3.80 \(\int \sin ^m(c+d x) (a+b \tan (c+d x))^2 \, dx\)

Optimal. Leaf size=179 \[ \frac{a^2 \cos (c+d x) \sin ^{m+1}(c+d x) \, _2F_1\left (\frac{1}{2},\frac{m+1}{2};\frac{m+3}{2};\sin ^2(c+d x)\right )}{d (m+1) \sqrt{\cos ^2(c+d x)}}+\frac{2 a b \sin ^{m+2}(c+d x) \, _2F_1\left (1,\frac{m+2}{2};\frac{m+4}{2};\sin ^2(c+d x)\right )}{d (m+2)}+\frac{b^2 \sqrt{\cos ^2(c+d x)} \sec (c+d x) \sin ^{m+3}(c+d x) \, _2F_1\left (\frac{3}{2},\frac{m+3}{2};\frac{m+5}{2};\sin ^2(c+d x)\right )}{d (m+3)} \]

[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]^(1 + m))/(d*(1 + m
)*Sqrt[Cos[c + d*x]^2]) + (2*a*b*Hypergeometric2F1[1, (2 + m)/2, (4 + m)/2, Sin[c + d*x]^2]*Sin[c + d*x]^(2 +
m))/(d*(2 + m)) + (b^2*Sqrt[Cos[c + d*x]^2]*Hypergeometric2F1[3/2, (3 + m)/2, (5 + m)/2, Sin[c + d*x]^2]*Sec[c
+ d*x]*Sin[c + d*x]^(3 + m))/(d*(3 + m))

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Rubi [A]  time = 0.268018, antiderivative size = 179, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {4401, 2643, 2564, 364, 2577} \[ \frac{a^2 \cos (c+d x) \sin ^{m+1}(c+d x) \, _2F_1\left (\frac{1}{2},\frac{m+1}{2};\frac{m+3}{2};\sin ^2(c+d x)\right )}{d (m+1) \sqrt{\cos ^2(c+d x)}}+\frac{2 a b \sin ^{m+2}(c+d x) \, _2F_1\left (1,\frac{m+2}{2};\frac{m+4}{2};\sin ^2(c+d x)\right )}{d (m+2)}+\frac{b^2 \sqrt{\cos ^2(c+d x)} \sec (c+d x) \sin ^{m+3}(c+d x) \, _2F_1\left (\frac{3}{2},\frac{m+3}{2};\frac{m+5}{2};\sin ^2(c+d x)\right )}{d (m+3)} \]

Antiderivative was successfully verified.

[In]

Int[Sin[c + d*x]^m*(a + b*Tan[c + d*x])^2,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]^(1 + m))/(d*(1 + m
)*Sqrt[Cos[c + d*x]^2]) + (2*a*b*Hypergeometric2F1[1, (2 + m)/2, (4 + m)/2, Sin[c + d*x]^2]*Sin[c + d*x]^(2 +
m))/(d*(2 + m)) + (b^2*Sqrt[Cos[c + d*x]^2]*Hypergeometric2F1[3/2, (3 + m)/2, (5 + m)/2, Sin[c + d*x]^2]*Sec[c
+ d*x]*Sin[c + d*x]^(3 + m))/(d*(3 + m))

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 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 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 \sin ^m(c+d x) (a+b \tan (c+d x))^2 \, dx &=\int \left (a^2 \sin ^m(c+d x)+2 a b \sec (c+d x) \sin ^{1+m}(c+d x)+b^2 \sec ^2(c+d x) \sin ^{2+m}(c+d x)\right ) \, dx\\ &=a^2 \int \sin ^m(c+d x) \, dx+(2 a b) \int \sec (c+d x) \sin ^{1+m}(c+d x) \, dx+b^2 \int \sec ^2(c+d x) \sin ^{2+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 ^{1+m}(c+d x)}{d (1+m) \sqrt{\cos ^2(c+d x)}}+\frac{b^2 \sqrt{\cos ^2(c+d x)} \, _2F_1\left (\frac{3}{2},\frac{3+m}{2};\frac{5+m}{2};\sin ^2(c+d x)\right ) \sec (c+d x) \sin ^{3+m}(c+d x)}{d (3+m)}+\frac{(2 a b) \operatorname{Subst}\left (\int \frac{x^{1+m}}{1-x^2} \, dx,x,\sin (c+d x)\right )}{d}\\ &=\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 ^{1+m}(c+d x)}{d (1+m) \sqrt{\cos ^2(c+d x)}}+\frac{2 a b \, _2F_1\left (1,\frac{2+m}{2};\frac{4+m}{2};\sin ^2(c+d x)\right ) \sin ^{2+m}(c+d x)}{d (2+m)}+\frac{b^2 \sqrt{\cos ^2(c+d x)} \, _2F_1\left (\frac{3}{2},\frac{3+m}{2};\frac{5+m}{2};\sin ^2(c+d x)\right ) \sec (c+d x) \sin ^{3+m}(c+d x)}{d (3+m)}\\ \end{align*}

Mathematica [A]  time = 1.19385, size = 166, normalized size = 0.93 \[ \frac{\sin ^{m+1}(c+d x) \left (\frac{a^2 \sqrt{\cos ^2(c+d x)} \sec (c+d x) \, _2F_1\left (\frac{1}{2},\frac{m+1}{2};\frac{m+3}{2};\sin ^2(c+d x)\right )}{m+1}+\frac{b \sin (c+d x) \left (2 a (m+3) \, _2F_1\left (1,\frac{m+2}{2};\frac{m+4}{2};\sin ^2(c+d x)\right )+b (m+2) \sqrt{\cos ^2(c+d x)} \tan (c+d x) \, _2F_1\left (\frac{3}{2},\frac{m+3}{2};\frac{m+5}{2};\sin ^2(c+d x)\right )\right )}{(m+2) (m+3)}\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sin[c + d*x]^(1 + m)*((a^2*Sqrt[Cos[c + d*x]^2]*Hypergeometric2F1[1/2, (1 + m)/2, (3 + m)/2, Sin[c + d*x]^2]*
Sec[c + d*x])/(1 + m) + (b*Sin[c + d*x]*(2*a*(3 + m)*Hypergeometric2F1[1, (2 + m)/2, (4 + m)/2, Sin[c + d*x]^2
] + b*(2 + m)*Sqrt[Cos[c + d*x]^2]*Hypergeometric2F1[3/2, (3 + m)/2, (5 + m)/2, Sin[c + d*x]^2]*Tan[c + d*x]))
/((2 + m)*(3 + m))))/d

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Maple [F]  time = 0.248, size = 0, normalized size = 0. \begin{align*} \int \left ( \sin \left ( dx+c \right ) \right ) ^{m} \left ( a+b\tan \left ( dx+c \right ) \right ) ^{2}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^m*(a+b*tan(d*x+c))^2,x, algorithm="maxima")

[Out]

integrate((b*tan(d*x + c) + a)^2*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} \tan \left (d x + c\right )^{2} + 2 \, a b \tan \left (d x + c\right ) + a^{2}\right )} \sin \left (d x + c\right )^{m}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^m*(a+b*tan(d*x+c))^2,x, algorithm="fricas")

[Out]

integral((b^2*tan(d*x + c)^2 + 2*a*b*tan(d*x + c) + a^2)*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(sin(d*x+c)**m*(a+b*tan(d*x+c))**2,x)

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^m*(a+b*tan(d*x+c))^2,x, algorithm="giac")

[Out]

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