3.357 \(\int \frac {(a+a \sin (c+d x))^m}{\sqrt {e \cos (c+d x)}} \, dx\)

Optimal. Leaf size=86 \[ -\frac {a 2^{m+\frac {5}{4}} \sqrt {e \cos (c+d x)} (\sin (c+d x)+1)^{\frac {3}{4}-m} (a \sin (c+d x)+a)^{m-1} \, _2F_1\left (\frac {1}{4},\frac {3}{4}-m;\frac {5}{4};\frac {1}{2} (1-\sin (c+d x))\right )}{d e} \]

[Out]

-2^(5/4+m)*a*hypergeom([1/4, 3/4-m],[5/4],1/2-1/2*sin(d*x+c))*(1+sin(d*x+c))^(3/4-m)*(a+a*sin(d*x+c))^(-1+m)*(
e*cos(d*x+c))^(1/2)/d/e

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Rubi [A]  time = 0.09, antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {2689, 70, 69} \[ -\frac {a 2^{m+\frac {5}{4}} \sqrt {e \cos (c+d x)} (\sin (c+d x)+1)^{\frac {3}{4}-m} (a \sin (c+d x)+a)^{m-1} \, _2F_1\left (\frac {1}{4},\frac {3}{4}-m;\frac {5}{4};\frac {1}{2} (1-\sin (c+d x))\right )}{d e} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sin[c + d*x])^m/Sqrt[e*Cos[c + d*x]],x]

[Out]

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

Rule 69

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*Hypergeometric2F1[
-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b*(m + 1)*(b/(b*c - a*d))^n), x] /; FreeQ[{a, b, c, d, m, n}
, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] ||  !(Ra
tionalQ[n] && GtQ[-(d/(b*c - a*d)), 0]))

Rule 70

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Dist[(c + d*x)^FracPart[n]/((b/(b*c - a*d)
)^IntPart[n]*((b*(c + d*x))/(b*c - a*d))^FracPart[n]), Int[(a + b*x)^m*Simp[(b*c)/(b*c - a*d) + (b*d*x)/(b*c -
a*d), x]^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] &&
(RationalQ[m] ||  !SimplerQ[n + 1, m + 1])

Rule 2689

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

Rubi steps

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

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Mathematica [A]  time = 0.08, size = 83, normalized size = 0.97 \[ -\frac {2^{m+\frac {5}{4}} \sqrt {e \cos (c+d x)} (\sin (c+d x)+1)^{-m-\frac {1}{4}} (a (\sin (c+d x)+1))^m \, _2F_1\left (\frac {1}{4},\frac {3}{4}-m;\frac {5}{4};\frac {1}{2} (1-\sin (c+d x))\right )}{d e} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sin[c + d*x])^m/Sqrt[e*Cos[c + d*x]],x]

[Out]

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

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fricas [F]  time = 0.47, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {e \cos \left (d x + c\right )} {\left (a \sin \left (d x + c\right ) + a\right )}^{m}}{e \cos \left (d x + c\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(d*x+c))^m/(e*cos(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

integral(sqrt(e*cos(d*x + c))*(a*sin(d*x + c) + a)^m/(e*cos(d*x + c)), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (a \sin \left (d x + c\right ) + a\right )}^{m}}{\sqrt {e \cos \left (d x + c\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(d*x+c))^m/(e*cos(d*x+c))^(1/2),x, algorithm="giac")

[Out]

integrate((a*sin(d*x + c) + a)^m/sqrt(e*cos(d*x + c)), x)

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maple [F]  time = 0.19, size = 0, normalized size = 0.00 \[ \int \frac {\left (a +a \sin \left (d x +c \right )\right )^{m}}{\sqrt {e \cos \left (d x +c \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sin(d*x+c))^m/(e*cos(d*x+c))^(1/2),x)

[Out]

int((a+a*sin(d*x+c))^m/(e*cos(d*x+c))^(1/2),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (a \sin \left (d x + c\right ) + a\right )}^{m}}{\sqrt {e \cos \left (d x + c\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(d*x+c))^m/(e*cos(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

integrate((a*sin(d*x + c) + a)^m/sqrt(e*cos(d*x + c)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (a+a\,\sin \left (c+d\,x\right )\right )}^m}{\sqrt {e\,\cos \left (c+d\,x\right )}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*sin(c + d*x))^m/(e*cos(c + d*x))^(1/2),x)

[Out]

int((a + a*sin(c + d*x))^m/(e*cos(c + d*x))^(1/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a \left (\sin {\left (c + d x \right )} + 1\right )\right )^{m}}{\sqrt {e \cos {\left (c + d x \right )}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(d*x+c))**m/(e*cos(d*x+c))**(1/2),x)

[Out]

Integral((a*(sin(c + d*x) + 1))**m/sqrt(e*cos(c + d*x)), x)

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