3.100 \(\int (a+a \sin (e+f x))^{5/2} \tan ^2(e+f x) \, dx\)

Optimal. Leaf size=118 \[ \frac {124 a^3 \cos (e+f x)}{15 f \sqrt {a \sin (e+f x)+a}}+\frac {31 a^2 \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{15 f}-\frac {2 \sec (e+f x) (a \sin (e+f x)+a)^{7/2}}{5 a f}+\frac {9 \sec (e+f x) (a \sin (e+f x)+a)^{5/2}}{5 f} \]

[Out]

9/5*sec(f*x+e)*(a+a*sin(f*x+e))^(5/2)/f-2/5*sec(f*x+e)*(a+a*sin(f*x+e))^(7/2)/a/f+124/15*a^3*cos(f*x+e)/f/(a+a
*sin(f*x+e))^(1/2)+31/15*a^2*cos(f*x+e)*(a+a*sin(f*x+e))^(1/2)/f

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Rubi [A]  time = 0.21, antiderivative size = 118, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {2713, 2855, 2647, 2646} \[ \frac {31 a^2 \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{15 f}+\frac {124 a^3 \cos (e+f x)}{15 f \sqrt {a \sin (e+f x)+a}}-\frac {2 \sec (e+f x) (a \sin (e+f x)+a)^{7/2}}{5 a f}+\frac {9 \sec (e+f x) (a \sin (e+f x)+a)^{5/2}}{5 f} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sin[e + f*x])^(5/2)*Tan[e + f*x]^2,x]

[Out]

(124*a^3*Cos[e + f*x])/(15*f*Sqrt[a + a*Sin[e + f*x]]) + (31*a^2*Cos[e + f*x]*Sqrt[a + a*Sin[e + f*x]])/(15*f)
+ (9*Sec[e + f*x]*(a + a*Sin[e + f*x])^(5/2))/(5*f) - (2*Sec[e + f*x]*(a + a*Sin[e + f*x])^(7/2))/(5*a*f)

Rule 2646

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(-2*b*Cos[c + d*x])/(d*Sqrt[a + b*Sin[c + d*
x]]), x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 2647

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(a + b*Sin[c + d*x])^(n -
1))/(d*n), x] + Dist[(a*(2*n - 1))/n, Int[(a + b*Sin[c + d*x])^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && Eq
Q[a^2 - b^2, 0] && IGtQ[n - 1/2, 0]

Rule 2713

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*tan[(e_.) + (f_.)*(x_)]^2, x_Symbol] :> -Simp[(a + b*Sin[e + f
*x])^(m + 1)/(b*f*m*Cos[e + f*x]), x] + Dist[1/(b*m), Int[((a + b*Sin[e + f*x])^m*(b*(m + 1) + a*Sin[e + f*x])
)/Cos[e + f*x]^2, x], x] /; FreeQ[{a, b, e, f, m}, x] && EqQ[a^2 - b^2, 0] &&  !IntegerQ[m] &&  !LtQ[m, 0]

Rule 2855

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> -Simp[((b*c + a*d)*(g*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^m)/(a*f*g*(p +
1)), x] + Dist[(b*(a*d*m + b*c*(m + p + 1)))/(a*g^2*(p + 1)), Int[(g*Cos[e + f*x])^(p + 2)*(a + b*Sin[e + f*x]
)^(m - 1), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && EqQ[a^2 - b^2, 0] && GtQ[m, -1] && LtQ[p, -1]

Rubi steps

\begin {align*} \int (a+a \sin (e+f x))^{5/2} \tan ^2(e+f x) \, dx &=-\frac {2 \sec (e+f x) (a+a \sin (e+f x))^{7/2}}{5 a f}+\frac {2 \int \sec ^2(e+f x) (a+a \sin (e+f x))^{5/2} \left (\frac {7 a}{2}+a \sin (e+f x)\right ) \, dx}{5 a}\\ &=\frac {9 \sec (e+f x) (a+a \sin (e+f x))^{5/2}}{5 f}-\frac {2 \sec (e+f x) (a+a \sin (e+f x))^{7/2}}{5 a f}-\frac {1}{10} (31 a) \int (a+a \sin (e+f x))^{3/2} \, dx\\ &=\frac {31 a^2 \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{15 f}+\frac {9 \sec (e+f x) (a+a \sin (e+f x))^{5/2}}{5 f}-\frac {2 \sec (e+f x) (a+a \sin (e+f x))^{7/2}}{5 a f}-\frac {1}{15} \left (62 a^2\right ) \int \sqrt {a+a \sin (e+f x)} \, dx\\ &=\frac {124 a^3 \cos (e+f x)}{15 f \sqrt {a+a \sin (e+f x)}}+\frac {31 a^2 \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{15 f}+\frac {9 \sec (e+f x) (a+a \sin (e+f x))^{5/2}}{5 f}-\frac {2 \sec (e+f x) (a+a \sin (e+f x))^{7/2}}{5 a f}\\ \end {align*}

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Mathematica [A]  time = 5.47, size = 60, normalized size = 0.51 \[ \frac {a^2 \sec (e+f x) \sqrt {a (\sin (e+f x)+1)} (-185 \sin (e+f x)+3 \sin (3 (e+f x))+22 \cos (2 (e+f x))+330)}{30 f} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sin[e + f*x])^(5/2)*Tan[e + f*x]^2,x]

[Out]

(a^2*Sec[e + f*x]*Sqrt[a*(1 + Sin[e + f*x])]*(330 + 22*Cos[2*(e + f*x)] - 185*Sin[e + f*x] + 3*Sin[3*(e + f*x)
]))/(30*f)

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fricas [A]  time = 0.43, size = 70, normalized size = 0.59 \[ \frac {2 \, {\left (11 \, a^{2} \cos \left (f x + e\right )^{2} + 77 \, a^{2} + {\left (3 \, a^{2} \cos \left (f x + e\right )^{2} - 47 \, a^{2}\right )} \sin \left (f x + e\right )\right )} \sqrt {a \sin \left (f x + e\right ) + a}}{15 \, f \cos \left (f x + e\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^(5/2)*tan(f*x+e)^2,x, algorithm="fricas")

[Out]

2/15*(11*a^2*cos(f*x + e)^2 + 77*a^2 + (3*a^2*cos(f*x + e)^2 - 47*a^2)*sin(f*x + e))*sqrt(a*sin(f*x + e) + a)/
(f*cos(f*x + e))

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giac [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((a+a*sin(f*x+e))^(5/2)*tan(f*x+e)^2,x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 0.49, size = 67, normalized size = 0.57 \[ -\frac {2 a^{3} \left (1+\sin \left (f x +e \right )\right ) \left (3 \left (\sin ^{3}\left (f x +e \right )\right )+11 \left (\sin ^{2}\left (f x +e \right )\right )+44 \sin \left (f x +e \right )-88\right )}{15 \cos \left (f x +e \right ) \sqrt {a +a \sin \left (f x +e \right )}\, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sin(f*x+e))^(5/2)*tan(f*x+e)^2,x)

[Out]

-2/15*a^3*(1+sin(f*x+e))*(3*sin(f*x+e)^3+11*sin(f*x+e)^2+44*sin(f*x+e)-88)/cos(f*x+e)/(a+a*sin(f*x+e))^(1/2)/f

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maxima [A]  time = 0.48, size = 191, normalized size = 1.62 \[ -\frac {8 \, {\left (22 \, a^{\frac {5}{2}} - \frac {22 \, a^{\frac {5}{2}} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {55 \, a^{\frac {5}{2}} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {50 \, a^{\frac {5}{2}} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {55 \, a^{\frac {5}{2}} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - \frac {22 \, a^{\frac {5}{2}} \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}} + \frac {22 \, a^{\frac {5}{2}} \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}}\right )}}{15 \, f {\left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )} {\left (\frac {\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1\right )}^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^(5/2)*tan(f*x+e)^2,x, algorithm="maxima")

[Out]

-8/15*(22*a^(5/2) - 22*a^(5/2)*sin(f*x + e)/(cos(f*x + e) + 1) + 55*a^(5/2)*sin(f*x + e)^2/(cos(f*x + e) + 1)^
2 - 50*a^(5/2)*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 55*a^(5/2)*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - 22*a^(5/
2)*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 22*a^(5/2)*sin(f*x + e)^6/(cos(f*x + e) + 1)^6)/(f*(sin(f*x + e)/(cos
(f*x + e) + 1) - 1)*(sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 1)^(5/2))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\mathrm {tan}\left (e+f\,x\right )}^2\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^{5/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(e + f*x)^2*(a + a*sin(e + f*x))^(5/2),x)

[Out]

int(tan(e + f*x)^2*(a + a*sin(e + f*x))^(5/2), 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((a+a*sin(f*x+e))**(5/2)*tan(f*x+e)**2,x)

[Out]

Timed out

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