3.3.100 \(\int \sec ^{10}(e+f x) (a+b \sin ^2(e+f x))^2 \, dx\) [300]

Optimal. Leaf size=106 \[ \frac {a^2 \tan (e+f x)}{f}+\frac {2 a (2 a+b) \tan ^3(e+f x)}{3 f}+\frac {\left (6 a^2+6 a b+b^2\right ) \tan ^5(e+f x)}{5 f}+\frac {2 (a+b) (2 a+b) \tan ^7(e+f x)}{7 f}+\frac {(a+b)^2 \tan ^9(e+f x)}{9 f} \]

[Out]

a^2*tan(f*x+e)/f+2/3*a*(2*a+b)*tan(f*x+e)^3/f+1/5*(6*a^2+6*a*b+b^2)*tan(f*x+e)^5/f+2/7*(a+b)*(2*a+b)*tan(f*x+e
)^7/f+1/9*(a+b)^2*tan(f*x+e)^9/f

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Rubi [A]
time = 0.06, antiderivative size = 106, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {3270, 380} \begin {gather*} \frac {\left (6 a^2+6 a b+b^2\right ) \tan ^5(e+f x)}{5 f}+\frac {a^2 \tan (e+f x)}{f}+\frac {(a+b)^2 \tan ^9(e+f x)}{9 f}+\frac {2 (a+b) (2 a+b) \tan ^7(e+f x)}{7 f}+\frac {2 a (2 a+b) \tan ^3(e+f x)}{3 f} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sec[e + f*x]^10*(a + b*Sin[e + f*x]^2)^2,x]

[Out]

(a^2*Tan[e + f*x])/f + (2*a*(2*a + b)*Tan[e + f*x]^3)/(3*f) + ((6*a^2 + 6*a*b + b^2)*Tan[e + f*x]^5)/(5*f) + (
2*(a + b)*(2*a + b)*Tan[e + f*x]^7)/(7*f) + ((a + b)^2*Tan[e + f*x]^9)/(9*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 ^{10}(e+f x) \left (a+b \sin ^2(e+f x)\right )^2 \, dx &=\frac {\text {Subst}\left (\int \left (1+x^2\right )^2 \left (a+(a+b) x^2\right )^2 \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {\text {Subst}\left (\int \left (a^2+2 a (2 a+b) x^2+\left (6 a^2+6 a b+b^2\right ) x^4+2 (a+b) (2 a+b) x^6+(a+b)^2 x^8\right ) \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {a^2 \tan (e+f x)}{f}+\frac {2 a (2 a+b) \tan ^3(e+f x)}{3 f}+\frac {\left (6 a^2+6 a b+b^2\right ) \tan ^5(e+f x)}{5 f}+\frac {2 (a+b) (2 a+b) \tan ^7(e+f x)}{7 f}+\frac {(a+b)^2 \tan ^9(e+f x)}{9 f}\\ \end {align*}

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Mathematica [A]
time = 0.34, size = 107, normalized size = 1.01 \begin {gather*} \frac {\sec ^9(e+f x) \left (252 \left (8 a^2+8 a b+3 b^2\right ) \sin (e+f x)+336 \left (4 a^2-a b-b^2\right ) \sin (3 (e+f x))+\left (16 a^2-4 a b+b^2\right ) (36 \sin (5 (e+f x))+9 \sin (7 (e+f x))+\sin (9 (e+f x)))\right )}{10080 f} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sec[e + f*x]^10*(a + b*Sin[e + f*x]^2)^2,x]

[Out]

(Sec[e + f*x]^9*(252*(8*a^2 + 8*a*b + 3*b^2)*Sin[e + f*x] + 336*(4*a^2 - a*b - b^2)*Sin[3*(e + f*x)] + (16*a^2
- 4*a*b + b^2)*(36*Sin[5*(e + f*x)] + 9*Sin[7*(e + f*x)] + Sin[9*(e + f*x)])))/(10080*f)

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Maple [A]
time = 0.38, size = 195, normalized size = 1.84

method	result	size
derivativedivides	\(\frac {-a^{2} \left (-\frac {128}{315}-\frac {\left (\sec ^{8}\left (f x +e \right )\right )}{9}-\frac {8 \left (\sec ^{6}\left (f x +e \right )\right )}{63}-\frac {16 \left (\sec ^{4}\left (f x +e \right )\right )}{105}-\frac {64 \left (\sec ^{2}\left (f x +e \right )\right )}{315}\right ) \tan \left (f x +e \right )+2 a b \left (\frac {\sin ^{3}\left (f x +e \right )}{9 \cos \left (f x +e \right )^{9}}+\frac {2 \left (\sin ^{3}\left (f x +e \right )\right )}{21 \cos \left (f x +e \right )^{7}}+\frac {8 \left (\sin ^{3}\left (f x +e \right )\right )}{105 \cos \left (f x +e \right )^{5}}+\frac {16 \left (\sin ^{3}\left (f x +e \right )\right )}{315 \cos \left (f x +e \right )^{3}}\right )+b^{2} \left (\frac {\sin ^{5}\left (f x +e \right )}{9 \cos \left (f x +e \right )^{9}}+\frac {4 \left (\sin ^{5}\left (f x +e \right )\right )}{63 \cos \left (f x +e \right )^{7}}+\frac {8 \left (\sin ^{5}\left (f x +e \right )\right )}{315 \cos \left (f x +e \right )^{5}}\right )}{f}\)	\(195\)
default	\(\frac {-a^{2} \left (-\frac {128}{315}-\frac {\left (\sec ^{8}\left (f x +e \right )\right )}{9}-\frac {8 \left (\sec ^{6}\left (f x +e \right )\right )}{63}-\frac {16 \left (\sec ^{4}\left (f x +e \right )\right )}{105}-\frac {64 \left (\sec ^{2}\left (f x +e \right )\right )}{315}\right ) \tan \left (f x +e \right )+2 a b \left (\frac {\sin ^{3}\left (f x +e \right )}{9 \cos \left (f x +e \right )^{9}}+\frac {2 \left (\sin ^{3}\left (f x +e \right )\right )}{21 \cos \left (f x +e \right )^{7}}+\frac {8 \left (\sin ^{3}\left (f x +e \right )\right )}{105 \cos \left (f x +e \right )^{5}}+\frac {16 \left (\sin ^{3}\left (f x +e \right )\right )}{315 \cos \left (f x +e \right )^{3}}\right )+b^{2} \left (\frac {\sin ^{5}\left (f x +e \right )}{9 \cos \left (f x +e \right )^{9}}+\frac {4 \left (\sin ^{5}\left (f x +e \right )\right )}{63 \cos \left (f x +e \right )^{7}}+\frac {8 \left (\sin ^{5}\left (f x +e \right )\right )}{315 \cos \left (f x +e \right )^{5}}\right )}{f}\)	\(195\)
risch	\(\frac {16 i \left (210 b^{2} {\mathrm e}^{12 i \left (f x +e \right )}-1260 a b \,{\mathrm e}^{10 i \left (f x +e \right )}-315 b^{2} {\mathrm e}^{10 i \left (f x +e \right )}+2016 a^{2} {\mathrm e}^{8 i \left (f x +e \right )}+756 a b \,{\mathrm e}^{8 i \left (f x +e \right )}+441 b^{2} {\mathrm e}^{8 i \left (f x +e \right )}+1344 a^{2} {\mathrm e}^{6 i \left (f x +e \right )}-336 a b \,{\mathrm e}^{6 i \left (f x +e \right )}-126 b^{2} {\mathrm e}^{6 i \left (f x +e \right )}+576 a^{2} {\mathrm e}^{4 i \left (f x +e \right )}-144 a b \,{\mathrm e}^{4 i \left (f x +e \right )}+36 b^{2} {\mathrm e}^{4 i \left (f x +e \right )}+144 a^{2} {\mathrm e}^{2 i \left (f x +e \right )}-36 a b \,{\mathrm e}^{2 i \left (f x +e \right )}+9 b^{2} {\mathrm e}^{2 i \left (f x +e \right )}+16 a^{2}-4 a b +b^{2}\right )}{315 f \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{9}}\)	\(238\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(f*x+e)^10*(a+b*sin(f*x+e)^2)^2,x,method=_RETURNVERBOSE)

[Out]

1/f*(-a^2*(-128/315-1/9*sec(f*x+e)^8-8/63*sec(f*x+e)^6-16/105*sec(f*x+e)^4-64/315*sec(f*x+e)^2)*tan(f*x+e)+2*a
*b*(1/9*sin(f*x+e)^3/cos(f*x+e)^9+2/21*sin(f*x+e)^3/cos(f*x+e)^7+8/105*sin(f*x+e)^3/cos(f*x+e)^5+16/315*sin(f*
x+e)^3/cos(f*x+e)^3)+b^2*(1/9*sin(f*x+e)^5/cos(f*x+e)^9+4/63*sin(f*x+e)^5/cos(f*x+e)^7+8/315*sin(f*x+e)^5/cos(
f*x+e)^5))

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Maxima [A]
time = 0.28, size = 108, normalized size = 1.02 \begin {gather*} \frac {35 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \tan \left (f x + e\right )^{9} + 90 \, {\left (2 \, a^{2} + 3 \, a b + b^{2}\right )} \tan \left (f x + e\right )^{7} + 63 \, {\left (6 \, a^{2} + 6 \, a b + b^{2}\right )} \tan \left (f x + e\right )^{5} + 210 \, {\left (2 \, a^{2} + a b\right )} \tan \left (f x + e\right )^{3} + 315 \, a^{2} \tan \left (f x + e\right )}{315 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)^10*(a+b*sin(f*x+e)^2)^2,x, algorithm="maxima")

[Out]

1/315*(35*(a^2 + 2*a*b + b^2)*tan(f*x + e)^9 + 90*(2*a^2 + 3*a*b + b^2)*tan(f*x + e)^7 + 63*(6*a^2 + 6*a*b + b
^2)*tan(f*x + e)^5 + 210*(2*a^2 + a*b)*tan(f*x + e)^3 + 315*a^2*tan(f*x + e))/f

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Fricas [A]
time = 0.41, size = 128, normalized size = 1.21 \begin {gather*} \frac {{\left (8 \, {\left (16 \, a^{2} - 4 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{8} + 4 \, {\left (16 \, a^{2} - 4 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{6} + 3 \, {\left (16 \, a^{2} - 4 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{4} + 10 \, {\left (4 \, a^{2} - a b - 5 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 35 \, a^{2} + 70 \, a b + 35 \, b^{2}\right )} \sin \left (f x + e\right )}{315 \, f \cos \left (f x + e\right )^{9}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)^10*(a+b*sin(f*x+e)^2)^2,x, algorithm="fricas")

[Out]

1/315*(8*(16*a^2 - 4*a*b + b^2)*cos(f*x + e)^8 + 4*(16*a^2 - 4*a*b + b^2)*cos(f*x + e)^6 + 3*(16*a^2 - 4*a*b +
b^2)*cos(f*x + e)^4 + 10*(4*a^2 - a*b - 5*b^2)*cos(f*x + e)^2 + 35*a^2 + 70*a*b + 35*b^2)*sin(f*x + e)/(f*cos
(f*x + e)^9)

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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(sec(f*x+e)**10*(a+b*sin(f*x+e)**2)**2,x)

[Out]

Timed out

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Giac [A]
time = 0.50, size = 168, normalized size = 1.58 \begin {gather*} \frac {35 \, a^{2} \tan \left (f x + e\right )^{9} + 70 \, a b \tan \left (f x + e\right )^{9} + 35 \, b^{2} \tan \left (f x + e\right )^{9} + 180 \, a^{2} \tan \left (f x + e\right )^{7} + 270 \, a b \tan \left (f x + e\right )^{7} + 90 \, b^{2} \tan \left (f x + e\right )^{7} + 378 \, a^{2} \tan \left (f x + e\right )^{5} + 378 \, a b \tan \left (f x + e\right )^{5} + 63 \, b^{2} \tan \left (f x + e\right )^{5} + 420 \, a^{2} \tan \left (f x + e\right )^{3} + 210 \, a b \tan \left (f x + e\right )^{3} + 315 \, a^{2} \tan \left (f x + e\right )}{315 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)^10*(a+b*sin(f*x+e)^2)^2,x, algorithm="giac")

[Out]

1/315*(35*a^2*tan(f*x + e)^9 + 70*a*b*tan(f*x + e)^9 + 35*b^2*tan(f*x + e)^9 + 180*a^2*tan(f*x + e)^7 + 270*a*
b*tan(f*x + e)^7 + 90*b^2*tan(f*x + e)^7 + 378*a^2*tan(f*x + e)^5 + 378*a*b*tan(f*x + e)^5 + 63*b^2*tan(f*x +
e)^5 + 420*a^2*tan(f*x + e)^3 + 210*a*b*tan(f*x + e)^3 + 315*a^2*tan(f*x + e))/f

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Mupad [B]
time = 14.22, size = 94, normalized size = 0.89 \begin {gather*} \frac {a^2\,\mathrm {tan}\left (e+f\,x\right )+\frac {{\mathrm {tan}\left (e+f\,x\right )}^9\,{\left (a+b\right )}^2}{9}+{\mathrm {tan}\left (e+f\,x\right )}^5\,\left (\frac {6\,a^2}{5}+\frac {6\,a\,b}{5}+\frac {b^2}{5}\right )+{\mathrm {tan}\left (e+f\,x\right )}^7\,\left (\frac {4\,a^2}{7}+\frac {6\,a\,b}{7}+\frac {2\,b^2}{7}\right )+\frac {2\,a\,{\mathrm {tan}\left (e+f\,x\right )}^3\,\left (2\,a+b\right )}{3}}{f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*sin(e + f*x)^2)^2/cos(e + f*x)^10,x)

[Out]

(a^2*tan(e + f*x) + (tan(e + f*x)^9*(a + b)^2)/9 + tan(e + f*x)^5*((6*a*b)/5 + (6*a^2)/5 + b^2/5) + tan(e + f*
x)^7*((6*a*b)/7 + (4*a^2)/7 + (2*b^2)/7) + (2*a*tan(e + f*x)^3*(2*a + b))/3)/f

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