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

Optimal. Leaf size=82 \[ -\frac {a^2 \csc ^2(e+f x) \sec (e+f x)}{2 f}+\frac {a (a+4 b) \sec (e+f x)}{2 f}-\frac {a (a+4 b) \tanh ^{-1}(\cos (e+f x))}{2 f}+\frac {b^2 \sec ^3(e+f x)}{3 f} \]

[Out]

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

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Rubi [A]  time = 0.11, antiderivative size = 82, 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 = {3664, 463, 459, 321, 207} \[ -\frac {a^2 \csc ^2(e+f x) \sec (e+f x)}{2 f}+\frac {a (a+4 b) \sec (e+f x)}{2 f}-\frac {a (a+4 b) \tanh ^{-1}(\cos (e+f x))}{2 f}+\frac {b^2 \sec ^3(e+f x)}{3 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 321

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

Rule 459

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

Rule 463

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

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Mathematica [B]  time = 6.13, size = 376, normalized size = 4.59 \[ \frac {\left (a^2+4 a b\right ) \log \left (\sin \left (\frac {1}{2} (e+f x)\right )\right )}{2 f}+\frac {\left (-a^2-4 a b\right ) \log \left (\cos \left (\frac {1}{2} (e+f x)\right )\right )}{2 f}-\frac {a^2 \csc ^2\left (\frac {1}{2} (e+f x)\right )}{8 f}+\frac {a^2 \sec ^2\left (\frac {1}{2} (e+f x)\right )}{8 f}+\frac {b^2 \left (-\sin \left (\frac {1}{2} (e+f x)\right )\right )-12 a b \sin \left (\frac {1}{2} (e+f x)\right )}{6 f \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )}+\frac {12 a b \sin \left (\frac {1}{2} (e+f x)\right )+b^2 \sin \left (\frac {1}{2} (e+f x)\right )}{6 f \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )}+\frac {b^2 \sin \left (\frac {1}{2} (e+f x)\right )}{6 f \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^3}+\frac {b^2}{12 f \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^2}+\frac {b^2}{12 f \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^2}-\frac {b^2 \sin \left (\frac {1}{2} (e+f x)\right )}{6 f \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/8*(a^2*Csc[(e + f*x)/2]^2)/f + ((-a^2 - 4*a*b)*Log[Cos[(e + f*x)/2]])/(2*f) + ((a^2 + 4*a*b)*Log[Sin[(e + f
*x)/2]])/(2*f) + (a^2*Sec[(e + f*x)/2]^2)/(8*f) + b^2/(12*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^2) + (b^2*Si
n[(e + f*x)/2])/(6*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^3) - (b^2*Sin[(e + f*x)/2])/(6*f*(Cos[(e + f*x)/2]
+ Sin[(e + f*x)/2])^3) + b^2/(12*f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^2) + (-12*a*b*Sin[(e + f*x)/2] - b^2*
Sin[(e + f*x)/2])/(6*f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])) + (12*a*b*Sin[(e + f*x)/2] + b^2*Sin[(e + f*x)/2
])/(6*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2]))

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fricas [B]  time = 0.44, size = 168, normalized size = 2.05 \[ \frac {6 \, {\left (a^{2} + 4 \, a b\right )} \cos \left (f x + e\right )^{4} - 4 \, {\left (6 \, a b - b^{2}\right )} \cos \left (f x + e\right )^{2} - 4 \, b^{2} - 3 \, {\left ({\left (a^{2} + 4 \, a b\right )} \cos \left (f x + e\right )^{5} - {\left (a^{2} + 4 \, a b\right )} \cos \left (f x + e\right )^{3}\right )} \log \left (\frac {1}{2} \, \cos \left (f x + e\right ) + \frac {1}{2}\right ) + 3 \, {\left ({\left (a^{2} + 4 \, a b\right )} \cos \left (f x + e\right )^{5} - {\left (a^{2} + 4 \, a b\right )} \cos \left (f x + e\right )^{3}\right )} \log \left (-\frac {1}{2} \, \cos \left (f x + e\right ) + \frac {1}{2}\right )}{12 \, {\left (f \cos \left (f x + e\right )^{5} - f \cos \left (f x + e\right )^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(6*(a^2 + 4*a*b)*cos(f*x + e)^4 - 4*(6*a*b - b^2)*cos(f*x + e)^2 - 4*b^2 - 3*((a^2 + 4*a*b)*cos(f*x + e)^
5 - (a^2 + 4*a*b)*cos(f*x + e)^3)*log(1/2*cos(f*x + e) + 1/2) + 3*((a^2 + 4*a*b)*cos(f*x + e)^5 - (a^2 + 4*a*b
)*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*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*((-2*(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)*1/16/(1-cos(f*x+exp(1)))*(1+cos(f*x+exp(1)))+(-6*((1-cos(f*x+exp(1))
)/(1+cos(f*x+exp(1))))^2*a*b-3*((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1))))^2*b^2+12*(1-cos(f*x+exp(1)))/(1+cos(f
*x+exp(1)))*a*b-6*a*b-b^2)*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+ex
p(1)))*a^2/16+(a^2+4*a*b)/8*ln(abs(1-cos(f*x+exp(1)))/abs(1+cos(f*x+exp(1)))))

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maple [A]  time = 0.74, size = 100, normalized size = 1.22 \[ -\frac {a^{2} \csc \left (f x +e \right ) \cot \left (f x +e \right )}{2 f}+\frac {a^{2} \ln \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right )}{2 f}+\frac {2 a b}{f \cos \left (f x +e \right )}+\frac {2 a b \ln \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right )}{f}+\frac {b^{2}}{3 f \cos \left (f x +e \right )^{3}} \]

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

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maxima [A]  time = 0.62, size = 111, normalized size = 1.35 \[ -\frac {3 \, {\left (a^{2} + 4 \, a b\right )} \log \left (\cos \left (f x + e\right ) + 1\right ) - 3 \, {\left (a^{2} + 4 \, a b\right )} \log \left (\cos \left (f x + e\right ) - 1\right ) - \frac {2 \, {\left (3 \, {\left (a^{2} + 4 \, a b\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (6 \, a b - b^{2}\right )} \cos \left (f x + e\right )^{2} - 2 \, b^{2}\right )}}{\cos \left (f x + e\right )^{5} - \cos \left (f x + e\right )^{3}}}{12 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(3*(a^2 + 4*a*b)*log(cos(f*x + e) + 1) - 3*(a^2 + 4*a*b)*log(cos(f*x + e) - 1) - 2*(3*(a^2 + 4*a*b)*cos(
f*x + e)^4 - 2*(6*a*b - 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 = 12.61, size = 188, normalized size = 2.29 \[ \frac {\ln \left (\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\right )\,\left (\frac {a^2}{2}+2\,b\,a\right )}{f}+\frac {a^2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2}{8\,f}-\frac {{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4\,\left (\frac {3\,a^2}{2}+32\,b\,a\right )-{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6\,\left (\frac {a^2}{2}+16\,a\,b+8\,b^2\right )-{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (\frac {3\,a^2}{2}+16\,a\,b+\frac {8\,b^2}{3}\right )+\frac {a^2}{2}}{f\,\left (-4\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^8+12\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6-12\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4+4\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(log(tan(e/2 + (f*x)/2))*(2*a*b + a^2/2))/f + (a^2*tan(e/2 + (f*x)/2)^2)/(8*f) - (tan(e/2 + (f*x)/2)^4*(32*a*b
+ (3*a^2)/2) - tan(e/2 + (f*x)/2)^6*(16*a*b + a^2/2 + 8*b^2) - tan(e/2 + (f*x)/2)^2*(16*a*b + (3*a^2)/2 + (8*
b^2)/3) + a^2/2)/(f*(4*tan(e/2 + (f*x)/2)^2 - 12*tan(e/2 + (f*x)/2)^4 + 12*tan(e/2 + (f*x)/2)^6 - 4*tan(e/2 +
(f*x)/2)^8))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a + b \tan ^{2}{\left (e + f x \right )}\right )^{2} \csc ^{3}{\left (e + f x \right )}\, dx \]

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]

Integral((a + b*tan(e + f*x)**2)**2*csc(e + f*x)**3, x)

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