3.193 \(\int \frac{\sec ^2(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=167 \[ \frac{35 \sec (c+d x)}{96 a^2 d \sqrt{a \sin (c+d x)+a}}-\frac{35 \tanh ^{-1}\left (\frac{\sqrt{a} \cos (c+d x)}{\sqrt{2} \sqrt{a \sin (c+d x)+a}}\right )}{128 \sqrt{2} a^{5/2} d}-\frac{35 \cos (c+d x)}{128 a d (a \sin (c+d x)+a)^{3/2}}-\frac{7 \sec (c+d x)}{48 a d (a \sin (c+d x)+a)^{3/2}}-\frac{\sec (c+d x)}{6 d (a \sin (c+d x)+a)^{5/2}} \]

[Out]

(-35*ArcTanh[(Sqrt[a]*Cos[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sin[c + d*x]])])/(128*Sqrt[2]*a^(5/2)*d) - Sec[c + d*x
]/(6*d*(a + a*Sin[c + d*x])^(5/2)) - (35*Cos[c + d*x])/(128*a*d*(a + a*Sin[c + d*x])^(3/2)) - (7*Sec[c + d*x])
/(48*a*d*(a + a*Sin[c + d*x])^(3/2)) + (35*Sec[c + d*x])/(96*a^2*d*Sqrt[a + a*Sin[c + d*x]])

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Rubi [A]  time = 0.230418, antiderivative size = 167, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used = {2681, 2687, 2650, 2649, 206} \[ \frac{35 \sec (c+d x)}{96 a^2 d \sqrt{a \sin (c+d x)+a}}-\frac{35 \tanh ^{-1}\left (\frac{\sqrt{a} \cos (c+d x)}{\sqrt{2} \sqrt{a \sin (c+d x)+a}}\right )}{128 \sqrt{2} a^{5/2} d}-\frac{35 \cos (c+d x)}{128 a d (a \sin (c+d x)+a)^{3/2}}-\frac{7 \sec (c+d x)}{48 a d (a \sin (c+d x)+a)^{3/2}}-\frac{\sec (c+d x)}{6 d (a \sin (c+d x)+a)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^2/(a + a*Sin[c + d*x])^(5/2),x]

[Out]

(-35*ArcTanh[(Sqrt[a]*Cos[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sin[c + d*x]])])/(128*Sqrt[2]*a^(5/2)*d) - Sec[c + d*x
]/(6*d*(a + a*Sin[c + d*x])^(5/2)) - (35*Cos[c + d*x])/(128*a*d*(a + a*Sin[c + d*x])^(3/2)) - (7*Sec[c + d*x])
/(48*a*d*(a + a*Sin[c + d*x])^(3/2)) + (35*Sec[c + d*x])/(96*a^2*d*Sqrt[a + a*Sin[c + d*x]])

Rule 2681

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

Rule 2687

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> -Simp[(b*(g*
Cos[e + f*x])^(p + 1))/(a*f*g*(p + 1)*Sqrt[a + b*Sin[e + f*x]]), x] + Dist[(a*(2*p + 1))/(2*g^2*(p + 1)), Int[
(g*Cos[e + f*x])^(p + 2)/(a + b*Sin[e + f*x])^(3/2), x], x] /; FreeQ[{a, b, e, f, g}, x] && EqQ[a^2 - b^2, 0]
&& LtQ[p, -1] && IntegerQ[2*p]

Rule 2650

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*Cos[c + d*x]*(a + b*Sin[c + d*x])^n)/(a*
d*(2*n + 1)), x] + Dist[(n + 1)/(a*(2*n + 1)), Int[(a + b*Sin[c + d*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d},
x] && EqQ[a^2 - b^2, 0] && LtQ[n, -1] && IntegerQ[2*n]

Rule 2649

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[-2/d, Subst[Int[1/(2*a - x^2), x], x, (b*C
os[c + d*x])/Sqrt[a + b*Sin[c + d*x]]], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{\sec ^2(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx &=-\frac{\sec (c+d x)}{6 d (a+a \sin (c+d x))^{5/2}}+\frac{7 \int \frac{\sec ^2(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx}{12 a}\\ &=-\frac{\sec (c+d x)}{6 d (a+a \sin (c+d x))^{5/2}}-\frac{7 \sec (c+d x)}{48 a d (a+a \sin (c+d x))^{3/2}}+\frac{35 \int \frac{\sec ^2(c+d x)}{\sqrt{a+a \sin (c+d x)}} \, dx}{96 a^2}\\ &=-\frac{\sec (c+d x)}{6 d (a+a \sin (c+d x))^{5/2}}-\frac{7 \sec (c+d x)}{48 a d (a+a \sin (c+d x))^{3/2}}+\frac{35 \sec (c+d x)}{96 a^2 d \sqrt{a+a \sin (c+d x)}}+\frac{35 \int \frac{1}{(a+a \sin (c+d x))^{3/2}} \, dx}{64 a}\\ &=-\frac{\sec (c+d x)}{6 d (a+a \sin (c+d x))^{5/2}}-\frac{35 \cos (c+d x)}{128 a d (a+a \sin (c+d x))^{3/2}}-\frac{7 \sec (c+d x)}{48 a d (a+a \sin (c+d x))^{3/2}}+\frac{35 \sec (c+d x)}{96 a^2 d \sqrt{a+a \sin (c+d x)}}+\frac{35 \int \frac{1}{\sqrt{a+a \sin (c+d x)}} \, dx}{256 a^2}\\ &=-\frac{\sec (c+d x)}{6 d (a+a \sin (c+d x))^{5/2}}-\frac{35 \cos (c+d x)}{128 a d (a+a \sin (c+d x))^{3/2}}-\frac{7 \sec (c+d x)}{48 a d (a+a \sin (c+d x))^{3/2}}+\frac{35 \sec (c+d x)}{96 a^2 d \sqrt{a+a \sin (c+d x)}}-\frac{35 \operatorname{Subst}\left (\int \frac{1}{2 a-x^2} \, dx,x,\frac{a \cos (c+d x)}{\sqrt{a+a \sin (c+d x)}}\right )}{128 a^2 d}\\ &=-\frac{35 \tanh ^{-1}\left (\frac{\sqrt{a} \cos (c+d x)}{\sqrt{2} \sqrt{a+a \sin (c+d x)}}\right )}{128 \sqrt{2} a^{5/2} d}-\frac{\sec (c+d x)}{6 d (a+a \sin (c+d x))^{5/2}}-\frac{35 \cos (c+d x)}{128 a d (a+a \sin (c+d x))^{3/2}}-\frac{7 \sec (c+d x)}{48 a d (a+a \sin (c+d x))^{3/2}}+\frac{35 \sec (c+d x)}{96 a^2 d \sqrt{a+a \sin (c+d x)}}\\ \end{align*}

Mathematica [C]  time = 0.395521, size = 284, normalized size = 1.7 \[ \frac{\frac{48 \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^5}{\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )}-57 \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^4+114 \sin \left (\frac{1}{2} (c+d x)\right ) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^3-44 \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2+88 \sin \left (\frac{1}{2} (c+d x)\right ) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+\frac{64 \sin \left (\frac{1}{2} (c+d x)\right )}{\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )}+(105+105 i) (-1)^{3/4} \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^5 \tanh ^{-1}\left (\left (\frac{1}{2}+\frac{i}{2}\right ) (-1)^{3/4} \left (\tan \left (\frac{1}{4} (c+d x)\right )-1\right )\right )-32}{384 d (a (\sin (c+d x)+1))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^2/(a + a*Sin[c + d*x])^(5/2),x]

[Out]

(-32 + (64*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]) + 88*Sin[(c + d*x)/2]*(Cos[(c + d*x)/2] + S
in[(c + d*x)/2]) - 44*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2 + 114*Sin[(c + d*x)/2]*(Cos[(c + d*x)/2] + Sin[(
c + d*x)/2])^3 - 57*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^4 + (105 + 105*I)*(-1)^(3/4)*ArcTanh[(1/2 + I/2)*(-1
)^(3/4)*(-1 + Tan[(c + d*x)/4])]*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^5 + (48*(Cos[(c + d*x)/2] + Sin[(c + d*
x)/2])^5)/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2]))/(384*d*(a*(1 + Sin[c + d*x]))^(5/2))

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Maple [A]  time = 0.187, size = 266, normalized size = 1.6 \begin{align*} -{\frac{1}{768\, \left ( 1+\sin \left ( dx+c \right ) \right ) ^{2}\cos \left ( dx+c \right ) d} \left ( \left ( 210\,{a}^{7/2}-105\,\sqrt{a-a\sin \left ( dx+c \right ) }\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{a-a\sin \left ( dx+c \right ) }\sqrt{2}}{\sqrt{a}}} \right ){a}^{3} \right ) \sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}+ \left ( -448\,{a}^{7/2}+420\,\sqrt{a-a\sin \left ( dx+c \right ) }\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{a-a\sin \left ( dx+c \right ) }\sqrt{2}}{\sqrt{a}}} \right ){a}^{3} \right ) \sin \left ( dx+c \right ) + \left ( 490\,{a}^{7/2}-315\,\sqrt{a-a\sin \left ( dx+c \right ) }\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{a-a\sin \left ( dx+c \right ) }\sqrt{2}}{\sqrt{a}}} \right ){a}^{3} \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}-320\,{a}^{7/2}+420\,\sqrt{a-a\sin \left ( dx+c \right ) }\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{a-a\sin \left ( dx+c \right ) }\sqrt{2}}{\sqrt{a}}} \right ){a}^{3} \right ){a}^{-{\frac{11}{2}}}{\frac{1}{\sqrt{a+a\sin \left ( dx+c \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^2/(a+a*sin(d*x+c))^(5/2),x)

[Out]

-1/768/a^(11/2)*((210*a^(7/2)-105*(a-a*sin(d*x+c))^(1/2)*2^(1/2)*arctanh(1/2*(a-a*sin(d*x+c))^(1/2)*2^(1/2)/a^
(1/2))*a^3)*sin(d*x+c)*cos(d*x+c)^2+(-448*a^(7/2)+420*(a-a*sin(d*x+c))^(1/2)*2^(1/2)*arctanh(1/2*(a-a*sin(d*x+
c))^(1/2)*2^(1/2)/a^(1/2))*a^3)*sin(d*x+c)+(490*a^(7/2)-315*(a-a*sin(d*x+c))^(1/2)*2^(1/2)*arctanh(1/2*(a-a*si
n(d*x+c))^(1/2)*2^(1/2)/a^(1/2))*a^3)*cos(d*x+c)^2-320*a^(7/2)+420*(a-a*sin(d*x+c))^(1/2)*2^(1/2)*arctanh(1/2*
(a-a*sin(d*x+c))^(1/2)*2^(1/2)/a^(1/2))*a^3)/(1+sin(d*x+c))^2/cos(d*x+c)/(a+a*sin(d*x+c))^(1/2)/d

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sec \left (d x + c\right )^{2}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2/(a+a*sin(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

integrate(sec(d*x + c)^2/(a*sin(d*x + c) + a)^(5/2), x)

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Fricas [A]  time = 2.44893, size = 755, normalized size = 4.52 \begin{align*} \frac{105 \, \sqrt{2}{\left (3 \, \cos \left (d x + c\right )^{3} +{\left (\cos \left (d x + c\right )^{3} - 4 \, \cos \left (d x + c\right )\right )} \sin \left (d x + c\right ) - 4 \, \cos \left (d x + c\right )\right )} \sqrt{a} \log \left (-\frac{a \cos \left (d x + c\right )^{2} - 2 \, \sqrt{2} \sqrt{a \sin \left (d x + c\right ) + a} \sqrt{a}{\left (\cos \left (d x + c\right ) - \sin \left (d x + c\right ) + 1\right )} + 3 \, a \cos \left (d x + c\right ) -{\left (a \cos \left (d x + c\right ) - 2 \, a\right )} \sin \left (d x + c\right ) + 2 \, a}{\cos \left (d x + c\right )^{2} -{\left (\cos \left (d x + c\right ) + 2\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 2}\right ) + 4 \,{\left (245 \, \cos \left (d x + c\right )^{2} + 7 \,{\left (15 \, \cos \left (d x + c\right )^{2} - 32\right )} \sin \left (d x + c\right ) - 160\right )} \sqrt{a \sin \left (d x + c\right ) + a}}{1536 \,{\left (3 \, a^{3} d \cos \left (d x + c\right )^{3} - 4 \, a^{3} d \cos \left (d x + c\right ) +{\left (a^{3} d \cos \left (d x + c\right )^{3} - 4 \, a^{3} d \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2/(a+a*sin(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

1/1536*(105*sqrt(2)*(3*cos(d*x + c)^3 + (cos(d*x + c)^3 - 4*cos(d*x + c))*sin(d*x + c) - 4*cos(d*x + c))*sqrt(
a)*log(-(a*cos(d*x + c)^2 - 2*sqrt(2)*sqrt(a*sin(d*x + c) + a)*sqrt(a)*(cos(d*x + c) - sin(d*x + c) + 1) + 3*a
*cos(d*x + c) - (a*cos(d*x + c) - 2*a)*sin(d*x + c) + 2*a)/(cos(d*x + c)^2 - (cos(d*x + c) + 2)*sin(d*x + c) -
cos(d*x + c) - 2)) + 4*(245*cos(d*x + c)^2 + 7*(15*cos(d*x + c)^2 - 32)*sin(d*x + c) - 160)*sqrt(a*sin(d*x +
c) + a))/(3*a^3*d*cos(d*x + c)^3 - 4*a^3*d*cos(d*x + c) + (a^3*d*cos(d*x + c)^3 - 4*a^3*d*cos(d*x + c))*sin(d*
x + c))

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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(sec(d*x+c)**2/(a+a*sin(d*x+c))**(5/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \mathit{sage}_{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2/(a+a*sin(d*x+c))^(5/2),x, algorithm="giac")

[Out]

sage2