3.144 \(\int (a+a \sec (c+d x))^n (e \sin (c+d x))^m \, dx\)

Optimal. Leaf size=130 \[ -\frac{e \cos (c+d x) (1-\cos (c+d x))^{\frac{1-m}{2}} (a \sec (c+d x)+a)^n (e \sin (c+d x))^{m-1} (\cos (c+d x)+1)^{\frac{1}{2} (-m-2 n+1)} F_1\left (1-n;\frac{1-m}{2},\frac{1}{2} (-m-2 n+1);2-n;\cos (c+d x),-\cos (c+d x)\right )}{d (1-n)} \]

[Out]

-((e*AppellF1[1 - n, (1 - m)/2, (1 - m - 2*n)/2, 2 - n, Cos[c + d*x], -Cos[c + d*x]]*(1 - Cos[c + d*x])^((1 -
m)/2)*Cos[c + d*x]*(1 + Cos[c + d*x])^((1 - m - 2*n)/2)*(a + a*Sec[c + d*x])^n*(e*Sin[c + d*x])^(-1 + m))/(d*(
1 - n)))

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Rubi [A]  time = 0.276536, antiderivative size = 130, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {3876, 2886, 135, 133} \[ -\frac{e \cos (c+d x) (1-\cos (c+d x))^{\frac{1-m}{2}} (a \sec (c+d x)+a)^n (e \sin (c+d x))^{m-1} (\cos (c+d x)+1)^{\frac{1}{2} (-m-2 n+1)} F_1\left (1-n;\frac{1-m}{2},\frac{1}{2} (-m-2 n+1);2-n;\cos (c+d x),-\cos (c+d x)\right )}{d (1-n)} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sec[c + d*x])^n*(e*Sin[c + d*x])^m,x]

[Out]

-((e*AppellF1[1 - n, (1 - m)/2, (1 - m - 2*n)/2, 2 - n, Cos[c + d*x], -Cos[c + d*x]]*(1 - Cos[c + d*x])^((1 -
m)/2)*Cos[c + d*x]*(1 + Cos[c + d*x])^((1 - m - 2*n)/2)*(a + a*Sec[c + d*x])^n*(e*Sin[c + d*x])^(-1 + m))/(d*(
1 - n)))

Rule 3876

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Dist[(Sin[
e + f*x]^FracPart[m]*(a + b*Csc[e + f*x])^FracPart[m])/(b + a*Sin[e + f*x])^FracPart[m], Int[((g*Cos[e + f*x])
^p*(b + a*Sin[e + f*x])^m)/Sin[e + f*x]^m, x], x] /; FreeQ[{a, b, e, f, g, m, p}, x] && (EqQ[a^2 - b^2, 0] ||
IntegersQ[2*m, p])

Rule 2886

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*
(x_)])^(m_), x_Symbol] :> Dist[(g*(g*Cos[e + f*x])^(p - 1))/(f*(a + b*Sin[e + f*x])^((p - 1)/2)*(a - b*Sin[e +
f*x])^((p - 1)/2)), Subst[Int[(d*x)^n*(a + b*x)^(m + (p - 1)/2)*(a - b*x)^((p - 1)/2), x], x, Sin[e + f*x]],
x] /; FreeQ[{a, b, d, e, f, m, n, p}, x] && EqQ[a^2 - b^2, 0] &&  !IntegerQ[m]

Rule 135

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_Symbol] :> Dist[(c^IntPart[n]*(c +
d*x)^FracPart[n])/(1 + (d*x)/c)^FracPart[n], Int[(b*x)^m*(1 + (d*x)/c)^n*(e + f*x)^p, x], x] /; FreeQ[{b, c, d
, e, f, m, n, p}, x] &&  !IntegerQ[m] &&  !IntegerQ[n] &&  !GtQ[c, 0]

Rule 133

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[(c^n*e^p*(b*x)^(m +
1)*AppellF1[m + 1, -n, -p, m + 2, -((d*x)/c), -((f*x)/e)])/(b*(m + 1)), x] /; FreeQ[{b, c, d, e, f, m, n, p},
x] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[c, 0] && (IntegerQ[p] || GtQ[e, 0])

Rubi steps

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

Mathematica [B]  time = 1.84603, size = 276, normalized size = 2.12 \[ \frac{4 (m+3) \sin \left (\frac{1}{2} (c+d x)\right ) \cos ^3\left (\frac{1}{2} (c+d x)\right ) (a (\sec (c+d x)+1))^n (e \sin (c+d x))^m F_1\left (\frac{m+1}{2};n,m+1;\frac{m+3}{2};\tan ^2\left (\frac{1}{2} (c+d x)\right ),-\tan ^2\left (\frac{1}{2} (c+d x)\right )\right )}{d (m+1) \left ((m+3) (\cos (c+d x)+1) F_1\left (\frac{m+1}{2};n,m+1;\frac{m+3}{2};\tan ^2\left (\frac{1}{2} (c+d x)\right ),-\tan ^2\left (\frac{1}{2} (c+d x)\right )\right )-4 \sin ^2\left (\frac{1}{2} (c+d x)\right ) \left ((m+1) F_1\left (\frac{m+3}{2};n,m+2;\frac{m+5}{2};\tan ^2\left (\frac{1}{2} (c+d x)\right ),-\tan ^2\left (\frac{1}{2} (c+d x)\right )\right )-n F_1\left (\frac{m+3}{2};n+1,m+1;\frac{m+5}{2};\tan ^2\left (\frac{1}{2} (c+d x)\right ),-\tan ^2\left (\frac{1}{2} (c+d x)\right )\right )\right )\right )} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + a*Sec[c + d*x])^n*(e*Sin[c + d*x])^m,x]

[Out]

(4*(3 + m)*AppellF1[(1 + m)/2, n, 1 + m, (3 + m)/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*Cos[(c + d*x)/2]^
3*(a*(1 + Sec[c + d*x]))^n*Sin[(c + d*x)/2]*(e*Sin[c + d*x])^m)/(d*(1 + m)*((3 + m)*AppellF1[(1 + m)/2, n, 1 +
m, (3 + m)/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*(1 + Cos[c + d*x]) - 4*((1 + m)*AppellF1[(3 + m)/2, n,
2 + m, (5 + m)/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] - n*AppellF1[(3 + m)/2, 1 + n, 1 + m, (5 + m)/2, T
an[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2])*Sin[(c + d*x)/2]^2))

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Maple [F]  time = 0.728, size = 0, normalized size = 0. \begin{align*} \int \left ( a+a\sec \left ( dx+c \right ) \right ) ^{n} \left ( e\sin \left ( dx+c \right ) \right ) ^{m}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sec(d*x+c))^n*(e*sin(d*x+c))^m,x)

[Out]

int((a+a*sec(d*x+c))^n*(e*sin(d*x+c))^m,x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (a \sec \left (d x + c\right ) + a\right )}^{n} \left (e \sin \left (d x + c\right )\right )^{m}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^n*(e*sin(d*x+c))^m,x, algorithm="maxima")

[Out]

integrate((a*sec(d*x + c) + a)^n*(e*sin(d*x + c))^m, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (a \sec \left (d x + c\right ) + a\right )}^{n} \left (e \sin \left (d x + c\right )\right )^{m}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^n*(e*sin(d*x+c))^m,x, algorithm="fricas")

[Out]

integral((a*sec(d*x + c) + a)^n*(e*sin(d*x + c))^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((a+a*sec(d*x+c))**n*(e*sin(d*x+c))**m,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (a \sec \left (d x + c\right ) + a\right )}^{n} \left (e \sin \left (d x + c\right )\right )^{m}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^n*(e*sin(d*x+c))^m,x, algorithm="giac")

[Out]

integrate((a*sec(d*x + c) + a)^n*(e*sin(d*x + c))^m, x)