3.258 \(\int (a+b \sec (c+d x))^2 (e \sin (c+d x))^m \, dx\)

Optimal. Leaf size=190 \[ \frac{a^2 \sin (c+d x) \cos (c+d x) (e \sin (c+d x))^m \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{m+1}{2},\frac{m+3}{2},\sin ^2(c+d x)\right )}{d (m+1) \sqrt{\cos ^2(c+d x)}}+\frac{2 a b (e \sin (c+d x))^{m+1} \text{Hypergeometric2F1}\left (1,\frac{m+1}{2},\frac{m+3}{2},\sin ^2(c+d x)\right )}{d e (m+1)}+\frac{b^2 \sqrt{\cos ^2(c+d x)} \tan (c+d x) (e \sin (c+d x))^m \text{Hypergeometric2F1}\left (\frac{3}{2},\frac{m+1}{2},\frac{m+3}{2},\sin ^2(c+d x)\right )}{d (m+1)} \]

[Out]

(a^2*Cos[c + d*x]*Hypergeometric2F1[1/2, (1 + m)/2, (3 + m)/2, Sin[c + d*x]^2]*Sin[c + d*x]*(e*Sin[c + d*x])^m
)/(d*(1 + m)*Sqrt[Cos[c + d*x]^2]) + (2*a*b*Hypergeometric2F1[1, (1 + m)/2, (3 + m)/2, Sin[c + d*x]^2]*(e*Sin[
c + d*x])^(1 + m))/(d*e*(1 + m)) + (b^2*Sqrt[Cos[c + d*x]^2]*Hypergeometric2F1[3/2, (1 + m)/2, (3 + m)/2, Sin[
c + d*x]^2]*(e*Sin[c + d*x])^m*Tan[c + d*x])/(d*(1 + m))

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Rubi [A]  time = 0.839372, antiderivative size = 190, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 8, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.348, Rules used = {3872, 2911, 2564, 364, 4398, 4401, 2643, 2577} \[ \frac{a^2 \sin (c+d x) \cos (c+d x) (e \sin (c+d x))^m \, _2F_1\left (\frac{1}{2},\frac{m+1}{2};\frac{m+3}{2};\sin ^2(c+d x)\right )}{d (m+1) \sqrt{\cos ^2(c+d x)}}+\frac{2 a b (e \sin (c+d x))^{m+1} \, _2F_1\left (1,\frac{m+1}{2};\frac{m+3}{2};\sin ^2(c+d x)\right )}{d e (m+1)}+\frac{b^2 \sqrt{\cos ^2(c+d x)} \tan (c+d x) (e \sin (c+d x))^m \, _2F_1\left (\frac{3}{2},\frac{m+1}{2};\frac{m+3}{2};\sin ^2(c+d x)\right )}{d (m+1)} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sec[c + d*x])^2*(e*Sin[c + d*x])^m,x]

[Out]

(a^2*Cos[c + d*x]*Hypergeometric2F1[1/2, (1 + m)/2, (3 + m)/2, Sin[c + d*x]^2]*Sin[c + d*x]*(e*Sin[c + d*x])^m
)/(d*(1 + m)*Sqrt[Cos[c + d*x]^2]) + (2*a*b*Hypergeometric2F1[1, (1 + m)/2, (3 + m)/2, Sin[c + d*x]^2]*(e*Sin[
c + d*x])^(1 + m))/(d*e*(1 + m)) + (b^2*Sqrt[Cos[c + d*x]^2]*Hypergeometric2F1[3/2, (1 + m)/2, (3 + m)/2, Sin[
c + d*x]^2]*(e*Sin[c + d*x])^m*Tan[c + d*x])/(d*(1 + m))

Rule 3872

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[((g*C
os[e + f*x])^p*(b + a*Sin[e + f*x])^m)/Sin[e + f*x]^m, x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]

Rule 2911

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

Rule 2564

Int[cos[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(a*f), Subst[Int[
x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Sin[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2] &&
!(IntegerQ[(m - 1)/2] && LtQ[0, m, n])

Rule 364

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a^p*(c*x)^(m + 1)*Hypergeometric2F1[-
p, (m + 1)/n, (m + 1)/n + 1, -((b*x^n)/a)])/(c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 4398

Int[(u_.)*((a_)*(v_))^(p_), x_Symbol] :> With[{uu = ActivateTrig[u], vv = ActivateTrig[v]}, Dist[(a^IntPart[p]
*(a*vv)^FracPart[p])/vv^FracPart[p], Int[uu*vv^p, x], x]] /; FreeQ[{a, p}, x] &&  !IntegerQ[p] &&  !InertTrigF
reeQ[v]

Rule 4401

Int[u_, x_Symbol] :> With[{v = ExpandTrig[u, x]}, Int[v, x] /; SumQ[v]] /;  !InertTrigFreeQ[u]

Rule 2643

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(Cos[c + d*x]*(b*Sin[c + d*x])^(n + 1)*Hypergeomet
ric2F1[1/2, (n + 1)/2, (n + 3)/2, Sin[c + d*x]^2])/(b*d*(n + 1)*Sqrt[Cos[c + d*x]^2]), x] /; FreeQ[{b, c, d, n
}, x] &&  !IntegerQ[2*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 (a+b \sec (c+d x))^2 (e \sin (c+d x))^m \, dx &=\int (-b-a \cos (c+d x))^2 \sec ^2(c+d x) (e \sin (c+d x))^m \, dx\\ &=(2 a b) \int \sec (c+d x) (e \sin (c+d x))^m \, dx+\int \left (b^2+a^2 \cos ^2(c+d x)\right ) \sec ^2(c+d x) (e \sin (c+d x))^m \, dx\\ &=\frac{(2 a b) \operatorname{Subst}\left (\int \frac{x^m}{1-\frac{x^2}{e^2}} \, dx,x,e \sin (c+d x)\right )}{d e}+\left (\sin ^{-m}(c+d x) (e \sin (c+d x))^m\right ) \int \left (b^2+a^2 \cos ^2(c+d x)\right ) \sec ^2(c+d x) \sin ^m(c+d x) \, dx\\ &=\frac{2 a b \, _2F_1\left (1,\frac{1+m}{2};\frac{3+m}{2};\sin ^2(c+d x)\right ) (e \sin (c+d x))^{1+m}}{d e (1+m)}+\left (\sin ^{-m}(c+d x) (e \sin (c+d x))^m\right ) \int \left (a^2 \sin ^m(c+d x)+b^2 \sec ^2(c+d x) \sin ^m(c+d x)\right ) \, dx\\ &=\frac{2 a b \, _2F_1\left (1,\frac{1+m}{2};\frac{3+m}{2};\sin ^2(c+d x)\right ) (e \sin (c+d x))^{1+m}}{d e (1+m)}+\left (a^2 \sin ^{-m}(c+d x) (e \sin (c+d x))^m\right ) \int \sin ^m(c+d x) \, dx+\left (b^2 \sin ^{-m}(c+d x) (e \sin (c+d x))^m\right ) \int \sec ^2(c+d x) \sin ^m(c+d x) \, dx\\ &=\frac{a^2 \cos (c+d x) \, _2F_1\left (\frac{1}{2},\frac{1+m}{2};\frac{3+m}{2};\sin ^2(c+d x)\right ) \sin (c+d x) (e \sin (c+d x))^m}{d (1+m) \sqrt{\cos ^2(c+d x)}}+\frac{2 a b \, _2F_1\left (1,\frac{1+m}{2};\frac{3+m}{2};\sin ^2(c+d x)\right ) (e \sin (c+d x))^{1+m}}{d e (1+m)}+\frac{b^2 \sqrt{\cos ^2(c+d x)} \, _2F_1\left (\frac{3}{2},\frac{1+m}{2};\frac{3+m}{2};\sin ^2(c+d x)\right ) (e \sin (c+d x))^m \tan (c+d x)}{d (1+m)}\\ \end{align*}

Mathematica [A]  time = 0.252578, size = 134, normalized size = 0.71 \[ \frac{(e \sin (c+d x))^m \left (\sqrt{\cos ^2(c+d x)} \tan (c+d x) \left (a^2 \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{m+1}{2},\frac{m+3}{2},\sin ^2(c+d x)\right )+b^2 \text{Hypergeometric2F1}\left (\frac{3}{2},\frac{m+1}{2},\frac{m+3}{2},\sin ^2(c+d x)\right )\right )+2 a b \sin (c+d x) \text{Hypergeometric2F1}\left (1,\frac{m+1}{2},\frac{m+3}{2},\sin ^2(c+d x)\right )\right )}{d (m+1)} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sec[c + d*x])^2*(e*Sin[c + d*x])^m,x]

[Out]

((e*Sin[c + d*x])^m*(2*a*b*Hypergeometric2F1[1, (1 + m)/2, (3 + m)/2, Sin[c + d*x]^2]*Sin[c + d*x] + Sqrt[Cos[
c + d*x]^2]*(a^2*Hypergeometric2F1[1/2, (1 + m)/2, (3 + m)/2, Sin[c + d*x]^2] + b^2*Hypergeometric2F1[3/2, (1
+ m)/2, (3 + m)/2, Sin[c + d*x]^2])*Tan[c + d*x]))/(d*(1 + m))

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Maple [F]  time = 1.267, size = 0, normalized size = 0. \begin{align*} \int \left ( a+b\sec \left ( dx+c \right ) \right ) ^{2} \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+b*sec(d*x+c))^2*(e*sin(d*x+c))^m,x)

[Out]

int((a+b*sec(d*x+c))^2*(e*sin(d*x+c))^m,x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \sec \left (d x + c\right ) + a\right )}^{2} \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+b*sec(d*x+c))^2*(e*sin(d*x+c))^m,x, algorithm="maxima")

[Out]

integrate((b*sec(d*x + c) + a)^2*(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 (b^{2} \sec \left (d x + c\right )^{2} + 2 \, a b \sec \left (d x + c\right ) + a^{2}\right )} \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+b*sec(d*x+c))^2*(e*sin(d*x+c))^m,x, algorithm="fricas")

[Out]

integral((b^2*sec(d*x + c)^2 + 2*a*b*sec(d*x + c) + a^2)*(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+b*sec(d*x+c))**2*(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 (b \sec \left (d x + c\right ) + a\right )}^{2} \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+b*sec(d*x+c))^2*(e*sin(d*x+c))^m,x, algorithm="giac")

[Out]

integrate((b*sec(d*x + c) + a)^2*(e*sin(d*x + c))^m, x)