3.123 \(\int (a+a \sin (e+f x))^3 (g \tan (e+f x))^p \, dx\)

Optimal. Leaf size=269 \[ \frac {3 a^3 (g \tan (e+f x))^{p+3} \, _2F_1\left (2,\frac {p+3}{2};\frac {p+5}{2};-\tan ^2(e+f x)\right )}{f g^3 (p+3)}+\frac {a^3 (g \tan (e+f x))^{p+1} \, _2F_1\left (1,\frac {p+1}{2};\frac {p+3}{2};-\tan ^2(e+f x)\right )}{f g (p+1)}+\frac {3 a^3 \sin (e+f x) \cos ^2(e+f x)^{\frac {p+1}{2}} (g \tan (e+f x))^{p+1} \, _2F_1\left (\frac {p+1}{2},\frac {p+2}{2};\frac {p+4}{2};\sin ^2(e+f x)\right )}{f g (p+2)}+\frac {a^3 \sin ^3(e+f x) \cos ^2(e+f x)^{\frac {p+1}{2}} (g \tan (e+f x))^{p+1} \, _2F_1\left (\frac {p+1}{2},\frac {p+4}{2};\frac {p+6}{2};\sin ^2(e+f x)\right )}{f g (p+4)} \]

[Out]

a^3*hypergeom([1, 1/2+1/2*p],[3/2+1/2*p],-tan(f*x+e)^2)*(g*tan(f*x+e))^(1+p)/f/g/(1+p)+3*a^3*(cos(f*x+e)^2)^(1
/2+1/2*p)*hypergeom([1+1/2*p, 1/2+1/2*p],[2+1/2*p],sin(f*x+e)^2)*sin(f*x+e)*(g*tan(f*x+e))^(1+p)/f/g/(2+p)+a^3
*(cos(f*x+e)^2)^(1/2+1/2*p)*hypergeom([2+1/2*p, 1/2+1/2*p],[3+1/2*p],sin(f*x+e)^2)*sin(f*x+e)^3*(g*tan(f*x+e))
^(1+p)/f/g/(4+p)+3*a^3*hypergeom([2, 3/2+1/2*p],[5/2+1/2*p],-tan(f*x+e)^2)*(g*tan(f*x+e))^(3+p)/f/g^3/(3+p)

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Rubi [A]  time = 0.35, antiderivative size = 269, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {2710, 3476, 364, 2602, 2577, 2591} \[ \frac {3 a^3 (g \tan (e+f x))^{p+3} \, _2F_1\left (2,\frac {p+3}{2};\frac {p+5}{2};-\tan ^2(e+f x)\right )}{f g^3 (p+3)}+\frac {a^3 (g \tan (e+f x))^{p+1} \, _2F_1\left (1,\frac {p+1}{2};\frac {p+3}{2};-\tan ^2(e+f x)\right )}{f g (p+1)}+\frac {a^3 \sin ^3(e+f x) \cos ^2(e+f x)^{\frac {p+1}{2}} (g \tan (e+f x))^{p+1} \, _2F_1\left (\frac {p+1}{2},\frac {p+4}{2};\frac {p+6}{2};\sin ^2(e+f x)\right )}{f g (p+4)}+\frac {3 a^3 \sin (e+f x) \cos ^2(e+f x)^{\frac {p+1}{2}} (g \tan (e+f x))^{p+1} \, _2F_1\left (\frac {p+1}{2},\frac {p+2}{2};\frac {p+4}{2};\sin ^2(e+f x)\right )}{f g (p+2)} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sin[e + f*x])^3*(g*Tan[e + f*x])^p,x]

[Out]

(a^3*Hypergeometric2F1[1, (1 + p)/2, (3 + p)/2, -Tan[e + f*x]^2]*(g*Tan[e + f*x])^(1 + p))/(f*g*(1 + p)) + (3*
a^3*(Cos[e + f*x]^2)^((1 + p)/2)*Hypergeometric2F1[(1 + p)/2, (2 + p)/2, (4 + p)/2, Sin[e + f*x]^2]*Sin[e + f*
x]*(g*Tan[e + f*x])^(1 + p))/(f*g*(2 + p)) + (a^3*(Cos[e + f*x]^2)^((1 + p)/2)*Hypergeometric2F1[(1 + p)/2, (4
+ p)/2, (6 + p)/2, Sin[e + f*x]^2]*Sin[e + f*x]^3*(g*Tan[e + f*x])^(1 + p))/(f*g*(4 + p)) + (3*a^3*Hypergeome
tric2F1[2, (3 + p)/2, (5 + p)/2, -Tan[e + f*x]^2]*(g*Tan[e + f*x])^(3 + p))/(f*g^3*(3 + p))

Rule 364

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a^p*(c*x)^(m + 1)*Hypergeometric2F1[-
p, (m + 1)/n, (m + 1)/n + 1, -((b*x^n)/a)])/(c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 2577

Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(b^(2*IntPart
[(n - 1)/2] + 1)*(b*Cos[e + f*x])^(2*FracPart[(n - 1)/2])*(a*Sin[e + f*x])^(m + 1)*Hypergeometric2F1[(1 + m)/2
, (1 - n)/2, (3 + m)/2, Sin[e + f*x]^2])/(a*f*(m + 1)*(Cos[e + f*x]^2)^FracPart[(n - 1)/2]), x] /; FreeQ[{a, b
, e, f, m, n}, x]

Rule 2591

Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> With[{ff = FreeFactors[Ta
n[e + f*x], x]}, Dist[(b*ff)/f, Subst[Int[(ff*x)^(m + n)/(b^2 + ff^2*x^2)^(m/2 + 1), x], x, (b*Tan[e + f*x])/f
f], x]] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2]

Rule 2602

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

Rule 2710

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((g_.)*tan[(e_.) + (f_.)*(x_)])^(p_.), x_Symbol] :> Int[Expan
dIntegrand[(g*Tan[e + f*x])^p, (a + b*Sin[e + f*x])^m, x], x] /; FreeQ[{a, b, e, f, g, p}, x] && EqQ[a^2 - b^2
, 0] && IGtQ[m, 0]

Rule 3476

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[b/d, Subst[Int[x^n/(b^2 + x^2), x], x, b*Tan[c + d
*x]], x] /; FreeQ[{b, c, d, n}, x] &&  !IntegerQ[n]

Rubi steps

\begin {align*} \int (a+a \sin (e+f x))^3 (g \tan (e+f x))^p \, dx &=\int \left (a^3 (g \tan (e+f x))^p+3 a^3 \sin (e+f x) (g \tan (e+f x))^p+3 a^3 \sin ^2(e+f x) (g \tan (e+f x))^p+a^3 \sin ^3(e+f x) (g \tan (e+f x))^p\right ) \, dx\\ &=a^3 \int (g \tan (e+f x))^p \, dx+a^3 \int \sin ^3(e+f x) (g \tan (e+f x))^p \, dx+\left (3 a^3\right ) \int \sin (e+f x) (g \tan (e+f x))^p \, dx+\left (3 a^3\right ) \int \sin ^2(e+f x) (g \tan (e+f x))^p \, dx\\ &=\frac {\left (a^3 g\right ) \operatorname {Subst}\left (\int \frac {x^p}{g^2+x^2} \, dx,x,g \tan (e+f x)\right )}{f}+\frac {\left (3 a^3 g\right ) \operatorname {Subst}\left (\int \frac {x^{2+p}}{\left (g^2+x^2\right )^2} \, dx,x,g \tan (e+f x)\right )}{f}+\frac {\left (a^3 \cos ^{1+p}(e+f x) \sin ^{-1-p}(e+f x) (g \tan (e+f x))^{1+p}\right ) \int \cos ^{-p}(e+f x) \sin ^{3+p}(e+f x) \, dx}{g}+\frac {\left (3 a^3 \cos ^{1+p}(e+f x) \sin ^{-1-p}(e+f x) (g \tan (e+f x))^{1+p}\right ) \int \cos ^{-p}(e+f x) \sin ^{1+p}(e+f x) \, dx}{g}\\ &=\frac {a^3 \, _2F_1\left (1,\frac {1+p}{2};\frac {3+p}{2};-\tan ^2(e+f x)\right ) (g \tan (e+f x))^{1+p}}{f g (1+p)}+\frac {3 a^3 \cos ^2(e+f x)^{\frac {1+p}{2}} \, _2F_1\left (\frac {1+p}{2},\frac {2+p}{2};\frac {4+p}{2};\sin ^2(e+f x)\right ) \sin (e+f x) (g \tan (e+f x))^{1+p}}{f g (2+p)}+\frac {a^3 \cos ^2(e+f x)^{\frac {1+p}{2}} \, _2F_1\left (\frac {1+p}{2},\frac {4+p}{2};\frac {6+p}{2};\sin ^2(e+f x)\right ) \sin ^3(e+f x) (g \tan (e+f x))^{1+p}}{f g (4+p)}+\frac {3 a^3 \, _2F_1\left (2,\frac {3+p}{2};\frac {5+p}{2};-\tan ^2(e+f x)\right ) (g \tan (e+f x))^{3+p}}{f g^3 (3+p)}\\ \end {align*}

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Mathematica [C]  time = 58.50, size = 5199, normalized size = 19.33 \[ \text {Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + a*Sin[e + f*x])^3*(g*Tan[e + f*x])^p,x]

[Out]

Result too large to show

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fricas [F]  time = 1.99, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-{\left (3 \, a^{3} \cos \left (f x + e\right )^{2} - 4 \, a^{3} + {\left (a^{3} \cos \left (f x + e\right )^{2} - 4 \, a^{3}\right )} \sin \left (f x + e\right )\right )} \left (g \tan \left (f x + e\right )\right )^{p}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^3*(g*tan(f*x+e))^p,x, algorithm="fricas")

[Out]

integral(-(3*a^3*cos(f*x + e)^2 - 4*a^3 + (a^3*cos(f*x + e)^2 - 4*a^3)*sin(f*x + e))*(g*tan(f*x + e))^p, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (a \sin \left (f x + e\right ) + a\right )}^{3} \left (g \tan \left (f x + e\right )\right )^{p}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^3*(g*tan(f*x+e))^p,x, algorithm="giac")

[Out]

integrate((a*sin(f*x + e) + a)^3*(g*tan(f*x + e))^p, x)

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maple [F]  time = 2.68, size = 0, normalized size = 0.00 \[ \int \left (a +a \sin \left (f x +e \right )\right )^{3} \left (g \tan \left (f x +e \right )\right )^{p}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sin(f*x+e))^3*(g*tan(f*x+e))^p,x)

[Out]

int((a+a*sin(f*x+e))^3*(g*tan(f*x+e))^p,x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (a \sin \left (f x + e\right ) + a\right )}^{3} \left (g \tan \left (f x + e\right )\right )^{p}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^3*(g*tan(f*x+e))^p,x, algorithm="maxima")

[Out]

integrate((a*sin(f*x + e) + a)^3*(g*tan(f*x + e))^p, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int {\left (g\,\mathrm {tan}\left (e+f\,x\right )\right )}^p\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^3 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((g*tan(e + f*x))^p*(a + a*sin(e + f*x))^3,x)

[Out]

int((g*tan(e + f*x))^p*(a + a*sin(e + f*x))^3, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ a^{3} \left (\int \left (g \tan {\left (e + f x \right )}\right )^{p}\, dx + \int 3 \left (g \tan {\left (e + f x \right )}\right )^{p} \sin {\left (e + f x \right )}\, dx + \int 3 \left (g \tan {\left (e + f x \right )}\right )^{p} \sin ^{2}{\left (e + f x \right )}\, dx + \int \left (g \tan {\left (e + f x \right )}\right )^{p} \sin ^{3}{\left (e + f x \right )}\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))**3*(g*tan(f*x+e))**p,x)

[Out]

a**3*(Integral((g*tan(e + f*x))**p, x) + Integral(3*(g*tan(e + f*x))**p*sin(e + f*x), x) + Integral(3*(g*tan(e
+ f*x))**p*sin(e + f*x)**2, x) + Integral((g*tan(e + f*x))**p*sin(e + f*x)**3, x))

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