3.372 \(\int (d \cos (e+f x))^m (a+b \sin ^2(e+f x))^p \, dx\)

Optimal. Leaf size=115 \[ \frac{d \sin (e+f x) \cos ^2(e+f x)^{\frac{1-m}{2}} (d \cos (e+f x))^{m-1} \left (a+b \sin ^2(e+f x)\right )^p \left (\frac{b \sin ^2(e+f x)}{a}+1\right )^{-p} F_1\left (\frac{1}{2};\frac{1-m}{2},-p;\frac{3}{2};\sin ^2(e+f x),-\frac{b \sin ^2(e+f x)}{a}\right )}{f} \]

[Out]

(d*AppellF1[1/2, (1 - m)/2, -p, 3/2, Sin[e + f*x]^2, -((b*Sin[e + f*x]^2)/a)]*(d*Cos[e + f*x])^(-1 + m)*(Cos[e
+ f*x]^2)^((1 - m)/2)*Sin[e + f*x]*(a + b*Sin[e + f*x]^2)^p)/(f*(1 + (b*Sin[e + f*x]^2)/a)^p)

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Rubi [A]  time = 0.0900169, antiderivative size = 115, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.12, Rules used = {3193, 430, 429} \[ \frac{d \sin (e+f x) \cos ^2(e+f x)^{\frac{1-m}{2}} (d \cos (e+f x))^{m-1} \left (a+b \sin ^2(e+f x)\right )^p \left (\frac{b \sin ^2(e+f x)}{a}+1\right )^{-p} F_1\left (\frac{1}{2};\frac{1-m}{2},-p;\frac{3}{2};\sin ^2(e+f x),-\frac{b \sin ^2(e+f x)}{a}\right )}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d*AppellF1[1/2, (1 - m)/2, -p, 3/2, Sin[e + f*x]^2, -((b*Sin[e + f*x]^2)/a)]*(d*Cos[e + f*x])^(-1 + m)*(Cos[e
+ f*x]^2)^((1 - m)/2)*Sin[e + f*x]*(a + b*Sin[e + f*x]^2)^p)/(f*(1 + (b*Sin[e + f*x]^2)/a)^p)

Rule 3193

Int[(cos[(e_.) + (f_.)*(x_)]*(d_.))^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.), x_Symbol] :> With[{ff
= FreeFactors[Sin[e + f*x], x]}, Dist[(ff*d^(2*IntPart[(m - 1)/2] + 1)*(d*Cos[e + f*x])^(2*FracPart[(m - 1)/2
]))/(f*(Cos[e + f*x]^2)^FracPart[(m - 1)/2]), Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*(a + b*ff^2*x^2)^p, x], x,
Sin[e + f*x]/ff], x]] /; FreeQ[{a, b, d, e, f, m, p}, x] &&  !IntegerQ[m]

Rule 430

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Dist[(a^IntPart[p]*(a + b*x^n)^F
racPart[p])/(1 + (b*x^n)/a)^FracPart[p], Int[(1 + (b*x^n)/a)^p*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, n,
p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[n, -1] &&  !(IntegerQ[p] || GtQ[a, 0])

Rule 429

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[a^p*c^q*x*AppellF1[1/n, -p,
-q, 1 + 1/n, -((b*x^n)/a), -((d*x^n)/c)], x] /; FreeQ[{a, b, c, d, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[n
, -1] && (IntegerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])

Rubi steps

\begin{align*} \int (d \cos (e+f x))^m \left (a+b \sin ^2(e+f x)\right )^p \, dx &=\frac{\left (d (d \cos (e+f x))^{2 \left (-\frac{1}{2}+\frac{m}{2}\right )} \cos ^2(e+f x)^{\frac{1}{2}-\frac{m}{2}}\right ) \operatorname{Subst}\left (\int \left (1-x^2\right )^{\frac{1}{2} (-1+m)} \left (a+b x^2\right )^p \, dx,x,\sin (e+f x)\right )}{f}\\ &=\frac{\left (d (d \cos (e+f x))^{2 \left (-\frac{1}{2}+\frac{m}{2}\right )} \cos ^2(e+f x)^{\frac{1}{2}-\frac{m}{2}} \left (a+b \sin ^2(e+f x)\right )^p \left (1+\frac{b \sin ^2(e+f x)}{a}\right )^{-p}\right ) \operatorname{Subst}\left (\int \left (1-x^2\right )^{\frac{1}{2} (-1+m)} \left (1+\frac{b x^2}{a}\right )^p \, dx,x,\sin (e+f x)\right )}{f}\\ &=\frac{d F_1\left (\frac{1}{2};\frac{1-m}{2},-p;\frac{3}{2};\sin ^2(e+f x),-\frac{b \sin ^2(e+f x)}{a}\right ) (d \cos (e+f x))^{-1+m} \cos ^2(e+f x)^{\frac{1-m}{2}} \sin (e+f x) \left (a+b \sin ^2(e+f x)\right )^p \left (1+\frac{b \sin ^2(e+f x)}{a}\right )^{-p}}{f}\\ \end{align*}

Mathematica [A]  time = 0.833344, size = 228, normalized size = 1.98 \[ \frac{3 a \tan (e+f x) (d \cos (e+f x))^m \left (a+b \sin ^2(e+f x)\right )^p F_1\left (\frac{1}{2};\frac{1-m}{2},-p;\frac{3}{2};\sin ^2(e+f x),-\frac{b \sin ^2(e+f x)}{a}\right )}{f \left (\sin ^2(e+f x) \left (2 b p F_1\left (\frac{3}{2};\frac{1-m}{2},1-p;\frac{5}{2};\sin ^2(e+f x),-\frac{b \sin ^2(e+f x)}{a}\right )-a (m-1) F_1\left (\frac{3}{2};\frac{3-m}{2},-p;\frac{5}{2};\sin ^2(e+f x),-\frac{b \sin ^2(e+f x)}{a}\right )\right )+3 a F_1\left (\frac{1}{2};\frac{1-m}{2},-p;\frac{3}{2};\sin ^2(e+f x),-\frac{b \sin ^2(e+f x)}{a}\right )\right )} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(d*Cos[e + f*x])^m*(a + b*Sin[e + f*x]^2)^p,x]

[Out]

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

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Maple [F]  time = 1.342, size = 0, normalized size = 0. \begin{align*} \int \left ( d\cos \left ( fx+e \right ) \right ) ^{m} \left ( a+b \left ( \sin \left ( fx+e \right ) \right ) ^{2} \right ) ^{p}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*cos(f*x+e))^m*(a+b*sin(f*x+e)^2)^p,x, algorithm="maxima")

[Out]

integrate((b*sin(f*x + e)^2 + a)^p*(d*cos(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 \cos \left (f x + e\right )^{2} + a + b\right )}^{p} \left (d \cos \left (f x + e\right )\right )^{m}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*cos(f*x+e))^m*(a+b*sin(f*x+e)^2)^p,x, algorithm="fricas")

[Out]

integral((-b*cos(f*x + e)^2 + a + b)^p*(d*cos(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*cos(f*x+e))**m*(a+b*sin(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 \sin \left (f x + e\right )^{2} + a\right )}^{p} \left (d \cos \left (f x + e\right )\right )^{m}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*cos(f*x+e))^m*(a+b*sin(f*x+e)^2)^p,x, algorithm="giac")

[Out]

integrate((b*sin(f*x + e)^2 + a)^p*(d*cos(f*x + e))^m, x)