3.1.93 \(\int \sin ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx\) [93]
Optimal. Leaf size=113 \[ \frac {\sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a-b+b \sec ^2(e+f x)}}\right )}{f}-\frac {\cos (e+f x) \sqrt {a-b+b \sec ^2(e+f x)}}{f}+\frac {\cos ^3(e+f x) \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}{3 (a-b) f} \]
[Out]
1/3*cos(f*x+e)^3*(a-b+b*sec(f*x+e)^2)^(3/2)/(a-b)/f+arctanh(sec(f*x+e)*b^(1/2)/(a-b+b*sec(f*x+e)^2)^(1/2))*b^(
1/2)/f-cos(f*x+e)*(a-b+b*sec(f*x+e)^2)^(1/2)/f
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Rubi [A]
time = 0.07, antiderivative size = 113, normalized size of antiderivative = 1.00, number of steps
used = 5, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3745, 462, 283,
223, 212} \begin {gather*} \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} \end {gather*}
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 212
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])
Rule 223
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 283
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 462
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 3745
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]
Rubi steps
\begin {align*} \int \sin ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx &=\frac {\text {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 {\text {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 \text {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 \text {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*}
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Mathematica [A]
time = 0.74, size = 170, normalized size = 1.50 \begin {gather*} \frac {\cos (e+f x) \left (6 \sqrt {2} (a-b) \sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {a+b+(a-b) \cos (2 (e+f x))}}{\sqrt {2} \sqrt {b}}\right )+\sqrt {a+b+(a-b) \cos (2 (e+f x))} (-5 a+7 b+(a-b) \cos (2 (e+f x)))\right ) \sqrt {(a+b+(a-b) \cos (2 (e+f x))) \sec ^2(e+f x)}}{6 \sqrt {2} (a-b) f \sqrt {a+b+(a-b) \cos (2 (e+f x))}} \end {gather*}
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] Leaf count of result is larger than twice the leaf count of optimal. \(4295\) vs.
\(2(101)=202\).
time = 0.35, size = 4296, normalized size = 38.02
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\(4296\) |
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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,method=_RETURNVERBOSE)
[Out]
-1/6/f*(cos(f*x+e)-1)^2*(-6*a^(5/2)*b^(1/2)*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)+12*a^(3/2)*b^(3/2)*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)-6*a^(1/2)*b^(
5/2)*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)-3*a^(7/2)*4^(1/2)*arctanh(1/8*(co
s(f*x+e)-1)*(4^(1/2)*cos(f*x+e)-4^(1/2)-2*cos(f*x+e)-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)*b^(1/2)*4^(1/2))*b+4*a^(5/2)*b^(1/2)*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)-8*a^(3/2)*b^(3/2)*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)-6*a^(1/
2)*4^(1/2)*b^(7/2)*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)+4*a^(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)+1)^2)^(3/2)+9*4^(1/2)*b^(7/2)*ln(-4*(cos(f*x+e)-1)*(cos(f*x+e)*a^(
1/2)*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(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/a^(1/2))*a-9*4^(1/2)*b^(7/2)*ln(-2*(cos(f*x+e)-
1)*(cos(f*x+e)*a^(1/2)*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(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/a^(1/2))*a+9*a^(5/2)*4^(1/2)*
arctanh(1/8*(cos(f*x+e)-1)*(4^(1/2)*cos(f*x+e)-4^(1/2)-2*cos(f*x+e)-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)*b^(1/2)*4^(1/2))*b^2-9*4^(1/2)*b^(5/2)*ln(-4*(cos(f*x+e)-1)*(cos(f*x+e)*a^(1/
2)*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(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/a^(1/2))*a^2+9*4^(1/2)*b^(5/2)*ln(-2*(cos(f*x+e)-
1)*(cos(f*x+e)*a^(1/2)*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(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/a^(1/2))*a^2-9*a^(3/2)*4^(1/2
)*arctanh(1/8*(cos(f*x+e)-1)*(4^(1/2)*cos(f*x+e)-4^(1/2)-2*cos(f*x+e)-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)*b^(1/2)*4^(1/2))*b^3+3*4^(1/2)*b^(3/2)*ln(-4*(cos(f*x+e)-1)*(cos(f*x+e)*a^(
1/2)*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(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/a^(1/2))*a^3-3*4^(1/2)*b^(3/2)*ln(-2*(cos(f*x+e
)-1)*(cos(f*x+e)*a^(1/2)*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(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/a^(1/2))*a^3+3*a^(1/2)*4^(1
/2)*arctanh(1/8*(cos(f*x+e)-1)*(4^(1/2)*cos(f*x+e)-4^(1/2)-2*cos(f*x+e)-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)*b^(1/2)*4^(1/2))*b^4-2*a^(5/2)*b^(1/2)*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)+4*a^(3/2)*b^(3/2)*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)-2*a^(1/2)*b^(5/2)*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)-8*
a^(5/2)*b^(1/2)*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)+3*a^(7/2)*4^(1/2)*b^(1
/2)*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)-12*a^(5/2)*4^(1/2)*b^(3/2)*((a*cos(f*x+e)^2-cos
(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)+15*a^(3/2)*4^(1/2)*b^(5/2)*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+
e)+1)^2)^(1/2)+16*a^(3/2)*b^(3/2)*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)-8*a^
(1/2)*b^(5/2)*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)-3*4^(1/2)*b^(9/2)*cos(f*
x+e)*ln(-4*(cos(f*x+e)-1)*(cos(f*x+e)*a^(1/2)*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(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/a^(1/2
))+3*4^(1/2)*b^(9/2)*cos(f*x+e)*ln(-2*(cos(f*x+e)-1)*(cos(f*x+e)*a^(1/2)*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(c
os(f*x+e)+1)^2)^(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/a^(1/2))-3*4^(1/2)*b^(9/2)*ln(-4*(cos(f*x+e)-1)*(cos(f*x+e)*a^(1/2)*((a*cos(f*x+e)^2-cos(
f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(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/a^(1/2))+3*4^(1/2)*b^(9/2)*ln(-2*(cos(f*x+e)-1)*(cos(f*x+e)*a^(1/2)*((a*c
os(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(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/a^(1/2))+4*a^(5/2)*b^(1/2)*((a*cos(f*x+e)^2-cos(f*x+e)^2*
b+b)/(cos(f*x+e)+1)^2)^(3/2)-8*a^(3/2)*b^(3/2)*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(3/2)+4*a^
(1/2)*b^(5/2)*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(3/2)-3*a^(7/2)*4^(1/2)*cos(f*x+e)*arctanh(
1/8*(cos(f*x+e)-1)*(4^(1/2)*cos(f*x+e)-4^(1/2)-2*cos(f*x+e)-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)*b^(1/2)*4^(1/2))*b-6*a...
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Maxima [A]
time = 0.52, size = 140, normalized size = 1.24 \begin {gather*} \frac {\frac {2 \, {\left (a - b + \frac {b}{\cos \left (f x + e\right )^{2}}\right )}^{\frac {3}{2}} \cos \left (f x + e\right )^{3}}{a - b} - 6 \, \sqrt {a - b + \frac {b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) - 3 \, \sqrt {b} \log \left (\frac {\sqrt {a - b + \frac {b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) - \sqrt {b}}{\sqrt {a - b + \frac {b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) + \sqrt {b}}\right )}{6 \, f} \end {gather*}
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]
1/6*(2*(a - b + b/cos(f*x + e)^2)^(3/2)*cos(f*x + e)^3/(a - b) - 6*sqrt(a - b + b/cos(f*x + e)^2)*cos(f*x + e)
- 3*sqrt(b)*log((sqrt(a - b + b/cos(f*x + e)^2)*cos(f*x + e) - sqrt(b))/(sqrt(a - b + b/cos(f*x + e)^2)*cos(f
*x + e) + sqrt(b))))/f
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Fricas [A]
time = 6.20, size = 293, normalized size = 2.59 \begin {gather*} \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 {gather*}
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)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
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] Leaf count of result is larger than twice the leaf count of optimal. 498 vs.
\(2 (107) = 214\).
time = 0.92, size = 498, normalized size = 4.41 \begin {gather*} \frac {1}{3} \, {\left (\frac {{\left (3 \, a b \arctan \left (\frac {\sqrt {b}}{\sqrt {-b}}\right ) - 3 \, 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}} - \frac {3 \, b \arctan \left (\frac {\sqrt {a \cos \left (f x + e\right )^{2} - b \cos \left (f x + e\right )^{2} + b}}{\sqrt {-b}}\right ) \mathrm {sgn}\left (f\right ) \mathrm {sgn}\left (\cos \left (f x + e\right )\right )}{\sqrt {-b} f^{2}} + \frac {{\left (a \cos \left (f x + e\right )^{2} - b \cos \left (f x + e\right )^{2} + b\right )}^{\frac {3}{2}} a^{2} f^{4} \mathrm {sgn}\left (f\right ) \mathrm {sgn}\left (\cos \left (f x + e\right )\right ) - 3 \, \sqrt {a \cos \left (f x + e\right )^{2} - b \cos \left (f x + e\right )^{2} + b} a^{3} f^{4} \mathrm {sgn}\left (f\right ) \mathrm {sgn}\left (\cos \left (f x + e\right )\right ) - 2 \, {\left (a \cos \left (f x + e\right )^{2} - b \cos \left (f x + e\right )^{2} + b\right )}^{\frac {3}{2}} a b f^{4} \mathrm {sgn}\left (f\right ) \mathrm {sgn}\left (\cos \left (f x + e\right )\right ) + 9 \, \sqrt {a \cos \left (f x + e\right )^{2} - b \cos \left (f x + e\right )^{2} + b} a^{2} b f^{4} \mathrm {sgn}\left (f\right ) \mathrm {sgn}\left (\cos \left (f x + e\right )\right ) + {\left (a \cos \left (f x + e\right )^{2} - b \cos \left (f x + e\right )^{2} + b\right )}^{\frac {3}{2}} b^{2} f^{4} \mathrm {sgn}\left (f\right ) \mathrm {sgn}\left (\cos \left (f x + e\right )\right ) - 9 \, \sqrt {a \cos \left (f x + e\right )^{2} - b \cos \left (f x + e\right )^{2} + b} a b^{2} f^{4} \mathrm {sgn}\left (f\right ) \mathrm {sgn}\left (\cos \left (f x + e\right )\right ) + 3 \, \sqrt {a \cos \left (f x + e\right )^{2} - b \cos \left (f x + e\right )^{2} + b} b^{3} f^{4} \mathrm {sgn}\left (f\right ) \mathrm {sgn}\left (\cos \left (f x + e\right )\right )}{a^{3} f^{6} - 3 \, a^{2} b f^{6} + 3 \, a b^{2} f^{6} - b^{3} f^{6}}\right )} {\left | f \right |} \end {gather*}
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*((3*a*b*arctan(sqrt(b)/sqrt(-b)) - 3*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) - 3*b*arctan(sqrt(a*cos(f*x + e)^2 - b*cos(f*x
+ e)^2 + b)/sqrt(-b))*sgn(f)*sgn(cos(f*x + e))/(sqrt(-b)*f^2) + ((a*cos(f*x + e)^2 - b*cos(f*x + e)^2 + b)^(3
/2)*a^2*f^4*sgn(f)*sgn(cos(f*x + e)) - 3*sqrt(a*cos(f*x + e)^2 - b*cos(f*x + e)^2 + b)*a^3*f^4*sgn(f)*sgn(cos(
f*x + e)) - 2*(a*cos(f*x + e)^2 - b*cos(f*x + e)^2 + b)^(3/2)*a*b*f^4*sgn(f)*sgn(cos(f*x + e)) + 9*sqrt(a*cos(
f*x + e)^2 - b*cos(f*x + e)^2 + b)*a^2*b*f^4*sgn(f)*sgn(cos(f*x + e)) + (a*cos(f*x + e)^2 - b*cos(f*x + e)^2 +
b)^(3/2)*b^2*f^4*sgn(f)*sgn(cos(f*x + e)) - 9*sqrt(a*cos(f*x + e)^2 - b*cos(f*x + e)^2 + b)*a*b^2*f^4*sgn(f)*
sgn(cos(f*x + e)) + 3*sqrt(a*cos(f*x + e)^2 - b*cos(f*x + e)^2 + b)*b^3*f^4*sgn(f)*sgn(cos(f*x + e)))/(a^3*f^6
- 3*a^2*b*f^6 + 3*a*b^2*f^6 - b^3*f^6))*abs(f)
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\sin \left (e+f\,x\right )}^3\,\sqrt {b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(sin(e + f*x)^3*(a + b*tan(e + f*x)^2)^(1/2),x)
[Out]
int(sin(e + f*x)^3*(a + b*tan(e + f*x)^2)^(1/2), x)
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