3.84 \(\int \csc ^3(e+f x) (a+b \sec ^2(e+f x))^{3/2} \, dx\)
Optimal. Leaf size=161 \[ \frac{b \sec (e+f x) \sqrt{a+b \sec ^2(e+f x)}}{f}+\frac{\sqrt{b} (3 a+4 b) \tanh ^{-1}\left (\frac{\sqrt{b} \sec (e+f x)}{\sqrt{a+b \sec ^2(e+f x)}}\right )}{2 f}-\frac{\sqrt{a+b} (a+4 b) \tanh ^{-1}\left (\frac{\sqrt{a+b} \sec (e+f x)}{\sqrt{a+b \sec ^2(e+f x)}}\right )}{2 f}-\frac{\cot (e+f x) \csc (e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2}}{2 f} \]
[Out]
(Sqrt[b]*(3*a + 4*b)*ArcTanh[(Sqrt[b]*Sec[e + f*x])/Sqrt[a + b*Sec[e + f*x]^2]])/(2*f) - (Sqrt[a + b]*(a + 4*b
)*ArcTanh[(Sqrt[a + b]*Sec[e + f*x])/Sqrt[a + b*Sec[e + f*x]^2]])/(2*f) + (b*Sec[e + f*x]*Sqrt[a + b*Sec[e + f
*x]^2])/f - (Cot[e + f*x]*Csc[e + f*x]*(a + b*Sec[e + f*x]^2)^(3/2))/(2*f)
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Rubi [A] time = 0.203121, antiderivative size = 161, normalized size of antiderivative = 1.,
number of steps used = 8, number of rules used = 8, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.32, Rules used =
{4134, 467, 528, 523, 217, 206, 377, 207} \[ \frac{b \sec (e+f x) \sqrt{a+b \sec ^2(e+f x)}}{f}+\frac{\sqrt{b} (3 a+4 b) \tanh ^{-1}\left (\frac{\sqrt{b} \sec (e+f x)}{\sqrt{a+b \sec ^2(e+f x)}}\right )}{2 f}-\frac{\sqrt{a+b} (a+4 b) \tanh ^{-1}\left (\frac{\sqrt{a+b} \sec (e+f x)}{\sqrt{a+b \sec ^2(e+f x)}}\right )}{2 f}-\frac{\cot (e+f x) \csc (e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2}}{2 f} \]
Antiderivative was successfully verified.
[In]
Int[Csc[e + f*x]^3*(a + b*Sec[e + f*x]^2)^(3/2),x]
[Out]
(Sqrt[b]*(3*a + 4*b)*ArcTanh[(Sqrt[b]*Sec[e + f*x])/Sqrt[a + b*Sec[e + f*x]^2]])/(2*f) - (Sqrt[a + b]*(a + 4*b
)*ArcTanh[(Sqrt[a + b]*Sec[e + f*x])/Sqrt[a + b*Sec[e + f*x]^2]])/(2*f) + (b*Sec[e + f*x]*Sqrt[a + b*Sec[e + f
*x]^2])/f - (Cot[e + f*x]*Csc[e + f*x]*(a + b*Sec[e + f*x]^2)^(3/2))/(2*f)
Rule 4134
Int[((a_) + (b_.)*((c_.)*sec[(e_.) + (f_.)*(x_)])^(n_))^(p_.)*sin[(e_.) + (f_.)*(x_)]^(m_.), x_Symbol] :> With
[{ff = FreeFactors[Cos[e + f*x], x]}, Dist[1/(f*ff^m), Subst[Int[((-1 + ff^2*x^2)^((m - 1)/2)*(a + b*(c*ff*x)^
n)^p)/x^(m + 1), x], x, Sec[e + f*x]/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && IntegerQ[(m - 1)/2] && (Gt
Q[m, 0] || EqQ[n, 2] || EqQ[n, 4])
Rule 467
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)/(b*n*(p + 1)), x] - Dist[e^n/(b*n*(p + 1)), Int[(e*x)^
(m - n)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q - 1)*Simp[c*(m - n + 1) + d*(m + n*(q - 1) + 1)*x^n, x], x], x] /;
FreeQ[{a, b, c, d, e}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[q, 0] && GtQ[m - n + 1, 0] &
& IntBinomialQ[a, b, c, d, e, m, n, p, q, x]
Rule 528
Int[((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)), x_Symbol] :> Simp[
(f*x*(a + b*x^n)^(p + 1)*(c + d*x^n)^q)/(b*(n*(p + q + 1) + 1)), x] + Dist[1/(b*(n*(p + q + 1) + 1)), Int[(a +
b*x^n)^p*(c + d*x^n)^(q - 1)*Simp[c*(b*e - a*f + b*e*n*(p + q + 1)) + (d*(b*e - a*f) + f*n*q*(b*c - a*d) + b*
d*e*n*(p + q + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && GtQ[q, 0] && NeQ[n*(p + q + 1) + 1
, 0]
Rule 523
Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*Sqrt[(c_) + (d_.)*(x_)^(n_)]), x_Symbol] :> Dist[f/b, I
nt[1/Sqrt[c + d*x^n], x], x] + Dist[(b*e - a*f)/b, Int[1/((a + b*x^n)*Sqrt[c + d*x^n]), x], x] /; FreeQ[{a, b,
c, d, e, f, n}, x]
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])
Rule 377
Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]
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])
Rubi steps
\begin{align*} \int \csc ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^2 \left (a+b x^2\right )^{3/2}}{\left (-1+x^2\right )^2} \, dx,x,\sec (e+f x)\right )}{f}\\ &=-\frac{\cot (e+f x) \csc (e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2}}{2 f}+\frac{\operatorname{Subst}\left (\int \frac{\sqrt{a+b x^2} \left (a+4 b x^2\right )}{-1+x^2} \, dx,x,\sec (e+f x)\right )}{2 f}\\ &=\frac{b \sec (e+f x) \sqrt{a+b \sec ^2(e+f x)}}{f}-\frac{\cot (e+f x) \csc (e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2}}{2 f}+\frac{\operatorname{Subst}\left (\int \frac{2 a (a+2 b)+2 b (3 a+4 b) x^2}{\left (-1+x^2\right ) \sqrt{a+b x^2}} \, dx,x,\sec (e+f x)\right )}{4 f}\\ &=\frac{b \sec (e+f x) \sqrt{a+b \sec ^2(e+f x)}}{f}-\frac{\cot (e+f x) \csc (e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2}}{2 f}+\frac{((a+b) (a+4 b)) \operatorname{Subst}\left (\int \frac{1}{\left (-1+x^2\right ) \sqrt{a+b x^2}} \, dx,x,\sec (e+f x)\right )}{2 f}+\frac{(b (3 a+4 b)) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x^2}} \, dx,x,\sec (e+f x)\right )}{2 f}\\ &=\frac{b \sec (e+f x) \sqrt{a+b \sec ^2(e+f x)}}{f}-\frac{\cot (e+f x) \csc (e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2}}{2 f}+\frac{((a+b) (a+4 b)) \operatorname{Subst}\left (\int \frac{1}{-1-(-a-b) x^2} \, dx,x,\frac{\sec (e+f x)}{\sqrt{a+b \sec ^2(e+f x)}}\right )}{2 f}+\frac{(b (3 a+4 b)) \operatorname{Subst}\left (\int \frac{1}{1-b x^2} \, dx,x,\frac{\sec (e+f x)}{\sqrt{a+b \sec ^2(e+f x)}}\right )}{2 f}\\ &=\frac{\sqrt{b} (3 a+4 b) \tanh ^{-1}\left (\frac{\sqrt{b} \sec (e+f x)}{\sqrt{a+b \sec ^2(e+f x)}}\right )}{2 f}-\frac{\sqrt{a+b} (a+4 b) \tanh ^{-1}\left (\frac{\sqrt{a+b} \sec (e+f x)}{\sqrt{a+b \sec ^2(e+f x)}}\right )}{2 f}+\frac{b \sec (e+f x) \sqrt{a+b \sec ^2(e+f x)}}{f}-\frac{\cot (e+f x) \csc (e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2}}{2 f}\\ \end{align*}
Mathematica [A] time = 1.4986, size = 202, normalized size = 1.25 \[ -\frac{\csc ^2(e+f x) \sec (e+f x) \sqrt{a+b \sec ^2(e+f x)} \left (\sqrt{2} \sqrt{a \cos (2 (e+f x))+a+2 b} ((a+2 b) \cos (2 (e+f x))+a)-\sqrt{b} (3 a+4 b) \sin ^2(2 (e+f x)) \tanh ^{-1}\left (\frac{\sqrt{-a \sin ^2(e+f x)+a+b}}{\sqrt{b}}\right )+\sqrt{a+b} (a+4 b) \sin ^2(2 (e+f x)) \tanh ^{-1}\left (\frac{\sqrt{-a \sin ^2(e+f x)+a+b}}{\sqrt{a+b}}\right )\right )}{4 \sqrt{2} f \sqrt{a \cos (2 (e+f x))+a+2 b}} \]
Antiderivative was successfully verified.
[In]
Integrate[Csc[e + f*x]^3*(a + b*Sec[e + f*x]^2)^(3/2),x]
[Out]
-(Csc[e + f*x]^2*Sec[e + f*x]*Sqrt[a + b*Sec[e + f*x]^2]*(Sqrt[2]*Sqrt[a + 2*b + a*Cos[2*(e + f*x)]]*(a + (a +
2*b)*Cos[2*(e + f*x)]) - Sqrt[b]*(3*a + 4*b)*ArcTanh[Sqrt[a + b - a*Sin[e + f*x]^2]/Sqrt[b]]*Sin[2*(e + f*x)]
^2 + Sqrt[a + b]*(a + 4*b)*ArcTanh[Sqrt[a + b - a*Sin[e + f*x]^2]/Sqrt[a + b]]*Sin[2*(e + f*x)]^2))/(4*Sqrt[2]
*f*Sqrt[a + 2*b + a*Cos[2*(e + f*x)]])
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Maple [B] time = 0.378, size = 5178, normalized size = 32.2 \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)^3*(a+b*sec(f*x+e)^2)^(3/2),x)
[Out]
result too large to display
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Maxima [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 )^{3}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(csc(f*x+e)^3*(a+b*sec(f*x+e)^2)^(3/2),x, algorithm="maxima")
[Out]
integrate((b*sec(f*x + e)^2 + a)^(3/2)*csc(f*x + e)^3, x)
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Fricas [A] time = 1.12478, size = 2477, normalized size = 15.39 \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)^3*(a+b*sec(f*x+e)^2)^(3/2),x, algorithm="fricas")
[Out]
[1/4*(((a + 4*b)*cos(f*x + e)^3 - (a + 4*b)*cos(f*x + e))*sqrt(a + b)*log(2*(a*cos(f*x + e)^2 - 2*sqrt(a + b)*
sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e) + a + 2*b)/(cos(f*x + e)^2 - 1)) + ((3*a + 4*b)*cos(f
*x + e)^3 - (3*a + 4*b)*cos(f*x + e))*sqrt(b)*log((a*cos(f*x + e)^2 + 2*sqrt(b)*sqrt((a*cos(f*x + e)^2 + b)/co
s(f*x + e)^2)*cos(f*x + e) + 2*b)/cos(f*x + e)^2) + 2*((a + 2*b)*cos(f*x + e)^2 - b)*sqrt((a*cos(f*x + e)^2 +
b)/cos(f*x + e)^2))/(f*cos(f*x + e)^3 - f*cos(f*x + e)), 1/4*(2*((a + 4*b)*cos(f*x + e)^3 - (a + 4*b)*cos(f*x
+ e))*sqrt(-a - b)*arctan(sqrt(-a - b)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e)/(a + b)) + ((3
*a + 4*b)*cos(f*x + e)^3 - (3*a + 4*b)*cos(f*x + e))*sqrt(b)*log((a*cos(f*x + e)^2 + 2*sqrt(b)*sqrt((a*cos(f*x
+ e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e) + 2*b)/cos(f*x + e)^2) + 2*((a + 2*b)*cos(f*x + e)^2 - b)*sqrt((a*co
s(f*x + e)^2 + b)/cos(f*x + e)^2))/(f*cos(f*x + e)^3 - f*cos(f*x + e)), -1/4*(2*((3*a + 4*b)*cos(f*x + e)^3 -
(3*a + 4*b)*cos(f*x + e))*sqrt(-b)*arctan(sqrt(-b)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e)/b)
- ((a + 4*b)*cos(f*x + e)^3 - (a + 4*b)*cos(f*x + e))*sqrt(a + b)*log(2*(a*cos(f*x + e)^2 - 2*sqrt(a + b)*sqr
t((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e) + a + 2*b)/(cos(f*x + e)^2 - 1)) - 2*((a + 2*b)*cos(f*x
+ e)^2 - b)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/(f*cos(f*x + e)^3 - f*cos(f*x + e)), 1/2*(((a + 4*b)*
cos(f*x + e)^3 - (a + 4*b)*cos(f*x + e))*sqrt(-a - b)*arctan(sqrt(-a - b)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x
+ e)^2)*cos(f*x + e)/(a + b)) - ((3*a + 4*b)*cos(f*x + e)^3 - (3*a + 4*b)*cos(f*x + e))*sqrt(-b)*arctan(sqrt(-
b)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e)/b) + ((a + 2*b)*cos(f*x + e)^2 - b)*sqrt((a*cos(f*
x + e)^2 + b)/cos(f*x + e)^2))/(f*cos(f*x + e)^3 - f*cos(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)**3*(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 )^{3}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(csc(f*x+e)^3*(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)^3, x)