3.3.58 \(\int (a+b \sec (c+d x))^2 (e \sin (c+d x))^m \, dx\) [258]

Optimal. Leaf size=190 \[ \frac {a^2 \cos (c+d x) \, _2F_1\left (\frac {1}{2},\frac {1+m}{2};\frac {3+m}{2};\sin ^2(c+d x)\right ) \sin (c+d x) (e \sin (c+d x))^m}{d (1+m) \sqrt {\cos ^2(c+d x)}}+\frac {2 a b \, _2F_1\left (1,\frac {1+m}{2};\frac {3+m}{2};\sin ^2(c+d x)\right ) (e \sin (c+d x))^{1+m}}{d e (1+m)}+\frac {b^2 \sqrt {\cos ^2(c+d x)} \, _2F_1\left (\frac {3}{2},\frac {1+m}{2};\frac {3+m}{2};\sin ^2(c+d x)\right ) (e \sin (c+d x))^m \tan (c+d x)}{d (1+m)} \]

[Out]

2*a*b*hypergeom([1, 1/2+1/2*m],[3/2+1/2*m],sin(d*x+c)^2)*(e*sin(d*x+c))^(1+m)/d/e/(1+m)+a^2*cos(d*x+c)*hyperge
om([1/2, 1/2+1/2*m],[3/2+1/2*m],sin(d*x+c)^2)*sin(d*x+c)*(e*sin(d*x+c))^m/d/(1+m)/(cos(d*x+c)^2)^(1/2)+b^2*hyp
ergeom([3/2, 1/2+1/2*m],[3/2+1/2*m],sin(d*x+c)^2)*(e*sin(d*x+c))^m*(cos(d*x+c)^2)^(1/2)*tan(d*x+c)/d/(1+m)

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Rubi [A]
time = 0.60, antiderivative size = 190, normalized size of antiderivative = 1.00, 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 = {3957, 2990, 2644, 371, 4483, 4486, 2722, 2657} \begin {gather*} \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)} \end {gather*}

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 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 2644

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 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 2722

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]*((b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1
)*Sqrt[Cos[c + d*x]^2]))*Hypergeometric2F1[1/2, (n + 1)/2, (n + 3)/2, Sin[c + d*x]^2], x] /; FreeQ[{b, c, d, n
}, x] &&  !IntegerQ[2*n]

Rule 2990

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 3957

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[(g*Co
s[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 4483

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 4486

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

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) \text {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*}

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Mathematica [A]
time = 0.17, size = 134, normalized size = 0.71 \begin {gather*} \frac {(e \sin (c+d x))^m \left (2 a b \, _2F_1\left (1,\frac {1+m}{2};\frac {3+m}{2};\sin ^2(c+d x)\right ) \sin (c+d x)+\sqrt {\cos ^2(c+d x)} \left (a^2 \, _2F_1\left (\frac {1}{2},\frac {1+m}{2};\frac {3+m}{2};\sin ^2(c+d x)\right )+b^2 \, _2F_1\left (\frac {3}{2},\frac {1+m}{2};\frac {3+m}{2};\sin ^2(c+d x)\right )\right ) \tan (c+d x)\right )}{d (1+m)} \end {gather*}

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 = 0.12, size = 0, normalized size = 0.00 \[\int \left (a +b \sec \left (d x +c \right )\right )^{2} \left (e \sin \left (d x +c \right )\right )^{m}\, dx\]

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 = 0.00, 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((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 = 0.00, 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((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]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (e \sin {\left (c + d x \right )}\right )^{m} \left (a + b \sec {\left (c + d x \right )}\right )^{2}\, dx \end {gather*}

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]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\left (e\,\sin \left (c+d\,x\right )\right )}^m\,{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^2 \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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