3.2.64 \(\int (d \sin (e+f x))^m (b (c \tan (e+f x))^n)^p \, dx\) [164]

Optimal. Leaf size=98 \[ \frac {\cos ^2(e+f x)^{\frac {1}{2} (1+n p)} \, _2F_1\left (\frac {1}{2} (1+n p),\frac {1}{2} (1+m+n p);\frac {1}{2} (3+m+n p);\sin ^2(e+f x)\right ) (d \sin (e+f x))^m \tan (e+f x) \left (b (c \tan (e+f x))^n\right )^p}{f (1+m+n p)} \]

[Out]

(cos(f*x+e)^2)^(1/2*n*p+1/2)*hypergeom([1/2*n*p+1/2, 1/2*n*p+1/2*m+1/2],[1/2*n*p+1/2*m+3/2],sin(f*x+e)^2)*(d*s
in(f*x+e))^m*tan(f*x+e)*(b*(c*tan(f*x+e))^n)^p/f/(n*p+m+1)

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Rubi [A]
time = 0.11, antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {3740, 2682, 2657} \begin {gather*} \frac {\tan (e+f x) (d \sin (e+f x))^m \cos ^2(e+f x)^{\frac {1}{2} (n p+1)} \left (b (c \tan (e+f x))^n\right )^p \, _2F_1\left (\frac {1}{2} (n p+1),\frac {1}{2} (m+n p+1);\frac {1}{2} (m+n p+3);\sin ^2(e+f x)\right )}{f (m+n p+1)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((Cos[e + f*x]^2)^((1 + n*p)/2)*Hypergeometric2F1[(1 + n*p)/2, (1 + m + n*p)/2, (3 + m + n*p)/2, Sin[e + f*x]^
2]*(d*Sin[e + f*x])^m*Tan[e + f*x]*(b*(c*Tan[e + f*x])^n)^p)/(f*(1 + m + 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 (d \sin (e+f x))^m \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 (d \sin (e+f x))^m (c \tan (e+f x))^{n p} \, dx\\ &=\left (d \cos ^{n p}(e+f x) \sin (e+f x) (d \sin (e+f x))^{-1-n p} \left (b (c \tan (e+f x))^n\right )^p\right ) \int \cos ^{-n p}(e+f x) (d \sin (e+f x))^{m+n p} \, 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} (1+m+n p);\frac {1}{2} (3+m+n p);\sin ^2(e+f x)\right ) (d \sin (e+f x))^m \tan (e+f x) \left (b (c \tan (e+f x))^n\right )^p}{f (1+m+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 = 1.32, size = 295, normalized size = 3.01 \begin {gather*} \frac {(3+m+n p) F_1\left (\frac {1}{2} (1+m+n p);n p,1+m;\frac {1}{2} (3+m+n p);\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right ) \sin (e+f x) (d \sin (e+f x))^m \left (b (c \tan (e+f x))^n\right )^p}{f (1+m+n p) \left ((3+m+n p) F_1\left (\frac {1}{2} (1+m+n p);n p,1+m;\frac {1}{2} (3+m+n p);\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )-2 \left ((1+m) F_1\left (\frac {1}{2} (3+m+n p);n p,2+m;\frac {1}{2} (5+m+n p);\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 (\frac {1}{2} (3+m+n p);1+n p,1+m;\frac {1}{2} (5+m+n p);\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )\right ) \tan ^2\left (\frac {1}{2} (e+f x)\right )\right )} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((3 + m + n*p)*AppellF1[(1 + m + n*p)/2, n*p, 1 + m, (3 + m + n*p)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]
*Sin[e + f*x]*(d*Sin[e + f*x])^m*(b*(c*Tan[e + f*x])^n)^p)/(f*(1 + m + n*p)*((3 + m + n*p)*AppellF1[(1 + m + n
*p)/2, n*p, 1 + m, (3 + m + n*p)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] - 2*((1 + m)*AppellF1[(3 + m + n*
p)/2, n*p, 2 + m, (5 + m + n*p)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] - n*p*AppellF1[(3 + m + n*p)/2, 1
+ n*p, 1 + m, (5 + m + n*p)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2])*Tan[(e + f*x)/2]^2))

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Maple [F]
time = 0.22, size = 0, normalized size = 0.00 \[\int \left (d \sin \left (f x +e \right )\right )^{m} \left (b \left (c \tan \left (f x +e \right )\right )^{n}\right )^{p}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

[Out]

integral(((c*tan(f*x + e))^n*b)^p*(d*sin(f*x + e))^m, 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} \left (d \sin {\left (e + f x \right )}\right )^{m}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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

[Out]

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

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