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

Optimal. Leaf size=92 \[ \frac{\tan (e+f x) \cos ^2(e+f x)^{p+\frac{1}{2}} \left (b \tan ^2(e+f x)\right )^p (d \sin (e+f x))^m \text{Hypergeometric2F1}\left (\frac{1}{2} (2 p+1),\frac{1}{2} (m+2 p+1),\frac{1}{2} (m+2 p+3),\sin ^2(e+f x)\right )}{f (m+2 p+1)} \]

[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))

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Rubi [A]  time = 0.153309, antiderivative size = 92, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {3658, 2602, 2577} \[ \frac{\tan (e+f x) \cos ^2(e+f x)^{p+\frac{1}{2}} \left (b \tan ^2(e+f x)\right )^p (d \sin (e+f x))^m \, _2F_1\left (\frac{1}{2} (2 p+1),\frac{1}{2} (m+2 p+1);\frac{1}{2} (m+2 p+3);\sin ^2(e+f x)\right )}{f (m+2 p+1)} \]

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 3658

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
]])

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 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]

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*}

Mathematica [C]  time = 2.07811, size = 292, normalized size = 3.17 \[ \frac{(m+2 p+3) \sin (e+f x) \left (b \tan ^2(e+f x)\right )^p (d \sin (e+f x))^m F_1\left (\frac{m}{2}+p+\frac{1}{2};2 p,m+1;\frac{m}{2}+p+\frac{3}{2};\tan ^2\left (\frac{1}{2} (e+f x)\right ),-\tan ^2\left (\frac{1}{2} (e+f x)\right )\right )}{f (m+2 p+1) \left ((m+2 p+3) F_1\left (\frac{m}{2}+p+\frac{1}{2};2 p,m+1;\frac{m}{2}+p+\frac{3}{2};\tan ^2\left (\frac{1}{2} (e+f x)\right ),-\tan ^2\left (\frac{1}{2} (e+f x)\right )\right )-2 \tan ^2\left (\frac{1}{2} (e+f x)\right ) \left ((m+1) F_1\left (\frac{m}{2}+p+\frac{3}{2};2 p,m+2;\frac{m}{2}+p+\frac{5}{2};\tan ^2\left (\frac{1}{2} (e+f x)\right ),-\tan ^2\left (\frac{1}{2} (e+f x)\right )\right )-2 p F_1\left (\frac{m}{2}+p+\frac{3}{2};2 p+1,m+1;\frac{m}{2}+p+\frac{5}{2};\tan ^2\left (\frac{1}{2} (e+f x)\right ),-\tan ^2\left (\frac{1}{2} (e+f x)\right )\right )\right )\right )} \]

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 = 0.777, size = 0, normalized size = 0. \begin{align*} \int \left ( d\sin \left ( fx+e \right ) \right ) ^{m} \left ( b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{p}\, dx \end{align*}

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., size = 0, normalized size = 0. \begin{align*} \int \left (b \tan \left (f x + e\right )^{2}\right )^{p} \left (d \sin \left (f x + e\right )\right )^{m}\,{d x} \end{align*}

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., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\left (b \tan \left (f x + e\right )^{2}\right )^{p} \left (d \sin \left (f x + e\right )\right )^{m}, x\right ) \end{align*}

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

[Out]

Timed out

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

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)