3.146 \(\int \sec (c+d x) (a+a \sin (c+d x))^{7/2} \, dx\)

Optimal. Leaf size=110 \[ \frac {8 \sqrt {2} a^{7/2} \tanh ^{-1}\left (\frac {\sqrt {a \sin (c+d x)+a}}{\sqrt {2} \sqrt {a}}\right )}{d}-\frac {8 a^3 \sqrt {a \sin (c+d x)+a}}{d}-\frac {4 a^2 (a \sin (c+d x)+a)^{3/2}}{3 d}-\frac {2 a (a \sin (c+d x)+a)^{5/2}}{5 d} \]

[Out]

-4/3*a^2*(a+a*sin(d*x+c))^(3/2)/d-2/5*a*(a+a*sin(d*x+c))^(5/2)/d+8*a^(7/2)*arctanh(1/2*(a+a*sin(d*x+c))^(1/2)*
2^(1/2)/a^(1/2))*2^(1/2)/d-8*a^3*(a+a*sin(d*x+c))^(1/2)/d

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Rubi [A]  time = 0.09, antiderivative size = 110, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {2667, 50, 63, 206} \[ -\frac {8 a^3 \sqrt {a \sin (c+d x)+a}}{d}-\frac {4 a^2 (a \sin (c+d x)+a)^{3/2}}{3 d}+\frac {8 \sqrt {2} a^{7/2} \tanh ^{-1}\left (\frac {\sqrt {a \sin (c+d x)+a}}{\sqrt {2} \sqrt {a}}\right )}{d}-\frac {2 a (a \sin (c+d x)+a)^{5/2}}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(8*Sqrt[2]*a^(7/2)*ArcTanh[Sqrt[a + a*Sin[c + d*x]]/(Sqrt[2]*Sqrt[a])])/d - (8*a^3*Sqrt[a + a*Sin[c + d*x]])/d
- (4*a^2*(a + a*Sin[c + d*x])^(3/2))/(3*d) - (2*a*(a + a*Sin[c + d*x])^(5/2))/(5*d)

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

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 2667

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x]
&& IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && (GeQ[p, -1] ||  !IntegerQ[m + 1/2])

Rubi steps

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

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Mathematica [A]  time = 0.35, size = 85, normalized size = 0.77 \[ \frac {120 \sqrt {2} a^{7/2} \tanh ^{-1}\left (\frac {\sqrt {a (\sin (c+d x)+1)}}{\sqrt {2} \sqrt {a}}\right )-2 a^3 \left (3 \sin ^2(c+d x)+16 \sin (c+d x)+73\right ) \sqrt {a (\sin (c+d x)+1)}}{15 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(120*Sqrt[2]*a^(7/2)*ArcTanh[Sqrt[a*(1 + Sin[c + d*x])]/(Sqrt[2]*Sqrt[a])] - 2*a^3*Sqrt[a*(1 + Sin[c + d*x])]*
(73 + 16*Sin[c + d*x] + 3*Sin[c + d*x]^2))/(15*d)

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fricas [A]  time = 0.74, size = 102, normalized size = 0.93 \[ \frac {2 \, {\left (30 \, \sqrt {2} a^{\frac {7}{2}} \log \left (-\frac {a \sin \left (d x + c\right ) + 2 \, \sqrt {2} \sqrt {a \sin \left (d x + c\right ) + a} \sqrt {a} + 3 \, a}{\sin \left (d x + c\right ) - 1}\right ) + {\left (3 \, a^{3} \cos \left (d x + c\right )^{2} - 16 \, a^{3} \sin \left (d x + c\right ) - 76 \, a^{3}\right )} \sqrt {a \sin \left (d x + c\right ) + a}\right )}}{15 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*(30*sqrt(2)*a^(7/2)*log(-(a*sin(d*x + c) + 2*sqrt(2)*sqrt(a*sin(d*x + c) + a)*sqrt(a) + 3*a)/(sin(d*x + c
) - 1)) + (3*a^3*cos(d*x + c)^2 - 16*a^3*sin(d*x + c) - 76*a^3)*sqrt(a*sin(d*x + c) + a))/d

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

[Out]

Timed out

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maple [A]  time = 0.17, size = 83, normalized size = 0.75 \[ -\frac {2 a \left (\frac {\left (a +a \sin \left (d x +c \right )\right )^{\frac {5}{2}}}{5}+\frac {2 \left (a +a \sin \left (d x +c \right )\right )^{\frac {3}{2}} a}{3}+4 a^{2} \sqrt {a +a \sin \left (d x +c \right )}-4 a^{\frac {5}{2}} \sqrt {2}\, \arctanh \left (\frac {\sqrt {a +a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*a*(1/5*(a+a*sin(d*x+c))^(5/2)+2/3*(a+a*sin(d*x+c))^(3/2)*a+4*a^2*(a+a*sin(d*x+c))^(1/2)-4*a^(5/2)*2^(1/2)*a
rctanh(1/2*(a+a*sin(d*x+c))^(1/2)*2^(1/2)/a^(1/2)))/d

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maxima [A]  time = 0.81, size = 115, normalized size = 1.05 \[ -\frac {2 \, {\left (30 \, \sqrt {2} a^{\frac {9}{2}} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {a \sin \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {a \sin \left (d x + c\right ) + a}}\right ) + 3 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {5}{2}} a^{2} + 10 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}} a^{3} + 60 \, \sqrt {a \sin \left (d x + c\right ) + a} a^{4}\right )}}{15 \, a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/15*(30*sqrt(2)*a^(9/2)*log(-(sqrt(2)*sqrt(a) - sqrt(a*sin(d*x + c) + a))/(sqrt(2)*sqrt(a) + sqrt(a*sin(d*x
+ c) + a))) + 3*(a*sin(d*x + c) + a)^(5/2)*a^2 + 10*(a*sin(d*x + c) + a)^(3/2)*a^3 + 60*sqrt(a*sin(d*x + c) +
a)*a^4)/(a*d)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{7/2}}{\cos \left (c+d\,x\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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