3.91 \(\int \csc ^4(e+f x) (a+b \sec ^2(e+f x))^{3/2} \, dx\)

Optimal. Leaf size=172 \[ \frac{b (3 a+5 b) \tan (e+f x) \sqrt{a+b \tan ^2(e+f x)+b}}{2 f (a+b)}+\frac{\sqrt{b} (3 a+5 b) \tanh ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b \tan ^2(e+f x)+b}}\right )}{2 f}-\frac{\cot ^3(e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{5/2}}{3 f (a+b)}-\frac{(3 a+5 b) \cot (e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}{3 f (a+b)} \]

[Out]

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

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Rubi [A]  time = 0.152781, antiderivative size = 172, 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, 453, 277, 195, 217, 206} \[ \frac{b (3 a+5 b) \tan (e+f x) \sqrt{a+b \tan ^2(e+f x)+b}}{2 f (a+b)}+\frac{\sqrt{b} (3 a+5 b) \tanh ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b \tan ^2(e+f x)+b}}\right )}{2 f}-\frac{\cot ^3(e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{5/2}}{3 f (a+b)}-\frac{(3 a+5 b) \cot (e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}{3 f (a+b)} \]

Antiderivative was successfully verified.

[In]

Int[Csc[e + f*x]^4*(a + b*Sec[e + f*x]^2)^(3/2),x]

[Out]

(Sqrt[b]*(3*a + 5*b)*ArcTanh[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b + b*Tan[e + f*x]^2]])/(2*f) + (b*(3*a + 5*b)*Ta
n[e + f*x]*Sqrt[a + b + b*Tan[e + f*x]^2])/(2*(a + b)*f) - ((3*a + 5*b)*Cot[e + f*x]*(a + b + b*Tan[e + f*x]^2
)^(3/2))/(3*(a + b)*f) - (Cot[e + f*x]^3*(a + b + b*Tan[e + f*x]^2)^(5/2))/(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 453

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[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*e^n*(m + 1)), In
t[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[n] ||
GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1])) &&  !ILtQ[p, -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 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

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

Mathematica [C]  time = 8.62628, size = 369, normalized size = 2.15 \[ \frac{\sqrt{2} e^{i (e+f x)} \cos ^3(e+f x) \sqrt{4 b+a e^{-2 i (e+f x)} \left (1+e^{2 i (e+f x)}\right )^2} \left (-\frac{i \left (4 a \left (-4 e^{2 i (e+f x)}+e^{4 i (e+f x)}+1\right ) \left (1+e^{2 i (e+f x)}\right )^2+b \left (-20 e^{2 i (e+f x)}-22 e^{4 i (e+f x)}-20 e^{6 i (e+f x)}+15 e^{8 i (e+f x)}+15\right )\right )}{\left (-1+e^{2 i (e+f x)}\right )^3 \left (1+e^{2 i (e+f x)}\right )^2}-\frac{3 \sqrt{b} (3 a+5 b) \log \left (\frac{4 i f \sqrt{a \left (1+e^{2 i (e+f x)}\right )^2+4 b e^{2 i (e+f x)}}-4 \sqrt{b} f \left (-1+e^{2 i (e+f x)}\right )}{1+e^{2 i (e+f x)}}\right )}{\sqrt{a \left (1+e^{2 i (e+f x)}\right )^2+4 b e^{2 i (e+f x)}}}\right ) \left (a+b \sec ^2(e+f x)\right )^{3/2}}{3 f (a \cos (2 e+2 f x)+a+2 b)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[e + f*x]^4*(a + b*Sec[e + f*x]^2)^(3/2),x]

[Out]

(Sqrt[2]*E^(I*(e + f*x))*Sqrt[4*b + (a*(1 + E^((2*I)*(e + f*x)))^2)/E^((2*I)*(e + f*x))]*Cos[e + f*x]^3*(((-I)
*(4*a*(1 + E^((2*I)*(e + f*x)))^2*(1 - 4*E^((2*I)*(e + f*x)) + E^((4*I)*(e + f*x))) + b*(15 - 20*E^((2*I)*(e +
f*x)) - 22*E^((4*I)*(e + f*x)) - 20*E^((6*I)*(e + f*x)) + 15*E^((8*I)*(e + f*x)))))/((-1 + E^((2*I)*(e + f*x)
))^3*(1 + E^((2*I)*(e + f*x)))^2) - (3*Sqrt[b]*(3*a + 5*b)*Log[(-4*Sqrt[b]*(-1 + E^((2*I)*(e + f*x)))*f + (4*I
)*Sqrt[4*b*E^((2*I)*(e + f*x)) + a*(1 + E^((2*I)*(e + f*x)))^2]*f)/(1 + E^((2*I)*(e + f*x)))])/Sqrt[4*b*E^((2*
I)*(e + f*x)) + a*(1 + E^((2*I)*(e + f*x)))^2])*(a + b*Sec[e + f*x]^2)^(3/2))/(3*f*(a + 2*b + a*Cos[2*e + 2*f*
x])^(3/2))

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Maple [C]  time = 0.396, size = 3925, normalized size = 22.8 \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)^(3/2),x)

[Out]

1/6/f/((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*(18*cos(f*x+e)^5*sin(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+30*cos(f*x+e)^5*sin(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^2-9*cos(f*x+e)^5*sin(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*co
s(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-15*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))*cos(f*x+e)^5*sin(f*x+e
)*b^2+18*cos(f*x+e)^4*sin(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+30*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*c
os(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))*cos(f*x+e)^4*sin(f*x+e)*b^2-9*cos(f*x+e)^4*sin(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-15*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))*cos(f*x+e)^4*sin(f*x+e)*b^2-18*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*co
s(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-30*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+9*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+15*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))*b^2-18*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-30*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+9*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+15*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+4*cos(f
*x+e)^6*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*a^2+15*cos(f*x+e)^6*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*a*
b-6*cos(f*x+e)^4*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*a^2-16*cos(f*x+e)^4*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))
^(1/2)*a*b+15*cos(f*x+e)^4*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*b^2-3*cos(f*x+e)^2*((2*I*a^(1/2)*b^(1/2)+a-
b)/(a+b))^(1/2)*a*b-20*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)*b^2)*cos(f*x+e)*((b+a*cos(f*x+e)^2)/cos(f*x+e)^2)^(3/2)/(b+a*cos(f*x+e)^2)^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)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 4.95812, size = 1180, normalized size = 6.86 \begin{align*} \left [\frac{3 \,{\left ({\left (3 \, a + 5 \, b\right )} \cos \left (f x + e\right )^{3} -{\left (3 \, a + 5 \, b\right )} \cos \left (f x + e\right )\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 (4 \, a + 15 \, b\right )} \cos \left (f x + e\right )^{4} - 2 \,{\left (3 \, a + 10 \, b\right )} \cos \left (f x + e\right )^{2} + 3 \, b\right )} \sqrt{\frac{a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{24 \,{\left (f \cos \left (f x + e\right )^{3} - f \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}, \frac{3 \,{\left ({\left (3 \, a + 5 \, b\right )} \cos \left (f x + e\right )^{3} -{\left (3 \, a + 5 \, b\right )} \cos \left (f x + e\right )\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 (4 \, a + 15 \, b\right )} \cos \left (f x + e\right )^{4} - 2 \,{\left (3 \, a + 10 \, b\right )} \cos \left (f x + e\right )^{2} + 3 \, b\right )} \sqrt{\frac{a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{12 \,{\left (f \cos \left (f x + e\right )^{3} - f \cos \left (f x + e\right )\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)^(3/2),x, algorithm="fricas")

[Out]

[1/24*(3*((3*a + 5*b)*cos(f*x + e)^3 - (3*a + 5*b)*cos(f*x + e))*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*((4*a + 15*b)*cos(f*x + e)^4 -
2*(3*a + 10*b)*cos(f*x + e)^2 + 3*b)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/((f*cos(f*x + e)^3 - f*cos(f
*x + e))*sin(f*x + e)), 1/12*(3*((3*a + 5*b)*cos(f*x + e)^3 - (3*a + 5*b)*cos(f*x + e))*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*((4*a + 15*b)*cos(f*x + e)^4 - 2*(3*a + 10*b)*cos(f*x + e)^2 + 3
*b)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/((f*cos(f*x + e)^3 - f*cos(f*x + e))*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)**(3/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \sec \left (f x + e\right )^{2} + a\right )}^{\frac{3}{2}} \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)^(3/2),x, algorithm="giac")

[Out]

integrate((b*sec(f*x + e)^2 + a)^(3/2)*csc(f*x + e)^4, x)