3.124 \(\int \sec ^4(c+d x) (a+a \sin (c+d x))^{3/2} \, dx\)

Optimal. Leaf size=107 \[ -\frac{a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a} \cos (c+d x)}{\sqrt{2} \sqrt{a \sin (c+d x)+a}}\right )}{2 \sqrt{2} d}+\frac{\sec ^3(c+d x) (a \sin (c+d x)+a)^{3/2}}{3 d}+\frac{a \sec (c+d x) \sqrt{a \sin (c+d x)+a}}{2 d} \]

[Out]

-(a^(3/2)*ArcTanh[(Sqrt[a]*Cos[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sin[c + d*x]])])/(2*Sqrt[2]*d) + (a*Sec[c + d*x]*
Sqrt[a + a*Sin[c + d*x]])/(2*d) + (Sec[c + d*x]^3*(a + a*Sin[c + d*x])^(3/2))/(3*d)

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Rubi [A]  time = 0.134985, antiderivative size = 107, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {2675, 2649, 206} \[ -\frac{a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a} \cos (c+d x)}{\sqrt{2} \sqrt{a \sin (c+d x)+a}}\right )}{2 \sqrt{2} d}+\frac{\sec ^3(c+d x) (a \sin (c+d x)+a)^{3/2}}{3 d}+\frac{a \sec (c+d x) \sqrt{a \sin (c+d x)+a}}{2 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^4*(a + a*Sin[c + d*x])^(3/2),x]

[Out]

-(a^(3/2)*ArcTanh[(Sqrt[a]*Cos[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sin[c + d*x]])])/(2*Sqrt[2]*d) + (a*Sec[c + d*x]*
Sqrt[a + a*Sin[c + d*x]])/(2*d) + (Sec[c + d*x]^3*(a + a*Sin[c + d*x])^(3/2))/(3*d)

Rule 2675

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*(p + 1)), x] + Dist[(a*(m + p + 1))/(g^2*(p + 1)), Int[(
g*Cos[e + f*x])^(p + 2)*(a + b*Sin[e + f*x])^(m - 1), x], x] /; FreeQ[{a, b, e, f, g}, x] && EqQ[a^2 - b^2, 0]
&& GtQ[m, 0] && LeQ[p, -2*m] && IntegersQ[m + 1/2, 2*p]

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 \sec ^4(c+d x) (a+a \sin (c+d x))^{3/2} \, dx &=\frac{\sec ^3(c+d x) (a+a \sin (c+d x))^{3/2}}{3 d}+\frac{1}{2} a \int \sec ^2(c+d x) \sqrt{a+a \sin (c+d x)} \, dx\\ &=\frac{a \sec (c+d x) \sqrt{a+a \sin (c+d x)}}{2 d}+\frac{\sec ^3(c+d x) (a+a \sin (c+d x))^{3/2}}{3 d}+\frac{1}{4} a^2 \int \frac{1}{\sqrt{a+a \sin (c+d x)}} \, dx\\ &=\frac{a \sec (c+d x) \sqrt{a+a \sin (c+d x)}}{2 d}+\frac{\sec ^3(c+d x) (a+a \sin (c+d x))^{3/2}}{3 d}-\frac{a^2 \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 )}{2 d}\\ &=-\frac{a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a} \cos (c+d x)}{\sqrt{2} \sqrt{a+a \sin (c+d x)}}\right )}{2 \sqrt{2} d}+\frac{a \sec (c+d x) \sqrt{a+a \sin (c+d x)}}{2 d}+\frac{\sec ^3(c+d x) (a+a \sin (c+d x))^{3/2}}{3 d}\\ \end{align*}

Mathematica [C]  time = 0.382373, size = 130, normalized size = 1.21 \[ \frac{\left (\frac{1}{12}+\frac{i}{12}\right ) a \sec ^3(c+d x) \sqrt{a (\sin (c+d x)+1)} \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2 \left (6 (-1)^{3/4} \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^3 \tanh ^{-1}\left (\left (\frac{1}{2}+\frac{i}{2}\right ) (-1)^{3/4} \left (\tan \left (\frac{1}{4} (c+d x)\right )-1\right )\right )-(1-i) (3 \sin (c+d x)-5)\right )}{d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^4*(a + a*Sin[c + d*x])^(3/2),x]

[Out]

((1/12 + I/12)*a*Sec[c + d*x]^3*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2*Sqrt[a*(1 + Sin[c + d*x])]*(6*(-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])^3 - (1 - I)*(
-5 + 3*Sin[c + d*x])))/d

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Maple [A]  time = 0.135, size = 107, normalized size = 1. \begin{align*}{\frac{1+\sin \left ( dx+c \right ) }{ \left ( 12\,\sin \left ( dx+c \right ) -12 \right ) \cos \left ( dx+c \right ) d} \left ( 6\,{a}^{7/2}\sin \left ( dx+c \right ) -10\,{a}^{7/2}+3\,\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{a-a\sin \left ( dx+c \right ) }\sqrt{2}}{\sqrt{a}}} \right ){a}^{2} \left ( a-a\sin \left ( dx+c \right ) \right ) ^{3/2} \right ){a}^{-{\frac{3}{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)^4*(a+a*sin(d*x+c))^(3/2),x)

[Out]

1/12/a^(3/2)*(1+sin(d*x+c))/(sin(d*x+c)-1)*(6*a^(7/2)*sin(d*x+c)-10*a^(7/2)+3*2^(1/2)*arctanh(1/2*(a-a*sin(d*x
+c))^(1/2)*2^(1/2)/a^(1/2))*a^2*(a-a*sin(d*x+c))^(3/2))/cos(d*x+c)/(a+a*sin(d*x+c))^(1/2)/d

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Maxima [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)^4*(a+a*sin(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [B]  time = 1.76247, size = 585, normalized size = 5.47 \begin{align*} \frac{3 \,{\left (\sqrt{2} a \cos \left (d x + c\right ) \sin \left (d x + c\right ) - \sqrt{2} a \cos \left (d x + c\right )\right )} \sqrt{a} \log \left (-\frac{a \cos \left (d x + c\right )^{2} - 2 \, \sqrt{a \sin \left (d x + c\right ) + a}{\left (\sqrt{2} \cos \left (d x + c\right ) - \sqrt{2} \sin \left (d x + c\right ) + \sqrt{2}\right )} \sqrt{a} + 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 (3 \, a \sin \left (d x + c\right ) - 5 \, a\right )} \sqrt{a \sin \left (d x + c\right ) + a}}{24 \,{\left (d \cos \left (d x + c\right ) \sin \left (d x + c\right ) - d \cos \left (d x + c\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*(a+a*sin(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

1/24*(3*(sqrt(2)*a*cos(d*x + c)*sin(d*x + c) - sqrt(2)*a*cos(d*x + c))*sqrt(a)*log(-(a*cos(d*x + c)^2 - 2*sqrt
(a*sin(d*x + c) + a)*(sqrt(2)*cos(d*x + c) - sqrt(2)*sin(d*x + c) + sqrt(2))*sqrt(a) + 3*a*cos(d*x + c) - (a*c
os(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*(3*a*sin(d*x + c) - 5*a)*sqrt(a*sin(d*x + c) + a))/(d*cos(d*x + c)*sin(d*x + c) - d*cos(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)**4*(a+a*sin(d*x+c))**(3/2),x)

[Out]

Timed out

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Giac [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)^4*(a+a*sin(d*x+c))^(3/2),x, algorithm="giac")

[Out]

Timed out