3.9.35 \(\int \frac {(c (d \sin (e+f x))^p)^n}{a+b \sin (e+f x)} \, dx\) [835]

Optimal. Leaf size=204 \[ \frac {b F_1\left (\frac {1}{2};-\frac {n p}{2},1;\frac {3}{2};\cos ^2(e+f x),-\frac {b^2 \cos ^2(e+f x)}{a^2-b^2}\right ) \cos (e+f x) \sin ^2(e+f x)^{-\frac {n p}{2}} \left (c (d \sin (e+f x))^p\right )^n}{\left (a^2-b^2\right ) f}-\frac {a F_1\left (\frac {1}{2};\frac {1}{2} (1-n p),1;\frac {3}{2};\cos ^2(e+f x),-\frac {b^2 \cos ^2(e+f x)}{a^2-b^2}\right ) \cot (e+f x) \sin ^2(e+f x)^{\frac {1}{2} (1-n p)} \left (c (d \sin (e+f x))^p\right )^n}{\left (a^2-b^2\right ) f} \]

[Out]

b*AppellF1(1/2,-1/2*n*p,1,3/2,cos(f*x+e)^2,-b^2*cos(f*x+e)^2/(a^2-b^2))*cos(f*x+e)*(c*(d*sin(f*x+e))^p)^n/(a^2
-b^2)/f/((sin(f*x+e)^2)^(1/2*n*p))-a*AppellF1(1/2,-1/2*n*p+1/2,1,3/2,cos(f*x+e)^2,-b^2*cos(f*x+e)^2/(a^2-b^2))
*cot(f*x+e)*(sin(f*x+e)^2)^(-1/2*n*p+1/2)*(c*(d*sin(f*x+e))^p)^n/(a^2-b^2)/f

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Rubi [A]
time = 0.23, antiderivative size = 204, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {2905, 2902, 3268, 440} \begin {gather*} \frac {b \cos (e+f x) \sin ^2(e+f x)^{-\frac {n p}{2}} \left (c (d \sin (e+f x))^p\right )^n F_1\left (\frac {1}{2};-\frac {n p}{2},1;\frac {3}{2};\cos ^2(e+f x),-\frac {b^2 \cos ^2(e+f x)}{a^2-b^2}\right )}{f \left (a^2-b^2\right )}-\frac {a \cot (e+f x) \sin ^2(e+f x)^{\frac {1}{2} (1-n p)} \left (c (d \sin (e+f x))^p\right )^n F_1\left (\frac {1}{2};\frac {1}{2} (1-n p),1;\frac {3}{2};\cos ^2(e+f x),-\frac {b^2 \cos ^2(e+f x)}{a^2-b^2}\right )}{f \left (a^2-b^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(c*(d*Sin[e + f*x])^p)^n/(a + b*Sin[e + f*x]),x]

[Out]

(b*AppellF1[1/2, -1/2*(n*p), 1, 3/2, Cos[e + f*x]^2, -((b^2*Cos[e + f*x]^2)/(a^2 - b^2))]*Cos[e + f*x]*(c*(d*S
in[e + f*x])^p)^n)/((a^2 - b^2)*f*(Sin[e + f*x]^2)^((n*p)/2)) - (a*AppellF1[1/2, (1 - n*p)/2, 1, 3/2, Cos[e +
f*x]^2, -((b^2*Cos[e + f*x]^2)/(a^2 - b^2))]*Cot[e + f*x]*(Sin[e + f*x]^2)^((1 - n*p)/2)*(c*(d*Sin[e + f*x])^p
)^n)/((a^2 - b^2)*f)

Rule 440

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])

Rule 2902

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

Rule 2905

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

Rule 3268

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

Rubi steps

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

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(1808\) vs. \(2(204)=408\).
time = 15.73, size = 1808, normalized size = 8.86 \begin {gather*} \frac {\sec ^2(e+f x)^{\frac {n p}{2}} \left (c (d \sin (e+f x))^p\right )^n \tan (e+f x) \left (\frac {\tan (e+f x)}{\sqrt {\sec ^2(e+f x)}}\right )^{n p} \left (\left (a^2-b^2\right ) (1+n p) F_1\left (1+\frac {n p}{2};\frac {1}{2} (-1+n p),1;2+\frac {n p}{2};-\tan ^2(e+f x),\left (-1+\frac {b^2}{a^2}\right ) \tan ^2(e+f x)\right ) \tan (e+f x)+a \left (b (2+n p) F_1\left (\frac {1}{2} (1+n p);\frac {n p}{2},1;\frac {1}{2} (3+n p);-\tan ^2(e+f x),\frac {\left (-a^2+b^2\right ) \tan ^2(e+f x)}{a^2}\right )-a (1+n p) \, _2F_1\left (1+\frac {n p}{2},\frac {1}{2} (1+n p);2+\frac {n p}{2};-\tan ^2(e+f x)\right ) \tan (e+f x)\right )\right )}{a^2 b f (1+n p) (2+n p) (a+b \sin (e+f x)) \left (\frac {\sec ^2(e+f x)^{1+\frac {n p}{2}} \left (\frac {\tan (e+f x)}{\sqrt {\sec ^2(e+f x)}}\right )^{n p} \left (\left (a^2-b^2\right ) (1+n p) F_1\left (1+\frac {n p}{2};\frac {1}{2} (-1+n p),1;2+\frac {n p}{2};-\tan ^2(e+f x),\left (-1+\frac {b^2}{a^2}\right ) \tan ^2(e+f x)\right ) \tan (e+f x)+a \left (b (2+n p) F_1\left (\frac {1}{2} (1+n p);\frac {n p}{2},1;\frac {1}{2} (3+n p);-\tan ^2(e+f x),\frac {\left (-a^2+b^2\right ) \tan ^2(e+f x)}{a^2}\right )-a (1+n p) \, _2F_1\left (1+\frac {n p}{2},\frac {1}{2} (1+n p);2+\frac {n p}{2};-\tan ^2(e+f x)\right ) \tan (e+f x)\right )\right )}{a^2 b (1+n p) (2+n p)}+\frac {n p \sec ^2(e+f x)^{\frac {n p}{2}} \tan ^2(e+f x) \left (\frac {\tan (e+f x)}{\sqrt {\sec ^2(e+f x)}}\right )^{n p} \left (\left (a^2-b^2\right ) (1+n p) F_1\left (1+\frac {n p}{2};\frac {1}{2} (-1+n p),1;2+\frac {n p}{2};-\tan ^2(e+f x),\left (-1+\frac {b^2}{a^2}\right ) \tan ^2(e+f x)\right ) \tan (e+f x)+a \left (b (2+n p) F_1\left (\frac {1}{2} (1+n p);\frac {n p}{2},1;\frac {1}{2} (3+n p);-\tan ^2(e+f x),\frac {\left (-a^2+b^2\right ) \tan ^2(e+f x)}{a^2}\right )-a (1+n p) \, _2F_1\left (1+\frac {n p}{2},\frac {1}{2} (1+n p);2+\frac {n p}{2};-\tan ^2(e+f x)\right ) \tan (e+f x)\right )\right )}{a^2 b (1+n p) (2+n p)}+\frac {n p \sec ^2(e+f x)^{\frac {n p}{2}} \tan (e+f x) \left (\frac {\tan (e+f x)}{\sqrt {\sec ^2(e+f x)}}\right )^{-1+n p} \left (\sqrt {\sec ^2(e+f x)}-\frac {\tan ^2(e+f x)}{\sqrt {\sec ^2(e+f x)}}\right ) \left (\left (a^2-b^2\right ) (1+n p) F_1\left (1+\frac {n p}{2};\frac {1}{2} (-1+n p),1;2+\frac {n p}{2};-\tan ^2(e+f x),\left (-1+\frac {b^2}{a^2}\right ) \tan ^2(e+f x)\right ) \tan (e+f x)+a \left (b (2+n p) F_1\left (\frac {1}{2} (1+n p);\frac {n p}{2},1;\frac {1}{2} (3+n p);-\tan ^2(e+f x),\frac {\left (-a^2+b^2\right ) \tan ^2(e+f x)}{a^2}\right )-a (1+n p) \, _2F_1\left (1+\frac {n p}{2},\frac {1}{2} (1+n p);2+\frac {n p}{2};-\tan ^2(e+f x)\right ) \tan (e+f x)\right )\right )}{a^2 b (1+n p) (2+n p)}+\frac {\sec ^2(e+f x)^{\frac {n p}{2}} \tan (e+f x) \left (\frac {\tan (e+f x)}{\sqrt {\sec ^2(e+f x)}}\right )^{n p} \left (\left (a^2-b^2\right ) (1+n p) F_1\left (1+\frac {n p}{2};\frac {1}{2} (-1+n p),1;2+\frac {n p}{2};-\tan ^2(e+f x),\left (-1+\frac {b^2}{a^2}\right ) \tan ^2(e+f x)\right ) \sec ^2(e+f x)+\left (a^2-b^2\right ) (1+n p) \tan (e+f x) \left (\frac {2 \left (-1+\frac {b^2}{a^2}\right ) \left (1+\frac {n p}{2}\right ) F_1\left (2+\frac {n p}{2};\frac {1}{2} (-1+n p),2;3+\frac {n p}{2};-\tan ^2(e+f x),\left (-1+\frac {b^2}{a^2}\right ) \tan ^2(e+f x)\right ) \sec ^2(e+f x) \tan (e+f x)}{2+\frac {n p}{2}}-\frac {\left (1+\frac {n p}{2}\right ) (-1+n p) F_1\left (2+\frac {n p}{2};1+\frac {1}{2} (-1+n p),1;3+\frac {n p}{2};-\tan ^2(e+f x),\left (-1+\frac {b^2}{a^2}\right ) \tan ^2(e+f x)\right ) \sec ^2(e+f x) \tan (e+f x)}{2+\frac {n p}{2}}\right )+a \left (-a (1+n p) \, _2F_1\left (1+\frac {n p}{2},\frac {1}{2} (1+n p);2+\frac {n p}{2};-\tan ^2(e+f x)\right ) \sec ^2(e+f x)+b (2+n p) \left (\frac {2 \left (-a^2+b^2\right ) (1+n p) F_1\left (1+\frac {1}{2} (1+n p);\frac {n p}{2},2;1+\frac {1}{2} (3+n p);-\tan ^2(e+f x),\frac {\left (-a^2+b^2\right ) \tan ^2(e+f x)}{a^2}\right ) \sec ^2(e+f x) \tan (e+f x)}{a^2 (3+n p)}-\frac {n p (1+n p) F_1\left (1+\frac {1}{2} (1+n p);1+\frac {n p}{2},1;1+\frac {1}{2} (3+n p);-\tan ^2(e+f x),\frac {\left (-a^2+b^2\right ) \tan ^2(e+f x)}{a^2}\right ) \sec ^2(e+f x) \tan (e+f x)}{3+n p}\right )-2 a \left (1+\frac {n p}{2}\right ) (1+n p) \sec ^2(e+f x) \left (-\, _2F_1\left (1+\frac {n p}{2},\frac {1}{2} (1+n p);2+\frac {n p}{2};-\tan ^2(e+f x)\right )+\left (1+\tan ^2(e+f x)\right )^{\frac {1}{2} (-1-n p)}\right )\right )\right )}{a^2 b (1+n p) (2+n p)}\right )} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[(c*(d*Sin[e + f*x])^p)^n/(a + b*Sin[e + f*x]),x]

[Out]

((Sec[e + f*x]^2)^((n*p)/2)*(c*(d*Sin[e + f*x])^p)^n*Tan[e + f*x]*(Tan[e + f*x]/Sqrt[Sec[e + f*x]^2])^(n*p)*((
a^2 - b^2)*(1 + n*p)*AppellF1[1 + (n*p)/2, (-1 + n*p)/2, 1, 2 + (n*p)/2, -Tan[e + f*x]^2, (-1 + b^2/a^2)*Tan[e
+ f*x]^2]*Tan[e + f*x] + a*(b*(2 + n*p)*AppellF1[(1 + n*p)/2, (n*p)/2, 1, (3 + n*p)/2, -Tan[e + f*x]^2, ((-a^
2 + b^2)*Tan[e + f*x]^2)/a^2] - a*(1 + n*p)*Hypergeometric2F1[1 + (n*p)/2, (1 + n*p)/2, 2 + (n*p)/2, -Tan[e +
f*x]^2]*Tan[e + f*x])))/(a^2*b*f*(1 + n*p)*(2 + n*p)*(a + b*Sin[e + f*x])*(((Sec[e + f*x]^2)^(1 + (n*p)/2)*(Ta
n[e + f*x]/Sqrt[Sec[e + f*x]^2])^(n*p)*((a^2 - b^2)*(1 + n*p)*AppellF1[1 + (n*p)/2, (-1 + n*p)/2, 1, 2 + (n*p)
/2, -Tan[e + f*x]^2, (-1 + b^2/a^2)*Tan[e + f*x]^2]*Tan[e + f*x] + a*(b*(2 + n*p)*AppellF1[(1 + n*p)/2, (n*p)/
2, 1, (3 + n*p)/2, -Tan[e + f*x]^2, ((-a^2 + b^2)*Tan[e + f*x]^2)/a^2] - a*(1 + n*p)*Hypergeometric2F1[1 + (n*
p)/2, (1 + n*p)/2, 2 + (n*p)/2, -Tan[e + f*x]^2]*Tan[e + f*x])))/(a^2*b*(1 + n*p)*(2 + n*p)) + (n*p*(Sec[e + f
*x]^2)^((n*p)/2)*Tan[e + f*x]^2*(Tan[e + f*x]/Sqrt[Sec[e + f*x]^2])^(n*p)*((a^2 - b^2)*(1 + n*p)*AppellF1[1 +
(n*p)/2, (-1 + n*p)/2, 1, 2 + (n*p)/2, -Tan[e + f*x]^2, (-1 + b^2/a^2)*Tan[e + f*x]^2]*Tan[e + f*x] + a*(b*(2
+ n*p)*AppellF1[(1 + n*p)/2, (n*p)/2, 1, (3 + n*p)/2, -Tan[e + f*x]^2, ((-a^2 + b^2)*Tan[e + f*x]^2)/a^2] - a*
(1 + n*p)*Hypergeometric2F1[1 + (n*p)/2, (1 + n*p)/2, 2 + (n*p)/2, -Tan[e + f*x]^2]*Tan[e + f*x])))/(a^2*b*(1
+ n*p)*(2 + n*p)) + (n*p*(Sec[e + f*x]^2)^((n*p)/2)*Tan[e + f*x]*(Tan[e + f*x]/Sqrt[Sec[e + f*x]^2])^(-1 + n*p
)*(Sqrt[Sec[e + f*x]^2] - Tan[e + f*x]^2/Sqrt[Sec[e + f*x]^2])*((a^2 - b^2)*(1 + n*p)*AppellF1[1 + (n*p)/2, (-
1 + n*p)/2, 1, 2 + (n*p)/2, -Tan[e + f*x]^2, (-1 + b^2/a^2)*Tan[e + f*x]^2]*Tan[e + f*x] + a*(b*(2 + n*p)*Appe
llF1[(1 + n*p)/2, (n*p)/2, 1, (3 + n*p)/2, -Tan[e + f*x]^2, ((-a^2 + b^2)*Tan[e + f*x]^2)/a^2] - a*(1 + n*p)*H
ypergeometric2F1[1 + (n*p)/2, (1 + n*p)/2, 2 + (n*p)/2, -Tan[e + f*x]^2]*Tan[e + f*x])))/(a^2*b*(1 + n*p)*(2 +
n*p)) + ((Sec[e + f*x]^2)^((n*p)/2)*Tan[e + f*x]*(Tan[e + f*x]/Sqrt[Sec[e + f*x]^2])^(n*p)*((a^2 - b^2)*(1 +
n*p)*AppellF1[1 + (n*p)/2, (-1 + n*p)/2, 1, 2 + (n*p)/2, -Tan[e + f*x]^2, (-1 + b^2/a^2)*Tan[e + f*x]^2]*Sec[e
+ f*x]^2 + (a^2 - b^2)*(1 + n*p)*Tan[e + f*x]*((2*(-1 + b^2/a^2)*(1 + (n*p)/2)*AppellF1[2 + (n*p)/2, (-1 + n*
p)/2, 2, 3 + (n*p)/2, -Tan[e + f*x]^2, (-1 + b^2/a^2)*Tan[e + f*x]^2]*Sec[e + f*x]^2*Tan[e + f*x])/(2 + (n*p)/
2) - ((1 + (n*p)/2)*(-1 + n*p)*AppellF1[2 + (n*p)/2, 1 + (-1 + n*p)/2, 1, 3 + (n*p)/2, -Tan[e + f*x]^2, (-1 +
b^2/a^2)*Tan[e + f*x]^2]*Sec[e + f*x]^2*Tan[e + f*x])/(2 + (n*p)/2)) + a*(-(a*(1 + n*p)*Hypergeometric2F1[1 +
(n*p)/2, (1 + n*p)/2, 2 + (n*p)/2, -Tan[e + f*x]^2]*Sec[e + f*x]^2) + b*(2 + n*p)*((2*(-a^2 + b^2)*(1 + n*p)*A
ppellF1[1 + (1 + n*p)/2, (n*p)/2, 2, 1 + (3 + n*p)/2, -Tan[e + f*x]^2, ((-a^2 + b^2)*Tan[e + f*x]^2)/a^2]*Sec[
e + f*x]^2*Tan[e + f*x])/(a^2*(3 + n*p)) - (n*p*(1 + n*p)*AppellF1[1 + (1 + n*p)/2, 1 + (n*p)/2, 1, 1 + (3 + n
*p)/2, -Tan[e + f*x]^2, ((-a^2 + b^2)*Tan[e + f*x]^2)/a^2]*Sec[e + f*x]^2*Tan[e + f*x])/(3 + n*p)) - 2*a*(1 +
(n*p)/2)*(1 + n*p)*Sec[e + f*x]^2*(-Hypergeometric2F1[1 + (n*p)/2, (1 + n*p)/2, 2 + (n*p)/2, -Tan[e + f*x]^2]
+ (1 + Tan[e + f*x]^2)^((-1 - n*p)/2)))))/(a^2*b*(1 + n*p)*(2 + n*p))))

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Maple [F]
time = 0.11, size = 0, normalized size = 0.00 \[\int \frac {\left (c \left (d \sin \left (f x +e \right )\right )^{p}\right )^{n}}{a +b \sin \left (f x +e \right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*(d*sin(f*x+e))^p)^n/(a+b*sin(f*x+e)),x)

[Out]

int((c*(d*sin(f*x+e))^p)^n/(a+b*sin(f*x+e)),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((c*(d*sin(f*x+e))^p)^n/(a+b*sin(f*x+e)),x, algorithm="maxima")

[Out]

integrate(((d*sin(f*x + e))^p*c)^n/(b*sin(f*x + e) + a), 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((c*(d*sin(f*x+e))^p)^n/(a+b*sin(f*x+e)),x, algorithm="fricas")

[Out]

integral(((d*sin(f*x + e))^p*c)^n/(b*sin(f*x + e) + a), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (c \left (d \sin {\left (e + f x \right )}\right )^{p}\right )^{n}}{a + b \sin {\left (e + f x \right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*(d*sin(f*x+e))**p)**n/(a+b*sin(f*x+e)),x)

[Out]

Integral((c*(d*sin(e + f*x))**p)**n/(a + b*sin(e + f*x)), 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((c*(d*sin(f*x+e))^p)^n/(a+b*sin(f*x+e)),x, algorithm="giac")

[Out]

integrate(((d*sin(f*x + e))^p*c)^n/(b*sin(f*x + e) + a), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*(d*sin(e + f*x))^p)^n/(a + b*sin(e + f*x)),x)

[Out]

int((c*(d*sin(e + f*x))^p)^n/(a + b*sin(e + f*x)), x)

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