3.2.50 \(\int \frac {\csc ^4(e+f x)}{(a+b \tan ^2(e+f x))^{5/2}} \, dx\) [150]

Optimal. Leaf size=146 \[ -\frac {(a-2 b) \cot (e+f x)}{a^2 f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac {\cot ^3(e+f x)}{3 a f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac {4 (a-2 b) b \tan (e+f x)}{3 a^3 f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac {8 (a-2 b) b \tan (e+f x)}{3 a^4 f \sqrt {a+b \tan ^2(e+f x)}} \]

[Out]

-8/3*(a-2*b)*b*tan(f*x+e)/a^4/f/(a+b*tan(f*x+e)^2)^(1/2)-(a-2*b)*cot(f*x+e)/a^2/f/(a+b*tan(f*x+e)^2)^(3/2)-1/3
*cot(f*x+e)^3/a/f/(a+b*tan(f*x+e)^2)^(3/2)-4/3*(a-2*b)*b*tan(f*x+e)/a^3/f/(a+b*tan(f*x+e)^2)^(3/2)

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Rubi [A]
time = 0.09, antiderivative size = 146, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3744, 464, 277, 198, 197} \begin {gather*} -\frac {8 b (a-2 b) \tan (e+f x)}{3 a^4 f \sqrt {a+b \tan ^2(e+f x)}}-\frac {4 b (a-2 b) \tan (e+f x)}{3 a^3 f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac {(a-2 b) \cot (e+f x)}{a^2 f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac {\cot ^3(e+f x)}{3 a f \left (a+b \tan ^2(e+f x)\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Csc[e + f*x]^4/(a + b*Tan[e + f*x]^2)^(5/2),x]

[Out]

-(((a - 2*b)*Cot[e + f*x])/(a^2*f*(a + b*Tan[e + f*x]^2)^(3/2))) - Cot[e + f*x]^3/(3*a*f*(a + b*Tan[e + f*x]^2
)^(3/2)) - (4*(a - 2*b)*b*Tan[e + f*x])/(3*a^3*f*(a + b*Tan[e + f*x]^2)^(3/2)) - (8*(a - 2*b)*b*Tan[e + f*x])/
(3*a^4*f*Sqrt[a + b*Tan[e + f*x]^2])

Rule 197

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x*((a + b*x^n)^(p + 1)/a), x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 0]

Rule 198

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^n)^(p + 1)/(a*n*(p + 1))), x] + Dist[(n*(p
+ 1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, n, p}, x] && ILtQ[Simplify[1/n + p +
1], 0] && NeQ[p, -1]

Rule 277

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

Rule 464

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 3744

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

Rubi steps

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

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Mathematica [A]
time = 0.78, size = 140, normalized size = 0.96 \begin {gather*} \frac {\sqrt {(a+b+(a-b) \cos (2 (e+f x))) \sec ^2(e+f x)} \left (-\cot (e+f x) \left (2 a-8 b+a \csc ^2(e+f x)\right )+\frac {2 b \left (-3 a^2+2 a b+4 b^2+\left (-3 a^2+7 a b-4 b^2\right ) \cos (2 (e+f x))\right ) \sin (2 (e+f x))}{(a+b+(a-b) \cos (2 (e+f x)))^2}\right )}{3 \sqrt {2} a^4 f} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Csc[e + f*x]^4/(a + b*Tan[e + f*x]^2)^(5/2),x]

[Out]

(Sqrt[(a + b + (a - b)*Cos[2*(e + f*x)])*Sec[e + f*x]^2]*(-(Cot[e + f*x]*(2*a - 8*b + a*Csc[e + f*x]^2)) + (2*
b*(-3*a^2 + 2*a*b + 4*b^2 + (-3*a^2 + 7*a*b - 4*b^2)*Cos[2*(e + f*x)])*Sin[2*(e + f*x)])/(a + b + (a - b)*Cos[
2*(e + f*x)])^2))/(3*Sqrt[2]*a^4*f)

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Maple [A]
time = 0.55, size = 245, normalized size = 1.68

method	result	size
default	\(\frac {\left (2 \left (\cos ^{6}\left (f x +e \right )\right ) a^{3}-18 \left (\cos ^{6}\left (f x +e \right )\right ) a^{2} b +32 \left (\cos ^{6}\left (f x +e \right )\right ) a \,b^{2}-16 \left (\cos ^{6}\left (f x +e \right )\right ) b^{3}-3 \left (\cos ^{4}\left (f x +e \right )\right ) a^{3}+30 \left (\cos ^{4}\left (f x +e \right )\right ) a^{2} b -72 \left (\cos ^{4}\left (f x +e \right )\right ) a \,b^{2}+48 \left (\cos ^{4}\left (f x +e \right )\right ) b^{3}-12 \left (\cos ^{2}\left (f x +e \right )\right ) a^{2} b +48 \left (\cos ^{2}\left (f x +e \right )\right ) a \,b^{2}-48 \left (\cos ^{2}\left (f x +e \right )\right ) b^{3}-8 a \,b^{2}+16 b^{3}\right ) \left (\cos ^{5}\left (f x +e \right )\right ) \left (\frac {a \left (\cos ^{2}\left (f x +e \right )\right )-\left (\cos ^{2}\left (f x +e \right )\right ) b +b}{\cos \left (f x +e \right )^{2}}\right )^{\frac {5}{2}}}{3 f \left (a \left (\cos ^{2}\left (f x +e \right )\right )-\left (\cos ^{2}\left (f x +e \right )\right ) b +b \right )^{4} \sin \left (f x +e \right )^{3} a^{4}}\)	\(245\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(f*x+e)^4/(a+b*tan(f*x+e)^2)^(5/2),x,method=_RETURNVERBOSE)

[Out]

1/3/f/(a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)^4*(2*cos(f*x+e)^6*a^3-18*cos(f*x+e)^6*a^2*b+32*cos(f*x+e)^6*a*b^2-16*c
os(f*x+e)^6*b^3-3*cos(f*x+e)^4*a^3+30*cos(f*x+e)^4*a^2*b-72*cos(f*x+e)^4*a*b^2+48*cos(f*x+e)^4*b^3-12*cos(f*x+
e)^2*a^2*b+48*cos(f*x+e)^2*a*b^2-48*cos(f*x+e)^2*b^3-8*a*b^2+16*b^3)*cos(f*x+e)^5*((a*cos(f*x+e)^2-cos(f*x+e)^
2*b+b)/cos(f*x+e)^2)^(5/2)/sin(f*x+e)^3/a^4

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Maxima [A]
time = 0.29, size = 209, normalized size = 1.43 \begin {gather*} -\frac {\frac {8 \, b \tan \left (f x + e\right )}{\sqrt {b \tan \left (f x + e\right )^{2} + a} a^{3}} + \frac {4 \, b \tan \left (f x + e\right )}{{\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} a^{2}} - \frac {16 \, b^{2} \tan \left (f x + e\right )}{\sqrt {b \tan \left (f x + e\right )^{2} + a} a^{4}} - \frac {8 \, b^{2} \tan \left (f x + e\right )}{{\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} a^{3}} + \frac {3}{{\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} a \tan \left (f x + e\right )} - \frac {6 \, b}{{\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} a^{2} \tan \left (f x + e\right )} + \frac {1}{{\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} a \tan \left (f x + e\right )^{3}}}{3 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^4/(a+b*tan(f*x+e)^2)^(5/2),x, algorithm="maxima")

[Out]

-1/3*(8*b*tan(f*x + e)/(sqrt(b*tan(f*x + e)^2 + a)*a^3) + 4*b*tan(f*x + e)/((b*tan(f*x + e)^2 + a)^(3/2)*a^2)
- 16*b^2*tan(f*x + e)/(sqrt(b*tan(f*x + e)^2 + a)*a^4) - 8*b^2*tan(f*x + e)/((b*tan(f*x + e)^2 + a)^(3/2)*a^3)
+ 3/((b*tan(f*x + e)^2 + a)^(3/2)*a*tan(f*x + e)) - 6*b/((b*tan(f*x + e)^2 + a)^(3/2)*a^2*tan(f*x + e)) + 1/(
(b*tan(f*x + e)^2 + a)^(3/2)*a*tan(f*x + e)^3))/f

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Fricas [A]
time = 95.74, size = 250, normalized size = 1.71 \begin {gather*} -\frac {{\left (2 \, {\left (a^{3} - 9 \, a^{2} b + 16 \, a b^{2} - 8 \, b^{3}\right )} \cos \left (f x + e\right )^{7} - 3 \, {\left (a^{3} - 10 \, a^{2} b + 24 \, a b^{2} - 16 \, b^{3}\right )} \cos \left (f x + e\right )^{5} - 12 \, {\left (a^{2} b - 4 \, a b^{2} + 4 \, b^{3}\right )} \cos \left (f x + e\right )^{3} - 8 \, {\left (a b^{2} - 2 \, b^{3}\right )} \cos \left (f x + e\right )\right )} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{3 \, {\left ({\left (a^{6} - 2 \, a^{5} b + a^{4} b^{2}\right )} f \cos \left (f x + e\right )^{6} - a^{4} b^{2} f - {\left (a^{6} - 4 \, a^{5} b + 3 \, a^{4} b^{2}\right )} f \cos \left (f x + e\right )^{4} - {\left (2 \, a^{5} b - 3 \, a^{4} b^{2}\right )} f \cos \left (f x + e\right )^{2}\right )} \sin \left (f x + e\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^4/(a+b*tan(f*x+e)^2)^(5/2),x, algorithm="fricas")

[Out]

-1/3*(2*(a^3 - 9*a^2*b + 16*a*b^2 - 8*b^3)*cos(f*x + e)^7 - 3*(a^3 - 10*a^2*b + 24*a*b^2 - 16*b^3)*cos(f*x + e
)^5 - 12*(a^2*b - 4*a*b^2 + 4*b^3)*cos(f*x + e)^3 - 8*(a*b^2 - 2*b^3)*cos(f*x + e))*sqrt(((a - b)*cos(f*x + e)
^2 + b)/cos(f*x + e)^2)/(((a^6 - 2*a^5*b + a^4*b^2)*f*cos(f*x + e)^6 - a^4*b^2*f - (a^6 - 4*a^5*b + 3*a^4*b^2)
*f*cos(f*x + e)^4 - (2*a^5*b - 3*a^4*b^2)*f*cos(f*x + e)^2)*sin(f*x + e))

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\csc ^{4}{\left (e + f x \right )}}{\left (a + b \tan ^{2}{\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)**4/(a+b*tan(f*x+e)**2)**(5/2),x)

[Out]

Integral(csc(e + f*x)**4/(a + b*tan(e + f*x)**2)**(5/2), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^4/(a+b*tan(f*x+e)^2)^(5/2),x, algorithm="giac")

[Out]

integrate(csc(f*x + e)^4/(b*tan(f*x + e)^2 + a)^(5/2), x)

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Mupad [F(-1)]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \text {Hanged} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sin(e + f*x)^4*(a + b*tan(e + f*x)^2)^(5/2)),x)

[Out]

\text{Hanged}

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