3.80 \(\int (a+b \sec ^2(e+f x))^{3/2} \sin ^5(e+f x) \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.179943, 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 = {4134, 462, 453, 277, 195, 217, 206} \[ \frac{b (3 a-4 b) \sec (e+f x) \sqrt{a+b \sec ^2(e+f x)}}{2 a f}-\frac{\cos ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^{5/2}}{5 a f}+\frac{2 \cos ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^{5/2}}{3 a f}-\frac{(3 a-4 b) \cos (e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2}}{3 a f}+\frac{\sqrt{b} (3 a-4 b) \tanh ^{-1}\left (\frac{\sqrt{b} \sec (e+f x)}{\sqrt{a+b \sec ^2(e+f x)}}\right )}{2 f} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sec[e + f*x]^2)^(3/2)*Sin[e + f*x]^5,x]

[Out]

((3*a - 4*b)*Sqrt[b]*ArcTanh[(Sqrt[b]*Sec[e + f*x])/Sqrt[a + b*Sec[e + f*x]^2]])/(2*f) + ((3*a - 4*b)*b*Sec[e
+ f*x]*Sqrt[a + b*Sec[e + f*x]^2])/(2*a*f) - ((3*a - 4*b)*Cos[e + f*x]*(a + b*Sec[e + f*x]^2)^(3/2))/(3*a*f) +
(2*Cos[e + f*x]^3*(a + b*Sec[e + f*x]^2)^(5/2))/(3*a*f) - (Cos[e + f*x]^5*(a + b*Sec[e + f*x]^2)^(5/2))/(5*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 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 \left (a+b \sec ^2(e+f x)\right )^{3/2} \sin ^5(e+f x) \, 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,\sec (e+f x)\right )}{f}\\ &=-\frac{\cos ^5(e+f x) \left (a+b \sec ^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,\sec (e+f x)\right )}{5 a f}\\ &=\frac{2 \cos ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^{5/2}}{3 a f}-\frac{\cos ^5(e+f x) \left (a+b \sec ^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,\sec (e+f x)\right )}{3 a f}\\ &=-\frac{(3 a-4 b) \cos (e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2}}{3 a f}+\frac{2 \cos ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^{5/2}}{3 a f}-\frac{\cos ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^{5/2}}{5 a f}+\frac{((3 a-4 b) b) \operatorname{Subst}\left (\int \sqrt{a+b x^2} \, dx,x,\sec (e+f x)\right )}{a f}\\ &=\frac{(3 a-4 b) b \sec (e+f x) \sqrt{a+b \sec ^2(e+f x)}}{2 a f}-\frac{(3 a-4 b) \cos (e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2}}{3 a f}+\frac{2 \cos ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^{5/2}}{3 a f}-\frac{\cos ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^{5/2}}{5 a f}+\frac{((3 a-4 b) b) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x^2}} \, dx,x,\sec (e+f x)\right )}{2 f}\\ &=\frac{(3 a-4 b) b \sec (e+f x) \sqrt{a+b \sec ^2(e+f x)}}{2 a f}-\frac{(3 a-4 b) \cos (e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2}}{3 a f}+\frac{2 \cos ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^{5/2}}{3 a f}-\frac{\cos ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^{5/2}}{5 a f}+\frac{((3 a-4 b) 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{(3 a-4 b) \sqrt{b} \tanh ^{-1}\left (\frac{\sqrt{b} \sec (e+f x)}{\sqrt{a+b \sec ^2(e+f x)}}\right )}{2 f}+\frac{(3 a-4 b) b \sec (e+f x) \sqrt{a+b \sec ^2(e+f x)}}{2 a f}-\frac{(3 a-4 b) \cos (e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2}}{3 a f}+\frac{2 \cos ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^{5/2}}{3 a f}-\frac{\cos ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^{5/2}}{5 a f}\\ \end{align*}

Mathematica [A]  time = 1.47831, size = 188, normalized size = 0.96 \[ \frac{\sqrt{2} \cos ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \left (-5 (3 a-4 b) \left (\sqrt{-a \sin ^2(e+f x)+a+b} \left (-a \sin ^2(e+f x)+a+4 b\right )-3 b^{3/2} \tanh ^{-1}\left (\frac{\sqrt{-a \sin ^2(e+f x)+a+b}}{\sqrt{b}}\right )\right )-\frac{6 b \left (-a \sin ^2(e+f x)+a+b\right )^{5/2}}{a}+15 \sec ^2(e+f x) \left (-a \sin ^2(e+f x)+a+b\right )^{5/2}\right )}{15 b f (a \cos (2 (e+f x))+a+2 b)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sec[e + f*x]^2)^(3/2)*Sin[e + f*x]^5,x]

[Out]

(Sqrt[2]*Cos[e + f*x]^3*(a + b*Sec[e + f*x]^2)^(3/2)*((-6*b*(a + b - a*Sin[e + f*x]^2)^(5/2))/a + 15*Sec[e + f
*x]^2*(a + b - a*Sin[e + f*x]^2)^(5/2) - 5*(3*a - 4*b)*(-3*b^(3/2)*ArcTanh[Sqrt[a + b - a*Sin[e + f*x]^2]/Sqrt
[b]] + Sqrt[a + b - a*Sin[e + f*x]^2]*(a + 4*b - a*Sin[e + f*x]^2))))/(15*b*f*(a + 2*b + a*Cos[2*(e + f*x)])^(
3/2))

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Maple [B]  time = 0.624, size = 2537, normalized size = 12.9 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*sec(f*x+e)^2)^(3/2)*sin(f*x+e)^5,x)

[Out]

1/60/f/a/b^(1/2)/(a+b)^(9/2)*(-1+cos(f*x+e))^3*(45*cos(f*x+e)^2*b^(5/2)*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)*a^5-45*cos(f*x+e)^2*b^(5/2)*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)*a^5+225*cos(f*x+e)^2*b^(11/2)*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)*a^2-225*cos(f*x+e)^2*b^(11/2)*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)*a^2+315*cos(f*x+e)^2*b^(9/2)*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)*a^3-315*cos(f*x+e)^2*b^(9/2)*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)*a^3+195*cos(f*x+e)^2*b^(7/2)*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)*a^4-195*cos(f*x+e)^2*b^(7/2)*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)*a^4+60*cos(f*x+e)^2*b^(13/2)*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)*a-60*cos(f*x+e)^2*b^(13/2)*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)/si
n(f*x+e)^2)*a-15*(a+b)^(7/2)*b^(5/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a-15*((b+a*cos(f*x+e)^2)/(1+c
os(f*x+e))^2)^(1/2)*b^(3/2)*(a+b)^(7/2)*a^2+6*cos(f*x+e)^3*(a+b)^(7/2)*b^(7/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+
e))^2)^(1/2)+6*cos(f*x+e)^2*b^(7/2)*(a+b)^(7/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)+6*cos(f*x+e)^7*b^(
3/2)*(a+b)^(7/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a^2+6*(a+b)^(7/2)*cos(f*x+e)^7*a^3*b^(1/2)*((b+a*
cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)+6*(a+b)^(7/2)*cos(f*x+e)^6*a^2*b^(3/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e)
)^2)^(1/2)+12*cos(f*x+e)^5*b^(5/2)*(a+b)^(7/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a+6*cos(f*x+e)^6*b^
(1/2)*(a+b)^(7/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a^3-8*cos(f*x+e)^5*b^(3/2)*(a+b)^(7/2)*((b+a*cos
(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a^2+12*cos(f*x+e)^4*b^(5/2)*(a+b)^(7/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^
2)^(1/2)*a-20*(a+b)^(7/2)*cos(f*x+e)^5*a^3*b^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-8*cos(f*x+e)^4*
b^(3/2)*(a+b)^(7/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a^2-74*cos(f*x+e)^3*b^(5/2)*(a+b)^(7/2)*((b+a*
cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a-20*cos(f*x+e)^4*b^(1/2)*(a+b)^(7/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))
^2)^(1/2)*a^3-50*cos(f*x+e)^3*b^(3/2)*(a+b)^(7/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a^2-74*cos(f*x+e
)^2*b^(5/2)*(a+b)^(7/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a+30*cos(f*x+e)^3*(a+b)^(7/2)*a^3*b^(1/2)*
((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-50*cos(f*x+e)^2*b^(3/2)*(a+b)^(7/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x
+e))^2)^(1/2)*a^2-15*cos(f*x+e)*b^(5/2)*(a+b)^(7/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a+30*cos(f*x+e
)^2*b^(1/2)*(a+b)^(7/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a^3-45*cos(f*x+e)^2*(a+b)^(7/2)*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^3*b+15*cos(f*x+e)^2*(a+b)^(7/2)*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^2*b^2+60*cos
(f*x+e)^2*(a+b)^(7/2)*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^3-15*cos(f*x+e)*b^(3/2)*(a+b)^(7/2)*((b+a*cos(f*
x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a^2)*cos(f*x+e)*((b+a*cos(f*x+e)^2)/cos(f*x+e)^2)^(3/2)*4^(1/2)/((b+a*cos(f*x+
e)^2)/(1+cos(f*x+e))^2)^(3/2)/sin(f*x+e)^6

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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((a+b*sec(f*x+e)^2)^(3/2)*sin(f*x+e)^5,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.72373, size = 876, normalized size = 4.47 \begin{align*} \left [-\frac{15 \,{\left (3 \, a^{2} - 4 \, a b\right )} \sqrt{b} \cos \left (f x + e\right ) \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 (6 \, a^{2} \cos \left (f x + e\right )^{6} - 4 \,{\left (5 \, a^{2} - 3 \, a b\right )} \cos \left (f x + e\right )^{4} + 2 \,{\left (15 \, a^{2} - 40 \, a b + 3 \, b^{2}\right )} \cos \left (f x + e\right )^{2} - 15 \, a b\right )} \sqrt{\frac{a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{60 \, a f \cos \left (f x + e\right )}, -\frac{15 \,{\left (3 \, a^{2} - 4 \, a b\right )} \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 ) \cos \left (f x + e\right ) +{\left (6 \, a^{2} \cos \left (f x + e\right )^{6} - 4 \,{\left (5 \, a^{2} - 3 \, a b\right )} \cos \left (f x + e\right )^{4} + 2 \,{\left (15 \, a^{2} - 40 \, a b + 3 \, b^{2}\right )} \cos \left (f x + e\right )^{2} - 15 \, a b\right )} \sqrt{\frac{a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{30 \, a f \cos \left (f x + e\right )}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(f*x+e)^2)^(3/2)*sin(f*x+e)^5,x, algorithm="fricas")

[Out]

[-1/60*(15*(3*a^2 - 4*a*b)*sqrt(b)*cos(f*x + e)*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*(6*a^2*cos(f*x + e)^6 - 4*(5*a^2 - 3*a*b)*cos(f*x + e)
^4 + 2*(15*a^2 - 40*a*b + 3*b^2)*cos(f*x + e)^2 - 15*a*b)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/(a*f*co
s(f*x + e)), -1/30*(15*(3*a^2 - 4*a*b)*sqrt(-b)*arctan(sqrt(-b)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*co
s(f*x + e)/b)*cos(f*x + e) + (6*a^2*cos(f*x + e)^6 - 4*(5*a^2 - 3*a*b)*cos(f*x + e)^4 + 2*(15*a^2 - 40*a*b + 3
*b^2)*cos(f*x + e)^2 - 15*a*b)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/(a*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((a+b*sec(f*x+e)**2)**(3/2)*sin(f*x+e)**5,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}} \sin \left (f x + e\right )^{5}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(f*x+e)^2)^(3/2)*sin(f*x+e)^5,x, algorithm="giac")

[Out]

integrate((b*sec(f*x + e)^2 + a)^(3/2)*sin(f*x + e)^5, x)