3.3.92 \(\int \sec ^8(e+f x) (a+b \sin ^2(e+f x)) \, dx\) [292]

Optimal. Leaf size=72 \[ \frac {a \tan (e+f x)}{f}+\frac {(3 a+b) \tan ^3(e+f x)}{3 f}+\frac {(3 a+2 b) \tan ^5(e+f x)}{5 f}+\frac {(a+b) \tan ^7(e+f x)}{7 f} \]

[Out]

a*tan(f*x+e)/f+1/3*(3*a+b)*tan(f*x+e)^3/f+1/5*(3*a+2*b)*tan(f*x+e)^5/f+1/7*(a+b)*tan(f*x+e)^7/f

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Rubi [A]
time = 0.04, antiderivative size = 72, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {3270, 380} \begin {gather*} \frac {(a+b) \tan ^7(e+f x)}{7 f}+\frac {(3 a+2 b) \tan ^5(e+f x)}{5 f}+\frac {(3 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]^8*(a + b*Sin[e + f*x]^2),x]

[Out]

(a*Tan[e + f*x])/f + ((3*a + b)*Tan[e + f*x]^3)/(3*f) + ((3*a + 2*b)*Tan[e + f*x]^5)/(5*f) + ((a + b)*Tan[e +
f*x]^7)/(7*f)

Rule 380

Int[((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x^n
)^p*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[p, 0] && IGtQ[q, 0]

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 ^8(e+f x) \left (a+b \sin ^2(e+f x)\right ) \, dx &=\frac {\text {Subst}\left (\int \left (1+x^2\right )^2 \left (a+(a+b) x^2\right ) \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {\text {Subst}\left (\int \left (a+(3 a+b) x^2+(3 a+2 b) x^4+(a+b) x^6\right ) \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {a \tan (e+f x)}{f}+\frac {(3 a+b) \tan ^3(e+f x)}{3 f}+\frac {(3 a+2 b) \tan ^5(e+f x)}{5 f}+\frac {(a+b) \tan ^7(e+f x)}{7 f}\\ \end {align*}

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Mathematica [A]
time = 0.23, size = 86, normalized size = 1.19 \begin {gather*} \frac {\tan (e+f x) \left (105 a-8 b-4 b \sec ^2(e+f x)-3 b \sec ^4(e+f x)+15 b \sec ^6(e+f x)+105 a \tan ^2(e+f x)+63 a \tan ^4(e+f x)+15 a \tan ^6(e+f x)\right )}{105 f} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sec[e + f*x]^8*(a + b*Sin[e + f*x]^2),x]

[Out]

(Tan[e + f*x]*(105*a - 8*b - 4*b*Sec[e + f*x]^2 - 3*b*Sec[e + f*x]^4 + 15*b*Sec[e + f*x]^6 + 105*a*Tan[e + f*x
]^2 + 63*a*Tan[e + f*x]^4 + 15*a*Tan[e + f*x]^6))/(105*f)

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Maple [A]
time = 0.29, size = 104, normalized size = 1.44

method	result	size
derivativedivides	\(\frac {-a \left (-\frac {16}{35}-\frac {\left (\sec ^{6}\left (f x +e \right )\right )}{7}-\frac {6 \left (\sec ^{4}\left (f x +e \right )\right )}{35}-\frac {8 \left (\sec ^{2}\left (f x +e \right )\right )}{35}\right ) \tan \left (f x +e \right )+b \left (\frac {\sin ^{3}\left (f x +e \right )}{7 \cos \left (f x +e \right )^{7}}+\frac {4 \left (\sin ^{3}\left (f x +e \right )\right )}{35 \cos \left (f x +e \right )^{5}}+\frac {8 \left (\sin ^{3}\left (f x +e \right )\right )}{105 \cos \left (f x +e \right )^{3}}\right )}{f}\)	\(104\)
default	\(\frac {-a \left (-\frac {16}{35}-\frac {\left (\sec ^{6}\left (f x +e \right )\right )}{7}-\frac {6 \left (\sec ^{4}\left (f x +e \right )\right )}{35}-\frac {8 \left (\sec ^{2}\left (f x +e \right )\right )}{35}\right ) \tan \left (f x +e \right )+b \left (\frac {\sin ^{3}\left (f x +e \right )}{7 \cos \left (f x +e \right )^{7}}+\frac {4 \left (\sin ^{3}\left (f x +e \right )\right )}{35 \cos \left (f x +e \right )^{5}}+\frac {8 \left (\sin ^{3}\left (f x +e \right )\right )}{105 \cos \left (f x +e \right )^{3}}\right )}{f}\)	\(104\)
risch	\(-\frac {16 i \left (70 b \,{\mathrm e}^{8 i \left (f x +e \right )}-210 a \,{\mathrm e}^{6 i \left (f x +e \right )}-35 b \,{\mathrm e}^{6 i \left (f x +e \right )}-126 a \,{\mathrm e}^{4 i \left (f x +e \right )}+21 \,{\mathrm e}^{4 i \left (f x +e \right )} b -42 a \,{\mathrm e}^{2 i \left (f x +e \right )}+7 \,{\mathrm e}^{2 i \left (f x +e \right )} b -6 a +b \right )}{105 f \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{7}}\)	\(109\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(f*x+e)^8*(a+b*sin(f*x+e)^2),x,method=_RETURNVERBOSE)

[Out]

1/f*(-a*(-16/35-1/7*sec(f*x+e)^6-6/35*sec(f*x+e)^4-8/35*sec(f*x+e)^2)*tan(f*x+e)+b*(1/7*sin(f*x+e)^3/cos(f*x+e
)^7+4/35*sin(f*x+e)^3/cos(f*x+e)^5+8/105*sin(f*x+e)^3/cos(f*x+e)^3))

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Maxima [A]
time = 0.29, size = 64, normalized size = 0.89 \begin {gather*} \frac {15 \, {\left (a + b\right )} \tan \left (f x + e\right )^{7} + 21 \, {\left (3 \, a + 2 \, b\right )} \tan \left (f x + e\right )^{5} + 35 \, {\left (3 \, a + b\right )} \tan \left (f x + e\right )^{3} + 105 \, a \tan \left (f x + e\right )}{105 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)^8*(a+b*sin(f*x+e)^2),x, algorithm="maxima")

[Out]

1/105*(15*(a + b)*tan(f*x + e)^7 + 21*(3*a + 2*b)*tan(f*x + e)^5 + 35*(3*a + b)*tan(f*x + e)^3 + 105*a*tan(f*x
+ e))/f

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Fricas [A]
time = 0.37, size = 77, normalized size = 1.07 \begin {gather*} \frac {{\left (8 \, {\left (6 \, a - b\right )} \cos \left (f x + e\right )^{6} + 4 \, {\left (6 \, a - b\right )} \cos \left (f x + e\right )^{4} + 3 \, {\left (6 \, a - b\right )} \cos \left (f x + e\right )^{2} + 15 \, a + 15 \, b\right )} \sin \left (f x + e\right )}{105 \, f \cos \left (f x + e\right )^{7}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)^8*(a+b*sin(f*x+e)^2),x, algorithm="fricas")

[Out]

1/105*(8*(6*a - b)*cos(f*x + e)^6 + 4*(6*a - b)*cos(f*x + e)^4 + 3*(6*a - b)*cos(f*x + e)^2 + 15*a + 15*b)*sin
(f*x + e)/(f*cos(f*x + e)^7)

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)**8*(a+b*sin(f*x+e)**2),x)

[Out]

Exception raised: SystemError >> excessive stack use: stack is 4370 deep

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Giac [A]
time = 0.51, size = 88, normalized size = 1.22 \begin {gather*} \frac {15 \, a \tan \left (f x + e\right )^{7} + 15 \, b \tan \left (f x + e\right )^{7} + 63 \, a \tan \left (f x + e\right )^{5} + 42 \, b \tan \left (f x + e\right )^{5} + 105 \, a \tan \left (f x + e\right )^{3} + 35 \, b \tan \left (f x + e\right )^{3} + 105 \, a \tan \left (f x + e\right )}{105 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)^8*(a+b*sin(f*x+e)^2),x, algorithm="giac")

[Out]

1/105*(15*a*tan(f*x + e)^7 + 15*b*tan(f*x + e)^7 + 63*a*tan(f*x + e)^5 + 42*b*tan(f*x + e)^5 + 105*a*tan(f*x +
e)^3 + 35*b*tan(f*x + e)^3 + 105*a*tan(f*x + e))/f

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Mupad [B]
time = 13.84, size = 59, normalized size = 0.82 \begin {gather*} \frac {\left (\frac {a}{7}+\frac {b}{7}\right )\,{\mathrm {tan}\left (e+f\,x\right )}^7+\left (\frac {3\,a}{5}+\frac {2\,b}{5}\right )\,{\mathrm {tan}\left (e+f\,x\right )}^5+\left (a+\frac {b}{3}\right )\,{\mathrm {tan}\left (e+f\,x\right )}^3+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)^8,x)

[Out]

(tan(e + f*x)^5*((3*a)/5 + (2*b)/5) + tan(e + f*x)^7*(a/7 + b/7) + a*tan(e + f*x) + tan(e + f*x)^3*(a + b/3))/
f

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