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

Optimal. Leaf size=105 \[ \frac{\sqrt{b} \tanh ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b \tan ^2(e+f x)+b}}\right )}{f}-\frac{\cot ^3(e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}{3 f (a+b)}-\frac{\cot (e+f x) \sqrt{a+b \tan ^2(e+f x)+b}}{f} \]

[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)

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Rubi [A]  time = 0.0971405, antiderivative size = 105, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {4132, 451, 277, 217, 206} \[ \frac{\sqrt{b} \tanh ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b \tan ^2(e+f x)+b}}\right )}{f}-\frac{\cot ^3(e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}{3 f (a+b)}-\frac{\cot (e+f x) \sqrt{a+b \tan ^2(e+f x)+b}}{f} \]

Antiderivative was successfully verified.

[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 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 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 ^4(e+f x) \sqrt{a+b \sec ^2(e+f x)} \, dx &=\frac{\operatorname{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{\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{\cot ^3(e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}{3 (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{\cot ^3(e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}{3 (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{\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]  time = 3.8762, size = 285, normalized size = 2.71 \[ -\frac{\sqrt{2} \cot (e+f x) \csc ^2(e+f x) \left (1-\frac{a \sin ^2(e+f x)}{a+b}\right ) \sqrt{a+b \sec ^2(e+f x)} \left (\frac{4 b \tan ^2(e+f x) \sec ^2(e+f x) \left (-a \sin ^2(e+f x)+a+b\right )^2 \sqrt{\frac{a+b \sec ^2(e+f x)}{a+b}} \text{Hypergeometric2F1}\left (2,2,\frac{3}{2},-\frac{b \tan ^2(e+f x)}{a+b}\right )}{(a+b)^2}+\left (2 a \sin ^2(e+f x)+a+b\right ) \left (\sqrt{\frac{a+b \sec ^2(e+f x)}{a+b}}+\sqrt{-\frac{b \tan ^2(e+f x)}{a+b}} \sin ^{-1}\left (\sqrt{-\frac{b \tan ^2(e+f x)}{a+b}}\right )\right )\right )}{3 f \sqrt{-a \sin ^2(e+f x)+a+b} \sqrt{a \cos (2 e+2 f x)+a+2 b} \sqrt{\frac{a+b \sec ^2(e+f x)}{a+b}}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-(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*Hyperg
eometric2F1[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))])))/(3*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])

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Maple [C]  time = 0.415, size = 3847, normalized size = 36.6 \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)^4*(a+b*sec(f*x+e)^2)^(1/2),x)

[Out]

-1/3/f/((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)/(a+b)*(3*sin(f*x+e)*cos(f*x+e)^3*2^(1/2)*(1/(a+b)*(I*cos(f*x+e)
*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)+a*cos(f*x+e)+b)/(1+cos(f*x+e)))^(1/2)*(-2/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/
2)-I*a^(1/2)*b^(1/2)-a*cos(f*x+e)-b)/(1+cos(f*x+e)))^(1/2)*EllipticF((-1+cos(f*x+e))*((2*I*a^(1/2)*b^(1/2)+a-b
)/(a+b))^(1/2)/sin(f*x+e),(-(4*I*a^(3/2)*b^(1/2)-4*I*a^(1/2)*b^(3/2)-a^2+6*a*b-b^2)/(a+b)^2)^(1/2))*a*b+3*sin(
f*x+e)*cos(f*x+e)^3*2^(1/2)*(1/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)+a*cos(f*x+e)+b)/(1+cos(f*
x+e)))^(1/2)*(-2/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)-a*cos(f*x+e)-b)/(1+cos(f*x+e)))^(1/2)*E
llipticF((-1+cos(f*x+e))*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)/sin(f*x+e),(-(4*I*a^(3/2)*b^(1/2)-4*I*a^(1/2)
*b^(3/2)-a^2+6*a*b-b^2)/(a+b)^2)^(1/2))*b^2-6*sin(f*x+e)*cos(f*x+e)^3*2^(1/2)*(1/(a+b)*(I*cos(f*x+e)*a^(1/2)*b
^(1/2)-I*a^(1/2)*b^(1/2)+a*cos(f*x+e)+b)/(1+cos(f*x+e)))^(1/2)*(-2/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/
2)*b^(1/2)-a*cos(f*x+e)-b)/(1+cos(f*x+e)))^(1/2)*EllipticPi((-1+cos(f*x+e))*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^
(1/2)/sin(f*x+e),1/(2*I*a^(1/2)*b^(1/2)+a-b)*(a+b),(-(2*I*a^(1/2)*b^(1/2)-a+b)/(a+b))^(1/2)/((2*I*a^(1/2)*b^(1
/2)+a-b)/(a+b))^(1/2))*a*b-6*sin(f*x+e)*cos(f*x+e)^3*2^(1/2)*(1/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*
b^(1/2)+a*cos(f*x+e)+b)/(1+cos(f*x+e)))^(1/2)*(-2/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)-a*cos(
f*x+e)-b)/(1+cos(f*x+e)))^(1/2)*EllipticPi((-1+cos(f*x+e))*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)/sin(f*x+e),
1/(2*I*a^(1/2)*b^(1/2)+a-b)*(a+b),(-(2*I*a^(1/2)*b^(1/2)-a+b)/(a+b))^(1/2)/((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(
1/2))*b^2+3*sin(f*x+e)*cos(f*x+e)^2*2^(1/2)*(1/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)+a*cos(f*x
+e)+b)/(1+cos(f*x+e)))^(1/2)*(-2/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)-a*cos(f*x+e)-b)/(1+cos(
f*x+e)))^(1/2)*EllipticF((-1+cos(f*x+e))*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)/sin(f*x+e),(-(4*I*a^(3/2)*b^(
1/2)-4*I*a^(1/2)*b^(3/2)-a^2+6*a*b-b^2)/(a+b)^2)^(1/2))*a*b+3*sin(f*x+e)*cos(f*x+e)^2*2^(1/2)*(1/(a+b)*(I*cos(
f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)+a*cos(f*x+e)+b)/(1+cos(f*x+e)))^(1/2)*(-2/(a+b)*(I*cos(f*x+e)*a^(1/2)
*b^(1/2)-I*a^(1/2)*b^(1/2)-a*cos(f*x+e)-b)/(1+cos(f*x+e)))^(1/2)*EllipticF((-1+cos(f*x+e))*((2*I*a^(1/2)*b^(1/
2)+a-b)/(a+b))^(1/2)/sin(f*x+e),(-(4*I*a^(3/2)*b^(1/2)-4*I*a^(1/2)*b^(3/2)-a^2+6*a*b-b^2)/(a+b)^2)^(1/2))*b^2-
6*sin(f*x+e)*cos(f*x+e)^2*2^(1/2)*(1/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)+a*cos(f*x+e)+b)/(1+
cos(f*x+e)))^(1/2)*(-2/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)-a*cos(f*x+e)-b)/(1+cos(f*x+e)))^(
1/2)*EllipticPi((-1+cos(f*x+e))*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)/sin(f*x+e),1/(2*I*a^(1/2)*b^(1/2)+a-b)
*(a+b),(-(2*I*a^(1/2)*b^(1/2)-a+b)/(a+b))^(1/2)/((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2))*a*b-6*sin(f*x+e)*cos(
f*x+e)^2*2^(1/2)*(1/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)+a*cos(f*x+e)+b)/(1+cos(f*x+e)))^(1/2
)*(-2/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)-a*cos(f*x+e)-b)/(1+cos(f*x+e)))^(1/2)*EllipticPi((
-1+cos(f*x+e))*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)/sin(f*x+e),1/(2*I*a^(1/2)*b^(1/2)+a-b)*(a+b),(-(2*I*a^(
1/2)*b^(1/2)-a+b)/(a+b))^(1/2)/((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2))*b^2-3*sin(f*x+e)*cos(f*x+e)*2^(1/2)*(1
/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)+a*cos(f*x+e)+b)/(1+cos(f*x+e)))^(1/2)*(-2/(a+b)*(I*cos(
f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)-a*cos(f*x+e)-b)/(1+cos(f*x+e)))^(1/2)*EllipticF((-1+cos(f*x+e))*((2*I
*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)/sin(f*x+e),(-(4*I*a^(3/2)*b^(1/2)-4*I*a^(1/2)*b^(3/2)-a^2+6*a*b-b^2)/(a+b)^
2)^(1/2))*a*b-3*sin(f*x+e)*cos(f*x+e)*2^(1/2)*(1/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)+a*cos(f
*x+e)+b)/(1+cos(f*x+e)))^(1/2)*(-2/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)-a*cos(f*x+e)-b)/(1+co
s(f*x+e)))^(1/2)*EllipticF((-1+cos(f*x+e))*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)/sin(f*x+e),(-(4*I*a^(3/2)*b
^(1/2)-4*I*a^(1/2)*b^(3/2)-a^2+6*a*b-b^2)/(a+b)^2)^(1/2))*b^2+6*sin(f*x+e)*cos(f*x+e)*2^(1/2)*(1/(a+b)*(I*cos(
f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)+a*cos(f*x+e)+b)/(1+cos(f*x+e)))^(1/2)*(-2/(a+b)*(I*cos(f*x+e)*a^(1/2)
*b^(1/2)-I*a^(1/2)*b^(1/2)-a*cos(f*x+e)-b)/(1+cos(f*x+e)))^(1/2)*EllipticPi((-1+cos(f*x+e))*((2*I*a^(1/2)*b^(1
/2)+a-b)/(a+b))^(1/2)/sin(f*x+e),1/(2*I*a^(1/2)*b^(1/2)+a-b)*(a+b),(-(2*I*a^(1/2)*b^(1/2)-a+b)/(a+b))^(1/2)/((
2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2))*a*b+6*sin(f*x+e)*cos(f*x+e)*2^(1/2)*(1/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1
/2)-I*a^(1/2)*b^(1/2)+a*cos(f*x+e)+b)/(1+cos(f*x+e)))^(1/2)*(-2/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*
b^(1/2)-a*cos(f*x+e)-b)/(1+cos(f*x+e)))^(1/2)*EllipticPi((-1+cos(f*x+e))*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/
2)/sin(f*x+e),1/(2*I*a^(1/2)*b^(1/2)+a-b)*(a+b),(-(2*I*a^(1/2)*b^(1/2)-a+b)/(a+b))^(1/2)/((2*I*a^(1/2)*b^(1/2)
+a-b)/(a+b))^(1/2))*b^2-3*2^(1/2)*(1/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)+a*cos(f*x+e)+b)/(1+
cos(f*x+e)))^(1/2)*(-2/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)-a*cos(f*x+e)-b)/(1+cos(f*x+e)))^(
1/2)*EllipticF((-1+cos(f*x+e))*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)/sin(f*x+e),(-(4*I*a^(3/2)*b^(1/2)-4*I*a
^(1/2)*b^(3/2)-a^2+6*a*b-b^2)/(a+b)^2)^(1/2))*a*b*sin(f*x+e)-3*2^(1/2)*(1/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-
I*a^(1/2)*b^(1/2)+a*cos(f*x+e)+b)/(1+cos(f*x+e)))^(1/2)*(-2/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1
/2)-a*cos(f*x+e)-b)/(1+cos(f*x+e)))^(1/2)*EllipticF((-1+cos(f*x+e))*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)/si
n(f*x+e),(-(4*I*a^(3/2)*b^(1/2)-4*I*a^(1/2)*b^(3/2)-a^2+6*a*b-b^2)/(a+b)^2)^(1/2))*b^2*sin(f*x+e)+6*2^(1/2)*(1
/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)+a*cos(f*x+e)+b)/(1+cos(f*x+e)))^(1/2)*(-2/(a+b)*(I*cos(
f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)-a*cos(f*x+e)-b)/(1+cos(f*x+e)))^(1/2)*EllipticPi((-1+cos(f*x+e))*((2*
I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)/sin(f*x+e),1/(2*I*a^(1/2)*b^(1/2)+a-b)*(a+b),(-(2*I*a^(1/2)*b^(1/2)-a+b)/(
a+b))^(1/2)/((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2))*a*b*sin(f*x+e)+6*2^(1/2)*(1/(a+b)*(I*cos(f*x+e)*a^(1/2)*b
^(1/2)-I*a^(1/2)*b^(1/2)+a*cos(f*x+e)+b)/(1+cos(f*x+e)))^(1/2)*(-2/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/
2)*b^(1/2)-a*cos(f*x+e)-b)/(1+cos(f*x+e)))^(1/2)*EllipticPi((-1+cos(f*x+e))*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^
(1/2)/sin(f*x+e),1/(2*I*a^(1/2)*b^(1/2)+a-b)*(a+b),(-(2*I*a^(1/2)*b^(1/2)-a+b)/(a+b))^(1/2)/((2*I*a^(1/2)*b^(1
/2)+a-b)/(a+b))^(1/2))*b^2*sin(f*x+e)-2*cos(f*x+e)^4*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*a^2-3*cos(f*x+e)^
4*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*a*b+3*cos(f*x+e)^2*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*a^2+2*cos
(f*x+e)^2*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*a*b-3*cos(f*x+e)^2*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*b
^2+3*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*a*b+4*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*b^2)*cos(f*x+e)*((b
+a*cos(f*x+e)^2)/cos(f*x+e)^2)^(1/2)/(b+a*cos(f*x+e)^2)/sin(f*x+e)^3

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.27468, size = 1092, normalized size = 10.4 \begin{align*} \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 ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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))]

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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)**4*(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 )^{4}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)