3.3.90 \(\int \sec ^4(e+f x) (a+b \sin ^2(e+f x)) \, dx\) [290]

Optimal. Leaf size=30 \[ \frac {a \tan (e+f x)}{f}+\frac {(a+b) \tan ^3(e+f x)}{3 f} \]

[Out]

a*tan(f*x+e)/f+1/3*(a+b)*tan(f*x+e)^3/f

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Rubi [A]
time = 0.02, antiderivative size = 30, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.048, Rules used = {3270} \begin {gather*} \frac {(a+b) \tan ^3(e+f x)}{3 f}+\frac {a \tan (e+f x)}{f} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sec[e + f*x]^4*(a + b*Sin[e + f*x]^2),x]

[Out]

(a*Tan[e + f*x])/f + ((a + b)*Tan[e + f*x]^3)/(3*f)

Rule 3270

Int[cos[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.), x_Symbol] :> With[{ff = FreeF
actors[Tan[e + f*x], x]}, Dist[ff/f, Subst[Int[(a + (a + b)*ff^2*x^2)^p/(1 + ff^2*x^2)^(m/2 + p + 1), x], x, T
an[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x] && IntegerQ[m/2] && IntegerQ[p]

Rubi steps

\begin {align*} \int \sec ^4(e+f x) \left (a+b \sin ^2(e+f x)\right ) \, dx &=\frac {\text {Subst}\left (\int \left (a+(a+b) x^2\right ) \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {a \tan (e+f x)}{f}+\frac {(a+b) \tan ^3(e+f x)}{3 f}\\ \end {align*}

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Mathematica [A]
time = 0.06, size = 41, normalized size = 1.37 \begin {gather*} \frac {b \tan ^3(e+f x)}{3 f}+\frac {a \left (\tan (e+f x)+\frac {1}{3} \tan ^3(e+f x)\right )}{f} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sec[e + f*x]^4*(a + b*Sin[e + f*x]^2),x]

[Out]

(b*Tan[e + f*x]^3)/(3*f) + (a*(Tan[e + f*x] + Tan[e + f*x]^3/3))/f

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Maple [A]
time = 0.28, size = 46, normalized size = 1.53

method	result	size
derivativedivides	\(\frac {-a \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (f x +e \right )\right )}{3}\right ) \tan \left (f x +e \right )+\frac {b \left (\sin ^{3}\left (f x +e \right )\right )}{3 \cos \left (f x +e \right )^{3}}}{f}\)	\(46\)
default	\(\frac {-a \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (f x +e \right )\right )}{3}\right ) \tan \left (f x +e \right )+\frac {b \left (\sin ^{3}\left (f x +e \right )\right )}{3 \cos \left (f x +e \right )^{3}}}{f}\)	\(46\)
risch	\(-\frac {2 i \left (3 \,{\mathrm e}^{4 i \left (f x +e \right )} b -6 a \,{\mathrm e}^{2 i \left (f x +e \right )}-2 a +b \right )}{3 f \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{3}}\)	\(49\)
norman	\(\frac {-\frac {2 a \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{f}-\frac {2 a \left (\tan ^{9}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{f}-\frac {8 \left (a +b \right ) \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{3 f}-\frac {8 \left (a +b \right ) \left (\tan ^{7}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{3 f}-\frac {4 \left (4 b +a \right ) \left (\tan ^{5}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{3 f}}{\left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{3} \left (1+\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )^{2}}\)	\(124\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(f*x+e)^4*(a+b*sin(f*x+e)^2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.28, size = 29, normalized size = 0.97 \begin {gather*} \frac {{\left (a + b\right )} \tan \left (f x + e\right )^{3} + 3 \, a \tan \left (f x + e\right )}{3 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)^4*(a+b*sin(f*x+e)^2),x, algorithm="maxima")

[Out]

1/3*((a + b)*tan(f*x + e)^3 + 3*a*tan(f*x + e))/f

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Fricas [A]
time = 0.39, size = 38, normalized size = 1.27 \begin {gather*} \frac {{\left ({\left (2 \, a - b\right )} \cos \left (f x + e\right )^{2} + a + b\right )} \sin \left (f x + e\right )}{3 \, f \cos \left (f x + e\right )^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)^4*(a+b*sin(f*x+e)^2),x, algorithm="fricas")

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (a + b \sin ^{2}{\left (e + f x \right )}\right ) \sec ^{4}{\left (e + f x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)**4*(a+b*sin(f*x+e)**2),x)

[Out]

Integral((a + b*sin(e + f*x)**2)*sec(e + f*x)**4, x)

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Giac [A]
time = 0.45, size = 38, normalized size = 1.27 \begin {gather*} \frac {a \tan \left (f x + e\right )^{3} + b \tan \left (f x + e\right )^{3} + 3 \, a \tan \left (f x + e\right )}{3 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)^4*(a+b*sin(f*x+e)^2),x, algorithm="giac")

[Out]

1/3*(a*tan(f*x + e)^3 + b*tan(f*x + e)^3 + 3*a*tan(f*x + e))/f

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Mupad [B]
time = 14.16, size = 31, normalized size = 1.03 \begin {gather*} \frac {{\mathrm {tan}\left (e+f\,x\right )}^3\,\left (\frac {a}{3}+\frac {b}{3}\right )}{f}+\frac {a\,\mathrm {tan}\left (e+f\,x\right )}{f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*sin(e + f*x)^2)/cos(e + f*x)^4,x)

[Out]

(tan(e + f*x)^3*(a/3 + b/3))/f + (a*tan(e + f*x))/f

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