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

Optimal. Leaf size=130 \[ -\frac {\sqrt {b} (a-b)^{3/2} \tan ^{-1}\left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a-b}}\right )}{a^3 f}-\frac {(5 a-4 b) \cot (e+f x) \csc (e+f x)}{8 a^2 f}-\frac {\left (3 a^2-12 a b+8 b^2\right ) \tanh ^{-1}(\cos (e+f x))}{8 a^3 f}-\frac {\cot ^3(e+f x) \csc (e+f x)}{4 a f} \]

[Out]

-1/8*(3*a^2-12*a*b+8*b^2)*arctanh(cos(f*x+e))/a^3/f-1/8*(5*a-4*b)*cot(f*x+e)*csc(f*x+e)/a^2/f-1/4*cot(f*x+e)^3
*csc(f*x+e)/a/f-(a-b)^(3/2)*arctan(sec(f*x+e)*b^(1/2)/(a-b)^(1/2))*b^(1/2)/a^3/f

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Rubi [A]  time = 0.18, antiderivative size = 130, normalized size of antiderivative = 1.00, number of steps used = 6, 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} \[ -\frac {\left (3 a^2-12 a b+8 b^2\right ) \tanh ^{-1}(\cos (e+f x))}{8 a^3 f}-\frac {(5 a-4 b) \cot (e+f x) \csc (e+f x)}{8 a^2 f}-\frac {\sqrt {b} (a-b)^{3/2} \tan ^{-1}\left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a-b}}\right )}{a^3 f}-\frac {\cot ^3(e+f x) \csc (e+f x)}{4 a f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((a - b)^(3/2)*Sqrt[b]*ArcTan[(Sqrt[b]*Sec[e + f*x])/Sqrt[a - b]])/(a^3*f)) - ((3*a^2 - 12*a*b + 8*b^2)*ArcT
anh[Cos[e + f*x]])/(8*a^3*f) - ((5*a - 4*b)*Cot[e + f*x]*Csc[e + f*x])/(8*a^2*f) - (Cot[e + f*x]^3*Csc[e + f*x
])/(4*a*f)

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)}{a+b \tan ^2(e+f x)} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {x^4}{\left (-1+x^2\right )^3 \left (a-b+b x^2\right )} \, dx,x,\sec (e+f x)\right )}{f}\\ &=-\frac {\cot ^3(e+f x) \csc (e+f x)}{4 a f}-\frac {\operatorname {Subst}\left (\int \frac {-a+b+(-4 a+3 b) x^2}{\left (-1+x^2\right )^2 \left (a-b+b x^2\right )} \, dx,x,\sec (e+f x)\right )}{4 a f}\\ &=-\frac {(5 a-4 b) \cot (e+f x) \csc (e+f x)}{8 a^2 f}-\frac {\cot ^3(e+f x) \csc (e+f x)}{4 a f}-\frac {\operatorname {Subst}\left (\int \frac {-(3 a-4 b) (a-b)+(5 a-4 b) b x^2}{\left (-1+x^2\right ) \left (a-b+b x^2\right )} \, dx,x,\sec (e+f x)\right )}{8 a^2 f}\\ &=-\frac {(5 a-4 b) \cot (e+f x) \csc (e+f x)}{8 a^2 f}-\frac {\cot ^3(e+f x) \csc (e+f x)}{4 a f}-\frac {\left ((a-b)^2 b\right ) \operatorname {Subst}\left (\int \frac {1}{a-b+b x^2} \, dx,x,\sec (e+f x)\right )}{a^3 f}+\frac {\left (3 a^2-12 a b+8 b^2\right ) \operatorname {Subst}\left (\int \frac {1}{-1+x^2} \, dx,x,\sec (e+f x)\right )}{8 a^3 f}\\ &=-\frac {(a-b)^{3/2} \sqrt {b} \tan ^{-1}\left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a-b}}\right )}{a^3 f}-\frac {\left (3 a^2-12 a b+8 b^2\right ) \tanh ^{-1}(\cos (e+f x))}{8 a^3 f}-\frac {(5 a-4 b) \cot (e+f x) \csc (e+f x)}{8 a^2 f}-\frac {\cot ^3(e+f x) \csc (e+f x)}{4 a f}\\ \end {align*}

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Mathematica [B]  time = 6.26, size = 326, normalized size = 2.51 \[ \frac {\sqrt {b} (a-b)^{3/2} \tan ^{-1}\left (\frac {\sec \left (\frac {1}{2} (e+f x)\right ) \left (\sqrt {a-b} \cos \left (\frac {1}{2} (e+f x)\right )-\sqrt {a} \sin \left (\frac {1}{2} (e+f x)\right )\right )}{\sqrt {b}}\right )}{a^3 f}+\frac {\sqrt {b} (a-b)^{3/2} \tan ^{-1}\left (\frac {\sec \left (\frac {1}{2} (e+f x)\right ) \left (\sqrt {a-b} \cos \left (\frac {1}{2} (e+f x)\right )+\sqrt {a} \sin \left (\frac {1}{2} (e+f x)\right )\right )}{\sqrt {b}}\right )}{a^3 f}+\frac {(4 b-3 a) \csc ^2\left (\frac {1}{2} (e+f x)\right )}{32 a^2 f}+\frac {(3 a-4 b) \sec ^2\left (\frac {1}{2} (e+f x)\right )}{32 a^2 f}+\frac {\left (3 a^2-12 a b+8 b^2\right ) \log \left (\sin \left (\frac {1}{2} (e+f x)\right )\right )}{8 a^3 f}+\frac {\left (-3 a^2+12 a b-8 b^2\right ) \log \left (\cos \left (\frac {1}{2} (e+f x)\right )\right )}{8 a^3 f}-\frac {\csc ^4\left (\frac {1}{2} (e+f x)\right )}{64 a f}+\frac {\sec ^4\left (\frac {1}{2} (e+f x)\right )}{64 a f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a - b)^(3/2)*Sqrt[b]*ArcTan[(Sec[(e + f*x)/2]*(Sqrt[a - b]*Cos[(e + f*x)/2] - Sqrt[a]*Sin[(e + f*x)/2]))/Sqr
t[b]])/(a^3*f) + ((a - b)^(3/2)*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]])/(a^3*f) + ((-3*a + 4*b)*Csc[(e + f*x)/2]^2)/(32*a^2*f) - Csc[(e + f*x)/2]^4/(64*a*f) +
((-3*a^2 + 12*a*b - 8*b^2)*Log[Cos[(e + f*x)/2]])/(8*a^3*f) + ((3*a^2 - 12*a*b + 8*b^2)*Log[Sin[(e + f*x)/2]]
)/(8*a^3*f) + ((3*a - 4*b)*Sec[(e + f*x)/2]^2)/(32*a^2*f) + Sec[(e + f*x)/2]^4/(64*a*f)

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fricas [B]  time = 0.57, size = 630, normalized size = 4.85 \[ \left [\frac {2 \, {\left (3 \, a^{2} - 4 \, a b\right )} \cos \left (f x + e\right )^{3} - 8 \, {\left ({\left (a - b\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (a - b\right )} \cos \left (f x + e\right )^{2} + a - b\right )} \sqrt {-a b + b^{2}} \log \left (\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} - 2 \, \sqrt {-a b + b^{2}} \cos \left (f x + e\right ) - b}{{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}\right ) - 2 \, {\left (5 \, a^{2} - 4 \, a b\right )} \cos \left (f x + e\right ) - {\left ({\left (3 \, a^{2} - 12 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (3 \, a^{2} - 12 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 3 \, a^{2} - 12 \, a b + 8 \, b^{2}\right )} \log \left (\frac {1}{2} \, \cos \left (f x + e\right ) + \frac {1}{2}\right ) + {\left ({\left (3 \, a^{2} - 12 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (3 \, a^{2} - 12 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 3 \, a^{2} - 12 \, a b + 8 \, b^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (f x + e\right ) + \frac {1}{2}\right )}{16 \, {\left (a^{3} f \cos \left (f x + e\right )^{4} - 2 \, a^{3} f \cos \left (f x + e\right )^{2} + a^{3} f\right )}}, \frac {2 \, {\left (3 \, a^{2} - 4 \, a b\right )} \cos \left (f x + e\right )^{3} + 16 \, {\left ({\left (a - b\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (a - b\right )} \cos \left (f x + e\right )^{2} + a - b\right )} \sqrt {a b - b^{2}} \arctan \left (\frac {\sqrt {a b - b^{2}} \cos \left (f x + e\right )}{b}\right ) - 2 \, {\left (5 \, a^{2} - 4 \, a b\right )} \cos \left (f x + e\right ) - {\left ({\left (3 \, a^{2} - 12 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (3 \, a^{2} - 12 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 3 \, a^{2} - 12 \, a b + 8 \, b^{2}\right )} \log \left (\frac {1}{2} \, \cos \left (f x + e\right ) + \frac {1}{2}\right ) + {\left ({\left (3 \, a^{2} - 12 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (3 \, a^{2} - 12 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 3 \, a^{2} - 12 \, a b + 8 \, b^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (f x + e\right ) + \frac {1}{2}\right )}{16 \, {\left (a^{3} f \cos \left (f x + e\right )^{4} - 2 \, a^{3} f \cos \left (f x + e\right )^{2} + a^{3} f\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/16*(2*(3*a^2 - 4*a*b)*cos(f*x + e)^3 - 8*((a - b)*cos(f*x + e)^4 - 2*(a - b)*cos(f*x + e)^2 + a - b)*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)) -
2*(5*a^2 - 4*a*b)*cos(f*x + e) - ((3*a^2 - 12*a*b + 8*b^2)*cos(f*x + e)^4 - 2*(3*a^2 - 12*a*b + 8*b^2)*cos(f*x
+ e)^2 + 3*a^2 - 12*a*b + 8*b^2)*log(1/2*cos(f*x + e) + 1/2) + ((3*a^2 - 12*a*b + 8*b^2)*cos(f*x + e)^4 - 2*(
3*a^2 - 12*a*b + 8*b^2)*cos(f*x + e)^2 + 3*a^2 - 12*a*b + 8*b^2)*log(-1/2*cos(f*x + e) + 1/2))/(a^3*f*cos(f*x
+ e)^4 - 2*a^3*f*cos(f*x + e)^2 + a^3*f), 1/16*(2*(3*a^2 - 4*a*b)*cos(f*x + e)^3 + 16*((a - b)*cos(f*x + e)^4
- 2*(a - b)*cos(f*x + e)^2 + a - b)*sqrt(a*b - b^2)*arctan(sqrt(a*b - b^2)*cos(f*x + e)/b) - 2*(5*a^2 - 4*a*b)
*cos(f*x + e) - ((3*a^2 - 12*a*b + 8*b^2)*cos(f*x + e)^4 - 2*(3*a^2 - 12*a*b + 8*b^2)*cos(f*x + e)^2 + 3*a^2 -
12*a*b + 8*b^2)*log(1/2*cos(f*x + e) + 1/2) + ((3*a^2 - 12*a*b + 8*b^2)*cos(f*x + e)^4 - 2*(3*a^2 - 12*a*b +
8*b^2)*cos(f*x + e)^2 + 3*a^2 - 12*a*b + 8*b^2)*log(-1/2*cos(f*x + e) + 1/2))/(a^3*f*cos(f*x + e)^4 - 2*a^3*f*
cos(f*x + e)^2 + a^3*f)]

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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),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*((32*((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1))))^2*a+256*(1-
cos(f*x+exp(1)))/(1+cos(f*x+exp(1)))*a-256*(1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1)))*b)*1/4096/a^2+(-18*((1-cos(
f*x+exp(1)))/(1+cos(f*x+exp(1))))^2*a^2+72*((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1))))^2*a*b-48*((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+8*(1-cos(f*x+exp(1)))/(1+cos(f*x
+exp(1)))*a*b-a^2)*1/128/a^3/((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1))))^2+(-2*a^2*b+4*a*b^2-2*b^3)*1/4/a^3/sqrt
(-b^2+a*b)*atan((-a*cos(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
^2-12*a*b+8*b^2)*1/32/a^3*ln(abs(1-cos(f*x+exp(1)))/abs(1+cos(f*x+exp(1)))))

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maple [B]  time = 0.60, size = 344, normalized size = 2.65 \[ \frac {b \arctan \left (\frac {\left (a -b \right ) \cos \left (f x +e \right )}{\sqrt {\left (a -b \right ) b}}\right )}{f a \sqrt {\left (a -b \right ) b}}-\frac {2 b^{2} \arctan \left (\frac {\left (a -b \right ) \cos \left (f x +e \right )}{\sqrt {\left (a -b \right ) b}}\right )}{f \,a^{2} \sqrt {\left (a -b \right ) b}}+\frac {b^{3} \arctan \left (\frac {\left (a -b \right ) \cos \left (f x +e \right )}{\sqrt {\left (a -b \right ) b}}\right )}{f \,a^{3} \sqrt {\left (a -b \right ) b}}-\frac {1}{16 f a \left (-1+\cos \left (f x +e \right )\right )^{2}}+\frac {3}{16 f a \left (-1+\cos \left (f x +e \right )\right )}-\frac {b}{4 f \,a^{2} \left (-1+\cos \left (f x +e \right )\right )}+\frac {3 \ln \left (-1+\cos \left (f x +e \right )\right )}{16 f a}-\frac {3 \ln \left (-1+\cos \left (f x +e \right )\right ) b}{4 f \,a^{2}}+\frac {\ln \left (-1+\cos \left (f x +e \right )\right ) b^{2}}{2 f \,a^{3}}+\frac {1}{16 f a \left (1+\cos \left (f x +e \right )\right )^{2}}+\frac {3}{16 f a \left (1+\cos \left (f x +e \right )\right )}-\frac {b}{4 f \,a^{2} \left (1+\cos \left (f x +e \right )\right )}-\frac {3 \ln \left (1+\cos \left (f x +e \right )\right )}{16 f a}+\frac {3 \ln \left (1+\cos \left (f x +e \right )\right ) b}{4 f \,a^{2}}-\frac {\ln \left (1+\cos \left (f x +e \right )\right ) b^{2}}{2 f \,a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/f/a*b/((a-b)*b)^(1/2)*arctan((a-b)*cos(f*x+e)/((a-b)*b)^(1/2))-2/f*b^2/a^2/((a-b)*b)^(1/2)*arctan((a-b)*cos(
f*x+e)/((a-b)*b)^(1/2))+1/f*b^3/a^3/((a-b)*b)^(1/2)*arctan((a-b)*cos(f*x+e)/((a-b)*b)^(1/2))-1/16/f/a/(-1+cos(
f*x+e))^2+3/16/f/a/(-1+cos(f*x+e))-1/4/f/a^2/(-1+cos(f*x+e))*b+3/16/f/a*ln(-1+cos(f*x+e))-3/4/f/a^2*ln(-1+cos(
f*x+e))*b+1/2/f/a^3*ln(-1+cos(f*x+e))*b^2+1/16/f/a/(1+cos(f*x+e))^2+3/16/f/a/(1+cos(f*x+e))-1/4/f/a^2/(1+cos(f
*x+e))*b-3/16/f/a*ln(1+cos(f*x+e))+3/4/f/a^2*ln(1+cos(f*x+e))*b-1/2/f/a^3*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),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 = 14.72, size = 740, normalized size = 5.69 \[ \frac {a^2\,\left (\frac {3\,\cos \left (3\,e+3\,f\,x\right )}{4}-\frac {11\,\cos \left (e+f\,x\right )}{4}+\frac {9\,\ln \left (\frac {\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}\right )}{8}-\frac {3\,\cos \left (2\,e+2\,f\,x\right )\,\ln \left (\frac {\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}\right )}{2}+\frac {3\,\cos \left (4\,e+4\,f\,x\right )\,\ln \left (\frac {\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}\right )}{8}\right )+3\,b^2\,\ln \left (\frac {\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}\right )-a\,\left (b\,\cos \left (3\,e+3\,f\,x\right )-b\,\cos \left (e+f\,x\right )+\frac {9\,b\,\ln \left (\frac {\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}\right )}{2}-6\,b\,\cos \left (2\,e+2\,f\,x\right )\,\ln \left (\frac {\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}\right )+\frac {3\,b\,\cos \left (4\,e+4\,f\,x\right )\,\ln \left (\frac {\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}\right )}{2}\right )-4\,b^2\,\cos \left (2\,e+2\,f\,x\right )\,\ln \left (\frac {\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}\right )+b^2\,\cos \left (4\,e+4\,f\,x\right )\,\ln \left (\frac {\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}\right )+3\,\sqrt {b}\,\mathrm {atan}\left (\frac {a^4\,\cos \left (e+f\,x\right )-a^3\,b-3\,a\,b^3+b^4\,\cos \left (e+f\,x\right )+b^4+3\,a^2\,b^2+6\,a^2\,b^2\,\cos \left (e+f\,x\right )-4\,a\,b^3\,\cos \left (e+f\,x\right )-4\,a^3\,b\,\cos \left (e+f\,x\right )}{2\,\sqrt {b}\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,{\left (a-b\right )}^{7/2}}\right )\,{\left (a-b\right )}^{3/2}-4\,\sqrt {b}\,\mathrm {atan}\left (\frac {a^4\,\cos \left (e+f\,x\right )-a^3\,b-3\,a\,b^3+b^4\,\cos \left (e+f\,x\right )+b^4+3\,a^2\,b^2+6\,a^2\,b^2\,\cos \left (e+f\,x\right )-4\,a\,b^3\,\cos \left (e+f\,x\right )-4\,a^3\,b\,\cos \left (e+f\,x\right )}{2\,\sqrt {b}\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,{\left (a-b\right )}^{7/2}}\right )\,\cos \left (2\,e+2\,f\,x\right )\,{\left (a-b\right )}^{3/2}+\sqrt {b}\,\mathrm {atan}\left (\frac {a^4\,\cos \left (e+f\,x\right )-a^3\,b-3\,a\,b^3+b^4\,\cos \left (e+f\,x\right )+b^4+3\,a^2\,b^2+6\,a^2\,b^2\,\cos \left (e+f\,x\right )-4\,a\,b^3\,\cos \left (e+f\,x\right )-4\,a^3\,b\,\cos \left (e+f\,x\right )}{2\,\sqrt {b}\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,{\left (a-b\right )}^{7/2}}\right )\,\cos \left (4\,e+4\,f\,x\right )\,{\left (a-b\right )}^{3/2}}{3\,a^3\,f-4\,a^3\,f\,\cos \left (2\,e+2\,f\,x\right )+a^3\,f\,\cos \left (4\,e+4\,f\,x\right )} \]

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)),x)

[Out]

(a^2*((3*cos(3*e + 3*f*x))/4 - (11*cos(e + f*x))/4 + (9*log(sin(e/2 + (f*x)/2)/cos(e/2 + (f*x)/2)))/8 - (3*cos
(2*e + 2*f*x)*log(sin(e/2 + (f*x)/2)/cos(e/2 + (f*x)/2)))/2 + (3*cos(4*e + 4*f*x)*log(sin(e/2 + (f*x)/2)/cos(e
/2 + (f*x)/2)))/8) + 3*b^2*log(sin(e/2 + (f*x)/2)/cos(e/2 + (f*x)/2)) - a*(b*cos(3*e + 3*f*x) - b*cos(e + f*x)
+ (9*b*log(sin(e/2 + (f*x)/2)/cos(e/2 + (f*x)/2)))/2 - 6*b*cos(2*e + 2*f*x)*log(sin(e/2 + (f*x)/2)/cos(e/2 +
(f*x)/2)) + (3*b*cos(4*e + 4*f*x)*log(sin(e/2 + (f*x)/2)/cos(e/2 + (f*x)/2)))/2) - 4*b^2*cos(2*e + 2*f*x)*log(
sin(e/2 + (f*x)/2)/cos(e/2 + (f*x)/2)) + b^2*cos(4*e + 4*f*x)*log(sin(e/2 + (f*x)/2)/cos(e/2 + (f*x)/2)) + 3*b
^(1/2)*atan((a^4*cos(e + f*x) - a^3*b - 3*a*b^3 + b^4*cos(e + f*x) + b^4 + 3*a^2*b^2 + 6*a^2*b^2*cos(e + f*x)
- 4*a*b^3*cos(e + f*x) - 4*a^3*b*cos(e + f*x))/(2*b^(1/2)*cos(e/2 + (f*x)/2)^2*(a - b)^(7/2)))*(a - b)^(3/2) -
4*b^(1/2)*atan((a^4*cos(e + f*x) - a^3*b - 3*a*b^3 + b^4*cos(e + f*x) + b^4 + 3*a^2*b^2 + 6*a^2*b^2*cos(e + f
*x) - 4*a*b^3*cos(e + f*x) - 4*a^3*b*cos(e + f*x))/(2*b^(1/2)*cos(e/2 + (f*x)/2)^2*(a - b)^(7/2)))*cos(2*e + 2
*f*x)*(a - b)^(3/2) + b^(1/2)*atan((a^4*cos(e + f*x) - a^3*b - 3*a*b^3 + b^4*cos(e + f*x) + b^4 + 3*a^2*b^2 +
6*a^2*b^2*cos(e + f*x) - 4*a*b^3*cos(e + f*x) - 4*a^3*b*cos(e + f*x))/(2*b^(1/2)*cos(e/2 + (f*x)/2)^2*(a - b)^
(7/2)))*cos(4*e + 4*f*x)*(a - b)^(3/2))/(3*a^3*f - 4*a^3*f*cos(2*e + 2*f*x) + a^3*f*cos(4*e + 4*f*x))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\csc ^{5}{\left (e + f x \right )}}{a + b \tan ^{2}{\left (e + f x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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