3.203 \(\int \frac{a+a \sin (c+d x)}{(e \cos (c+d x))^{7/2}} \, dx\)

Optimal. Leaf size=126 \[ \frac{6 a \sin (c+d x)}{5 d e^3 \sqrt{e \cos (c+d x)}}-\frac{6 a E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{e \cos (c+d x)}}{5 d e^4 \sqrt{\cos (c+d x)}}+\frac{2 a}{5 d e (e \cos (c+d x))^{5/2}}+\frac{2 a \sin (c+d x)}{5 d e (e \cos (c+d x))^{5/2}} \]

[Out]

(2*a)/(5*d*e*(e*Cos[c + d*x])^(5/2)) - (6*a*Sqrt[e*Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2])/(5*d*e^4*Sqrt[Cos[
c + d*x]]) + (2*a*Sin[c + d*x])/(5*d*e*(e*Cos[c + d*x])^(5/2)) + (6*a*Sin[c + d*x])/(5*d*e^3*Sqrt[e*Cos[c + d*
x]])

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Rubi [A]  time = 0.0863084, antiderivative size = 126, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {2669, 2636, 2640, 2639} \[ \frac{6 a \sin (c+d x)}{5 d e^3 \sqrt{e \cos (c+d x)}}-\frac{6 a E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{e \cos (c+d x)}}{5 d e^4 \sqrt{\cos (c+d x)}}+\frac{2 a}{5 d e (e \cos (c+d x))^{5/2}}+\frac{2 a \sin (c+d x)}{5 d e (e \cos (c+d x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sin[c + d*x])/(e*Cos[c + d*x])^(7/2),x]

[Out]

(2*a)/(5*d*e*(e*Cos[c + d*x])^(5/2)) - (6*a*Sqrt[e*Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2])/(5*d*e^4*Sqrt[Cos[
c + d*x]]) + (2*a*Sin[c + d*x])/(5*d*e*(e*Cos[c + d*x])^(5/2)) + (6*a*Sin[c + d*x])/(5*d*e^3*Sqrt[e*Cos[c + d*
x]])

Rule 2669

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

Rule 2636

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)), x] + Dist[(n + 2)/(b^2*(n + 1)), Int[(b*Sin[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1
] && IntegerQ[2*n]

Rule 2640

Int[Sqrt[(b_)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[b*Sin[c + d*x]]/Sqrt[Sin[c + d*x]], Int[Sqrt[Si
n[c + d*x]], x], x] /; FreeQ[{b, c, d}, x]

Rule 2639

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticE[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ[{
c, d}, x]

Rubi steps

\begin{align*} \int \frac{a+a \sin (c+d x)}{(e \cos (c+d x))^{7/2}} \, dx &=\frac{2 a}{5 d e (e \cos (c+d x))^{5/2}}+a \int \frac{1}{(e \cos (c+d x))^{7/2}} \, dx\\ &=\frac{2 a}{5 d e (e \cos (c+d x))^{5/2}}+\frac{2 a \sin (c+d x)}{5 d e (e \cos (c+d x))^{5/2}}+\frac{(3 a) \int \frac{1}{(e \cos (c+d x))^{3/2}} \, dx}{5 e^2}\\ &=\frac{2 a}{5 d e (e \cos (c+d x))^{5/2}}+\frac{2 a \sin (c+d x)}{5 d e (e \cos (c+d x))^{5/2}}+\frac{6 a \sin (c+d x)}{5 d e^3 \sqrt{e \cos (c+d x)}}-\frac{(3 a) \int \sqrt{e \cos (c+d x)} \, dx}{5 e^4}\\ &=\frac{2 a}{5 d e (e \cos (c+d x))^{5/2}}+\frac{2 a \sin (c+d x)}{5 d e (e \cos (c+d x))^{5/2}}+\frac{6 a \sin (c+d x)}{5 d e^3 \sqrt{e \cos (c+d x)}}-\frac{\left (3 a \sqrt{e \cos (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \, dx}{5 e^4 \sqrt{\cos (c+d x)}}\\ &=\frac{2 a}{5 d e (e \cos (c+d x))^{5/2}}-\frac{6 a \sqrt{e \cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d e^4 \sqrt{\cos (c+d x)}}+\frac{2 a \sin (c+d x)}{5 d e (e \cos (c+d x))^{5/2}}+\frac{6 a \sin (c+d x)}{5 d e^3 \sqrt{e \cos (c+d x)}}\\ \end{align*}

Mathematica [C]  time = 1.22272, size = 144, normalized size = 1.14 \[ \frac{2 a e^{i (c+d x)} \left (i \sqrt{1+e^{2 i (c+d x)}} \left (e^{i (c+d x)}-i\right )^2 \, _2F_1\left (\frac{1}{2},\frac{3}{4};\frac{7}{4};-e^{2 i (c+d x)}\right )-6 e^{i (c+d x)}-3 i e^{2 i (c+d x)}+i\right )}{5 d e^3 \left (e^{i (c+d x)}-i\right )^2 \sqrt{e \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sin[c + d*x])/(e*Cos[c + d*x])^(7/2),x]

[Out]

(2*a*E^(I*(c + d*x))*(I - 6*E^(I*(c + d*x)) - (3*I)*E^((2*I)*(c + d*x)) + I*(-I + E^(I*(c + d*x)))^2*Sqrt[1 +
E^((2*I)*(c + d*x))]*Hypergeometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))]))/(5*d*e^3*(-I + E^(I*(c + d*x)))^
2*Sqrt[e*Cos[c + d*x]])

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Maple [B]  time = 0.962, size = 304, normalized size = 2.4 \begin{align*} -{\frac{2\,a}{5\,d{e}^{3}} \left ( 12\,{\it EllipticE} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) \sqrt{2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}} \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}-24\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{6}\cos \left ( 1/2\,dx+c/2 \right ) -12\,{\it EllipticE} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) \sqrt{2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}} \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+24\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}\cos \left ( 1/2\,dx+c/2 \right ) +3\,\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}{\it EllipticE} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) -8\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}\cos \left ( 1/2\,dx+c/2 \right ) -\sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) \left ( 4\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}-4\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+1 \right ) ^{-1} \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}{\frac{1}{\sqrt{-2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}e+e}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sin(d*x+c))/(e*cos(d*x+c))^(7/2),x)

[Out]

-2/5/(4*sin(1/2*d*x+1/2*c)^4-4*sin(1/2*d*x+1/2*c)^2+1)/sin(1/2*d*x+1/2*c)/(-2*sin(1/2*d*x+1/2*c)^2*e+e)^(1/2)/
e^3*(12*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*si
n(1/2*d*x+1/2*c)^4-24*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)-12*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(
1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*sin(1/2*d*x+1/2*c)^2+24*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*
x+1/2*c)+3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))
-8*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-sin(1/2*d*x+1/2*c))*a/d

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{a \sin \left (d x + c\right ) + a}{\left (e \cos \left (d x + c\right )\right )^{\frac{7}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(d*x+c))/(e*cos(d*x+c))^(7/2),x, algorithm="maxima")

[Out]

integrate((a*sin(d*x + c) + a)/(e*cos(d*x + c))^(7/2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{e \cos \left (d x + c\right )}{\left (a \sin \left (d x + c\right ) + a\right )}}{e^{4} \cos \left (d x + c\right )^{4}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(d*x+c))/(e*cos(d*x+c))^(7/2),x, algorithm="fricas")

[Out]

integral(sqrt(e*cos(d*x + c))*(a*sin(d*x + c) + a)/(e^4*cos(d*x + c)^4), 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((a+a*sin(d*x+c))/(e*cos(d*x+c))**(7/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{a \sin \left (d x + c\right ) + a}{\left (e \cos \left (d x + c\right )\right )^{\frac{7}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(d*x+c))/(e*cos(d*x+c))^(7/2),x, algorithm="giac")

[Out]

integrate((a*sin(d*x + c) + a)/(e*cos(d*x + c))^(7/2), x)