} \[ \int \csc ^4(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx=\frac {\sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)+b}}\right )}{f}-\frac {\cot ^3(e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}{3 f (a+b)}-\frac {\cot (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{f} \]

[In]

Int[Csc[e + f*x]^4*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 - (Cot[e + f*x]^3*(a + b + b*Tan[e + f*x]^2)^(3/2))/(3*(a + b)*f)

Rule 212

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

Rule 223

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 283

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 462

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 4217

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]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {\left (1+x^2\right ) \sqrt {a+b+b x^2}}{x^4} \, dx,x,\tan (e+f x)\right )}{f} \\ & = -\frac {\cot ^3(e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}{3 (a+b) f}+\frac {\text {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 {\cot ^3(e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}{3 (a+b) f}+\frac {b \text {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 {\cot ^3(e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}{3 (a+b) f}+\frac {b \text {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} \text {arctanh}\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 {\cot ^3(e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}{3 (a+b) f} \\ \end{align*}

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 2.76 (sec) , antiderivative size = 285, normalized size of antiderivative = 2.71 \[ \int \csc ^4(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx=-\frac {\sqrt {2} \cot (e+f x) \csc ^2(e+f x) \sqrt {a+b \sec ^2(e+f x)} \left (1-\frac {a \sin ^2(e+f x)}{a+b}\right ) \left (\frac {4 b \operatorname {Hypergeometric2F1}\left (2,2,\frac {3}{2},-\frac {b \tan ^2(e+f x)}{a+b}\right ) \sec ^2(e+f x) \sqrt {\frac {a+b \sec ^2(e+f x)}{a+b}} \left (a+b-a \sin ^2(e+f x)\right )^2 \tan ^2(e+f x)}{(a+b)^2}+\left (a+b+2 a \sin ^2(e+f x)\right ) \left (\sqrt {\frac {a+b \sec ^2(e+f x)}{a+b}}+\arcsin \left (\sqrt {-\frac {b \tan ^2(e+f x)}{a+b}}\right ) \sqrt {-\frac {b \tan ^2(e+f x)}{a+b}}\right )\right )}{3 f \sqrt {a+2 b+a \cos (2 e+2 f x)} \sqrt {\frac {a+b \sec ^2(e+f x)}{a+b}} \sqrt {a+b-a \sin ^2(e+f x)}} \]

[In]

Integrate[Csc[e + f*x]^4*Sqrt[a + b*Sec[e + f*x]^2],x]

[Out]

-1/3*(Sqrt[2]*Cot[e + f*x]*Csc[e + f*x]^2*Sqrt[a + b*Sec[e + f*x]^2]*(1 - (a*Sin[e + f*x]^2)/(a + b))*((4*b*Hy
pergeometric2F1[2, 2, 3/2, -((b*Tan[e + f*x]^2)/(a + b))]*Sec[e + f*x]^2*Sqrt[(a + b*Sec[e + f*x]^2)/(a + b)]*
(a + b - a*Sin[e + f*x]^2)^2*Tan[e + f*x]^2)/(a + b)^2 + (a + b + 2*a*Sin[e + f*x]^2)*(Sqrt[(a + b*Sec[e + f*x
]^2)/(a + b)] + ArcSin[Sqrt[-((b*Tan[e + f*x]^2)/(a + b))]]*Sqrt[-((b*Tan[e + f*x]^2)/(a + b))])))/(f*Sqrt[a +
2*b + a*Cos[2*e + 2*f*x]]*Sqrt[(a + b*Sec[e + f*x]^2)/(a + b)]*Sqrt[a + b - a*Sin[e + f*x]^2])

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1036\) vs. \(2(93)=186\).

Time = 7.32 (sec) , antiderivative size = 1037, normalized size of antiderivative = 9.88

method	result	size
default	\(\text {Expression too large to display}\)	\(1037\)

[In]

int(csc(f*x+e)^4*(a+b*sec(f*x+e)^2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

-1/6/f/(a+b)*(3*sin(f*x+e)*b^(3/2)*ln(4*(((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^(1/2)*cos(f*x+e)+b^(1/2
)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-sin(f*x+e)*a-a-b)/(sin(f*x+e)+1))*cos(f*x+e)+3*sin(f*x+e)*b^(3/2
)*ln(-4*(((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^(1/2)*cos(f*x+e)+b^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x
+e))^2)^(1/2)-sin(f*x+e)*a+a+b)/(sin(f*x+e)-1))*cos(f*x+e)-3*sin(f*x+e)*b^(3/2)*ln(4*(((b+a*cos(f*x+e)^2)/(1+c
os(f*x+e))^2)^(1/2)*b^(1/2)*cos(f*x+e)+b^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-sin(f*x+e)*a-a-b)/(
sin(f*x+e)+1))-3*sin(f*x+e)*b^(3/2)*ln(-4*(((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^(1/2)*cos(f*x+e)+b^(1
/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-sin(f*x+e)*a+a+b)/(sin(f*x+e)-1))+3*sin(f*x+e)*b^(1/2)*a*ln(4*
(((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^(1/2)*cos(f*x+e)+b^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^
(1/2)-sin(f*x+e)*a-a-b)/(sin(f*x+e)+1))*cos(f*x+e)+3*sin(f*x+e)*b^(1/2)*a*ln(-4*(((b+a*cos(f*x+e)^2)/(1+cos(f*
x+e))^2)^(1/2)*b^(1/2)*cos(f*x+e)+b^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-sin(f*x+e)*a+a+b)/(sin(f
*x+e)-1))*cos(f*x+e)-3*sin(f*x+e)*b^(1/2)*a*ln(4*(((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^(1/2)*cos(f*x+
e)+b^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-sin(f*x+e)*a-a-b)/(sin(f*x+e)+1))-3*sin(f*x+e)*b^(1/2)*
a*ln(-4*(((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^(1/2)*cos(f*x+e)+b^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x
+e))^2)^(1/2)-sin(f*x+e)*a+a+b)/(sin(f*x+e)-1))-4*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e)^2*a-6
*cos(f*x+e)^2*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b+6*a*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)+8*
((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b)*(a+b*sec(f*x+e)^2)^(1/2)/((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^
(1/2)*cot(f*x+e)*csc(f*x+e)^2

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 202 vs. \(2 (93) = 186\).

Time = 0.47 (sec) , antiderivative size = 436, normalized size of antiderivative = 4.15 \[ \int \csc ^4(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx=\left [\frac {3 \, {\left ({\left (a + b\right )} \cos \left (f x + e\right )^{2} - a - b\right )} \sqrt {b} \log \left (\frac {{\left (a^{2} - 6 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{4} + 8 \, {\left (a b - b^{2}\right )} \cos \left (f x + e\right )^{2} + 4 \, {\left ({\left (a - b\right )} \cos \left (f x + e\right )^{3} + 2 \, b \cos \left (f x + e\right )\right )} \sqrt {b} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \sin \left (f x + e\right ) + 8 \, b^{2}}{\cos \left (f x + e\right )^{4}}\right ) \sin \left (f x + e\right ) - 4 \, {\left ({\left (2 \, a + 3 \, b\right )} \cos \left (f x + e\right )^{3} - {\left (3 \, a + 4 \, b\right )} \cos \left (f x + e\right )\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{12 \, {\left ({\left (a + b\right )} f \cos \left (f x + e\right )^{2} - {\left (a + b\right )} f\right )} \sin \left (f x + e\right )}, \frac {3 \, {\left ({\left (a + b\right )} \cos \left (f x + e\right )^{2} - a - b\right )} \sqrt {-b} \arctan \left (-\frac {{\left ({\left (a - b\right )} \cos \left (f x + e\right )^{3} + 2 \, b \cos \left (f x + e\right )\right )} \sqrt {-b} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{2 \, {\left (a b \cos \left (f x + e\right )^{2} + b^{2}\right )} \sin \left (f x + e\right )}\right ) \sin \left (f x + e\right ) - 2 \, {\left ({\left (2 \, a + 3 \, b\right )} \cos \left (f x + e\right )^{3} - {\left (3 \, a + 4 \, b\right )} \cos \left (f x + e\right )\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{6 \, {\left ({\left (a + b\right )} f \cos \left (f x + e\right )^{2} - {\left (a + b\right )} f\right )} \sin \left (f x + e\right )}\right ] \]

[In]

integrate(csc(f*x+e)^4*(a+b*sec(f*x+e)^2)^(1/2),x, algorithm="fricas")

[Out]

[1/12*(3*((a + b)*cos(f*x + e)^2 - a - b)*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*cos(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*((2*a + 3*b)*cos(f*x + e)^3 - (3*a + 4*b)*cos(f*x + e)
)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/(((a + b)*f*cos(f*x + e)^2 - (a + b)*f)*sin(f*x + e)), 1/6*(3*(
(a + b)*cos(f*x + e)^2 - a - b)*sqrt(-b)*arctan(-1/2*((a - b)*cos(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*((2*a + 3*
b)*cos(f*x + e)^3 - (3*a + 4*b)*cos(f*x + e))*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/(((a + b)*f*cos(f*x
+ e)^2 - (a + b)*f)*sin(f*x + e))]

Sympy [F]

\[ \int \csc ^4(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx=\int \sqrt {a + b \sec ^{2}{\left (e + f x \right )}} \csc ^{4}{\left (e + f x \right )}\, dx \]

[In]

integrate(csc(f*x+e)**4*(a+b*sec(f*x+e)**2)**(1/2),x)

[Out]

Integral(sqrt(a + b*sec(e + f*x)**2)*csc(e + f*x)**4, x)

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.78 \[ \int \csc ^4(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx=\frac {3 \, \sqrt {b} \operatorname {arsinh}\left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right ) - \frac {3 \, \sqrt {b \tan \left (f x + e\right )^{2} + a + b}}{\tan \left (f x + e\right )} - \frac {{\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{\frac {3}{2}}}{{\left (a + b\right )} \tan \left (f x + e\right )^{3}}}{3 \, f} \]

[In]

integrate(csc(f*x+e)^4*(a+b*sec(f*x+e)^2)^(1/2),x, algorithm="maxima")

[Out]

1/3*(3*sqrt(b)*arcsinh(b*tan(f*x + e)/sqrt((a + b)*b)) - 3*sqrt(b*tan(f*x + e)^2 + a + b)/tan(f*x + e) - (b*ta
n(f*x + e)^2 + a + b)^(3/2)/((a + b)*tan(f*x + e)^3))/f

Giac [F]

\[ \int \csc ^4(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx=\int { \sqrt {b \sec \left (f x + e\right )^{2} + a} \csc \left (f x + e\right )^{4} \,d x } \]

[In]

integrate(csc(f*x+e)^4*(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)^4, x)

Mupad [F(-1)]

Timed out. \[ \int \csc ^4(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx=\int \frac {\sqrt {a+\frac {b}{{\cos \left (e+f\,x\right )}^2}}}{{\sin \left (e+f\,x\right )}^4} \,d x \]

[In]

int((a + b/cos(e + f*x)^2)^(1/2)/sin(e + f*x)^4,x)

[Out]

int((a + b/cos(e + f*x)^2)^(1/2)/sin(e + f*x)^4, x)