3.115 \(\int \csc ^6(e+f x) (a+b \tan ^2(e+f x))^{3/2} \, dx\)

Optimal. Leaf size=196 \[ \frac{b (3 a+4 b) \tan (e+f x) \sqrt{a+b \tan ^2(e+f x)}}{2 a f}+\frac{\sqrt{b} (3 a+4 b) \tanh ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b \tan ^2(e+f x)}}\right )}{2 f}-\frac{\cot ^5(e+f x) \left (a+b \tan ^2(e+f x)\right )^{5/2}}{5 a f}-\frac{2 \cot ^3(e+f x) \left (a+b \tan ^2(e+f x)\right )^{5/2}}{3 a f}-\frac{(3 a+4 b) \cot (e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2}}{3 a f} \]

[Out]

(Sqrt[b]*(3*a + 4*b)*ArcTanh[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b*Tan[e + f*x]^2]])/(2*f) + (b*(3*a + 4*b)*Tan[e
+ f*x]*Sqrt[a + b*Tan[e + f*x]^2])/(2*a*f) - ((3*a + 4*b)*Cot[e + f*x]*(a + b*Tan[e + f*x]^2)^(3/2))/(3*a*f) -
(2*Cot[e + f*x]^3*(a + b*Tan[e + f*x]^2)^(5/2))/(3*a*f) - (Cot[e + f*x]^5*(a + b*Tan[e + f*x]^2)^(5/2))/(5*a*
f)

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Rubi [A]  time = 0.170512, antiderivative size = 196, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.28, Rules used = {3663, 462, 453, 277, 195, 217, 206} \[ \frac{b (3 a+4 b) \tan (e+f x) \sqrt{a+b \tan ^2(e+f x)}}{2 a f}+\frac{\sqrt{b} (3 a+4 b) \tanh ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b \tan ^2(e+f x)}}\right )}{2 f}-\frac{\cot ^5(e+f x) \left (a+b \tan ^2(e+f x)\right )^{5/2}}{5 a f}-\frac{2 \cot ^3(e+f x) \left (a+b \tan ^2(e+f x)\right )^{5/2}}{3 a f}-\frac{(3 a+4 b) \cot (e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2}}{3 a f} \]

Antiderivative was successfully verified.

[In]

Int[Csc[e + f*x]^6*(a + b*Tan[e + f*x]^2)^(3/2),x]

[Out]

(Sqrt[b]*(3*a + 4*b)*ArcTanh[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b*Tan[e + f*x]^2]])/(2*f) + (b*(3*a + 4*b)*Tan[e
+ f*x]*Sqrt[a + b*Tan[e + f*x]^2])/(2*a*f) - ((3*a + 4*b)*Cot[e + f*x]*(a + b*Tan[e + f*x]^2)^(3/2))/(3*a*f) -
(2*Cot[e + f*x]^3*(a + b*Tan[e + f*x]^2)^(5/2))/(3*a*f) - (Cot[e + f*x]^5*(a + b*Tan[e + f*x]^2)^(5/2))/(5*a*
f)

Rule 3663

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]

Rule 462

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

Rule 453

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 277

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

Rule 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 206

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

Rubi steps

\begin{align*} \int \csc ^6(e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (1+x^2\right )^2 \left (a+b x^2\right )^{3/2}}{x^6} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac{\cot ^5(e+f x) \left (a+b \tan ^2(e+f x)\right )^{5/2}}{5 a f}+\frac{\operatorname{Subst}\left (\int \frac{\left (10 a+5 a x^2\right ) \left (a+b x^2\right )^{3/2}}{x^4} \, dx,x,\tan (e+f x)\right )}{5 a f}\\ &=-\frac{2 \cot ^3(e+f x) \left (a+b \tan ^2(e+f x)\right )^{5/2}}{3 a f}-\frac{\cot ^5(e+f x) \left (a+b \tan ^2(e+f x)\right )^{5/2}}{5 a f}+\frac{(3 a+4 b) \operatorname{Subst}\left (\int \frac{\left (a+b x^2\right )^{3/2}}{x^2} \, dx,x,\tan (e+f x)\right )}{3 a f}\\ &=-\frac{(3 a+4 b) \cot (e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2}}{3 a f}-\frac{2 \cot ^3(e+f x) \left (a+b \tan ^2(e+f x)\right )^{5/2}}{3 a f}-\frac{\cot ^5(e+f x) \left (a+b \tan ^2(e+f x)\right )^{5/2}}{5 a f}+\frac{(b (3 a+4 b)) \operatorname{Subst}\left (\int \sqrt{a+b x^2} \, dx,x,\tan (e+f x)\right )}{a f}\\ &=\frac{b (3 a+4 b) \tan (e+f x) \sqrt{a+b \tan ^2(e+f x)}}{2 a f}-\frac{(3 a+4 b) \cot (e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2}}{3 a f}-\frac{2 \cot ^3(e+f x) \left (a+b \tan ^2(e+f x)\right )^{5/2}}{3 a f}-\frac{\cot ^5(e+f x) \left (a+b \tan ^2(e+f x)\right )^{5/2}}{5 a f}+\frac{(b (3 a+4 b)) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x^2}} \, dx,x,\tan (e+f x)\right )}{2 f}\\ &=\frac{b (3 a+4 b) \tan (e+f x) \sqrt{a+b \tan ^2(e+f x)}}{2 a f}-\frac{(3 a+4 b) \cot (e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2}}{3 a f}-\frac{2 \cot ^3(e+f x) \left (a+b \tan ^2(e+f x)\right )^{5/2}}{3 a f}-\frac{\cot ^5(e+f x) \left (a+b \tan ^2(e+f x)\right )^{5/2}}{5 a f}+\frac{(b (3 a+4 b)) \operatorname{Subst}\left (\int \frac{1}{1-b x^2} \, dx,x,\frac{\tan (e+f x)}{\sqrt{a+b \tan ^2(e+f x)}}\right )}{2 f}\\ &=\frac{\sqrt{b} (3 a+4 b) \tanh ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b \tan ^2(e+f x)}}\right )}{2 f}+\frac{b (3 a+4 b) \tan (e+f x) \sqrt{a+b \tan ^2(e+f x)}}{2 a f}-\frac{(3 a+4 b) \cot (e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2}}{3 a f}-\frac{2 \cot ^3(e+f x) \left (a+b \tan ^2(e+f x)\right )^{5/2}}{3 a f}-\frac{\cot ^5(e+f x) \left (a+b \tan ^2(e+f x)\right )^{5/2}}{5 a f}\\ \end{align*}

Mathematica [C]  time = 2.04711, size = 213, normalized size = 1.09 \[ \frac{\sqrt{\sec ^2(e+f x) ((a-b) \cos (2 (e+f x))+a+b)} \left (\frac{15 \sqrt{2} (3 a+4 b) \cot (e+f x) \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{\frac{\csc ^2(e+f x) ((a-b) \cos (2 (e+f x))+a+b)}{b}}}{\sqrt{2}}\right ),1\right )}{\sqrt{\frac{\csc ^2(e+f x) ((a-b) \cos (2 (e+f x))+a+b)}{b}}}-\frac{2 \left (8 a^2+34 a b+3 b^2\right ) \cot (e+f x)}{a}-4 (2 a+3 b) \cot (e+f x) \csc ^2(e+f x)-6 a \cot (e+f x) \csc ^4(e+f x)+15 b \tan (e+f x)\right )}{30 \sqrt{2} f} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[e + f*x]^6*(a + b*Tan[e + f*x]^2)^(3/2),x]

[Out]

(Sqrt[(a + b + (a - b)*Cos[2*(e + f*x)])*Sec[e + f*x]^2]*((-2*(8*a^2 + 34*a*b + 3*b^2)*Cot[e + f*x])/a - 4*(2*
a + 3*b)*Cot[e + f*x]*Csc[e + f*x]^2 - 6*a*Cot[e + f*x]*Csc[e + f*x]^4 + (15*Sqrt[2]*(3*a + 4*b)*Cot[e + f*x]*
EllipticF[ArcSin[Sqrt[((a + b + (a - b)*Cos[2*(e + f*x)])*Csc[e + f*x]^2)/b]/Sqrt[2]], 1])/Sqrt[((a + b + (a -
b)*Cos[2*(e + f*x)])*Csc[e + f*x]^2)/b] + 15*b*Tan[e + f*x]))/(30*Sqrt[2]*f)

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Maple [C]  time = 0.868, size = 6988, normalized size = 35.7 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(f*x+e)^6*(a+b*tan(f*x+e)^2)^(3/2),x)

[Out]

result too large to display

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^6*(a+b*tan(f*x+e)^2)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 26.6989, size = 1624, normalized size = 8.29 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^6*(a+b*tan(f*x+e)^2)^(3/2),x, algorithm="fricas")

[Out]

[1/120*(15*((3*a^2 + 4*a*b)*cos(f*x + e)^5 - 2*(3*a^2 + 4*a*b)*cos(f*x + e)^3 + (3*a^2 + 4*a*b)*cos(f*x + e))*
sqrt(b)*log(((a^2 - 8*a*b + 8*b^2)*cos(f*x + e)^4 + 8*(a*b - 2*b^2)*cos(f*x + e)^2 + 4*((a - 2*b)*cos(f*x + e)
^3 + 2*b*cos(f*x + e))*sqrt(b)*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*sin(f*x + e) + 8*b^2)/cos(f*x
+ e)^4)*sin(f*x + e) - 4*((16*a^2 + 83*a*b + 6*b^2)*cos(f*x + e)^6 - (40*a^2 + 193*a*b + 12*b^2)*cos(f*x + e)
^4 + (30*a^2 + 125*a*b + 6*b^2)*cos(f*x + e)^2 - 15*a*b)*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/((
a*f*cos(f*x + e)^5 - 2*a*f*cos(f*x + e)^3 + a*f*cos(f*x + e))*sin(f*x + e)), -1/60*(15*((3*a^2 + 4*a*b)*cos(f*
x + e)^5 - 2*(3*a^2 + 4*a*b)*cos(f*x + e)^3 + (3*a^2 + 4*a*b)*cos(f*x + e))*sqrt(-b)*arctan(1/2*((a - 2*b)*cos
(f*x + e)^3 + 2*b*cos(f*x + e))*sqrt(-b)*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2)/(((a*b - b^2)*cos(f
*x + e)^2 + b^2)*sin(f*x + e)))*sin(f*x + e) + 2*((16*a^2 + 83*a*b + 6*b^2)*cos(f*x + e)^6 - (40*a^2 + 193*a*b
+ 12*b^2)*cos(f*x + e)^4 + (30*a^2 + 125*a*b + 6*b^2)*cos(f*x + e)^2 - 15*a*b)*sqrt(((a - b)*cos(f*x + e)^2 +
b)/cos(f*x + e)^2))/((a*f*cos(f*x + e)^5 - 2*a*f*cos(f*x + e)^3 + a*f*cos(f*x + e))*sin(f*x + e))]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)**6*(a+b*tan(f*x+e)**2)**(3/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac{3}{2}} \csc \left (f x + e\right )^{6}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^6*(a+b*tan(f*x+e)^2)^(3/2),x, algorithm="giac")

[Out]

integrate((b*tan(f*x + e)^2 + a)^(3/2)*csc(f*x + e)^6, x)