3.2.71 \(\int \sin (e+f x) (b (c \tan (e+f x))^n)^p \, dx\) [171]

Optimal. Leaf size=91 \[ \frac {\cos ^2(e+f x)^{\frac {1}{2} (1+n p)} \, _2F_1\left (\frac {1}{2} (1+n p),\frac {1}{2} (2+n p);\frac {1}{2} (4+n p);\sin ^2(e+f x)\right ) \sin (e+f x) \tan (e+f x) \left (b (c \tan (e+f x))^n\right )^p}{f (2+n p)} \]

[Out]

(cos(f*x+e)^2)^(1/2*n*p+1/2)*hypergeom([1/2*n*p+1, 1/2*n*p+1/2],[1/2*n*p+2],sin(f*x+e)^2)*sin(f*x+e)*tan(f*x+e
)*(b*(c*tan(f*x+e))^n)^p/f/(n*p+2)

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Rubi [A]
time = 0.07, antiderivative size = 91, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3740, 2682, 2657} \begin {gather*} \frac {\sin (e+f x) \tan (e+f x) \cos ^2(e+f x)^{\frac {1}{2} (n p+1)} \, _2F_1\left (\frac {1}{2} (n p+1),\frac {1}{2} (n p+2);\frac {1}{2} (n p+4);\sin ^2(e+f x)\right ) \left (b (c \tan (e+f x))^n\right )^p}{f (n p+2)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sin[e + f*x]*(b*(c*Tan[e + f*x])^n)^p,x]

[Out]

((Cos[e + f*x]^2)^((1 + n*p)/2)*Hypergeometric2F1[(1 + n*p)/2, (2 + n*p)/2, (4 + n*p)/2, Sin[e + f*x]^2]*Sin[e
+ f*x]*Tan[e + f*x]*(b*(c*Tan[e + f*x])^n)^p)/(f*(2 + n*p))

Rule 2657

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)/(a*f*(m + 1)*(Cos[e + f*x]^
2)^FracPart[(n - 1)/2]))*Hypergeometric2F1[(1 + m)/2, (1 - n)/2, (3 + m)/2, Sin[e + f*x]^2], x] /; FreeQ[{a, b
, e, f, m, n}, x]

Rule 2682

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 3740

Int[(u_.)*((b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_), x_Symbol] :> Dist[b^IntPart[p]*((b*(c*Tan[e + f*x
])^n)^FracPart[p]/(c*Tan[e + f*x])^(n*FracPart[p])), Int[ActivateTrig[u]*(c*Tan[e + f*x])^(n*p), x], x] /; Fre
eQ[{b, c, e, f, n, p}, x] &&  !IntegerQ[p] &&  !IntegerQ[n] && (EqQ[u, 1] || MatchQ[u, ((d_.)*(trig_)[e + f*x]
)^(m_.) /; FreeQ[{d, m}, x] && MemberQ[{sin, cos, tan, cot, sec, csc}, trig]])

Rubi steps

\begin {align*} \int \sin (e+f x) \left (b (c \tan (e+f x))^n\right )^p \, dx &=\left ((c \tan (e+f x))^{-n p} \left (b (c \tan (e+f x))^n\right )^p\right ) \int \sin (e+f x) (c \tan (e+f x))^{n p} \, dx\\ &=\left (\cos ^{n p}(e+f x) \sin ^{-n p}(e+f x) \left (b (c \tan (e+f x))^n\right )^p\right ) \int \cos ^{-n p}(e+f x) \sin ^{1+n p}(e+f x) \, dx\\ &=\frac {\cos ^2(e+f x)^{\frac {1}{2} (1+n p)} \, _2F_1\left (\frac {1}{2} (1+n p),\frac {1}{2} (2+n p);\frac {1}{2} (4+n p);\sin ^2(e+f x)\right ) \sin (e+f x) \tan (e+f x) \left (b (c \tan (e+f x))^n\right )^p}{f (2+n p)}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.
time = 0.90, size = 284, normalized size = 3.12 \begin {gather*} \frac {8 (4+n p) F_1\left (1+\frac {n p}{2};n p,2;2+\frac {n p}{2};\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right ) \cos ^4\left (\frac {1}{2} (e+f x)\right ) \sin ^2\left (\frac {1}{2} (e+f x)\right ) \left (b (c \tan (e+f x))^n\right )^p}{f (2+n p) \left (2 (4+n p) F_1\left (1+\frac {n p}{2};n p,2;2+\frac {n p}{2};\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right ) \cos ^2\left (\frac {1}{2} (e+f x)\right )+2 \left (2 F_1\left (2+\frac {n p}{2};n p,3;3+\frac {n p}{2};\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )-n p F_1\left (2+\frac {n p}{2};1+n p,2;3+\frac {n p}{2};\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )\right ) (-1+\cos (e+f x))\right )} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[Sin[e + f*x]*(b*(c*Tan[e + f*x])^n)^p,x]

[Out]

(8*(4 + n*p)*AppellF1[1 + (n*p)/2, n*p, 2, 2 + (n*p)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Cos[(e + f*x)
/2]^4*Sin[(e + f*x)/2]^2*(b*(c*Tan[e + f*x])^n)^p)/(f*(2 + n*p)*(2*(4 + n*p)*AppellF1[1 + (n*p)/2, n*p, 2, 2 +
(n*p)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Cos[(e + f*x)/2]^2 + 2*(2*AppellF1[2 + (n*p)/2, n*p, 3, 3 +
(n*p)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] - n*p*AppellF1[2 + (n*p)/2, 1 + n*p, 2, 3 + (n*p)/2, Tan[(e
+ f*x)/2]^2, -Tan[(e + f*x)/2]^2])*(-1 + Cos[e + f*x])))

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Maple [F]
time = 0.01, size = 0, normalized size = 0.00 \[\int \sin \left (f x +e \right ) \left (b \left (c \tan \left (f x +e \right )\right )^{n}\right )^{p}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(f*x+e)*(b*(c*tan(f*x+e))^n)^p,x)

[Out]

int(sin(f*x+e)*(b*(c*tan(f*x+e))^n)^p,x)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)*(b*(c*tan(f*x+e))^n)^p,x, algorithm="maxima")

[Out]

integrate(((c*tan(f*x + e))^n*b)^p*sin(f*x + e), x)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)*(b*(c*tan(f*x+e))^n)^p,x, algorithm="fricas")

[Out]

integral(((c*tan(f*x + e))^n*b)^p*sin(f*x + e), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (b \left (c \tan {\left (e + f x \right )}\right )^{n}\right )^{p} \sin {\left (e + f x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)*(b*(c*tan(f*x+e))**n)**p,x)

[Out]

Integral((b*(c*tan(e + f*x))**n)**p*sin(e + f*x), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)*(b*(c*tan(f*x+e))^n)^p,x, algorithm="giac")

[Out]

integrate(((c*tan(f*x + e))^n*b)^p*sin(f*x + e), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \sin \left (e+f\,x\right )\,{\left (b\,{\left (c\,\mathrm {tan}\left (e+f\,x\right )\right )}^n\right )}^p \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(e + f*x)*(b*(c*tan(e + f*x))^n)^p,x)

[Out]

int(sin(e + f*x)*(b*(c*tan(e + f*x))^n)^p, x)

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