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

Optimal. Leaf size=209 \[ \frac{b (3 a+7 b) \tan (e+f x) \sqrt{a+b \tan ^2(e+f x)+b}}{2 f (a+b)}+\frac{\sqrt{b} (3 a+7 b) \tanh ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b \tan ^2(e+f x)+b}}\right )}{2 f}-\frac{\cot ^5(e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{5/2}}{5 f (a+b)}-\frac{2 \cot ^3(e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{5/2}}{3 f (a+b)}-\frac{(3 a+7 b) \cot (e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}{3 f (a+b)} \]

[Out]

(Sqrt[b]*(3*a + 7*b)*ArcTanh[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b + b*Tan[e + f*x]^2]])/(2*f) + (b*(3*a + 7*b)*Ta
n[e + f*x]*Sqrt[a + b + b*Tan[e + f*x]^2])/(2*(a + b)*f) - ((3*a + 7*b)*Cot[e + f*x]*(a + b + b*Tan[e + f*x]^2
)^(3/2))/(3*(a + b)*f) - (2*Cot[e + f*x]^3*(a + b + b*Tan[e + f*x]^2)^(5/2))/(3*(a + b)*f) - (Cot[e + f*x]^5*(
a + b + b*Tan[e + f*x]^2)^(5/2))/(5*(a + b)*f)

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Rubi [A]  time = 0.20454, antiderivative size = 209, 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 = {4132, 462, 453, 277, 195, 217, 206} \[ \frac{b (3 a+7 b) \tan (e+f x) \sqrt{a+b \tan ^2(e+f x)+b}}{2 f (a+b)}+\frac{\sqrt{b} (3 a+7 b) \tanh ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b \tan ^2(e+f x)+b}}\right )}{2 f}-\frac{\cot ^5(e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{5/2}}{5 f (a+b)}-\frac{2 \cot ^3(e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{5/2}}{3 f (a+b)}-\frac{(3 a+7 b) \cot (e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}{3 f (a+b)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[b]*(3*a + 7*b)*ArcTanh[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b + b*Tan[e + f*x]^2]])/(2*f) + (b*(3*a + 7*b)*Ta
n[e + f*x]*Sqrt[a + b + b*Tan[e + f*x]^2])/(2*(a + b)*f) - ((3*a + 7*b)*Cot[e + f*x]*(a + b + b*Tan[e + f*x]^2
)^(3/2))/(3*(a + b)*f) - (2*Cot[e + f*x]^3*(a + b + b*Tan[e + f*x]^2)^(5/2))/(3*(a + b)*f) - (Cot[e + f*x]^5*(
a + b + b*Tan[e + f*x]^2)^(5/2))/(5*(a + b)*f)

Rule 4132

Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*sin[(e_.) + (f_.)*(x_)]^(m_), x_Symbol] :> With[{ff = Fr
eeFactors[Tan[e + f*x], x]}, Dist[ff^(m + 1)/f, Subst[Int[(x^m*ExpandToSum[a + b*(1 + ff^2*x^2)^(n/2), x]^p)/(
1 + ff^2*x^2)^(m/2 + 1), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[m/2] && Integer
Q[n/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 \sec ^2(e+f x)\right )^{3/2} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (1+x^2\right )^2 \left (a+b+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+b \tan ^2(e+f x)\right )^{5/2}}{5 (a+b) f}+\frac{\operatorname{Subst}\left (\int \frac{\left (a+b+b x^2\right )^{3/2} \left (10 (a+b)+5 (a+b) x^2\right )}{x^4} \, dx,x,\tan (e+f x)\right )}{5 (a+b) f}\\ &=-\frac{2 \cot ^3(e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{5/2}}{3 (a+b) f}-\frac{\cot ^5(e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{5/2}}{5 (a+b) f}+\frac{(3 a+7 b) \operatorname{Subst}\left (\int \frac{\left (a+b+b x^2\right )^{3/2}}{x^2} \, dx,x,\tan (e+f x)\right )}{3 (a+b) f}\\ &=-\frac{(3 a+7 b) \cot (e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}{3 (a+b) f}-\frac{2 \cot ^3(e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{5/2}}{3 (a+b) f}-\frac{\cot ^5(e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{5/2}}{5 (a+b) f}+\frac{(b (3 a+7 b)) \operatorname{Subst}\left (\int \sqrt{a+b+b x^2} \, dx,x,\tan (e+f x)\right )}{(a+b) f}\\ &=\frac{b (3 a+7 b) \tan (e+f x) \sqrt{a+b+b \tan ^2(e+f x)}}{2 (a+b) f}-\frac{(3 a+7 b) \cot (e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}{3 (a+b) f}-\frac{2 \cot ^3(e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{5/2}}{3 (a+b) f}-\frac{\cot ^5(e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{5/2}}{5 (a+b) f}+\frac{(b (3 a+7 b)) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b+b x^2}} \, dx,x,\tan (e+f x)\right )}{2 f}\\ &=\frac{b (3 a+7 b) \tan (e+f x) \sqrt{a+b+b \tan ^2(e+f x)}}{2 (a+b) f}-\frac{(3 a+7 b) \cot (e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}{3 (a+b) f}-\frac{2 \cot ^3(e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{5/2}}{3 (a+b) f}-\frac{\cot ^5(e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{5/2}}{5 (a+b) f}+\frac{(b (3 a+7 b)) \operatorname{Subst}\left (\int \frac{1}{1-b x^2} \, dx,x,\frac{\tan (e+f x)}{\sqrt{a+b+b \tan ^2(e+f x)}}\right )}{2 f}\\ &=\frac{\sqrt{b} (3 a+7 b) \tanh ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b+b \tan ^2(e+f x)}}\right )}{2 f}+\frac{b (3 a+7 b) \tan (e+f x) \sqrt{a+b+b \tan ^2(e+f x)}}{2 (a+b) f}-\frac{(3 a+7 b) \cot (e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}{3 (a+b) f}-\frac{2 \cot ^3(e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{5/2}}{3 (a+b) f}-\frac{\cot ^5(e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{5/2}}{5 (a+b) f}\\ \end{align*}

Mathematica [C]  time = 10.4456, size = 512, normalized size = 2.45 \[ \frac{\sqrt{2} e^{i (e+f x)} \cos ^3(e+f x) \sqrt{4 b+a e^{-2 i (e+f x)} \left (1+e^{2 i (e+f x)}\right )^2} \left (-\frac{i \left (16 a^2 \left (1+e^{2 i (e+f x)}\right )^2 \left (-6 e^{2 i (e+f x)}+16 e^{4 i (e+f x)}-6 e^{6 i (e+f x)}+e^{8 i (e+f x)}+1\right )+a b \left (-402 e^{2 i (e+f x)}+317 e^{4 i (e+f x)}+708 e^{6 i (e+f x)}+317 e^{8 i (e+f x)}-402 e^{10 i (e+f x)}+115 e^{12 i (e+f x)}+115\right )+b^2 \left (-350 e^{2 i (e+f x)}+231 e^{4 i (e+f x)}+412 e^{6 i (e+f x)}+231 e^{8 i (e+f x)}-350 e^{10 i (e+f x)}+105 e^{12 i (e+f x)}+105\right )\right )}{(a+b) \left (-1+e^{2 i (e+f x)}\right )^5 \left (1+e^{2 i (e+f x)}\right )^2}-\frac{15 \sqrt{b} (3 a+7 b) \log \left (\frac{4 i f \sqrt{a \left (1+e^{2 i (e+f x)}\right )^2+4 b e^{2 i (e+f x)}}-4 \sqrt{b} f \left (-1+e^{2 i (e+f x)}\right )}{1+e^{2 i (e+f x)}}\right )}{\sqrt{a \left (1+e^{2 i (e+f x)}\right )^2+4 b e^{2 i (e+f x)}}}\right ) \left (a+b \sec ^2(e+f x)\right )^{3/2}}{15 f (a \cos (2 (e+f x))+a+2 b)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[2]*E^(I*(e + f*x))*Sqrt[4*b + (a*(1 + E^((2*I)*(e + f*x)))^2)/E^((2*I)*(e + f*x))]*Cos[e + f*x]^3*(((-I)
*(16*a^2*(1 + E^((2*I)*(e + f*x)))^2*(1 - 6*E^((2*I)*(e + f*x)) + 16*E^((4*I)*(e + f*x)) - 6*E^((6*I)*(e + f*x
)) + E^((8*I)*(e + f*x))) + b^2*(105 - 350*E^((2*I)*(e + f*x)) + 231*E^((4*I)*(e + f*x)) + 412*E^((6*I)*(e + f
*x)) + 231*E^((8*I)*(e + f*x)) - 350*E^((10*I)*(e + f*x)) + 105*E^((12*I)*(e + f*x))) + a*b*(115 - 402*E^((2*I
)*(e + f*x)) + 317*E^((4*I)*(e + f*x)) + 708*E^((6*I)*(e + f*x)) + 317*E^((8*I)*(e + f*x)) - 402*E^((10*I)*(e
+ f*x)) + 115*E^((12*I)*(e + f*x)))))/((a + b)*(-1 + E^((2*I)*(e + f*x)))^5*(1 + E^((2*I)*(e + f*x)))^2) - (15
*Sqrt[b]*(3*a + 7*b)*Log[(-4*Sqrt[b]*(-1 + E^((2*I)*(e + f*x)))*f + (4*I)*Sqrt[4*b*E^((2*I)*(e + f*x)) + a*(1
+ E^((2*I)*(e + f*x)))^2]*f)/(1 + E^((2*I)*(e + f*x)))])/Sqrt[4*b*E^((2*I)*(e + f*x)) + a*(1 + E^((2*I)*(e + f
*x)))^2])*(a + b*Sec[e + f*x]^2)^(3/2))/(15*f*(a + 2*b + a*Cos[2*(e + f*x)])^(3/2))

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Maple [C]  time = 0.951, size = 8726, normalized size = 41.8 \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*sec(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*sec(f*x+e)^2)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 20.3657, size = 1732, 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*sec(f*x+e)^2)^(3/2),x, algorithm="fricas")

[Out]

[1/120*(15*((3*a^2 + 10*a*b + 7*b^2)*cos(f*x + e)^5 - 2*(3*a^2 + 10*a*b + 7*b^2)*cos(f*x + e)^3 + (3*a^2 + 10*
a*b + 7*b^2)*cos(f*x + e))*sqrt(b)*log(((a^2 - 6*a*b + b^2)*cos(f*x + e)^4 + 8*(a*b - b^2)*cos(f*x + e)^2 + 4*
((a - b)*cos(f*x + e)^3 + 2*b*cos(f*x + e))*sqrt(b)*sqrt((a*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 + 115*a*b + 105*b^2)*cos(f*x + e)^6 - (40*a^2 + 273*a*b + 24
5*b^2)*cos(f*x + e)^4 + (30*a^2 + 185*a*b + 161*b^2)*cos(f*x + e)^2 - 15*a*b - 15*b^2)*sqrt((a*cos(f*x + e)^2
+ b)/cos(f*x + e)^2))/(((a + b)*f*cos(f*x + e)^5 - 2*(a + b)*f*cos(f*x + e)^3 + (a + b)*f*cos(f*x + e))*sin(f*
x + e)), 1/60*(15*((3*a^2 + 10*a*b + 7*b^2)*cos(f*x + e)^5 - 2*(3*a^2 + 10*a*b + 7*b^2)*cos(f*x + e)^3 + (3*a^
2 + 10*a*b + 7*b^2)*cos(f*x + e))*sqrt(-b)*arctan(-1/2*((a - b)*cos(f*x + e)^3 + 2*b*cos(f*x + e))*sqrt(-b)*sq
rt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)/((a*b*cos(f*x + e)^2 + b^2)*sin(f*x + e)))*sin(f*x + e) - 2*((16*a^2
+ 115*a*b + 105*b^2)*cos(f*x + e)^6 - (40*a^2 + 273*a*b + 245*b^2)*cos(f*x + e)^4 + (30*a^2 + 185*a*b + 161*b
^2)*cos(f*x + e)^2 - 15*a*b - 15*b^2)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/(((a + b)*f*cos(f*x + e)^5
- 2*(a + b)*f*cos(f*x + e)^3 + (a + b)*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*sec(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 \sec \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*sec(f*x+e)^2)^(3/2),x, algorithm="giac")

[Out]

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