3.72 \(\int \frac {\csc ^3(e+f x)}{(a+b \tan ^2(e+f x))^2} \, dx\)

Optimal. Leaf size=147 \[ -\frac {(a-4 b) \tanh ^{-1}(\cos (e+f x))}{2 a^3 f}-\frac {\sqrt {b} (3 a-4 b) \tan ^{-1}\left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a-b}}\right )}{2 a^3 f \sqrt {a-b}}-\frac {b \sec (e+f x)}{a^2 f \left (a+b \sec ^2(e+f x)-b\right )}-\frac {\cot (e+f x) \csc (e+f x)}{2 a f \left (a+b \sec ^2(e+f x)-b\right )} \]

[Out]

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

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Rubi [A]  time = 0.18, antiderivative size = 147, 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, 471, 527, 522, 207, 205} \[ -\frac {b \sec (e+f x)}{a^2 f \left (a+b \sec ^2(e+f x)-b\right )}-\frac {(a-4 b) \tanh ^{-1}(\cos (e+f x))}{2 a^3 f}-\frac {\sqrt {b} (3 a-4 b) \tan ^{-1}\left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a-b}}\right )}{2 a^3 f \sqrt {a-b}}-\frac {\cot (e+f x) \csc (e+f x)}{2 a f \left (a+b \sec ^2(e+f x)-b\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((3*a - 4*b)*Sqrt[b]*ArcTan[(Sqrt[b]*Sec[e + f*x])/Sqrt[a - b]])/(2*a^3*Sqrt[a - b]*f) - ((a - 4*b)*ArcTanh[C
os[e + f*x]])/(2*a^3*f) - (Cot[e + f*x]*Csc[e + f*x])/(2*a*f*(a - b + b*Sec[e + f*x]^2)) - (b*Sec[e + f*x])/(a
^2*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 471

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(e^(n -
1)*(e*x)^(m - n + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(n*(b*c - a*d)*(p + 1)), x] - Dist[e^n/(n*(b*c -
a*d)*(p + 1)), Int[(e*x)^(m - n)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*(m - n + 1) + d*(m + n*(p + q + 1)
+ 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] && GeQ[n
, m - n + 1] && GtQ[m - n + 1, 0] && 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 ^3(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^2} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {x^2}{\left (-1+x^2\right )^2 \left (a-b+b x^2\right )^2} \, dx,x,\sec (e+f x)\right )}{f}\\ &=-\frac {\cot (e+f x) \csc (e+f x)}{2 a f \left (a-b+b \sec ^2(e+f x)\right )}+\frac {\operatorname {Subst}\left (\int \frac {a-b-3 b x^2}{\left (-1+x^2\right ) \left (a-b+b x^2\right )^2} \, dx,x,\sec (e+f x)\right )}{2 a f}\\ &=-\frac {\cot (e+f x) \csc (e+f x)}{2 a f \left (a-b+b \sec ^2(e+f x)\right )}-\frac {b \sec (e+f x)}{a^2 f \left (a-b+b \sec ^2(e+f x)\right )}+\frac {\operatorname {Subst}\left (\int \frac {2 (a-2 b) (a-b)-4 (a-b) b x^2}{\left (-1+x^2\right ) \left (a-b+b x^2\right )} \, dx,x,\sec (e+f x)\right )}{4 a^2 (a-b) f}\\ &=-\frac {\cot (e+f x) \csc (e+f x)}{2 a f \left (a-b+b \sec ^2(e+f x)\right )}-\frac {b \sec (e+f x)}{a^2 f \left (a-b+b \sec ^2(e+f x)\right )}+\frac {(a-4 b) \operatorname {Subst}\left (\int \frac {1}{-1+x^2} \, dx,x,\sec (e+f x)\right )}{2 a^3 f}-\frac {((3 a-4 b) b) \operatorname {Subst}\left (\int \frac {1}{a-b+b x^2} \, dx,x,\sec (e+f x)\right )}{2 a^3 f}\\ &=-\frac {(3 a-4 b) \sqrt {b} \tan ^{-1}\left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a-b}}\right )}{2 a^3 \sqrt {a-b} f}-\frac {(a-4 b) \tanh ^{-1}(\cos (e+f x))}{2 a^3 f}-\frac {\cot (e+f x) \csc (e+f x)}{2 a f \left (a-b+b \sec ^2(e+f x)\right )}-\frac {b \sec (e+f x)}{a^2 f \left (a-b+b \sec ^2(e+f x)\right )}\\ \end {align*}

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Mathematica [B]  time = 6.31, size = 325, normalized size = 2.21 \[ \frac {(a-4 b) \log \left (\sin \left (\frac {1}{2} (e+f x)\right )\right )}{2 a^3 f}+\frac {(4 b-a) \log \left (\cos \left (\frac {1}{2} (e+f x)\right )\right )}{2 a^3 f}-\frac {\sqrt {b} (3 a-4 b) \sqrt {a-b} \tan ^{-1}\left (\frac {\sec \left (\frac {1}{2} (e+f x)\right ) \left (\sqrt {a-b} \cos \left (\frac {1}{2} (e+f x)\right )-\sqrt {a} \sin \left (\frac {1}{2} (e+f x)\right )\right )}{\sqrt {b}}\right )}{2 a^3 f (b-a)}-\frac {\sqrt {b} (3 a-4 b) \sqrt {a-b} \tan ^{-1}\left (\frac {\sec \left (\frac {1}{2} (e+f x)\right ) \left (\sqrt {a-b} \cos \left (\frac {1}{2} (e+f x)\right )+\sqrt {a} \sin \left (\frac {1}{2} (e+f x)\right )\right )}{\sqrt {b}}\right )}{2 a^3 f (b-a)}-\frac {b \cos (e+f x)}{a^2 f (a \cos (2 (e+f x))+a-b \cos (2 (e+f x))+b)}-\frac {\csc ^2\left (\frac {1}{2} (e+f x)\right )}{8 a^2 f}+\frac {\sec ^2\left (\frac {1}{2} (e+f x)\right )}{8 a^2 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*((3*a - 4*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]])/(a^3*(-a + b)*f) - ((3*a - 4*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^3*(-a + b)*f) - (b*Cos[e + f*x])/(a^2*f*(a + b
+ a*Cos[2*(e + f*x)] - b*Cos[2*(e + f*x)])) - Csc[(e + f*x)/2]^2/(8*a^2*f) + ((-a + 4*b)*Log[Cos[(e + f*x)/2]]
)/(2*a^3*f) + ((a - 4*b)*Log[Sin[(e + f*x)/2]])/(2*a^3*f) + Sec[(e + f*x)/2]^2/(8*a^2*f)

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fricas [B]  time = 0.56, size = 672, normalized size = 4.57 \[ \left [\frac {2 \, {\left (a^{2} - 2 \, a b\right )} \cos \left (f x + e\right )^{3} + 4 \, a b \cos \left (f x + e\right ) - {\left ({\left (3 \, a^{2} - 7 \, a b + 4 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - {\left (3 \, a^{2} - 10 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{2} - 3 \, a b + 4 \, b^{2}\right )} \sqrt {-\frac {b}{a - b}} \log \left (-\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} - 2 \, {\left (a - b\right )} \sqrt {-\frac {b}{a - b}} \cos \left (f x + e\right ) - b}{{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}\right ) - {\left ({\left (a^{2} - 5 \, a b + 4 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - {\left (a^{2} - 6 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{2} - a b + 4 \, b^{2}\right )} \log \left (\frac {1}{2} \, \cos \left (f x + e\right ) + \frac {1}{2}\right ) + {\left ({\left (a^{2} - 5 \, a b + 4 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - {\left (a^{2} - 6 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{2} - a b + 4 \, b^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (f x + e\right ) + \frac {1}{2}\right )}{4 \, {\left ({\left (a^{4} - a^{3} b\right )} f \cos \left (f x + e\right )^{4} - a^{3} b f - {\left (a^{4} - 2 \, a^{3} b\right )} f \cos \left (f x + e\right )^{2}\right )}}, \frac {2 \, {\left (a^{2} - 2 \, a b\right )} \cos \left (f x + e\right )^{3} + 4 \, a b \cos \left (f x + e\right ) - 2 \, {\left ({\left (3 \, a^{2} - 7 \, a b + 4 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - {\left (3 \, a^{2} - 10 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{2} - 3 \, a b + 4 \, b^{2}\right )} \sqrt {\frac {b}{a - b}} \arctan \left (-\frac {{\left (a - b\right )} \sqrt {\frac {b}{a - b}} \cos \left (f x + e\right )}{b}\right ) - {\left ({\left (a^{2} - 5 \, a b + 4 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - {\left (a^{2} - 6 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{2} - a b + 4 \, b^{2}\right )} \log \left (\frac {1}{2} \, \cos \left (f x + e\right ) + \frac {1}{2}\right ) + {\left ({\left (a^{2} - 5 \, a b + 4 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - {\left (a^{2} - 6 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{2} - a b + 4 \, b^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (f x + e\right ) + \frac {1}{2}\right )}{4 \, {\left ({\left (a^{4} - a^{3} b\right )} f \cos \left (f x + e\right )^{4} - a^{3} b f - {\left (a^{4} - 2 \, a^{3} b\right )} f \cos \left (f x + e\right )^{2}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*(2*(a^2 - 2*a*b)*cos(f*x + e)^3 + 4*a*b*cos(f*x + e) - ((3*a^2 - 7*a*b + 4*b^2)*cos(f*x + e)^4 - (3*a^2 -
10*a*b + 8*b^2)*cos(f*x + e)^2 - 3*a*b + 4*b^2)*sqrt(-b/(a - b))*log(-((a - b)*cos(f*x + e)^2 - 2*(a - b)*sqr
t(-b/(a - b))*cos(f*x + e) - b)/((a - b)*cos(f*x + e)^2 + b)) - ((a^2 - 5*a*b + 4*b^2)*cos(f*x + e)^4 - (a^2 -
6*a*b + 8*b^2)*cos(f*x + e)^2 - a*b + 4*b^2)*log(1/2*cos(f*x + e) + 1/2) + ((a^2 - 5*a*b + 4*b^2)*cos(f*x + e
)^4 - (a^2 - 6*a*b + 8*b^2)*cos(f*x + e)^2 - a*b + 4*b^2)*log(-1/2*cos(f*x + e) + 1/2))/((a^4 - a^3*b)*f*cos(f
*x + e)^4 - a^3*b*f - (a^4 - 2*a^3*b)*f*cos(f*x + e)^2), 1/4*(2*(a^2 - 2*a*b)*cos(f*x + e)^3 + 4*a*b*cos(f*x +
e) - 2*((3*a^2 - 7*a*b + 4*b^2)*cos(f*x + e)^4 - (3*a^2 - 10*a*b + 8*b^2)*cos(f*x + e)^2 - 3*a*b + 4*b^2)*sqr
t(b/(a - b))*arctan(-(a - b)*sqrt(b/(a - b))*cos(f*x + e)/b) - ((a^2 - 5*a*b + 4*b^2)*cos(f*x + e)^4 - (a^2 -
6*a*b + 8*b^2)*cos(f*x + e)^2 - a*b + 4*b^2)*log(1/2*cos(f*x + e) + 1/2) + ((a^2 - 5*a*b + 4*b^2)*cos(f*x + e)
^4 - (a^2 - 6*a*b + 8*b^2)*cos(f*x + e)^2 - a*b + 4*b^2)*log(-1/2*cos(f*x + e) + 1/2))/((a^4 - a^3*b)*f*cos(f*
x + e)^4 - a^3*b*f - (a^4 - 2*a^3*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)^3/(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)))*1/16/a^2+(-2*((1-
cos(f*x+exp(1)))/(1+cos(f*x+exp(1))))^3*a^2+8*((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1))))^3*a*b+((1-cos(f*x+exp(
1)))/(1+cos(f*x+exp(1))))^2*a^2-16*((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1))))^2*b^2+4*(1-cos(f*x+exp(1)))/(1+co
s(f*x+exp(1)))*a^2-28*(1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1)))*a*b-3*a^2)*1/48/a^3/(((1-cos(f*x+exp(1)))/(1+cos
(f*x+exp(1))))^3*a-2*((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1))))^2*a+4*((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1))))
^2*b+(1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1)))*a)+(a-4*b)*1/8/a^3*ln(abs(1-cos(f*x+exp(1)))/abs(1+cos(f*x+exp(1)
)))+(-3*a*b+4*b^2)*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))))

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maple [A]  time = 0.65, size = 229, normalized size = 1.56 \[ -\frac {b \cos \left (f x +e \right )}{2 f \,a^{2} \left (a \left (\cos ^{2}\left (f x +e \right )\right )-\left (\cos ^{2}\left (f x +e \right )\right ) b +b \right )}+\frac {3 b \arctan \left (\frac {\left (a -b \right ) \cos \left (f x +e \right )}{\sqrt {\left (a -b \right ) b}}\right )}{2 f \,a^{2} \sqrt {\left (a -b \right ) b}}-\frac {2 b^{2} \arctan \left (\frac {\left (a -b \right ) \cos \left (f x +e \right )}{\sqrt {\left (a -b \right ) b}}\right )}{f \,a^{3} \sqrt {\left (a -b \right ) b}}+\frac {1}{4 f \,a^{2} \left (-1+\cos \left (f x +e \right )\right )}+\frac {\ln \left (-1+\cos \left (f x +e \right )\right )}{4 f \,a^{2}}-\frac {\ln \left (-1+\cos \left (f x +e \right )\right ) b}{f \,a^{3}}+\frac {1}{4 f \,a^{2} \left (1+\cos \left (f x +e \right )\right )}-\frac {\ln \left (1+\cos \left (f x +e \right )\right )}{4 f \,a^{2}}+\frac {\ln \left (1+\cos \left (f x +e \right )\right ) b}{f \,a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(f*x+e)^3/(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)+3/2/f*b/a^2/((a-b)*b)^(1/2)*arctan((a-b)*cos(f*x+e)/
((a-b)*b)^(1/2))-2/f*b^2/a^3/((a-b)*b)^(1/2)*arctan((a-b)*cos(f*x+e)/((a-b)*b)^(1/2))+1/4/f/a^2/(-1+cos(f*x+e)
)+1/4/f/a^2*ln(-1+cos(f*x+e))-1/f/a^3*ln(-1+cos(f*x+e))*b+1/4/f/a^2/(1+cos(f*x+e))-1/4/f/a^2*ln(1+cos(f*x+e))+
1/f/a^3*ln(1+cos(f*x+e))*b

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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)^3/(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 = 12.37, size = 917, normalized size = 6.24 \[ \frac {{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2}{8\,a^2\,f}-\frac {\frac {a}{2}-{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (a-6\,b\right )+\frac {{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4\,\left (a^2-8\,a\,b+16\,b^2\right )}{2\,a}}{f\,\left (4\,a^3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+4\,a^3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4\,\left (16\,a^2\,b-8\,a^3\right )\right )}+\frac {\ln \left (\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\right )\,\left (a-4\,b\right )}{2\,a^3\,f}+\frac {\sqrt {b}\,\mathrm {atan}\left (\frac {2\,a^2\,\left (\frac {\sqrt {b}\,\left (3\,a-4\,b\right )\,\left (12\,a^6\,b-106\,a^5\,b^2+240\,a^4\,b^3-160\,a^3\,b^4\right )}{2\,a^9\,\sqrt {a-b}}+\frac {b^{3/2}\,{\left (3\,a-4\,b\right )}^3\,\left (8\,a^{11}-32\,a^{10}\,b+32\,a^9\,b^2\right )}{32\,a^{15}\,{\left (a-b\right )}^{3/2}}\right )\,\left (a-b\right )\,\left (15\,a^4-182\,a^3\,b+648\,a^2\,b^2-864\,a\,b^3+384\,b^4\right )}{\left (9\,a^2\,b-24\,a\,b^2+16\,b^3\right )\,\left (4\,a^3-27\,a^2\,b+72\,a\,b^2-48\,b^3\right )}-\frac {4\,a^7\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,{\left (a-b\right )}^{3/2}\,\left (\frac {\left (\frac {4\,\left (9\,a^2\,b^2-24\,a\,b^3+16\,b^4\right )}{a^5}-\frac {b\,{\left (3\,a-4\,b\right )}^2\,\left (2\,a^8-46\,a^7\,b+344\,a^6\,b^2-672\,a^5\,b^3+384\,a^4\,b^4\right )}{4\,a^{11}\,\left (a-b\right )}\right )\,\left (a^4-31\,a^3\,b+180\,a^2\,b^2-336\,a\,b^3+192\,b^4\right )}{\sqrt {b}\,\left (b\,\left (27\,a^7+b\,\left (48\,a^5\,b-72\,a^6\right )\right )-4\,a^8\right )}+\frac {\left (\frac {\sqrt {b}\,\left (3\,a-4\,b\right )\,\left (9\,a^5\,b-78\,a^4\,b^2+268\,a^3\,b^3-384\,a^2\,b^4+192\,a\,b^5\right )}{a^8\,\sqrt {a-b}}-\frac {b^{3/2}\,{\left (3\,a-4\,b\right )}^3\,\left (-4\,a^{10}+104\,a^9\,b-288\,a^8\,b^2+192\,a^7\,b^3\right )}{16\,a^{14}\,{\left (a-b\right )}^{3/2}}\right )\,\left (15\,a^4-182\,a^3\,b+648\,a^2\,b^2-864\,a\,b^3+384\,b^4\right )}{2\,a^5\,\sqrt {a-b}\,\left (4\,a^3-27\,a^2\,b+72\,a\,b^2-48\,b^3\right )}\right )}{9\,a^2\,b-24\,a\,b^2+16\,b^3}+\frac {4\,a^7\,{\left (a-b\right )}^{3/2}\,\left (\frac {2\,\left (9\,a^3\,b^2-60\,a^2\,b^3+112\,a\,b^4-64\,b^5\right )}{a^6}+\frac {b\,{\left (3\,a-4\,b\right )}^2\,\left (-4\,a^9+56\,a^8\,b-160\,a^7\,b^2+128\,a^6\,b^3\right )}{8\,a^{12}\,\left (a-b\right )}\right )\,\left (a^4-31\,a^3\,b+180\,a^2\,b^2-336\,a\,b^3+192\,b^4\right )}{\sqrt {b}\,\left (b\,\left (27\,a^7+b\,\left (48\,a^5\,b-72\,a^6\right )\right )-4\,a^8\right )\,\left (9\,a^2\,b-24\,a\,b^2+16\,b^3\right )}\right )\,\left (3\,a-4\,b\right )}{2\,a^3\,f\,\sqrt {a-b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sin(e + f*x)^3*(a + b*tan(e + f*x)^2)^2),x)

[Out]

tan(e/2 + (f*x)/2)^2/(8*a^2*f) - (a/2 - tan(e/2 + (f*x)/2)^2*(a - 6*b) + (tan(e/2 + (f*x)/2)^4*(a^2 - 8*a*b +
16*b^2))/(2*a))/(f*(4*a^3*tan(e/2 + (f*x)/2)^2 + 4*a^3*tan(e/2 + (f*x)/2)^6 + tan(e/2 + (f*x)/2)^4*(16*a^2*b -
8*a^3))) + (log(tan(e/2 + (f*x)/2))*(a - 4*b))/(2*a^3*f) + (b^(1/2)*atan((2*a^2*((b^(1/2)*(3*a - 4*b)*(12*a^6
*b - 160*a^3*b^4 + 240*a^4*b^3 - 106*a^5*b^2))/(2*a^9*(a - b)^(1/2)) + (b^(3/2)*(3*a - 4*b)^3*(8*a^11 - 32*a^1
0*b + 32*a^9*b^2))/(32*a^15*(a - b)^(3/2)))*(a - b)*(15*a^4 - 182*a^3*b - 864*a*b^3 + 384*b^4 + 648*a^2*b^2))/
((9*a^2*b - 24*a*b^2 + 16*b^3)*(72*a*b^2 - 27*a^2*b + 4*a^3 - 48*b^3)) - (4*a^7*tan(e/2 + (f*x)/2)^2*(a - b)^(
3/2)*((((4*(16*b^4 - 24*a*b^3 + 9*a^2*b^2))/a^5 - (b*(3*a - 4*b)^2*(2*a^8 - 46*a^7*b + 384*a^4*b^4 - 672*a^5*b
^3 + 344*a^6*b^2))/(4*a^11*(a - b)))*(a^4 - 31*a^3*b - 336*a*b^3 + 192*b^4 + 180*a^2*b^2))/(b^(1/2)*(b*(27*a^7
+ b*(48*a^5*b - 72*a^6)) - 4*a^8)) + (((b^(1/2)*(3*a - 4*b)*(192*a*b^5 + 9*a^5*b - 384*a^2*b^4 + 268*a^3*b^3
- 78*a^4*b^2))/(a^8*(a - b)^(1/2)) - (b^(3/2)*(3*a - 4*b)^3*(104*a^9*b - 4*a^10 + 192*a^7*b^3 - 288*a^8*b^2))/
(16*a^14*(a - b)^(3/2)))*(15*a^4 - 182*a^3*b - 864*a*b^3 + 384*b^4 + 648*a^2*b^2))/(2*a^5*(a - b)^(1/2)*(72*a*
b^2 - 27*a^2*b + 4*a^3 - 48*b^3))))/(9*a^2*b - 24*a*b^2 + 16*b^3) + (4*a^7*(a - b)^(3/2)*((2*(112*a*b^4 - 64*b
^5 - 60*a^2*b^3 + 9*a^3*b^2))/a^6 + (b*(3*a - 4*b)^2*(56*a^8*b - 4*a^9 + 128*a^6*b^3 - 160*a^7*b^2))/(8*a^12*(
a - b)))*(a^4 - 31*a^3*b - 336*a*b^3 + 192*b^4 + 180*a^2*b^2))/(b^(1/2)*(b*(27*a^7 + b*(48*a^5*b - 72*a^6)) -
4*a^8)*(9*a^2*b - 24*a*b^2 + 16*b^3)))*(3*a - 4*b))/(2*a^3*f*(a - b)^(1/2))

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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*tan(f*x+e)**2)**2,x)

[Out]

Timed out

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