3.2.52 \(\int (d \sin (e+f x))^m (b \tan ^2(e+f x))^p \, dx\) [152]

Optimal. Leaf size=92 \[ \frac {\cos ^2(e+f x)^{\frac {1}{2}+p} \, _2F_1\left (\frac {1}{2} (1+2 p),\frac {1}{2} (1+m+2 p);\frac {1}{2} (3+m+2 p);\sin ^2(e+f x)\right ) (d \sin (e+f x))^m \tan (e+f x) \left (b \tan ^2(e+f x)\right )^p}{f (1+m+2 p)} \]

[Out]

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

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Rubi [A]
time = 0.10, antiderivative size = 92, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {3739, 2682, 2657} \begin {gather*} \frac {\tan (e+f x) \cos ^2(e+f x)^{p+\frac {1}{2}} \left (b \tan ^2(e+f x)\right )^p (d \sin (e+f x))^m \, _2F_1\left (\frac {1}{2} (2 p+1),\frac {1}{2} (m+2 p+1);\frac {1}{2} (m+2 p+3);\sin ^2(e+f x)\right )}{f (m+2 p+1)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(d*Sin[e + f*x])^m*(b*Tan[e + f*x]^2)^p,x]

[Out]

((Cos[e + f*x]^2)^(1/2 + p)*Hypergeometric2F1[(1 + 2*p)/2, (1 + m + 2*p)/2, (3 + m + 2*p)/2, Sin[e + f*x]^2]*(
d*Sin[e + f*x])^m*Tan[e + f*x]*(b*Tan[e + f*x]^2)^p)/(f*(1 + m + 2*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 3739

Int[(u_.)*((b_.)*tan[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Di
st[(b*ff^n)^IntPart[p]*((b*Tan[e + f*x]^n)^FracPart[p]/(Tan[e + f*x]/ff)^(n*FracPart[p])), Int[ActivateTrig[u]
*(Tan[e + f*x]/ff)^(n*p), x], x]] /; FreeQ[{b, 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 \tan ^2(e+f x)\right )^p \, dx &=\left (\tan ^{-2 p}(e+f x) \left (b \tan ^2(e+f x)\right )^p\right ) \int (d \sin (e+f x))^m \tan ^{2 p}(e+f x) \, dx\\ &=\left (d \cos ^{2 p}(e+f x) \sin (e+f x) (d \sin (e+f x))^{-1-2 p} \left (b \tan ^2(e+f x)\right )^p\right ) \int \cos ^{-2 p}(e+f x) (d \sin (e+f x))^{m+2 p} \, dx\\ &=\frac {\cos ^2(e+f x)^{\frac {1}{2}+p} \, _2F_1\left (\frac {1}{2} (1+2 p),\frac {1}{2} (1+m+2 p);\frac {1}{2} (3+m+2 p);\sin ^2(e+f x)\right ) (d \sin (e+f x))^m \tan (e+f x) \left (b \tan ^2(e+f x)\right )^p}{f (1+m+2 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 = 1.37, size = 292, normalized size = 3.17 \begin {gather*} \frac {(3+m+2 p) F_1\left (\frac {1}{2}+\frac {m}{2}+p;2 p,1+m;\frac {3}{2}+\frac {m}{2}+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 \tan ^2(e+f x)\right )^p}{f (1+m+2 p) \left ((3+m+2 p) F_1\left (\frac {1}{2}+\frac {m}{2}+p;2 p,1+m;\frac {3}{2}+\frac {m}{2}+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 {3}{2}+\frac {m}{2}+p;2 p,2+m;\frac {5}{2}+\frac {m}{2}+p;\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )-2 p F_1\left (\frac {3}{2}+\frac {m}{2}+p;1+2 p,1+m;\frac {5}{2}+\frac {m}{2}+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*Tan[e + f*x]^2)^p,x]

[Out]

((3 + m + 2*p)*AppellF1[1/2 + m/2 + p, 2*p, 1 + m, 3/2 + m/2 + p, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Sin
[e + f*x]*(d*Sin[e + f*x])^m*(b*Tan[e + f*x]^2)^p)/(f*(1 + m + 2*p)*((3 + m + 2*p)*AppellF1[1/2 + m/2 + p, 2*p
, 1 + m, 3/2 + m/2 + p, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] - 2*((1 + m)*AppellF1[3/2 + m/2 + p, 2*p, 2 +
m, 5/2 + m/2 + p, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] - 2*p*AppellF1[3/2 + m/2 + p, 1 + 2*p, 1 + m, 5/2
+ m/2 + p, 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 \[\int \left (d \sin \left (f x +e \right )\right )^{m} \left (b \left (\tan ^{2}\left (f x +e \right )\right )\right )^{p}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*sin(f*x+e))^m*(b*tan(f*x+e)^2)^p,x)

[Out]

int((d*sin(f*x+e))^m*(b*tan(f*x+e)^2)^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*tan(f*x+e)^2)^p,x, algorithm="maxima")

[Out]

integrate((b*tan(f*x + e)^2)^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*tan(f*x+e)^2)^p,x, algorithm="fricas")

[Out]

integral((b*tan(f*x + e)^2)^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 \tan ^{2}{\left (e + f x \right )}\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*tan(f*x+e)**2)**p,x)

[Out]

Integral((b*tan(e + f*x)**2)**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*tan(f*x+e)^2)^p,x, algorithm="giac")

[Out]

integrate((b*tan(f*x + e)^2)^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\,{\mathrm {tan}\left (e+f\,x\right )}^2\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*tan(e + f*x)^2)^p,x)

[Out]

int((d*sin(e + f*x))^m*(b*tan(e + f*x)^2)^p, x)

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