3.95 \(\int (a+a \sin (e+f x))^{3/2} \tan ^4(e+f x) \, dx\)

Optimal. Leaf size=167 \[ \frac{2 a^3 \cos ^3(e+f x)}{3 f (a \sin (e+f x)+a)^{3/2}}-\frac{4 a^2 \cos (e+f x)}{f \sqrt{a \sin (e+f x)+a}}-\frac{a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{2} \sqrt{a \sin (e+f x)+a}}\right )}{2 \sqrt{2} f}+\frac{\sec ^3(e+f x) (a \sin (e+f x)+a)^{3/2}}{3 f}-\frac{7 a \sec (e+f x) \sqrt{a \sin (e+f x)+a}}{2 f} \]

[Out]

-(a^(3/2)*ArcTanh[(Sqrt[a]*Cos[e + f*x])/(Sqrt[2]*Sqrt[a + a*Sin[e + f*x]])])/(2*Sqrt[2]*f) + (2*a^3*Cos[e + f
*x]^3)/(3*f*(a + a*Sin[e + f*x])^(3/2)) - (4*a^2*Cos[e + f*x])/(f*Sqrt[a + a*Sin[e + f*x]]) - (7*a*Sec[e + f*x
]*Sqrt[a + a*Sin[e + f*x]])/(2*f) + (Sec[e + f*x]^3*(a + a*Sin[e + f*x])^(3/2))/(3*f)

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Rubi [A]  time = 0.975016, antiderivative size = 195, normalized size of antiderivative = 1.17, number of steps used = 14, number of rules used = 9, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.391, Rules used = {2714, 2647, 2646, 4401, 2675, 2649, 206, 2878, 2855} \[ -\frac{8 a^2 \cos (e+f x)}{3 f \sqrt{a \sin (e+f x)+a}}-\frac{a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{2} \sqrt{a \sin (e+f x)+a}}\right )}{2 \sqrt{2} f}-\frac{2 a \cos (e+f x) \sqrt{a \sin (e+f x)+a}}{3 f}+\frac{4 \sec ^3(e+f x) (a \sin (e+f x)+a)^{5/2}}{a f}-\frac{23 \sec ^3(e+f x) (a \sin (e+f x)+a)^{3/2}}{3 f}+\frac{a \sec (e+f x) \sqrt{a \sin (e+f x)+a}}{2 f} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sin[e + f*x])^(3/2)*Tan[e + f*x]^4,x]

[Out]

-(a^(3/2)*ArcTanh[(Sqrt[a]*Cos[e + f*x])/(Sqrt[2]*Sqrt[a + a*Sin[e + f*x]])])/(2*Sqrt[2]*f) - (8*a^2*Cos[e + f
*x])/(3*f*Sqrt[a + a*Sin[e + f*x]]) - (2*a*Cos[e + f*x]*Sqrt[a + a*Sin[e + f*x]])/(3*f) + (a*Sec[e + f*x]*Sqrt
[a + a*Sin[e + f*x]])/(2*f) - (23*Sec[e + f*x]^3*(a + a*Sin[e + f*x])^(3/2))/(3*f) + (4*Sec[e + f*x]^3*(a + a*
Sin[e + f*x])^(5/2))/(a*f)

Rule 2714

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*tan[(e_.) + (f_.)*(x_)]^4, x_Symbol] :> Int[(a + b*Sin[e + f*x
])^m, x] - Int[((a + b*Sin[e + f*x])^m*(1 - 2*Sin[e + f*x]^2))/Cos[e + f*x]^4, x] /; FreeQ[{a, b, e, f, m}, x]
&& EqQ[a^2 - b^2, 0] && IntegerQ[m - 1/2]

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 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 4401

Int[u_, x_Symbol] :> With[{v = ExpandTrig[u, x]}, Int[v, x] /; SumQ[v]] /;  !InertTrigFreeQ[u]

Rule 2675

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

Rule 2649

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[-2/d, Subst[Int[1/(2*a - x^2), x], x, (b*C
os[c + d*x])/Sqrt[a + b*Sin[c + d*x]]], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

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 2878

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*sin[(e_.) + (f_.)*(x_)]^2*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)
, x_Symbol] :> -Simp[((g*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^(m + 1))/(b*f*g*(m + p + 2)), x] + Dist[1/
(b*(m + p + 2)), Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^m*(b*(m + 1) - a*(p + 1)*Sin[e + f*x]), x], x] /;
FreeQ[{a, b, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0] && NeQ[m + p + 2, 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))^{3/2} \tan ^4(e+f x) \, dx &=\int (a+a \sin (e+f x))^{3/2} \, dx-\int \sec ^4(e+f x) (a+a \sin (e+f x))^{3/2} \left (1-2 \sin ^2(e+f x)\right ) \, dx\\ &=-\frac{2 a \cos (e+f x) \sqrt{a+a \sin (e+f x)}}{3 f}+\frac{1}{3} (4 a) \int \sqrt{a+a \sin (e+f x)} \, dx-\int \left (\sec ^4(e+f x) (a (1+\sin (e+f x)))^{3/2}-2 \sec ^2(e+f x) (a (1+\sin (e+f x)))^{3/2} \tan ^2(e+f x)\right ) \, dx\\ &=-\frac{8 a^2 \cos (e+f x)}{3 f \sqrt{a+a \sin (e+f x)}}-\frac{2 a \cos (e+f x) \sqrt{a+a \sin (e+f x)}}{3 f}+2 \int \sec ^2(e+f x) (a (1+\sin (e+f x)))^{3/2} \tan ^2(e+f x) \, dx-\int \sec ^4(e+f x) (a (1+\sin (e+f x)))^{3/2} \, dx\\ &=-\frac{8 a^2 \cos (e+f x)}{3 f \sqrt{a+a \sin (e+f x)}}-\frac{2 a \cos (e+f x) \sqrt{a+a \sin (e+f x)}}{3 f}-\frac{\sec ^3(e+f x) (a+a \sin (e+f x))^{3/2}}{3 f}+\frac{4 \sec ^3(e+f x) (a+a \sin (e+f x))^{5/2}}{a f}-\frac{4 \int \sec ^4(e+f x) (a+a \sin (e+f x))^{3/2} \left (\frac{5 a}{2}+3 a \sin (e+f x)\right ) \, dx}{a}-\frac{1}{2} a \int \sec ^2(e+f x) \sqrt{a+a \sin (e+f x)} \, dx\\ &=-\frac{8 a^2 \cos (e+f x)}{3 f \sqrt{a+a \sin (e+f x)}}-\frac{2 a \cos (e+f x) \sqrt{a+a \sin (e+f x)}}{3 f}-\frac{a \sec (e+f x) \sqrt{a+a \sin (e+f x)}}{2 f}-\frac{23 \sec ^3(e+f x) (a+a \sin (e+f x))^{3/2}}{3 f}+\frac{4 \sec ^3(e+f x) (a+a \sin (e+f x))^{5/2}}{a f}+a \int \sec ^2(e+f x) \sqrt{a+a \sin (e+f x)} \, dx-\frac{1}{4} a^2 \int \frac{1}{\sqrt{a+a \sin (e+f x)}} \, dx\\ &=-\frac{8 a^2 \cos (e+f x)}{3 f \sqrt{a+a \sin (e+f x)}}-\frac{2 a \cos (e+f x) \sqrt{a+a \sin (e+f x)}}{3 f}+\frac{a \sec (e+f x) \sqrt{a+a \sin (e+f x)}}{2 f}-\frac{23 \sec ^3(e+f x) (a+a \sin (e+f x))^{3/2}}{3 f}+\frac{4 \sec ^3(e+f x) (a+a \sin (e+f x))^{5/2}}{a f}+\frac{1}{2} a^2 \int \frac{1}{\sqrt{a+a \sin (e+f x)}} \, dx+\frac{a^2 \operatorname{Subst}\left (\int \frac{1}{2 a-x^2} \, dx,x,\frac{a \cos (e+f x)}{\sqrt{a+a \sin (e+f x)}}\right )}{2 f}\\ &=\frac{a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{2} \sqrt{a+a \sin (e+f x)}}\right )}{2 \sqrt{2} f}-\frac{8 a^2 \cos (e+f x)}{3 f \sqrt{a+a \sin (e+f x)}}-\frac{2 a \cos (e+f x) \sqrt{a+a \sin (e+f x)}}{3 f}+\frac{a \sec (e+f x) \sqrt{a+a \sin (e+f x)}}{2 f}-\frac{23 \sec ^3(e+f x) (a+a \sin (e+f x))^{3/2}}{3 f}+\frac{4 \sec ^3(e+f x) (a+a \sin (e+f x))^{5/2}}{a f}-\frac{a^2 \operatorname{Subst}\left (\int \frac{1}{2 a-x^2} \, dx,x,\frac{a \cos (e+f x)}{\sqrt{a+a \sin (e+f x)}}\right )}{f}\\ &=-\frac{a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{2} \sqrt{a+a \sin (e+f x)}}\right )}{2 \sqrt{2} f}-\frac{8 a^2 \cos (e+f x)}{3 f \sqrt{a+a \sin (e+f x)}}-\frac{2 a \cos (e+f x) \sqrt{a+a \sin (e+f x)}}{3 f}+\frac{a \sec (e+f x) \sqrt{a+a \sin (e+f x)}}{2 f}-\frac{23 \sec ^3(e+f x) (a+a \sin (e+f x))^{3/2}}{3 f}+\frac{4 \sec ^3(e+f x) (a+a \sin (e+f x))^{5/2}}{a f}\\ \end{align*}

Mathematica [C]  time = 5.56233, size = 141, normalized size = 0.84 \[ \frac{a \sec ^3(e+f x) \sqrt{a (\sin (e+f x)+1)} \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^2 \left (54 \sin (e+f x)+\sin (3 (e+f x))+6 \cos (2 (e+f x))+(3+3 i) (-1)^{3/4} \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^3 \tanh ^{-1}\left (\left (\frac{1}{2}+\frac{i}{2}\right ) (-1)^{3/4} \left (\tan \left (\frac{1}{4} (e+f x)\right )-1\right )\right )-45\right )}{6 f} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sin[e + f*x])^(3/2)*Tan[e + f*x]^4,x]

[Out]

(a*Sec[e + f*x]^3*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^2*Sqrt[a*(1 + Sin[e + f*x])]*(-45 + 6*Cos[2*(e + f*x)]
+ (3 + 3*I)*(-1)^(3/4)*ArcTanh[(1/2 + I/2)*(-1)^(3/4)*(-1 + Tan[(e + f*x)/4])]*(Cos[(e + f*x)/2] - Sin[(e + f
*x)/2])^3 + 54*Sin[e + f*x] + Sin[3*(e + f*x)]))/(6*f)

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Maple [A]  time = 0.574, size = 139, normalized size = 0.8 \begin{align*}{\frac{1+\sin \left ( fx+e \right ) }{12\,a \left ( -1+\sin \left ( fx+e \right ) \right ) \cos \left ( fx+e \right ) f} \left ( 3\,{a}^{3/2}\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{a-a\sin \left ( fx+e \right ) }\sqrt{2}}{\sqrt{a}}} \right ) \left ( a-a\sin \left ( fx+e \right ) \right ) ^{3/2}-8\,{a}^{3}\sin \left ( fx+e \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}-24\,{a}^{3} \left ( \cos \left ( fx+e \right ) \right ) ^{2}-106\,{a}^{3}\sin \left ( fx+e \right ) +102\,{a}^{3} \right ){\frac{1}{\sqrt{a+a\sin \left ( fx+e \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sin(f*x+e))^(3/2)*tan(f*x+e)^4,x)

[Out]

1/12*(1+sin(f*x+e))/a/(-1+sin(f*x+e))*(3*a^(3/2)*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*(
a-a*sin(f*x+e))^(3/2)-8*a^3*sin(f*x+e)*cos(f*x+e)^2-24*a^3*cos(f*x+e)^2-106*a^3*sin(f*x+e)+102*a^3)/cos(f*x+e)
/(a+a*sin(f*x+e))^(1/2)/f

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Maxima [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^(3/2)*tan(f*x+e)^4,x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 1.90228, size = 648, normalized size = 3.88 \begin{align*} \frac{3 \,{\left (\sqrt{2} a \cos \left (f x + e\right ) \sin \left (f x + e\right ) - \sqrt{2} a \cos \left (f x + e\right )\right )} \sqrt{a} \log \left (-\frac{a \cos \left (f x + e\right )^{2} - 2 \, \sqrt{a \sin \left (f x + e\right ) + a}{\left (\sqrt{2} \cos \left (f x + e\right ) - \sqrt{2} \sin \left (f x + e\right ) + \sqrt{2}\right )} \sqrt{a} + 3 \, a \cos \left (f x + e\right ) -{\left (a \cos \left (f x + e\right ) - 2 \, a\right )} \sin \left (f x + e\right ) + 2 \, a}{\cos \left (f x + e\right )^{2} -{\left (\cos \left (f x + e\right ) + 2\right )} \sin \left (f x + e\right ) - \cos \left (f x + e\right ) - 2}\right ) - 4 \,{\left (12 \, a \cos \left (f x + e\right )^{2} +{\left (4 \, a \cos \left (f x + e\right )^{2} + 53 \, a\right )} \sin \left (f x + e\right ) - 51 \, a\right )} \sqrt{a \sin \left (f x + e\right ) + a}}{24 \,{\left (f \cos \left (f x + e\right ) \sin \left (f x + e\right ) - f \cos \left (f x + e\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^(3/2)*tan(f*x+e)^4,x, algorithm="fricas")

[Out]

1/24*(3*(sqrt(2)*a*cos(f*x + e)*sin(f*x + e) - sqrt(2)*a*cos(f*x + e))*sqrt(a)*log(-(a*cos(f*x + e)^2 - 2*sqrt
(a*sin(f*x + e) + a)*(sqrt(2)*cos(f*x + e) - sqrt(2)*sin(f*x + e) + sqrt(2))*sqrt(a) + 3*a*cos(f*x + e) - (a*c
os(f*x + e) - 2*a)*sin(f*x + e) + 2*a)/(cos(f*x + e)^2 - (cos(f*x + e) + 2)*sin(f*x + e) - cos(f*x + e) - 2))
- 4*(12*a*cos(f*x + e)^2 + (4*a*cos(f*x + e)^2 + 53*a)*sin(f*x + e) - 51*a)*sqrt(a*sin(f*x + e) + a))/(f*cos(f
*x + e)*sin(f*x + e) - f*cos(f*x + e))

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))**(3/2)*tan(f*x+e)**4,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac{3}{2}} \tan \left (f x + e\right )^{4}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^(3/2)*tan(f*x+e)^4,x, algorithm="giac")

[Out]

integrate((a*sin(f*x + e) + a)^(3/2)*tan(f*x + e)^4, x)