3.73 \(\int \frac {\csc ^5(e+f x)}{(a+b \tan ^2(e+f x))^2} \, dx\)

Optimal. Leaf size=210 \[ -\frac {3 \sqrt {b} (a-2 b) \sqrt {a-b} \tan ^{-1}\left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a-b}}\right )}{2 a^4 f}-\frac {3 b (3 a-4 b) \sec (e+f x)}{8 a^3 f \left (a+b \sec ^2(e+f x)-b\right )}-\frac {(5 a-6 b) \cot (e+f x) \csc (e+f x)}{8 a^2 f \left (a+b \sec ^2(e+f x)-b\right )}-\frac {3 \left (a^2-8 a b+8 b^2\right ) \tanh ^{-1}(\cos (e+f x))}{8 a^4 f}-\frac {\cot ^3(e+f x) \csc (e+f x)}{4 a f \left (a+b \sec ^2(e+f x)-b\right )} \]

[Out]

-3/8*(a^2-8*a*b+8*b^2)*arctanh(cos(f*x+e))/a^4/f-1/8*(5*a-6*b)*cot(f*x+e)*csc(f*x+e)/a^2/f/(a-b+b*sec(f*x+e)^2
)-1/4*cot(f*x+e)^3*csc(f*x+e)/a/f/(a-b+b*sec(f*x+e)^2)-3/8*(3*a-4*b)*b*sec(f*x+e)/a^3/f/(a-b+b*sec(f*x+e)^2)-3
/2*(a-2*b)*arctan(sec(f*x+e)*b^(1/2)/(a-b)^(1/2))*(a-b)^(1/2)*b^(1/2)/a^4/f

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Rubi [A]  time = 0.28, antiderivative size = 210, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3664, 470, 527, 522, 207, 205} \[ -\frac {3 \left (a^2-8 a b+8 b^2\right ) \tanh ^{-1}(\cos (e+f x))}{8 a^4 f}-\frac {3 b (3 a-4 b) \sec (e+f x)}{8 a^3 f \left (a+b \sec ^2(e+f x)-b\right )}-\frac {3 \sqrt {b} (a-2 b) \sqrt {a-b} \tan ^{-1}\left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a-b}}\right )}{2 a^4 f}-\frac {(5 a-6 b) \cot (e+f x) \csc (e+f x)}{8 a^2 f \left (a+b \sec ^2(e+f x)-b\right )}-\frac {\cot ^3(e+f x) \csc (e+f x)}{4 a f \left (a+b \sec ^2(e+f x)-b\right )} \]

Antiderivative was successfully verified.

[In]

Int[Csc[e + f*x]^5/(a + b*Tan[e + f*x]^2)^2,x]

[Out]

(-3*(a - 2*b)*Sqrt[a - b]*Sqrt[b]*ArcTan[(Sqrt[b]*Sec[e + f*x])/Sqrt[a - b]])/(2*a^4*f) - (3*(a^2 - 8*a*b + 8*
b^2)*ArcTanh[Cos[e + f*x]])/(8*a^4*f) - ((5*a - 6*b)*Cot[e + f*x]*Csc[e + f*x])/(8*a^2*f*(a - b + b*Sec[e + f*
x]^2)) - (Cot[e + f*x]^3*Csc[e + f*x])/(4*a*f*(a - b + b*Sec[e + f*x]^2)) - (3*(3*a - 4*b)*b*Sec[e + f*x])/(8*
a^3*f*(a - b + b*Sec[e + f*x]^2))

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 207

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

Rule 470

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

Rule 522

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*((c_) + (d_.)*(x_)^(n_))), x_Symbol] :> Dist[(b*e - a*f
)/(b*c - a*d), Int[1/(a + b*x^n), x], x] - Dist[(d*e - c*f)/(b*c - a*d), Int[1/(c + d*x^n), x], x] /; FreeQ[{a
, b, c, d, e, f, n}, x]

Rule 527

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)), x_Symbol] :> -Simp[
((b*e - a*f)*x*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(a*n*(b*c - a*d)*(p + 1)), x] + Dist[1/(a*n*(b*c - a*d
)*(p + 1)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*(b*e - a*f) + e*n*(b*c - a*d)*(p + 1) + d*(b*e - a*f)
*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, q}, x] && LtQ[p, -1]

Rule 3664

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

Rubi steps

\begin {align*} \int \frac {\csc ^5(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^2} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {x^4}{\left (-1+x^2\right )^3 \left (a-b+b x^2\right )^2} \, dx,x,\sec (e+f x)\right )}{f}\\ &=-\frac {\cot ^3(e+f x) \csc (e+f x)}{4 a f \left (a-b+b \sec ^2(e+f x)\right )}-\frac {\operatorname {Subst}\left (\int \frac {-a+b+(-4 a+5 b) x^2}{\left (-1+x^2\right )^2 \left (a-b+b x^2\right )^2} \, dx,x,\sec (e+f x)\right )}{4 a f}\\ &=-\frac {(5 a-6 b) \cot (e+f x) \csc (e+f x)}{8 a^2 f \left (a-b+b \sec ^2(e+f x)\right )}-\frac {\cot ^3(e+f x) \csc (e+f x)}{4 a f \left (a-b+b \sec ^2(e+f x)\right )}-\frac {\operatorname {Subst}\left (\int \frac {-3 (a-2 b) (a-b)+3 (5 a-6 b) b x^2}{\left (-1+x^2\right ) \left (a-b+b x^2\right )^2} \, dx,x,\sec (e+f x)\right )}{8 a^2 f}\\ &=-\frac {(5 a-6 b) \cot (e+f x) \csc (e+f x)}{8 a^2 f \left (a-b+b \sec ^2(e+f x)\right )}-\frac {\cot ^3(e+f x) \csc (e+f x)}{4 a f \left (a-b+b \sec ^2(e+f x)\right )}-\frac {3 (3 a-4 b) b \sec (e+f x)}{8 a^3 f \left (a-b+b \sec ^2(e+f x)\right )}-\frac {\operatorname {Subst}\left (\int \frac {-6 (a-4 b) (a-b)^2+6 (3 a-4 b) (a-b) b x^2}{\left (-1+x^2\right ) \left (a-b+b x^2\right )} \, dx,x,\sec (e+f x)\right )}{16 a^3 (a-b) f}\\ &=-\frac {(5 a-6 b) \cot (e+f x) \csc (e+f x)}{8 a^2 f \left (a-b+b \sec ^2(e+f x)\right )}-\frac {\cot ^3(e+f x) \csc (e+f x)}{4 a f \left (a-b+b \sec ^2(e+f x)\right )}-\frac {3 (3 a-4 b) b \sec (e+f x)}{8 a^3 f \left (a-b+b \sec ^2(e+f x)\right )}-\frac {(3 (a-2 b) (a-b) b) \operatorname {Subst}\left (\int \frac {1}{a-b+b x^2} \, dx,x,\sec (e+f x)\right )}{2 a^4 f}+\frac {\left (3 \left (a^2-8 a b+8 b^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{-1+x^2} \, dx,x,\sec (e+f x)\right )}{8 a^4 f}\\ &=-\frac {3 (a-2 b) \sqrt {a-b} \sqrt {b} \tan ^{-1}\left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a-b}}\right )}{2 a^4 f}-\frac {3 \left (a^2-8 a b+8 b^2\right ) \tanh ^{-1}(\cos (e+f x))}{8 a^4 f}-\frac {(5 a-6 b) \cot (e+f x) \csc (e+f x)}{8 a^2 f \left (a-b+b \sec ^2(e+f x)\right )}-\frac {\cot ^3(e+f x) \csc (e+f x)}{4 a f \left (a-b+b \sec ^2(e+f x)\right )}-\frac {3 (3 a-4 b) b \sec (e+f x)}{8 a^3 f \left (a-b+b \sec ^2(e+f x)\right )}\\ \end {align*}

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Mathematica [A]  time = 6.36, size = 392, normalized size = 1.87 \[ \frac {3 \sqrt {b} (a-2 b) \sqrt {a-b} \tan ^{-1}\left (\frac {\sec \left (\frac {1}{2} (e+f x)\right ) \left (\sqrt {a-b} \cos \left (\frac {1}{2} (e+f x)\right )-\sqrt {a} \sin \left (\frac {1}{2} (e+f x)\right )\right )}{\sqrt {b}}\right )}{2 a^4 f}+\frac {3 \sqrt {b} (a-2 b) \sqrt {a-b} \tan ^{-1}\left (\frac {\sec \left (\frac {1}{2} (e+f x)\right ) \left (\sqrt {a-b} \cos \left (\frac {1}{2} (e+f x)\right )+\sqrt {a} \sin \left (\frac {1}{2} (e+f x)\right )\right )}{\sqrt {b}}\right )}{2 a^4 f}+\frac {b^2 \cos (e+f x)-a b \cos (e+f x)}{a^3 f (a \cos (2 (e+f x))+a-b \cos (2 (e+f x))+b)}+\frac {(8 b-3 a) \csc ^2\left (\frac {1}{2} (e+f x)\right )}{32 a^3 f}+\frac {(3 a-8 b) \sec ^2\left (\frac {1}{2} (e+f x)\right )}{32 a^3 f}-\frac {\csc ^4\left (\frac {1}{2} (e+f x)\right )}{64 a^2 f}+\frac {\sec ^4\left (\frac {1}{2} (e+f x)\right )}{64 a^2 f}+\frac {3 \left (a^2-8 a b+8 b^2\right ) \log \left (\sin \left (\frac {1}{2} (e+f x)\right )\right )}{8 a^4 f}-\frac {3 \left (a^2-8 a b+8 b^2\right ) \log \left (\cos \left (\frac {1}{2} (e+f x)\right )\right )}{8 a^4 f} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[e + f*x]^5/(a + b*Tan[e + f*x]^2)^2,x]

[Out]

(3*(a - 2*b)*Sqrt[a - b]*Sqrt[b]*ArcTan[(Sec[(e + f*x)/2]*(Sqrt[a - b]*Cos[(e + f*x)/2] - Sqrt[a]*Sin[(e + f*x
)/2]))/Sqrt[b]])/(2*a^4*f) + (3*(a - 2*b)*Sqrt[a - b]*Sqrt[b]*ArcTan[(Sec[(e + f*x)/2]*(Sqrt[a - b]*Cos[(e + f
*x)/2] + Sqrt[a]*Sin[(e + f*x)/2]))/Sqrt[b]])/(2*a^4*f) + (-(a*b*Cos[e + f*x]) + b^2*Cos[e + f*x])/(a^3*f*(a +
b + a*Cos[2*(e + f*x)] - b*Cos[2*(e + f*x)])) + ((-3*a + 8*b)*Csc[(e + f*x)/2]^2)/(32*a^3*f) - Csc[(e + f*x)/
2]^4/(64*a^2*f) - (3*(a^2 - 8*a*b + 8*b^2)*Log[Cos[(e + f*x)/2]])/(8*a^4*f) + (3*(a^2 - 8*a*b + 8*b^2)*Log[Sin
[(e + f*x)/2]])/(8*a^4*f) + ((3*a - 8*b)*Sec[(e + f*x)/2]^2)/(32*a^3*f) + Sec[(e + f*x)/2]^4/(64*a^2*f)

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fricas [B]  time = 0.59, size = 1052, normalized size = 5.01 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^5/(a+b*tan(f*x+e)^2)^2,x, algorithm="fricas")

[Out]

[1/16*(6*(a^3 - 5*a^2*b + 4*a*b^2)*cos(f*x + e)^5 - 2*(5*a^3 - 24*a^2*b + 24*a*b^2)*cos(f*x + e)^3 - 12*((a^2
- 3*a*b + 2*b^2)*cos(f*x + e)^6 - (2*a^2 - 7*a*b + 6*b^2)*cos(f*x + e)^4 + (a^2 - 5*a*b + 6*b^2)*cos(f*x + e)^
2 + a*b - 2*b^2)*sqrt(-a*b + b^2)*log(((a - b)*cos(f*x + e)^2 - 2*sqrt(-a*b + b^2)*cos(f*x + e) - b)/((a - b)*
cos(f*x + e)^2 + b)) - 6*(3*a^2*b - 4*a*b^2)*cos(f*x + e) - 3*((a^3 - 9*a^2*b + 16*a*b^2 - 8*b^3)*cos(f*x + e)
^6 - (2*a^3 - 19*a^2*b + 40*a*b^2 - 24*b^3)*cos(f*x + e)^4 + a^2*b - 8*a*b^2 + 8*b^3 + (a^3 - 11*a^2*b + 32*a*
b^2 - 24*b^3)*cos(f*x + e)^2)*log(1/2*cos(f*x + e) + 1/2) + 3*((a^3 - 9*a^2*b + 16*a*b^2 - 8*b^3)*cos(f*x + e)
^6 - (2*a^3 - 19*a^2*b + 40*a*b^2 - 24*b^3)*cos(f*x + e)^4 + a^2*b - 8*a*b^2 + 8*b^3 + (a^3 - 11*a^2*b + 32*a*
b^2 - 24*b^3)*cos(f*x + e)^2)*log(-1/2*cos(f*x + e) + 1/2))/((a^5 - a^4*b)*f*cos(f*x + e)^6 + a^4*b*f - (2*a^5
- 3*a^4*b)*f*cos(f*x + e)^4 + (a^5 - 3*a^4*b)*f*cos(f*x + e)^2), 1/16*(6*(a^3 - 5*a^2*b + 4*a*b^2)*cos(f*x +
e)^5 - 2*(5*a^3 - 24*a^2*b + 24*a*b^2)*cos(f*x + e)^3 + 24*((a^2 - 3*a*b + 2*b^2)*cos(f*x + e)^6 - (2*a^2 - 7*
a*b + 6*b^2)*cos(f*x + e)^4 + (a^2 - 5*a*b + 6*b^2)*cos(f*x + e)^2 + a*b - 2*b^2)*sqrt(a*b - b^2)*arctan(sqrt(
a*b - b^2)*cos(f*x + e)/b) - 6*(3*a^2*b - 4*a*b^2)*cos(f*x + e) - 3*((a^3 - 9*a^2*b + 16*a*b^2 - 8*b^3)*cos(f*
x + e)^6 - (2*a^3 - 19*a^2*b + 40*a*b^2 - 24*b^3)*cos(f*x + e)^4 + a^2*b - 8*a*b^2 + 8*b^3 + (a^3 - 11*a^2*b +
32*a*b^2 - 24*b^3)*cos(f*x + e)^2)*log(1/2*cos(f*x + e) + 1/2) + 3*((a^3 - 9*a^2*b + 16*a*b^2 - 8*b^3)*cos(f*
x + e)^6 - (2*a^3 - 19*a^2*b + 40*a*b^2 - 24*b^3)*cos(f*x + e)^4 + a^2*b - 8*a*b^2 + 8*b^3 + (a^3 - 11*a^2*b +
32*a*b^2 - 24*b^3)*cos(f*x + e)^2)*log(-1/2*cos(f*x + e) + 1/2))/((a^5 - a^4*b)*f*cos(f*x + e)^6 + a^4*b*f -
(2*a^5 - 3*a^4*b)*f*cos(f*x + e)^4 + (a^5 - 3*a^4*b)*f*cos(f*x + e)^2)]

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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)^5/(a+b*tan(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*(((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1)))*a^2*b-3*(1-cos(f
*x+exp(1)))/(1+cos(f*x+exp(1)))*a*b^2+2*(1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1)))*b^3-a^2*b+a*b^2)*1/2/a^4/(((1-
cos(f*x+exp(1)))/(1+cos(f*x+exp(1))))^2*a-2*(1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1)))*a+4*(1-cos(f*x+exp(1)))/(1
+cos(f*x+exp(1)))*b+a)+(-18*((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1))))^2*a^2+144*((1-cos(f*x+exp(1)))/(1+cos(f*
x+exp(1))))^2*a*b-144*((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1))))^2*b^2-8*(1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1))
)*a^2+16*(1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1)))*a*b-a^2)*1/128/a^4/((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1))))^
2+(32*((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1))))^2*a^2+256*(1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1)))*a^2-512*(1-c
os(f*x+exp(1)))/(1+cos(f*x+exp(1)))*a*b)*1/4096/a^4+(-3*a^2*b+9*a*b^2-6*b^3)*1/4/a^4/sqrt(-b^2+a*b)*atan((-a*c
os(f*x+exp(1))+b*cos(f*x+exp(1))+b)/(sqrt(-b^2+a*b)*cos(f*x+exp(1))+sqrt(-b^2+a*b)))+(3*a^2-24*a*b+24*b^2)*1/3
2/a^4*ln(abs(1-cos(f*x+exp(1)))/abs(1+cos(f*x+exp(1)))))

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maple [B]  time = 0.57, size = 428, normalized size = 2.04 \[ -\frac {b \cos \left (f x +e \right )}{2 f \,a^{2} \left (a \left (\cos ^{2}\left (f x +e \right )\right )-\left (\cos ^{2}\left (f x +e \right )\right ) b +b \right )}+\frac {b^{2} \cos \left (f x +e \right )}{2 f \,a^{3} \left (a \left (\cos ^{2}\left (f x +e \right )\right )-\left (\cos ^{2}\left (f x +e \right )\right ) b +b \right )}+\frac {3 b \arctan \left (\frac {\left (a -b \right ) \cos \left (f x +e \right )}{\sqrt {\left (a -b \right ) b}}\right )}{2 f \,a^{2} \sqrt {\left (a -b \right ) b}}-\frac {9 b^{2} \arctan \left (\frac {\left (a -b \right ) \cos \left (f x +e \right )}{\sqrt {\left (a -b \right ) b}}\right )}{2 f \,a^{3} \sqrt {\left (a -b \right ) b}}+\frac {3 b^{3} \arctan \left (\frac {\left (a -b \right ) \cos \left (f x +e \right )}{\sqrt {\left (a -b \right ) b}}\right )}{f \,a^{4} \sqrt {\left (a -b \right ) b}}-\frac {1}{16 f \,a^{2} \left (-1+\cos \left (f x +e \right )\right )^{2}}+\frac {3}{16 f \,a^{2} \left (-1+\cos \left (f x +e \right )\right )}-\frac {b}{2 f \,a^{3} \left (-1+\cos \left (f x +e \right )\right )}+\frac {3 \ln \left (-1+\cos \left (f x +e \right )\right )}{16 f \,a^{2}}-\frac {3 \ln \left (-1+\cos \left (f x +e \right )\right ) b}{2 f \,a^{3}}+\frac {3 \ln \left (-1+\cos \left (f x +e \right )\right ) b^{2}}{2 f \,a^{4}}+\frac {1}{16 f \,a^{2} \left (1+\cos \left (f x +e \right )\right )^{2}}+\frac {3}{16 f \,a^{2} \left (1+\cos \left (f x +e \right )\right )}-\frac {b}{2 f \,a^{3} \left (1+\cos \left (f x +e \right )\right )}-\frac {3 \ln \left (1+\cos \left (f x +e \right )\right )}{16 f \,a^{2}}+\frac {3 \ln \left (1+\cos \left (f x +e \right )\right ) b}{2 f \,a^{3}}-\frac {3 \ln \left (1+\cos \left (f x +e \right )\right ) b^{2}}{2 f \,a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(f*x+e)^5/(a+b*tan(f*x+e)^2)^2,x)

[Out]

-1/2/f*b/a^2*cos(f*x+e)/(a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)+1/2/f*b^2/a^3*cos(f*x+e)/(a*cos(f*x+e)^2-cos(f*x+e)^
2*b+b)+3/2/f*b/a^2/((a-b)*b)^(1/2)*arctan((a-b)*cos(f*x+e)/((a-b)*b)^(1/2))-9/2/f*b^2/a^3/((a-b)*b)^(1/2)*arct
an((a-b)*cos(f*x+e)/((a-b)*b)^(1/2))+3/f*b^3/a^4/((a-b)*b)^(1/2)*arctan((a-b)*cos(f*x+e)/((a-b)*b)^(1/2))-1/16
/f/a^2/(-1+cos(f*x+e))^2+3/16/f/a^2/(-1+cos(f*x+e))-1/2/f/a^3/(-1+cos(f*x+e))*b+3/16/f/a^2*ln(-1+cos(f*x+e))-3
/2/f/a^3*ln(-1+cos(f*x+e))*b+3/2/f/a^4*ln(-1+cos(f*x+e))*b^2+1/16/f/a^2/(1+cos(f*x+e))^2+3/16/f/a^2/(1+cos(f*x
+e))-1/2/f/a^3/(1+cos(f*x+e))*b-3/16/f/a^2*ln(1+cos(f*x+e))+3/2/f/a^3*ln(1+cos(f*x+e))*b-3/2/f/a^4*ln(1+cos(f*
x+e))*b^2

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^5/(a+b*tan(f*x+e)^2)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(b-a>0)', see `assume?` for mor
e details)Is b-a positive or negative?

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mupad [B]  time = 12.31, size = 1113, normalized size = 5.30 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sin(e + f*x)^5*(a + b*tan(e + f*x)^2)^2),x)

[Out]

tan(e/2 + (f*x)/2)^4/(64*a^2*f) - (a^2/4 - tan(e/2 + (f*x)/2)^4*((15*a^2)/4 - 32*a*b + 32*b^2) + (3*a*tan(e/2
+ (f*x)/2)^2*(a - 2*b))/2 + (2*tan(e/2 + (f*x)/2)^6*(24*a*b^2 - 10*a^2*b + a^3 - 16*b^3))/a)/(f*(16*a^4*tan(e/
2 + (f*x)/2)^4 + 16*a^4*tan(e/2 + (f*x)/2)^8 + tan(e/2 + (f*x)/2)^6*(64*a^3*b - 32*a^4))) + (tan(e/2 + (f*x)/2
)^2*(a - 2*b))/(8*a^3*f) + (log(tan(e/2 + (f*x)/2))*(3*a^2 - 24*a*b + 24*b^2))/(8*a^4*f) + (3*atan((8*a^10*tan
(e/2 + (f*x)/2)^2*((((756*a*b^6 - 216*b^7 - 1026*a^2*b^5 + 675*a^3*b^4 - 216*a^4*b^3 + 27*a^5*b^2)/a^8 + (9*(a
- 2*b)^2*(a*b - b^2)*(180*a^10*b - 6*a^11 + 2304*a^6*b^5 - 5760*a^7*b^4 + 4944*a^8*b^3 - 1656*a^9*b^2))/(16*a
^16))*(960*a*b^4 - 38*a^4*b + a^5 - 384*b^5 - 840*a^2*b^3 + 300*a^3*b^2))/(2*a^5*(b*(a - b))^(3/2)*(a^4 - 12*a
^3*b - 96*a*b^3 + 48*b^4 + 60*a^2*b^2)) + (((27*(a - 2*b)^3*(a*b - b^2)^(3/2)*(416*a^12*b - 16*a^13 + 768*a^10
*b^3 - 1152*a^11*b^2))/(64*a^20) - (3*(a - 2*b)*(a*b - b^2)^(1/2)*(27*a^8*b + 1728*a^2*b^7 - 6048*a^3*b^6 + 83
52*a^4*b^5 - 5760*a^5*b^4 + 2070*a^6*b^3 - 369*a^7*b^2))/(4*a^12))*(4*a^4 - 60*a^3*b - 384*a*b^3 + 192*b^4 + 2
52*a^2*b^2))/(a^5*b*(144*a*b^4 - 13*a^4*b + a^5 - 48*b^5 - 156*a^2*b^3 + 72*a^3*b^2))))/(27*a^2 - 108*a*b + 10
8*b^2) + (8*a^5*((27*(a - 2*b)^3*(a*b - b^2)^(3/2)*(32*a^14 - 128*a^13*b + 128*a^12*b^2))/(128*a^21) + (3*(a -
2*b)*(a*b - b^2)^(1/2)*(36*a^9*b - 1440*a^4*b^6 + 4320*a^5*b^5 - 4824*a^6*b^4 + 2448*a^7*b^3 - 540*a^8*b^2))/
(8*a^13))*(4*a^4 - 60*a^3*b - 384*a*b^3 + 192*b^4 + 252*a^2*b^2))/(b*(27*a^2 - 108*a*b + 108*b^2)*(144*a*b^4 -
13*a^4*b + a^5 - 48*b^5 - 156*a^2*b^3 + 72*a^3*b^2)) - (4*a^5*((864*b^8 - 3456*a*b^7 + 5508*a^2*b^6 - 4428*a^
3*b^5 + 1863*a^4*b^4 - 378*a^5*b^3 + 27*a^6*b^2)/(2*a^9) - (9*(a - 2*b)^2*(a*b - b^2)*(12*a^12 - 240*a^11*b +
768*a^8*b^4 - 1536*a^9*b^3 + 1008*a^10*b^2))/(32*a^17))*(960*a*b^4 - 38*a^4*b + a^5 - 384*b^5 - 840*a^2*b^3 +
300*a^3*b^2))/((b*(a - b))^(3/2)*(27*a^2 - 108*a*b + 108*b^2)*(a^4 - 12*a^3*b - 96*a*b^3 + 48*b^4 + 60*a^2*b^2
)))*(a - 2*b)*(a*b - b^2)^(1/2))/(2*a^4*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)**5/(a+b*tan(f*x+e)**2)**2,x)

[Out]

Timed out

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