3.19 \(\int \csc ^3(e+f x) (a+b \sec ^2(e+f x))^2 \, dx\)

Optimal. Leaf size=104 \[ -\frac {\left (3 a^2+6 a b+5 b^2\right ) \cot (e+f x) \csc (e+f x)}{6 f}+\frac {b (6 a+5 b) \sec (e+f x)}{3 f}-\frac {(a+b) (a+5 b) \tanh ^{-1}(\cos (e+f x))}{2 f}+\frac {b^2 \csc ^2(e+f x) \sec ^3(e+f x)}{3 f} \]

[Out]

-1/2*(a+b)*(a+5*b)*arctanh(cos(f*x+e))/f-1/6*(3*a^2+6*a*b+5*b^2)*cot(f*x+e)*csc(f*x+e)/f+1/3*b*(6*a+5*b)*sec(f
*x+e)/f+1/3*b^2*csc(f*x+e)^2*sec(f*x+e)^3/f

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Rubi [A]  time = 0.11, antiderivative size = 104, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {4133, 462, 456, 453, 206} \[ -\frac {\left (3 a^2+6 a b+5 b^2\right ) \cot (e+f x) \csc (e+f x)}{6 f}+\frac {b (6 a+5 b) \sec (e+f x)}{3 f}-\frac {(a+b) (a+5 b) \tanh ^{-1}(\cos (e+f x))}{2 f}+\frac {b^2 \csc ^2(e+f x) \sec ^3(e+f x)}{3 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a + b)*(a + 5*b)*ArcTanh[Cos[e + f*x]])/(2*f) - ((3*a^2 + 6*a*b + 5*b^2)*Cot[e + f*x]*Csc[e + f*x])/(6*f) +
(b*(6*a + 5*b)*Sec[e + f*x])/(3*f) + (b^2*Csc[e + f*x]^2*Sec[e + f*x]^3)/(3*f)

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 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 456

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[((-a)^(m/2 - 1)*(b*c - a*d)*
x*(a + b*x^2)^(p + 1))/(2*b^(m/2 + 1)*(p + 1)), x] + Dist[1/(2*b^(m/2 + 1)*(p + 1)), Int[x^m*(a + b*x^2)^(p +
1)*ExpandToSum[2*b*(p + 1)*Together[(b^(m/2)*(c + d*x^2) - (-a)^(m/2 - 1)*(b*c - a*d)*x^(-m + 2))/(a + b*x^2)]
- ((-a)^(m/2 - 1)*(b*c - a*d))/x^m, x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] &
& ILtQ[m/2, 0] && (IntegerQ[p] || EqQ[m + 2*p + 1, 0])

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 4133

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

Rubi steps

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

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Mathematica [B]  time = 6.59, size = 1021, normalized size = 9.82 \[ \frac {\left (-a^2-2 b a-b^2\right ) \csc ^2\left (\frac {e}{2}+\frac {f x}{2}\right ) \left (b \sec ^2(e+f x)+a\right )^2 \cos ^4(e+f x)}{2 f (\cos (2 e+2 f x) a+a+2 b)^2}+\frac {\left (a^2+2 b a+b^2\right ) \sec ^2\left (\frac {e}{2}+\frac {f x}{2}\right ) \left (b \sec ^2(e+f x)+a\right )^2 \cos ^4(e+f x)}{2 f (\cos (2 e+2 f x) a+a+2 b)^2}-\frac {2 \left (a^2+6 b a+5 b^2\right ) \log \left (\cos \left (\frac {e}{2}+\frac {f x}{2}\right )\right ) \left (b \sec ^2(e+f x)+a\right )^2 \cos ^4(e+f x)}{f (\cos (2 e+2 f x) a+a+2 b)^2}+\frac {2 \left (a^2+6 b a+5 b^2\right ) \log \left (\sin \left (\frac {e}{2}+\frac {f x}{2}\right )\right ) \left (b \sec ^2(e+f x)+a\right )^2 \cos ^4(e+f x)}{f (\cos (2 e+2 f x) a+a+2 b)^2}+\frac {2 b (12 a+13 b) \sec (e) \left (b \sec ^2(e+f x)+a\right )^2 \cos ^4(e+f x)}{3 f (\cos (2 e+2 f x) a+a+2 b)^2}+\frac {2 \left (b \sec ^2(e+f x)+a\right )^2 \left (13 \sin \left (\frac {f x}{2}\right ) b^2+12 a \sin \left (\frac {f x}{2}\right ) b\right ) \cos ^4(e+f x)}{3 f (\cos (2 e+2 f x) a+a+2 b)^2 \left (\cos \left (\frac {e}{2}\right )-\sin \left (\frac {e}{2}\right )\right ) \left (\cos \left (\frac {e}{2}+\frac {f x}{2}\right )-\sin \left (\frac {e}{2}+\frac {f x}{2}\right )\right )}-\frac {2 \left (b \sec ^2(e+f x)+a\right )^2 \left (13 \sin \left (\frac {f x}{2}\right ) b^2+12 a \sin \left (\frac {f x}{2}\right ) b\right ) \cos ^4(e+f x)}{3 f (\cos (2 e+2 f x) a+a+2 b)^2 \left (\cos \left (\frac {e}{2}\right )+\sin \left (\frac {e}{2}\right )\right ) \left (\cos \left (\frac {e}{2}+\frac {f x}{2}\right )+\sin \left (\frac {e}{2}+\frac {f x}{2}\right )\right )}+\frac {\left (b \sec ^2(e+f x)+a\right )^2 \left (\cos \left (\frac {e}{2}\right ) b^2+\sin \left (\frac {e}{2}\right ) b^2\right ) \cos ^4(e+f x)}{3 f (\cos (2 e+2 f x) a+a+2 b)^2 \left (\cos \left (\frac {e}{2}\right )-\sin \left (\frac {e}{2}\right )\right ) \left (\cos \left (\frac {e}{2}+\frac {f x}{2}\right )-\sin \left (\frac {e}{2}+\frac {f x}{2}\right )\right )^2}+\frac {\left (b \sec ^2(e+f x)+a\right )^2 \left (b^2 \cos \left (\frac {e}{2}\right )-b^2 \sin \left (\frac {e}{2}\right )\right ) \cos ^4(e+f x)}{3 f (\cos (2 e+2 f x) a+a+2 b)^2 \left (\cos \left (\frac {e}{2}\right )+\sin \left (\frac {e}{2}\right )\right ) \left (\cos \left (\frac {e}{2}+\frac {f x}{2}\right )+\sin \left (\frac {e}{2}+\frac {f x}{2}\right )\right )^2}+\frac {2 b^2 \left (b \sec ^2(e+f x)+a\right )^2 \sin \left (\frac {f x}{2}\right ) \cos ^4(e+f x)}{3 f (\cos (2 e+2 f x) a+a+2 b)^2 \left (\cos \left (\frac {e}{2}\right )-\sin \left (\frac {e}{2}\right )\right ) \left (\cos \left (\frac {e}{2}+\frac {f x}{2}\right )-\sin \left (\frac {e}{2}+\frac {f x}{2}\right )\right )^3}-\frac {2 b^2 \left (b \sec ^2(e+f x)+a\right )^2 \sin \left (\frac {f x}{2}\right ) \cos ^4(e+f x)}{3 f (\cos (2 e+2 f x) a+a+2 b)^2 \left (\cos \left (\frac {e}{2}\right )+\sin \left (\frac {e}{2}\right )\right ) \left (\cos \left (\frac {e}{2}+\frac {f x}{2}\right )+\sin \left (\frac {e}{2}+\frac {f x}{2}\right )\right )^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-a^2 - 2*a*b - b^2)*Cos[e + f*x]^4*Csc[e/2 + (f*x)/2]^2*(a + b*Sec[e + f*x]^2)^2)/(2*f*(a + 2*b + a*Cos[2*e
+ 2*f*x])^2) - (2*(a^2 + 6*a*b + 5*b^2)*Cos[e + f*x]^4*Log[Cos[e/2 + (f*x)/2]]*(a + b*Sec[e + f*x]^2)^2)/(f*(a
+ 2*b + a*Cos[2*e + 2*f*x])^2) + (2*(a^2 + 6*a*b + 5*b^2)*Cos[e + f*x]^4*Log[Sin[e/2 + (f*x)/2]]*(a + b*Sec[e
+ f*x]^2)^2)/(f*(a + 2*b + a*Cos[2*e + 2*f*x])^2) + (2*b*(12*a + 13*b)*Cos[e + f*x]^4*Sec[e]*(a + b*Sec[e + f
*x]^2)^2)/(3*f*(a + 2*b + a*Cos[2*e + 2*f*x])^2) + ((a^2 + 2*a*b + b^2)*Cos[e + f*x]^4*Sec[e/2 + (f*x)/2]^2*(a
+ b*Sec[e + f*x]^2)^2)/(2*f*(a + 2*b + a*Cos[2*e + 2*f*x])^2) + (2*b^2*Cos[e + f*x]^4*(a + b*Sec[e + f*x]^2)^
2*Sin[(f*x)/2])/(3*f*(a + 2*b + a*Cos[2*e + 2*f*x])^2*(Cos[e/2] - Sin[e/2])*(Cos[e/2 + (f*x)/2] - Sin[e/2 + (f
*x)/2])^3) + (Cos[e + f*x]^4*(a + b*Sec[e + f*x]^2)^2*(b^2*Cos[e/2] + b^2*Sin[e/2]))/(3*f*(a + 2*b + a*Cos[2*e
+ 2*f*x])^2*(Cos[e/2] - Sin[e/2])*(Cos[e/2 + (f*x)/2] - Sin[e/2 + (f*x)/2])^2) + (2*Cos[e + f*x]^4*(a + b*Sec
[e + f*x]^2)^2*(12*a*b*Sin[(f*x)/2] + 13*b^2*Sin[(f*x)/2]))/(3*f*(a + 2*b + a*Cos[2*e + 2*f*x])^2*(Cos[e/2] -
Sin[e/2])*(Cos[e/2 + (f*x)/2] - Sin[e/2 + (f*x)/2])) - (2*b^2*Cos[e + f*x]^4*(a + b*Sec[e + f*x]^2)^2*Sin[(f*x
)/2])/(3*f*(a + 2*b + a*Cos[2*e + 2*f*x])^2*(Cos[e/2] + Sin[e/2])*(Cos[e/2 + (f*x)/2] + Sin[e/2 + (f*x)/2])^3)
+ (Cos[e + f*x]^4*(a + b*Sec[e + f*x]^2)^2*(b^2*Cos[e/2] - b^2*Sin[e/2]))/(3*f*(a + 2*b + a*Cos[2*e + 2*f*x])
^2*(Cos[e/2] + Sin[e/2])*(Cos[e/2 + (f*x)/2] + Sin[e/2 + (f*x)/2])^2) - (2*Cos[e + f*x]^4*(a + b*Sec[e + f*x]^
2)^2*(12*a*b*Sin[(f*x)/2] + 13*b^2*Sin[(f*x)/2]))/(3*f*(a + 2*b + a*Cos[2*e + 2*f*x])^2*(Cos[e/2] + Sin[e/2])*
(Cos[e/2 + (f*x)/2] + Sin[e/2 + (f*x)/2]))

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fricas [B]  time = 0.61, size = 193, normalized size = 1.86 \[ \frac {6 \, {\left (a^{2} + 6 \, a b + 5 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - 4 \, {\left (6 \, a b + 5 \, b^{2}\right )} \cos \left (f x + e\right )^{2} - 4 \, b^{2} - 3 \, {\left ({\left (a^{2} + 6 \, a b + 5 \, b^{2}\right )} \cos \left (f x + e\right )^{5} - {\left (a^{2} + 6 \, a b + 5 \, b^{2}\right )} \cos \left (f x + e\right )^{3}\right )} \log \left (\frac {1}{2} \, \cos \left (f x + e\right ) + \frac {1}{2}\right ) + 3 \, {\left ({\left (a^{2} + 6 \, a b + 5 \, b^{2}\right )} \cos \left (f x + e\right )^{5} - {\left (a^{2} + 6 \, a b + 5 \, b^{2}\right )} \cos \left (f x + e\right )^{3}\right )} \log \left (-\frac {1}{2} \, \cos \left (f x + e\right ) + \frac {1}{2}\right )}{12 \, {\left (f \cos \left (f x + e\right )^{5} - f \cos \left (f x + e\right )^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^3*(a+b*sec(f*x+e)^2)^2,x, algorithm="fricas")

[Out]

1/12*(6*(a^2 + 6*a*b + 5*b^2)*cos(f*x + e)^4 - 4*(6*a*b + 5*b^2)*cos(f*x + e)^2 - 4*b^2 - 3*((a^2 + 6*a*b + 5*
b^2)*cos(f*x + e)^5 - (a^2 + 6*a*b + 5*b^2)*cos(f*x + e)^3)*log(1/2*cos(f*x + e) + 1/2) + 3*((a^2 + 6*a*b + 5*
b^2)*cos(f*x + e)^5 - (a^2 + 6*a*b + 5*b^2)*cos(f*x + e)^3)*log(-1/2*cos(f*x + e) + 1/2))/(f*cos(f*x + e)^5 -
f*cos(f*x + e)^3)

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: NotImplementedError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^3*(a+b*sec(f*x+e)^2)^2,x, algorithm="giac")

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: Unable to check sign: (2*pi/x/2)>(-2*pi/
x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)2/f*((-10*(1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1)))*b^2-12*(1-co
s(f*x+exp(1)))/(1+cos(f*x+exp(1)))*b*a-2*(1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1)))*a^2-b^2-2*b*a-a^2)*1/16/(1-co
s(f*x+exp(1)))*(1+cos(f*x+exp(1)))+(-9*((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1))))^2*b^2-6*((1-cos(f*x+exp(1)))/
(1+cos(f*x+exp(1))))^2*b*a+12*(1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1)))*b^2+12*(1-cos(f*x+exp(1)))/(1+cos(f*x+ex
p(1)))*b*a-7*b^2-6*b*a)*1/3/((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1)))-1)^3+((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(
1)))*b^2+2*(1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1)))*b*a+(1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1)))*a^2)/16+(5*b^2+
6*b*a+a^2)/8*ln(abs(1-cos(f*x+exp(1)))/abs(1+cos(f*x+exp(1)))))

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maple [B]  time = 1.38, size = 195, normalized size = 1.88 \[ -\frac {a^{2} \csc \left (f x +e \right ) \cot \left (f x +e \right )}{2 f}+\frac {a^{2} \ln \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right )}{2 f}-\frac {a b}{f \sin \left (f x +e \right )^{2} \cos \left (f x +e \right )}+\frac {3 a b}{f \cos \left (f x +e \right )}+\frac {3 a b \ln \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right )}{f}+\frac {b^{2}}{3 f \sin \left (f x +e \right )^{2} \cos \left (f x +e \right )^{3}}-\frac {5 b^{2}}{6 f \sin \left (f x +e \right )^{2} \cos \left (f x +e \right )}+\frac {5 b^{2}}{2 f \cos \left (f x +e \right )}+\frac {5 b^{2} \ln \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right )}{2 f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(f*x+e)^3*(a+b*sec(f*x+e)^2)^2,x)

[Out]

-1/2/f*a^2*csc(f*x+e)*cot(f*x+e)+1/2/f*a^2*ln(csc(f*x+e)-cot(f*x+e))-1/f*a*b/sin(f*x+e)^2/cos(f*x+e)+3/f*a*b/c
os(f*x+e)+3/f*a*b*ln(csc(f*x+e)-cot(f*x+e))+1/3/f*b^2/sin(f*x+e)^2/cos(f*x+e)^3-5/6/f*b^2/sin(f*x+e)^2/cos(f*x
+e)+5/2/f*b^2/cos(f*x+e)+5/2/f*b^2*ln(csc(f*x+e)-cot(f*x+e))

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maxima [A]  time = 0.33, size = 126, normalized size = 1.21 \[ -\frac {3 \, {\left (a^{2} + 6 \, a b + 5 \, b^{2}\right )} \log \left (\cos \left (f x + e\right ) + 1\right ) - 3 \, {\left (a^{2} + 6 \, a b + 5 \, b^{2}\right )} \log \left (\cos \left (f x + e\right ) - 1\right ) - \frac {2 \, {\left (3 \, {\left (a^{2} + 6 \, a b + 5 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (6 \, a b + 5 \, b^{2}\right )} \cos \left (f x + e\right )^{2} - 2 \, b^{2}\right )}}{\cos \left (f x + e\right )^{5} - \cos \left (f x + e\right )^{3}}}{12 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^3*(a+b*sec(f*x+e)^2)^2,x, algorithm="maxima")

[Out]

-1/12*(3*(a^2 + 6*a*b + 5*b^2)*log(cos(f*x + e) + 1) - 3*(a^2 + 6*a*b + 5*b^2)*log(cos(f*x + e) - 1) - 2*(3*(a
^2 + 6*a*b + 5*b^2)*cos(f*x + e)^4 - 2*(6*a*b + 5*b^2)*cos(f*x + e)^2 - 2*b^2)/(cos(f*x + e)^5 - cos(f*x + e)^
3))/f

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mupad [B]  time = 4.29, size = 96, normalized size = 0.92 \[ \frac {\frac {b^2}{3}+{\cos \left (e+f\,x\right )}^2\,\left (\frac {5\,b^2}{3}+2\,a\,b\right )-{\cos \left (e+f\,x\right )}^4\,\left (\frac {a^2}{2}+3\,a\,b+\frac {5\,b^2}{2}\right )}{f\,\left ({\cos \left (e+f\,x\right )}^3-{\cos \left (e+f\,x\right )}^5\right )}-\frac {\mathrm {atanh}\left (\cos \left (e+f\,x\right )\right )\,\left (a+b\right )\,\left (a+5\,b\right )}{2\,f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b/cos(e + f*x)^2)^2/sin(e + f*x)^3,x)

[Out]

(b^2/3 + cos(e + f*x)^2*(2*a*b + (5*b^2)/3) - cos(e + f*x)^4*(3*a*b + a^2/2 + (5*b^2)/2))/(f*(cos(e + f*x)^3 -
cos(e + f*x)^5)) - (atanh(cos(e + f*x))*(a + b)*(a + 5*b))/(2*f)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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