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

Optimal. Leaf size=186 \[ \frac {b (3 a-5 b) \sec (e+f x) \sqrt {a+b \sec ^2(e+f x)-b}}{2 f (a-b)}+\frac {\cos ^3(e+f x) \left (a+b \sec ^2(e+f x)-b\right )^{5/2}}{3 f (a-b)}-\frac {(3 a-5 b) \cos (e+f x) \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}{3 f (a-b)}+\frac {\sqrt {b} (3 a-5 b) \tanh ^{-1}\left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a+b \sec ^2(e+f x)-b}}\right )}{2 f} \]

[Out]

-1/3*(3*a-5*b)*cos(f*x+e)*(a-b+b*sec(f*x+e)^2)^(3/2)/(a-b)/f+1/3*cos(f*x+e)^3*(a-b+b*sec(f*x+e)^2)^(5/2)/(a-b)
/f+1/2*(3*a-5*b)*arctanh(sec(f*x+e)*b^(1/2)/(a-b+b*sec(f*x+e)^2)^(1/2))*b^(1/2)/f+1/2*(3*a-5*b)*b*sec(f*x+e)*(
a-b+b*sec(f*x+e)^2)^(1/2)/(a-b)/f

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Rubi [A]  time = 0.16, antiderivative size = 186, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {3664, 453, 277, 195, 217, 206} \[ \frac {b (3 a-5 b) \sec (e+f x) \sqrt {a+b \sec ^2(e+f x)-b}}{2 f (a-b)}+\frac {\cos ^3(e+f x) \left (a+b \sec ^2(e+f x)-b\right )^{5/2}}{3 f (a-b)}-\frac {(3 a-5 b) \cos (e+f x) \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}{3 f (a-b)}+\frac {\sqrt {b} (3 a-5 b) \tanh ^{-1}\left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a+b \sec ^2(e+f x)-b}}\right )}{2 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 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 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 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 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]

Rubi steps

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

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Mathematica [A]  time = 1.89, size = 188, normalized size = 1.01 \[ \frac {\sec (e+f x) \sqrt {\sec ^2(e+f x) ((a-b) \cos (2 (e+f x))+a+b)} \left (\sqrt {(a-b) \cos (2 (e+f x))+a+b} (-8 (a-3 b) \cos (2 (e+f x))+(a-b) \cos (4 (e+f x))-9 a+37 b)+12 \sqrt {2} \sqrt {b} (3 a-5 b) \cos ^2(e+f x) \tanh ^{-1}\left (\frac {\sqrt {(a-b) \cos (2 (e+f x))+a+b}}{\sqrt {2} \sqrt {b}}\right )\right )}{24 \sqrt {2} f \sqrt {(a-b) \cos (2 (e+f x))+a+b}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((12*Sqrt[2]*(3*a - 5*b)*Sqrt[b]*ArcTanh[Sqrt[a + b + (a - b)*Cos[2*(e + f*x)]]/(Sqrt[2]*Sqrt[b])]*Cos[e + f*x
]^2 + Sqrt[a + b + (a - b)*Cos[2*(e + f*x)]]*(-9*a + 37*b - 8*(a - 3*b)*Cos[2*(e + f*x)] + (a - b)*Cos[4*(e +
f*x)]))*Sec[e + f*x]*Sqrt[(a + b + (a - b)*Cos[2*(e + f*x)])*Sec[e + f*x]^2])/(24*Sqrt[2]*f*Sqrt[a + b + (a -
b)*Cos[2*(e + f*x)]])

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fricas [A]  time = 0.82, size = 307, normalized size = 1.65 \[ \left [-\frac {3 \, {\left (3 \, a - 5 \, b\right )} \sqrt {b} \cos \left (f x + e\right ) \log \left (-\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} - 2 \, \sqrt {b} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) + 2 \, b}{\cos \left (f x + e\right )^{2}}\right ) - 2 \, {\left (2 \, {\left (a - b\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (3 \, a - 7 \, b\right )} \cos \left (f x + e\right )^{2} + 3 \, b\right )} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{12 \, f \cos \left (f x + e\right )}, -\frac {3 \, {\left (3 \, a - 5 \, b\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{b}\right ) \cos \left (f x + e\right ) - {\left (2 \, {\left (a - b\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (3 \, a - 7 \, b\right )} \cos \left (f x + e\right )^{2} + 3 \, b\right )} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{6 \, f \cos \left (f x + e\right )}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/12*(3*(3*a - 5*b)*sqrt(b)*cos(f*x + e)*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*(2*(a - b)*cos(f*x + e)^4 - 2*(3*a - 7*b)*cos(
f*x + e)^2 + 3*b)*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/(f*cos(f*x + e)), -1/6*(3*(3*a - 5*b)*sqr
t(-b)*arctan(sqrt(-b)*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e)/b)*cos(f*x + e) - (2*(a -
b)*cos(f*x + e)^4 - 2*(3*a - 7*b)*cos(f*x + e)^2 + 3*b)*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/(f
*cos(f*x + e))]

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: NotImplementedError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: Unable to check sign: (2*pi/x/2)>(-2*pi/
x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check si
gn: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/
x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)abs(f)*(-1/2*(b^2*sqrt(a*f^2*(-cos(f*x+exp(1))/f)^2-b*f^2*(-co
s(f*x+exp(1))/f)^2+b)*sign(cos(f*x+exp(1)))*sign(f)-a*b*sqrt(a*f^2*(-cos(f*x+exp(1))/f)^2-b*f^2*(-cos(f*x+exp(
1))/f)^2+b)*sign(cos(f*x+exp(1)))*sign(f))/f^2/(a*f^2*(-cos(f*x+exp(1))/f)^2-b*f^2*(-cos(f*x+exp(1))/f)^2)+(1/
3*f^4*sqrt(a*f^2*(-cos(f*x+exp(1))/f)^2-b*f^2*(-cos(f*x+exp(1))/f)^2+b)*(a*f^2*(-cos(f*x+exp(1))/f)^2-b*f^2*(-
cos(f*x+exp(1))/f)^2+b)*sign(cos(f*x+exp(1)))*sign(f)-a*f^4*sqrt(a*f^2*(-cos(f*x+exp(1))/f)^2-b*f^2*(-cos(f*x+
exp(1))/f)^2+b)*sign(cos(f*x+exp(1)))*sign(f)+2*b*f^4*sqrt(a*f^2*(-cos(f*x+exp(1))/f)^2-b*f^2*(-cos(f*x+exp(1)
)/f)^2+b)*sign(cos(f*x+exp(1)))*sign(f))/f^6-1/2*(-5*b^2*sign(cos(f*x+exp(1)))*sign(f)+3*a*b*sign(cos(f*x+exp(
1)))*sign(f))*atan(sqrt(a*f^2*(-cos(f*x+exp(1))/f)^2-b*f^2*(-cos(f*x+exp(1))/f)^2+b)/sqrt(-b))/sqrt(-b)/f^2)

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maple [B]  time = 0.80, size = 1104, normalized size = 5.94 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12/f*(-1+cos(f*x+e))^3*(2*cos(f*x+e)^5*a^(7/2)*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(1+cos(f*x+e))^2)^(1/2)*b
^(1/2)-2*cos(f*x+e)^5*a^(5/2)*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(1+cos(f*x+e))^2)^(1/2)*b^(3/2)+2*cos(f*x+e)^
4*a^(7/2)*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(1+cos(f*x+e))^2)^(1/2)*b^(1/2)-2*cos(f*x+e)^4*a^(5/2)*((a*cos(f*
x+e)^2-cos(f*x+e)^2*b+b)/(1+cos(f*x+e))^2)^(1/2)*b^(3/2)-6*cos(f*x+e)^3*a^(7/2)*((a*cos(f*x+e)^2-cos(f*x+e)^2*
b+b)/(1+cos(f*x+e))^2)^(1/2)*b^(1/2)+14*cos(f*x+e)^3*a^(5/2)*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(1+cos(f*x+e))
^2)^(1/2)*b^(3/2)+9*cos(f*x+e)^2*a^(7/2)*arctanh(1/8*(-1+cos(f*x+e))*(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)/(1+cos(f*x+e))^2)^(1/2)*b^(1/2)*4^(1/2))*b-6*cos(f*x+e)^2*a
^(7/2)*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(1+cos(f*x+e))^2)^(1/2)*b^(1/2)+14*cos(f*x+e)^2*a^(5/2)*((a*cos(f*x+
e)^2-cos(f*x+e)^2*b+b)/(1+cos(f*x+e))^2)^(1/2)*b^(3/2)-6*cos(f*x+e)^2*ln(-2*(-1+cos(f*x+e))*(a^(1/2)*cos(f*x+e
)*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(1+cos(f*x+e))^2)^(1/2)+((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(1+cos(f*x+e))
^2)^(1/2)*a^(1/2)-a*cos(f*x+e)+b*cos(f*x+e)+b)/sin(f*x+e)^2/a^(1/2))*b^(7/2)*a+6*cos(f*x+e)^2*ln(-4*(-1+cos(f*
x+e))*(a^(1/2)*cos(f*x+e)*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(1+cos(f*x+e))^2)^(1/2)+((a*cos(f*x+e)^2-cos(f*x+
e)^2*b+b)/(1+cos(f*x+e))^2)^(1/2)*a^(1/2)-a*cos(f*x+e)+b*cos(f*x+e)+b)/sin(f*x+e)^2/a^(1/2))*b^(7/2)*a-15*cos(
f*x+e)^2*a^(5/2)*arctanh(1/8*(-1+cos(f*x+e))*(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)/(1+cos(f*x+e))^2)^(1/2)*b^(1/2)*4^(1/2))*b^2+3*cos(f*x+e)*a^(5/2)*((a*cos(f*x+e)^2-
cos(f*x+e)^2*b+b)/(1+cos(f*x+e))^2)^(1/2)*b^(3/2)+3*a^(5/2)*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(1+cos(f*x+e))^
2)^(1/2)*b^(3/2))*cos(f*x+e)*4^(1/2)*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/cos(f*x+e)^2)^(3/2)/((a*cos(f*x+e)^2-c
os(f*x+e)^2*b+b)/(1+cos(f*x+e))^2)^(3/2)/sin(f*x+e)^6/a^(5/2)/b^(1/2)

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maxima [A]  time = 0.98, size = 296, normalized size = 1.59 \[ \frac {4 \, {\left (a - b + \frac {b}{\cos \left (f x + e\right )^{2}}\right )}^{\frac {3}{2}} \cos \left (f x + e\right )^{3} - 12 \, \sqrt {a - b + \frac {b}{\cos \left (f x + e\right )^{2}}} {\left (a - b\right )} \cos \left (f x + e\right ) + 12 \, \sqrt {a - b + \frac {b}{\cos \left (f x + e\right )^{2}}} b \cos \left (f x + e\right ) + 6 \, b^{\frac {3}{2}} \log \left (\frac {\sqrt {a - b + \frac {b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) - \sqrt {b}}{\sqrt {a - b + \frac {b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) + \sqrt {b}}\right ) + \frac {6 \, {\left (a b - b^{2}\right )} \sqrt {a - b + \frac {b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{{\left (a - b + \frac {b}{\cos \left (f x + e\right )^{2}}\right )} \cos \left (f x + e\right )^{2} - b} - \frac {9 \, {\left (a b - b^{2}\right )} \log \left (\frac {\sqrt {a - b + \frac {b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) - \sqrt {b}}{\sqrt {a - b + \frac {b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) + \sqrt {b}}\right )}{\sqrt {b}}}{12 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(4*(a - b + b/cos(f*x + e)^2)^(3/2)*cos(f*x + e)^3 - 12*sqrt(a - b + b/cos(f*x + e)^2)*(a - b)*cos(f*x +
e) + 12*sqrt(a - b + b/cos(f*x + e)^2)*b*cos(f*x + e) + 6*b^(3/2)*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))) + 6*(a*b - b^2)*sqrt(a - b + b/cos(f*
x + e)^2)*cos(f*x + e)/((a - b + b/cos(f*x + e)^2)*cos(f*x + e)^2 - b) - 9*(a*b - b^2)*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)))/sqrt(b))/f

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\sin \left (e+f\,x\right )}^3\,{\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a\right )}^{3/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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