3.126 \(\int \sec ^6(c+d x) (a+a \sin (c+d x))^{3/2} \, dx\)

Optimal. Leaf size=169 \[ -\frac {7 a^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a \sin (c+d x)+a}}\right )}{16 \sqrt {2} d}-\frac {7 a^3 \cos (c+d x)}{16 d (a \sin (c+d x)+a)^{3/2}}+\frac {7 a^2 \sec (c+d x)}{12 d \sqrt {a \sin (c+d x)+a}}+\frac {\sec ^5(c+d x) (a \sin (c+d x)+a)^{3/2}}{5 d}+\frac {7 a \sec ^3(c+d x) \sqrt {a \sin (c+d x)+a}}{30 d} \]

[Out]

-7/16*a^3*cos(d*x+c)/d/(a+a*sin(d*x+c))^(3/2)+1/5*sec(d*x+c)^5*(a+a*sin(d*x+c))^(3/2)/d-7/32*a^(3/2)*arctanh(1
/2*cos(d*x+c)*a^(1/2)*2^(1/2)/(a+a*sin(d*x+c))^(1/2))/d*2^(1/2)+7/12*a^2*sec(d*x+c)/d/(a+a*sin(d*x+c))^(1/2)+7
/30*a*sec(d*x+c)^3*(a+a*sin(d*x+c))^(1/2)/d

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Rubi [A]  time = 0.22, antiderivative size = 169, normalized size of antiderivative = 1.00, 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 = {2675, 2687, 2650, 2649, 206} \[ -\frac {7 a^3 \cos (c+d x)}{16 d (a \sin (c+d x)+a)^{3/2}}+\frac {7 a^2 \sec (c+d x)}{12 d \sqrt {a \sin (c+d x)+a}}-\frac {7 a^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a \sin (c+d x)+a}}\right )}{16 \sqrt {2} d}+\frac {\sec ^5(c+d x) (a \sin (c+d x)+a)^{3/2}}{5 d}+\frac {7 a \sec ^3(c+d x) \sqrt {a \sin (c+d x)+a}}{30 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^6*(a + a*Sin[c + d*x])^(3/2),x]

[Out]

(-7*a^(3/2)*ArcTanh[(Sqrt[a]*Cos[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sin[c + d*x]])])/(16*Sqrt[2]*d) - (7*a^3*Cos[c
+ d*x])/(16*d*(a + a*Sin[c + d*x])^(3/2)) + (7*a^2*Sec[c + d*x])/(12*d*Sqrt[a + a*Sin[c + d*x]]) + (7*a*Sec[c
+ d*x]^3*Sqrt[a + a*Sin[c + d*x]])/(30*d) + (Sec[c + d*x]^5*(a + a*Sin[c + d*x])^(3/2))/(5*d)

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])

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 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 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 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]

Rubi steps

\begin {align*} \int \sec ^6(c+d x) (a+a \sin (c+d x))^{3/2} \, dx &=\frac {\sec ^5(c+d x) (a+a \sin (c+d x))^{3/2}}{5 d}+\frac {1}{10} (7 a) \int \sec ^4(c+d x) \sqrt {a+a \sin (c+d x)} \, dx\\ &=\frac {7 a \sec ^3(c+d x) \sqrt {a+a \sin (c+d x)}}{30 d}+\frac {\sec ^5(c+d x) (a+a \sin (c+d x))^{3/2}}{5 d}+\frac {1}{12} \left (7 a^2\right ) \int \frac {\sec ^2(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx\\ &=\frac {7 a^2 \sec (c+d x)}{12 d \sqrt {a+a \sin (c+d x)}}+\frac {7 a \sec ^3(c+d x) \sqrt {a+a \sin (c+d x)}}{30 d}+\frac {\sec ^5(c+d x) (a+a \sin (c+d x))^{3/2}}{5 d}+\frac {1}{8} \left (7 a^3\right ) \int \frac {1}{(a+a \sin (c+d x))^{3/2}} \, dx\\ &=-\frac {7 a^3 \cos (c+d x)}{16 d (a+a \sin (c+d x))^{3/2}}+\frac {7 a^2 \sec (c+d x)}{12 d \sqrt {a+a \sin (c+d x)}}+\frac {7 a \sec ^3(c+d x) \sqrt {a+a \sin (c+d x)}}{30 d}+\frac {\sec ^5(c+d x) (a+a \sin (c+d x))^{3/2}}{5 d}+\frac {1}{32} \left (7 a^2\right ) \int \frac {1}{\sqrt {a+a \sin (c+d x)}} \, dx\\ &=-\frac {7 a^3 \cos (c+d x)}{16 d (a+a \sin (c+d x))^{3/2}}+\frac {7 a^2 \sec (c+d x)}{12 d \sqrt {a+a \sin (c+d x)}}+\frac {7 a \sec ^3(c+d x) \sqrt {a+a \sin (c+d x)}}{30 d}+\frac {\sec ^5(c+d x) (a+a \sin (c+d x))^{3/2}}{5 d}-\frac {\left (7 a^2\right ) \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 )}{16 d}\\ &=-\frac {7 a^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a+a \sin (c+d x)}}\right )}{16 \sqrt {2} d}-\frac {7 a^3 \cos (c+d x)}{16 d (a+a \sin (c+d x))^{3/2}}+\frac {7 a^2 \sec (c+d x)}{12 d \sqrt {a+a \sin (c+d x)}}+\frac {7 a \sec ^3(c+d x) \sqrt {a+a \sin (c+d x)}}{30 d}+\frac {\sec ^5(c+d x) (a+a \sin (c+d x))^{3/2}}{5 d}\\ \end {align*}

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Mathematica [C]  time = 0.43, size = 288, normalized size = 1.70 \[ \frac {(a (\sin (c+d x)+1))^{3/2} \left (30 \sin \left (\frac {1}{2} (c+d x)\right )+\frac {90 \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^2}{\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )}+\frac {40 \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^2}{\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^3}+\frac {24 \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^2}{\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^5}-15 \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )+(105+105 i) (-1)^{3/4} \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^2 \tanh ^{-1}\left (\left (\frac {1}{2}+\frac {i}{2}\right ) (-1)^{3/4} \left (\tan \left (\frac {1}{4} (c+d x)\right )-1\right )\right )\right )}{240 d \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^5} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^6*(a + a*Sin[c + d*x])^(3/2),x]

[Out]

((30*Sin[(c + d*x)/2] - 15*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]) + (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])^2 + (24*(Cos[(c + d*x)/2] + Sin[(c
+ d*x)/2])^2)/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^5 + (40*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2)/(Cos[(c
+ d*x)/2] - Sin[(c + d*x)/2])^3 + (90*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2)/(Cos[(c + d*x)/2] - Sin[(c + d*
x)/2]))*(a*(1 + Sin[c + d*x]))^(3/2))/(240*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^5)

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fricas [A]  time = 0.77, size = 248, normalized size = 1.47 \[ \frac {105 \, {\left (\sqrt {2} a \cos \left (d x + c\right )^{3} \sin \left (d x + c\right ) - \sqrt {2} a \cos \left (d x + c\right )^{3}\right )} \sqrt {a} \log \left (-\frac {a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {a \sin \left (d x + c\right ) + a} {\left (\sqrt {2} \cos \left (d x + c\right ) - \sqrt {2} \sin \left (d x + c\right ) + \sqrt {2}\right )} \sqrt {a} + 3 \, a \cos \left (d x + c\right ) - {\left (a \cos \left (d x + c\right ) - 2 \, a\right )} \sin \left (d x + c\right ) + 2 \, a}{\cos \left (d x + c\right )^{2} - {\left (\cos \left (d x + c\right ) + 2\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 2}\right ) - 4 \, {\left (175 \, a \cos \left (d x + c\right )^{2} - 21 \, {\left (5 \, a \cos \left (d x + c\right )^{2} - 4 \, a\right )} \sin \left (d x + c\right ) - 36 \, a\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{960 \, {\left (d \cos \left (d x + c\right )^{3} \sin \left (d x + c\right ) - d \cos \left (d x + c\right )^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^6*(a+a*sin(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

1/960*(105*(sqrt(2)*a*cos(d*x + c)^3*sin(d*x + c) - sqrt(2)*a*cos(d*x + c)^3)*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*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*(175*a*cos(d*x + c)^2 - 21*(5*a*cos(d*x + c)^2 - 4*a)*sin(d*x + c) - 36*a)*sqrt(a*sin(d*x + c) + a)
)/(d*cos(d*x + c)^3*sin(d*x + c) - d*cos(d*x + c)^3)

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^6*(a+a*sin(d*x+c))^(3/2),x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 0.25, size = 172, normalized size = 1.02 \[ -\frac {210 a^{\frac {7}{2}} \sin \left (d x +c \right ) \left (\cos ^{2}\left (d x +c \right )\right )+\left (105 \left (a -a \sin \left (d x +c \right )\right )^{\frac {5}{2}} \sqrt {2}\, \arctanh \left (\frac {\sqrt {a -a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a -168 a^{\frac {7}{2}}\right ) \sin \left (d x +c \right )-350 a^{\frac {7}{2}} \left (\cos ^{2}\left (d x +c \right )\right )+105 \left (a -a \sin \left (d x +c \right )\right )^{\frac {5}{2}} \sqrt {2}\, \arctanh \left (\frac {\sqrt {a -a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a +72 a^{\frac {7}{2}}}{480 a^{\frac {3}{2}} \left (\sin \left (d x +c \right )-1\right )^{2} \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^6*(a+a*sin(d*x+c))^(3/2),x)

[Out]

-1/480/a^(3/2)*(210*a^(7/2)*sin(d*x+c)*cos(d*x+c)^2+(105*(a-a*sin(d*x+c))^(5/2)*2^(1/2)*arctanh(1/2*(a-a*sin(d
*x+c))^(1/2)*2^(1/2)/a^(1/2))*a-168*a^(7/2))*sin(d*x+c)-350*a^(7/2)*cos(d*x+c)^2+105*(a-a*sin(d*x+c))^(5/2)*2^
(1/2)*arctanh(1/2*(a-a*sin(d*x+c))^(1/2)*2^(1/2)/a^(1/2))*a+72*a^(7/2))/(sin(d*x+c)-1)^2/cos(d*x+c)/(a+a*sin(d
*x+c))^(1/2)/d

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maxima [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^6*(a+a*sin(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

Timed out

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{3/2}}{{\cos \left (c+d\,x\right )}^6} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*sin(c + d*x))^(3/2)/cos(c + d*x)^6,x)

[Out]

int((a + a*sin(c + d*x))^(3/2)/cos(c + d*x)^6, x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**6*(a+a*sin(d*x+c))**(3/2),x)

[Out]

Timed out

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