3.93 \(\int \sin ^3(e+f x) \sqrt{a+b \tan ^2(e+f x)} \, dx\)

Optimal. Leaf size=113 \[ \frac{\cos ^3(e+f x) \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}{3 f (a-b)}-\frac{\cos (e+f x) \sqrt{a+b \sec ^2(e+f x)-b}}{f}+\frac{\sqrt{b} \tanh ^{-1}\left (\frac{\sqrt{b} \sec (e+f x)}{\sqrt{a+b \sec ^2(e+f x)-b}}\right )}{f} \]

[Out]

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

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Rubi [A]  time = 0.104813, antiderivative size = 113, 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 = {3664, 451, 277, 217, 206} \[ \frac{\cos ^3(e+f x) \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}{3 f (a-b)}-\frac{\cos (e+f x) \sqrt{a+b \sec ^2(e+f x)-b}}{f}+\frac{\sqrt{b} \tanh ^{-1}\left (\frac{\sqrt{b} \sec (e+f x)}{\sqrt{a+b \sec ^2(e+f x)-b}}\right )}{f} \]

Antiderivative was successfully verified.

[In]

Int[Sin[e + f*x]^3*Sqrt[a + b*Tan[e + f*x]^2],x]

[Out]

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

Rule 3664

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]^2)^(p_.), x_Symbol] :> With[{ff = Free
Factors[Sec[e + f*x], x]}, Dist[1/(f*ff^m), Subst[Int[((-1 + ff^2*x^2)^((m - 1)/2)*(a - b + b*ff^2*x^2)^p)/x^(
m + 1), x], x, Sec[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]

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

Mathematica [A]  time = 1.05357, size = 170, normalized size = 1.5 \[ \frac{\cos (e+f x) \sqrt{\sec ^2(e+f x) ((a-b) \cos (2 (e+f x))+a+b)} \left (\sqrt{(a-b) \cos (2 (e+f x))+a+b} ((a-b) \cos (2 (e+f x))-5 a+7 b)+6 \sqrt{2} \sqrt{b} (a-b) \tanh ^{-1}\left (\frac{\sqrt{(a-b) \cos (2 (e+f x))+a+b}}{\sqrt{2} \sqrt{b}}\right )\right )}{6 \sqrt{2} f (a-b) \sqrt{(a-b) \cos (2 (e+f x))+a+b}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[e + f*x]^3*Sqrt[a + b*Tan[e + f*x]^2],x]

[Out]

(Cos[e + f*x]*(6*Sqrt[2]*(a - b)*Sqrt[b]*ArcTanh[Sqrt[a + b + (a - b)*Cos[2*(e + f*x)]]/(Sqrt[2]*Sqrt[b])] + S
qrt[a + b + (a - b)*Cos[2*(e + f*x)]]*(-5*a + 7*b + (a - b)*Cos[2*(e + f*x)]))*Sqrt[(a + b + (a - b)*Cos[2*(e
+ f*x)])*Sec[e + f*x]^2])/(6*Sqrt[2]*(a - b)*f*Sqrt[a + b + (a - b)*Cos[2*(e + f*x)]])

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Maple [B]  time = 0.243, size = 4296, normalized size = 38. \begin{align*} \text{output 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*tan(f*x+e)^2)^(1/2),x)

[Out]

1/6/f/b^(1/2)/(a-b)^3/a^(1/2)*(cos(f*x+e)-1)^2*(3*ln(-4/a^(1/2)*(cos(f*x+e)-1)*(cos(f*x+e)*((a*cos(f*x+e)^2-co
s(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)-cos(f*x+e)*a+b*cos(f*x+e)+((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(
cos(f*x+e)+1)^2)^(1/2)*a^(1/2)+b)/sin(f*x+e)^2)*4^(1/2)*b^(9/2)-3*ln(-2/a^(1/2)*(cos(f*x+e)-1)*(cos(f*x+e)*((a
*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)-cos(f*x+e)*a+b*cos(f*x+e)+((a*cos(f*x+e)^2-cos
(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)+b)/sin(f*x+e)^2)*4^(1/2)*b^(9/2)-4*((a*cos(f*x+e)^2-cos(f*x+e)^
2*b+b)/(cos(f*x+e)+1)^2)^(3/2)*a^(5/2)*b^(1/2)+8*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(3/2)*a^
(3/2)*b^(3/2)-4*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(3/2)*a^(1/2)*b^(5/2)+6*cos(f*x+e)^2*((a*
cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(3/2)*a^(1/2)*b^(5/2)-4*cos(f*x+e)*((a*cos(f*x+e)^2-cos(f*x+e
)^2*b+b)/(cos(f*x+e)+1)^2)^(3/2)*a^(5/2)*b^(1/2)+8*cos(f*x+e)*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1
)^2)^(3/2)*a^(3/2)*b^(3/2)-4*cos(f*x+e)*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(3/2)*a^(1/2)*b^(
5/2)+6*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)*4^(1/2)*b^(7/2)-9*ln(-4/a^(1/2)*(cos
(f*x+e)-1)*(cos(f*x+e)*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)-cos(f*x+e)*a+b*cos(f
*x+e)+((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)+b)/sin(f*x+e)^2)*4^(1/2)*b^(7/2)*a+9*
ln(-2/a^(1/2)*(cos(f*x+e)-1)*(cos(f*x+e)*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)-co
s(f*x+e)*a+b*cos(f*x+e)+((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)+b)/sin(f*x+e)^2)*4^
(1/2)*b^(7/2)*a+3*arctanh(1/8*b^(1/2)*4^(1/2)*(cos(f*x+e)-1)*(cos(f*x+e)*4^(1/2)-2*cos(f*x+e)-4^(1/2)-2)/sin(f
*x+e)^2/((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2))*a^(7/2)*4^(1/2)*b+9*ln(-4/a^(1/2)*(cos(f*x
+e)-1)*(cos(f*x+e)*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)-cos(f*x+e)*a+b*cos(f*x+e
)+((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)+b)/sin(f*x+e)^2)*4^(1/2)*b^(5/2)*a^2-9*ln
(-2/a^(1/2)*(cos(f*x+e)-1)*(cos(f*x+e)*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)-cos(
f*x+e)*a+b*cos(f*x+e)+((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)+b)/sin(f*x+e)^2)*4^(1
/2)*b^(5/2)*a^2-9*arctanh(1/8*b^(1/2)*4^(1/2)*(cos(f*x+e)-1)*(cos(f*x+e)*4^(1/2)-2*cos(f*x+e)-4^(1/2)-2)/sin(f
*x+e)^2/((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2))*a^(5/2)*4^(1/2)*b^2-3*ln(-4/a^(1/2)*(cos(f
*x+e)-1)*(cos(f*x+e)*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)-cos(f*x+e)*a+b*cos(f*x
+e)+((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)+b)/sin(f*x+e)^2)*4^(1/2)*b^(3/2)*a^3+3*
ln(-2/a^(1/2)*(cos(f*x+e)-1)*(cos(f*x+e)*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)-co
s(f*x+e)*a+b*cos(f*x+e)+((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)+b)/sin(f*x+e)^2)*4^
(1/2)*b^(3/2)*a^3+9*arctanh(1/8*b^(1/2)*4^(1/2)*(cos(f*x+e)-1)*(cos(f*x+e)*4^(1/2)-2*cos(f*x+e)-4^(1/2)-2)/sin
(f*x+e)^2/((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2))*a^(3/2)*4^(1/2)*b^3-3*arctanh(1/8*b^(1/2
)*4^(1/2)*(cos(f*x+e)-1)*(cos(f*x+e)*4^(1/2)-2*cos(f*x+e)-4^(1/2)-2)/sin(f*x+e)^2/((a*cos(f*x+e)^2-cos(f*x+e)^
2*b+b)/(cos(f*x+e)+1)^2)^(1/2))*a^(1/2)*4^(1/2)*b^4-3*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/
2)*a^(7/2)*4^(1/2)*b^(1/2)+12*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(5/2)*4^(1/2)*b^(3/
2)-15*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(3/2)*4^(1/2)*b^(5/2)+2*cos(f*x+e)^4*((a*co
s(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(3/2)*a^(5/2)*b^(1/2)-4*cos(f*x+e)^4*((a*cos(f*x+e)^2-cos(f*x+e
)^2*b+b)/(cos(f*x+e)+1)^2)^(3/2)*a^(3/2)*b^(3/2)+2*cos(f*x+e)^4*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)
+1)^2)^(3/2)*a^(1/2)*b^(5/2)+8*cos(f*x+e)^3*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(3/2)*a^(5/2)
*b^(1/2)-16*cos(f*x+e)^3*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(3/2)*a^(3/2)*b^(3/2)+8*cos(f*x+
e)^3*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(3/2)*a^(1/2)*b^(5/2)+3*cos(f*x+e)*ln(-4/a^(1/2)*(co
s(f*x+e)-1)*(cos(f*x+e)*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)-cos(f*x+e)*a+b*cos(
f*x+e)+((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)+b)/sin(f*x+e)^2)*4^(1/2)*b^(9/2)-3*c
os(f*x+e)*ln(-2/a^(1/2)*(cos(f*x+e)-1)*(cos(f*x+e)*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*
a^(1/2)-cos(f*x+e)*a+b*cos(f*x+e)+((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)+b)/sin(f*
x+e)^2)*4^(1/2)*b^(9/2)+6*cos(f*x+e)^2*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(3/2)*a^(5/2)*b^(1
/2)-12*cos(f*x+e)^2*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(3/2)*a^(3/2)*b^(3/2)+6*cos(f*x+e)*((
a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)*4^(1/2)*b^(7/2)-9*cos(f*x+e)*ln(-4/a^(1/2)*(c
os(f*x+e)-1)*(cos(f*x+e)*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)-cos(f*x+e)*a+b*cos
(f*x+e)+((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)+b)/sin(f*x+e)^2)*4^(1/2)*b^(7/2)*a+
9*cos(f*x+e)*ln(-2/a^(1/2)*(cos(f*x+e)-1)*(cos(f*x+e)*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/
2)*a^(1/2)-cos(f*x+e)*a+b*cos(f*x+e)+((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)+b)/sin
(f*x+e)^2)*4^(1/2)*b^(7/2)*a+3*cos(f*x+e)*arctanh(1/8*b^(1/2)*4^(1/2)*(cos(f*x+e)-1)*(cos(f*x+e)*4^(1/2)-2*cos
(f*x+e)-4^(1/2)-2)/sin(f*x+e)^2/((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2))*a^(7/2)*4^(1/2)*b+
9*cos(f*x+e)*ln(-4/a^(1/2)*(cos(f*x+e)-1)*(cos(f*x+e)*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/
2)*a^(1/2)-cos(f*x+e)*a+b*cos(f*x+e)+((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)+b)/sin
(f*x+e)^2)*4^(1/2)*b^(5/2)*a^2-9*cos(f*x+e)*ln(-2/a^(1/2)*(cos(f*x+e)-1)*(cos(f*x+e)*((a*cos(f*x+e)^2-cos(f*x+
e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)-cos(f*x+e)*a+b*cos(f*x+e)+((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*
x+e)+1)^2)^(1/2)*a^(1/2)+b)/sin(f*x+e)^2)*4^(1/2)*b^(5/2)*a^2-9*cos(f*x+e)*arctanh(1/8*b^(1/2)*4^(1/2)*(cos(f*
x+e)-1)*(cos(f*x+e)*4^(1/2)-2*cos(f*x+e)-4^(1/2)-2)/sin(f*x+e)^2/((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e
)+1)^2)^(1/2))*a^(5/2)*4^(1/2)*b^2-3*cos(f*x+e)*ln(-4/a^(1/2)*(cos(f*x+e)-1)*(cos(f*x+e)*((a*cos(f*x+e)^2-cos(
f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)-cos(f*x+e)*a+b*cos(f*x+e)+((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(co
s(f*x+e)+1)^2)^(1/2)*a^(1/2)+b)/sin(f*x+e)^2)*4^(1/2)*b^(3/2)*a^3+3*cos(f*x+e)*ln(-2/a^(1/2)*(cos(f*x+e)-1)*(c
os(f*x+e)*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)-cos(f*x+e)*a+b*cos(f*x+e)+((a*cos
(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)+b)/sin(f*x+e)^2)*4^(1/2)*b^(3/2)*a^3+9*cos(f*x+e)*
arctanh(1/8*b^(1/2)*4^(1/2)*(cos(f*x+e)-1)*(cos(f*x+e)*4^(1/2)-2*cos(f*x+e)-4^(1/2)-2)/sin(f*x+e)^2/((a*cos(f*
x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2))*a^(3/2)*4^(1/2)*b^3-3*cos(f*x+e)*arctanh(1/8*b^(1/2)*4^(1/2)
*(cos(f*x+e)-1)*(cos(f*x+e)*4^(1/2)-2*cos(f*x+e)-4^(1/2)-2)/sin(f*x+e)^2/((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(c
os(f*x+e)+1)^2)^(1/2))*a^(1/2)*4^(1/2)*b^4-6*cos(f*x+e)*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(
1/2)*a^(7/2)*4^(1/2)*b^(1/2)+18*cos(f*x+e)*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(5/2)*
4^(1/2)*b^(3/2)-18*cos(f*x+e)*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(3/2)*4^(1/2)*b^(5/
2))*cos(f*x+e)*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/cos(f*x+e)^2)^(1/2)/((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(
f*x+e)+1)^2)^(1/2)/sin(f*x+e)^4

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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(sin(f*x+e)^3*(a+b*tan(f*x+e)^2)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 3.45313, size = 674, normalized size = 5.96 \begin{align*} \left [\frac{3 \,{\left (a - b\right )} \sqrt{b} \log \left (-\frac{{\left (a - b\right )} \cos \left (f x + e\right )^{2} + 2 \, \sqrt{b} \sqrt{\frac{{\left (a - b\right )} \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 ({\left (a - b\right )} \cos \left (f x + e\right )^{3} -{\left (3 \, a - 4 \, b\right )} \cos \left (f x + e\right )\right )} \sqrt{\frac{{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{6 \,{\left (a - b\right )} f}, -\frac{3 \,{\left (a - b\right )} \sqrt{-b} \arctan \left (\frac{\sqrt{-b} \sqrt{\frac{{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{b}\right ) -{\left ({\left (a - b\right )} \cos \left (f x + e\right )^{3} -{\left (3 \, a - 4 \, b\right )} \cos \left (f x + e\right )\right )} \sqrt{\frac{{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{3 \,{\left (a - b\right )} f}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^3*(a+b*tan(f*x+e)^2)^(1/2),x, algorithm="fricas")

[Out]

[1/6*(3*(a - b)*sqrt(b)*log(-((a - b)*cos(f*x + e)^2 + 2*sqrt(b)*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e
)^2)*cos(f*x + e) + 2*b)/cos(f*x + e)^2) + 2*((a - b)*cos(f*x + e)^3 - (3*a - 4*b)*cos(f*x + e))*sqrt(((a - b)
*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/((a - b)*f), -1/3*(3*(a - b)*sqrt(-b)*arctan(sqrt(-b)*sqrt(((a - b)*cos(
f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e)/b) - ((a - b)*cos(f*x + e)^3 - (3*a - 4*b)*cos(f*x + e))*sqrt(((a
- b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/((a - b)*f)]

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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(sin(f*x+e)**3*(a+b*tan(f*x+e)**2)**(1/2),x)

[Out]

Timed out

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Giac [B]  time = 1.65597, size = 410, normalized size = 3.63 \begin{align*} \frac{1}{3} \,{\left (\frac{{\left (6 \, a b \arctan \left (\frac{\sqrt{b}}{\sqrt{-b}}\right ) - 6 \, b^{2} \arctan \left (\frac{\sqrt{b}}{\sqrt{-b}}\right ) + 3 \, a \sqrt{-b} \sqrt{b} - 4 \, \sqrt{-b} b^{\frac{3}{2}}\right )} \mathrm{sgn}\left (f\right ) \mathrm{sgn}\left (\cos \left (f x + e\right )\right )}{a \sqrt{-b} f^{2} - \sqrt{-b} b f^{2}} - \sqrt{a \cos \left (f x + e\right )^{2} - b \cos \left (f x + e\right )^{2} + b}{\left (\frac{3 \, a \mathrm{sgn}\left (f\right ) \mathrm{sgn}\left (\cos \left (f x + e\right )\right ) - 4 \, b \mathrm{sgn}\left (f\right ) \mathrm{sgn}\left (\cos \left (f x + e\right )\right )}{a f^{2} - b f^{2}} - \frac{{\left (a f^{2} \mathrm{sgn}\left (f\right ) \mathrm{sgn}\left (\cos \left (f x + e\right )\right ) - b f^{2} \mathrm{sgn}\left (f\right ) \mathrm{sgn}\left (\cos \left (f x + e\right )\right )\right )} \cos \left (f x + e\right )^{2}}{{\left (a f^{2} - b f^{2}\right )} f^{2}}\right )} - \frac{6 \, b \arctan \left (\frac{\sqrt{a \cos \left (f x + e\right )^{2} - b \cos \left (f x + e\right )^{2} + b} + \frac{\sqrt{a f^{2} - b f^{2}} \cos \left (f x + e\right )}{f}}{\sqrt{-b}}\right ) \mathrm{sgn}\left (f\right ) \mathrm{sgn}\left (\cos \left (f x + e\right )\right )}{\sqrt{-b} f^{2}}\right )}{\left | f \right |} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^3*(a+b*tan(f*x+e)^2)^(1/2),x, algorithm="giac")

[Out]

1/3*((6*a*b*arctan(sqrt(b)/sqrt(-b)) - 6*b^2*arctan(sqrt(b)/sqrt(-b)) + 3*a*sqrt(-b)*sqrt(b) - 4*sqrt(-b)*b^(3
/2))*sgn(f)*sgn(cos(f*x + e))/(a*sqrt(-b)*f^2 - sqrt(-b)*b*f^2) - sqrt(a*cos(f*x + e)^2 - b*cos(f*x + e)^2 + b
)*((3*a*sgn(f)*sgn(cos(f*x + e)) - 4*b*sgn(f)*sgn(cos(f*x + e)))/(a*f^2 - b*f^2) - (a*f^2*sgn(f)*sgn(cos(f*x +
e)) - b*f^2*sgn(f)*sgn(cos(f*x + e)))*cos(f*x + e)^2/((a*f^2 - b*f^2)*f^2)) - 6*b*arctan((sqrt(a*cos(f*x + e)
^2 - b*cos(f*x + e)^2 + b) + sqrt(a*f^2 - b*f^2)*cos(f*x + e)/f)/sqrt(-b))*sgn(f)*sgn(cos(f*x + e))/(sqrt(-b)*
f^2))*abs(f)