3.358 \(\int \frac {(a+a \sin (c+d x))^m}{(e \cos (c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=82 \[ \frac {2^{m+\frac {3}{4}} (\sin (c+d x)+1)^{\frac {1}{4}-m} (a \sin (c+d x)+a)^m \, _2F_1\left (-\frac {1}{4},\frac {5}{4}-m;\frac {3}{4};\frac {1}{2} (1-\sin (c+d x))\right )}{d e \sqrt {e \cos (c+d x)}} \]

[Out]

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

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Rubi [A]  time = 0.10, antiderivative size = 82, 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} \[ \frac {2^{m+\frac {3}{4}} (\sin (c+d x)+1)^{\frac {1}{4}-m} (a \sin (c+d x)+a)^m \, _2F_1\left (-\frac {1}{4},\frac {5}{4}-m;\frac {3}{4};\frac {1}{2} (1-\sin (c+d x))\right )}{d e \sqrt {e \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sin[c + d*x])^m/(e*Cos[c + d*x])^(3/2),x]

[Out]

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

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

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Mathematica [A]  time = 0.10, size = 82, normalized size = 1.00 \[ \frac {2^{m+\frac {3}{4}} (\sin (c+d x)+1)^{\frac {1}{4}-m} (a (\sin (c+d x)+1))^m \, _2F_1\left (-\frac {1}{4},\frac {5}{4}-m;\frac {3}{4};\frac {1}{2} (1-\sin (c+d x))\right )}{d e \sqrt {e \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sin[c + d*x])^m/(e*Cos[c + d*x])^(3/2),x]

[Out]

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

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fricas [F]  time = 0.46, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {e \cos \left (d x + c\right )} {\left (a \sin \left (d x + c\right ) + a\right )}^{m}}{e^{2} \cos \left (d x + c\right )^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(d*x+c))^m/(e*cos(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (a \sin \left (d x + c\right ) + a\right )}^{m}}{\left (e \cos \left (d x + c\right )\right )^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(d*x+c))^m/(e*cos(d*x+c))^(3/2),x, algorithm="giac")

[Out]

integrate((a*sin(d*x + c) + a)^m/(e*cos(d*x + c))^(3/2), x)

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maple [F]  time = 0.18, size = 0, normalized size = 0.00 \[ \int \frac {\left (a +a \sin \left (d x +c \right )\right )^{m}}{\left (e \cos \left (d x +c \right )\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sin(d*x+c))^m/(e*cos(d*x+c))^(3/2),x)

[Out]

int((a+a*sin(d*x+c))^m/(e*cos(d*x+c))^(3/2),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (a \sin \left (d x + c\right ) + a\right )}^{m}}{\left (e \cos \left (d x + c\right )\right )^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(d*x+c))^m/(e*cos(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

integrate((a*sin(d*x + c) + a)^m/(e*cos(d*x + c))^(3/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (a+a\,\sin \left (c+d\,x\right )\right )}^m}{{\left (e\,\cos \left (c+d\,x\right )\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*sin(c + d*x))^m/(e*cos(c + d*x))^(3/2),x)

[Out]

int((a + a*sin(c + d*x))^m/(e*cos(c + d*x))^(3/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a \left (\sin {\left (c + d x \right )} + 1\right )\right )^{m}}{\left (e \cos {\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(d*x+c))**m/(e*cos(d*x+c))**(3/2),x)

[Out]

Integral((a*(sin(c + d*x) + 1))**m/(e*cos(c + d*x))**(3/2), x)

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