3.68 \(\int \sqrt{a+b \sec ^2(e+f x)} \sin ^3(e+f x) \, dx\)
Optimal. Leaf size=100 \[ \frac{\cos ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2}}{3 a f}-\frac{\cos (e+f x) \sqrt{a+b \sec ^2(e+f x)}}{f}+\frac{\sqrt{b} \tanh ^{-1}\left (\frac{\sqrt{b} \sec (e+f x)}{\sqrt{a+b \sec ^2(e+f x)}}\right )}{f} \]
[Out]
(Sqrt[b]*ArcTanh[(Sqrt[b]*Sec[e + f*x])/Sqrt[a + b*Sec[e + f*x]^2]])/f - (Cos[e + f*x]*Sqrt[a + b*Sec[e + f*x]
^2])/f + (Cos[e + f*x]^3*(a + b*Sec[e + f*x]^2)^(3/2))/(3*a*f)
________________________________________________________________________________________
Rubi [A] time = 0.0938918, antiderivative size = 100, normalized size of antiderivative = 1.,
number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used =
{4134, 451, 277, 217, 206} \[ \frac{\cos ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2}}{3 a f}-\frac{\cos (e+f x) \sqrt{a+b \sec ^2(e+f x)}}{f}+\frac{\sqrt{b} \tanh ^{-1}\left (\frac{\sqrt{b} \sec (e+f x)}{\sqrt{a+b \sec ^2(e+f x)}}\right )}{f} \]
Antiderivative was successfully verified.
[In]
Int[Sqrt[a + b*Sec[e + f*x]^2]*Sin[e + f*x]^3,x]
[Out]
(Sqrt[b]*ArcTanh[(Sqrt[b]*Sec[e + f*x])/Sqrt[a + b*Sec[e + f*x]^2]])/f - (Cos[e + f*x]*Sqrt[a + b*Sec[e + f*x]
^2])/f + (Cos[e + f*x]^3*(a + b*Sec[e + f*x]^2)^(3/2))/(3*a*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 451
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[d/e^n, Int[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a,
b, c, d, e, m, n, p}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n*(p + 1) + 1, 0] && (IntegerQ[n] || GtQ[e, 0]) && (
(GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -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 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 \sqrt{a+b \sec ^2(e+f x)} \sin ^3(e+f x) \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (-1+x^2\right ) \sqrt{a+b x^2}}{x^4} \, dx,x,\sec (e+f x)\right )}{f}\\ &=\frac{\cos ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2}}{3 a f}+\frac{\operatorname{Subst}\left (\int \frac{\sqrt{a+b x^2}}{x^2} \, dx,x,\sec (e+f x)\right )}{f}\\ &=-\frac{\cos (e+f x) \sqrt{a+b \sec ^2(e+f x)}}{f}+\frac{\cos ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2}}{3 a f}+\frac{b \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x^2}} \, dx,x,\sec (e+f x)\right )}{f}\\ &=-\frac{\cos (e+f x) \sqrt{a+b \sec ^2(e+f x)}}{f}+\frac{\cos ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2}}{3 a f}+\frac{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 )}{f}\\ &=\frac{\sqrt{b} \tanh ^{-1}\left (\frac{\sqrt{b} \sec (e+f x)}{\sqrt{a+b \sec ^2(e+f x)}}\right )}{f}-\frac{\cos (e+f x) \sqrt{a+b \sec ^2(e+f x)}}{f}+\frac{\cos ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2}}{3 a f}\\ \end{align*}
Mathematica [A] time = 0.379185, size = 120, normalized size = 1.2 \[ \frac{\sqrt{2} \cos (e+f x) \sqrt{a+b \sec ^2(e+f x)} \left (\sqrt{a \cos ^2(e+f x)+b} \left (a \cos ^2(e+f x)-3 a+b\right )+3 a \sqrt{b} \tanh ^{-1}\left (\frac{\sqrt{a \cos ^2(e+f x)+b}}{\sqrt{b}}\right )\right )}{3 a f \sqrt{a \cos (2 (e+f x))+a+2 b}} \]
Antiderivative was successfully verified.
[In]
Integrate[Sqrt[a + b*Sec[e + f*x]^2]*Sin[e + f*x]^3,x]
[Out]
(Sqrt[2]*Cos[e + f*x]*(3*a*Sqrt[b]*ArcTanh[Sqrt[b + a*Cos[e + f*x]^2]/Sqrt[b]] + Sqrt[b + a*Cos[e + f*x]^2]*(-
3*a + b + a*Cos[e + f*x]^2))*Sqrt[a + b*Sec[e + f*x]^2])/(3*a*f*Sqrt[a + 2*b + a*Cos[2*(e + f*x)]])
________________________________________________________________________________________
Maple [B] time = 0.384, size = 1525, normalized size = 15.3 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(sin(f*x+e)^3*(a+b*sec(f*x+e)^2)^(1/2),x)
[Out]
-1/6/f/b^(1/2)/(a+b)^(3/2)/a*(-1+cos(f*x+e))^2*(-2*(a+b)^(3/2)*cos(f*x+e)^4*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))
^2)^(3/2)*b^(1/2)-8*(a+b)^(3/2)*cos(f*x+e)^3*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(3/2)*b^(1/2)-6*(a+b)^(3/2)
*cos(f*x+e)^2*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(3/2)*b^(1/2)-3*4^(1/2)*cos(f*x+e)*ln(-2/(a+b)^(1/2)*(-1+c
os(f*x+e))*(cos(f*x+e)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(a+b)^(1/2)-a*cos(f*x+e)+((b+a*cos(f*x+e)^2
)/(1+cos(f*x+e))^2)^(1/2)*(a+b)^(1/2)+b)/sin(f*x+e)^2)*b^(5/2)*a+3*4^(1/2)*cos(f*x+e)*ln(-4/(a+b)^(1/2)*(-1+co
s(f*x+e))*(cos(f*x+e)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(a+b)^(1/2)-a*cos(f*x+e)+((b+a*cos(f*x+e)^2)
/(1+cos(f*x+e))^2)^(1/2)*(a+b)^(1/2)+b)/sin(f*x+e)^2)*b^(5/2)*a+6*(a+b)^(3/2)*4^(1/2)*cos(f*x+e)*((b+a*cos(f*x
+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^(1/2)*a-3*(a+b)^(3/2)*4^(1/2)*cos(f*x+e)*arctanh(1/8*b^(1/2)*4^(1/2)*(-1+cos(
f*x+e))*(cos(f*x+e)*4^(1/2)-2*cos(f*x+e)-4^(1/2)-2)/sin(f*x+e)^2/((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2))*
a*b-3*(a+b)^(3/2)*4^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^(3/2)+4*(a+b)^(3/2)*cos(f*x+e)*((b+a*c
os(f*x+e)^2)/(1+cos(f*x+e))^2)^(3/2)*b^(1/2)-3*4^(1/2)*cos(f*x+e)*ln(-2/(a+b)^(1/2)*(-1+cos(f*x+e))*(cos(f*x+e
)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(a+b)^(1/2)-a*cos(f*x+e)+((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(
1/2)*(a+b)^(1/2)+b)/sin(f*x+e)^2)*b^(3/2)*a^2+3*4^(1/2)*cos(f*x+e)*ln(-4/(a+b)^(1/2)*(-1+cos(f*x+e))*(cos(f*x+
e)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(a+b)^(1/2)-a*cos(f*x+e)+((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^
(1/2)*(a+b)^(1/2)+b)/sin(f*x+e)^2)*b^(3/2)*a^2-3*4^(1/2)*ln(-2/(a+b)^(1/2)*(-1+cos(f*x+e))*(cos(f*x+e)*((b+a*c
os(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(a+b)^(1/2)-a*cos(f*x+e)+((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(a+b
)^(1/2)+b)/sin(f*x+e)^2)*b^(5/2)*a+3*4^(1/2)*ln(-4/(a+b)^(1/2)*(-1+cos(f*x+e))*(cos(f*x+e)*((b+a*cos(f*x+e)^2)
/(1+cos(f*x+e))^2)^(1/2)*(a+b)^(1/2)-a*cos(f*x+e)+((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(a+b)^(1/2)+b)/s
in(f*x+e)^2)*b^(5/2)*a+3*(a+b)^(3/2)*4^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^(1/2)*a-3*(a+b)^(3/
2)*4^(1/2)*arctanh(1/8*b^(1/2)*4^(1/2)*(-1+cos(f*x+e))*(cos(f*x+e)*4^(1/2)-2*cos(f*x+e)-4^(1/2)-2)/sin(f*x+e)^
2/((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2))*a*b+4*(a+b)^(3/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(3/2)*b
^(1/2)-3*4^(1/2)*ln(-2/(a+b)^(1/2)*(-1+cos(f*x+e))*(cos(f*x+e)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(a+
b)^(1/2)-a*cos(f*x+e)+((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(a+b)^(1/2)+b)/sin(f*x+e)^2)*b^(3/2)*a^2+3*4
^(1/2)*ln(-4/(a+b)^(1/2)*(-1+cos(f*x+e))*(cos(f*x+e)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(a+b)^(1/2)-a
*cos(f*x+e)+((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(a+b)^(1/2)+b)/sin(f*x+e)^2)*b^(3/2)*a^2)*cos(f*x+e)*(
(b+a*cos(f*x+e)^2)/cos(f*x+e)^2)^(1/2)/((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)/sin(f*x+e)^4
________________________________________________________________________________________
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(sin(f*x+e)^3*(a+b*sec(f*x+e)^2)^(1/2),x, algorithm="maxima")
[Out]
Exception raised: ValueError
________________________________________________________________________________________
Fricas [A] time = 1.12116, size = 578, normalized size = 5.78 \begin{align*} \left [\frac{3 \, a \sqrt{b} \log \left (\frac{a \cos \left (f x + e\right )^{2} + 2 \, \sqrt{b} \sqrt{\frac{a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) + 2 \, b}{\cos \left (f x + e\right )^{2}}\right ) + 2 \,{\left (a \cos \left (f x + e\right )^{3} -{\left (3 \, a - b\right )} \cos \left (f x + e\right )\right )} \sqrt{\frac{a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{6 \, a f}, -\frac{3 \, a \sqrt{-b} \arctan \left (\frac{\sqrt{-b} \sqrt{\frac{a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{b}\right ) -{\left (a \cos \left (f x + e\right )^{3} -{\left (3 \, a - b\right )} \cos \left (f x + e\right )\right )} \sqrt{\frac{a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{3 \, a f}\right ] \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sin(f*x+e)^3*(a+b*sec(f*x+e)^2)^(1/2),x, algorithm="fricas")
[Out]
[1/6*(3*a*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*cos(f*x + e)^3 - (3*a - b)*cos(f*x + e))*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)
^2))/(a*f), -1/3*(3*a*sqrt(-b)*arctan(sqrt(-b)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e)/b) - (
a*cos(f*x + e)^3 - (3*a - b)*cos(f*x + e))*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/(a*f)]
________________________________________________________________________________________
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(sin(f*x+e)**3*(a+b*sec(f*x+e)**2)**(1/2),x)
[Out]
Timed out
________________________________________________________________________________________
Giac [A] time = 1.31934, size = 120, normalized size = 1.2 \begin{align*} -\frac{{\left (\frac{3 \, b \arctan \left (\frac{\sqrt{a \cos \left (f x + e\right )^{2} + b}}{\sqrt{-b}}\right )}{\sqrt{-b}} - \frac{{\left (a \cos \left (f x + e\right )^{2} + b\right )}^{\frac{3}{2}} a^{2} - 3 \, \sqrt{a \cos \left (f x + e\right )^{2} + b} a^{3}}{a^{3}}\right )} \mathrm{sgn}\left (\cos \left (f x + e\right )\right )}{3 \, f} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sin(f*x+e)^3*(a+b*sec(f*x+e)^2)^(1/2),x, algorithm="giac")
[Out]
-1/3*(3*b*arctan(sqrt(a*cos(f*x + e)^2 + b)/sqrt(-b))/sqrt(-b) - ((a*cos(f*x + e)^2 + b)^(3/2)*a^2 - 3*sqrt(a*
cos(f*x + e)^2 + b)*a^3)/a^3)*sgn(cos(f*x + e))/f