Optimal. Leaf size=420 \[ -\frac {\left (8 a^4 A-29 a^2 A b^2+15 A b^4+9 a^3 b B-3 a b^3 B\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{4 a^3 \left (a^2-b^2\right )^2 d}+\frac {\left (11 a^2 A b-5 A b^3-7 a^3 B+a b^2 B\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{4 a^2 \left (a^2-b^2\right )^2 d}-\frac {\left (35 a^4 A b-38 a^2 A b^3+15 A b^5-15 a^5 B+6 a^3 b^2 B-3 a b^4 B\right ) \Pi \left (\frac {2 b}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{4 a^3 (a-b)^2 (a+b)^3 d}+\frac {\left (8 a^4 A-29 a^2 A b^2+15 A b^4+9 a^3 b B-3 a b^3 B\right ) \sin (c+d x)}{4 a^3 \left (a^2-b^2\right )^2 d \sqrt {\cos (c+d x)}}+\frac {b (A b-a B) \sin (c+d x)}{2 a \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^2}+\frac {b \left (11 a^2 A b-5 A b^3-7 a^3 B+a b^2 B\right ) \sin (c+d x)}{4 a^2 \left (a^2-b^2\right )^2 d \sqrt {\cos (c+d x)} (a+b \cos (c+d x))} \]
[Out]
________________________________________________________________________________________
Rubi [A]
time = 0.96, antiderivative size = 420, normalized size of antiderivative = 1.00, number of steps
used = 8, number of rules used = 7, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.212, Rules used = {3079, 3134,
3138, 2719, 3081, 2720, 2884} \begin {gather*} \frac {b (A b-a B) \sin (c+d x)}{2 a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^2}+\frac {\left (-7 a^3 B+11 a^2 A b+a b^2 B-5 A b^3\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{4 a^2 d \left (a^2-b^2\right )^2}+\frac {b \left (-7 a^3 B+11 a^2 A b+a b^2 B-5 A b^3\right ) \sin (c+d x)}{4 a^2 d \left (a^2-b^2\right )^2 \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}-\frac {\left (8 a^4 A+9 a^3 b B-29 a^2 A b^2-3 a b^3 B+15 A b^4\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{4 a^3 d \left (a^2-b^2\right )^2}+\frac {\left (8 a^4 A+9 a^3 b B-29 a^2 A b^2-3 a b^3 B+15 A b^4\right ) \sin (c+d x)}{4 a^3 d \left (a^2-b^2\right )^2 \sqrt {\cos (c+d x)}}-\frac {\left (-15 a^5 B+35 a^4 A b+6 a^3 b^2 B-38 a^2 A b^3-3 a b^4 B+15 A b^5\right ) \Pi \left (\frac {2 b}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{4 a^3 d (a-b)^2 (a+b)^3} \end {gather*}
Antiderivative was successfully verified.
[In]
[Out]
Rule 2719
Rule 2720
Rule 2884
Rule 3079
Rule 3081
Rule 3134
Rule 3138
Rubi steps
\begin {align*} \int \frac {A+B \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^3} \, dx &=\frac {b (A b-a B) \sin (c+d x)}{2 a \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^2}+\frac {\int \frac {\frac {1}{2} \left (4 a^2 A-5 A b^2+a b B\right )-2 a (A b-a B) \cos (c+d x)+\frac {3}{2} b (A b-a B) \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^2} \, dx}{2 a \left (a^2-b^2\right )}\\ &=\frac {b (A b-a B) \sin (c+d x)}{2 a \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^2}+\frac {b \left (11 a^2 A b-5 A b^3-7 a^3 B+a b^2 B\right ) \sin (c+d x)}{4 a^2 \left (a^2-b^2\right )^2 d \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}+\frac {\int \frac {\frac {1}{4} \left (8 a^4 A-29 a^2 A b^2+15 A b^4+9 a^3 b B-3 a b^3 B\right )-a \left (4 a^2 A b-A b^3-2 a^3 B-a b^2 B\right ) \cos (c+d x)+\frac {1}{4} b \left (11 a^2 A b-5 A b^3-7 a^3 B+a b^2 B\right ) \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))} \, dx}{2 a^2 \left (a^2-b^2\right )^2}\\ &=\frac {\left (8 a^4 A-29 a^2 A b^2+15 A b^4+9 a^3 b B-3 a b^3 B\right ) \sin (c+d x)}{4 a^3 \left (a^2-b^2\right )^2 d \sqrt {\cos (c+d x)}}+\frac {b (A b-a B) \sin (c+d x)}{2 a \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^2}+\frac {b \left (11 a^2 A b-5 A b^3-7 a^3 B+a b^2 B\right ) \sin (c+d x)}{4 a^2 \left (a^2-b^2\right )^2 d \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}+\frac {\int \frac {\frac {1}{8} \left (-24 a^4 A b+33 a^2 A b^3-15 A b^5+8 a^5 B-5 a^3 b^2 B+3 a b^4 B\right )-\frac {1}{2} a \left (2 a^4 A-10 a^2 A b^2+5 A b^4+4 a^3 b B-a b^3 B\right ) \cos (c+d x)-\frac {1}{8} b \left (8 a^4 A-29 a^2 A b^2+15 A b^4+9 a^3 b B-3 a b^3 B\right ) \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{a^3 \left (a^2-b^2\right )^2}\\ &=\frac {\left (8 a^4 A-29 a^2 A b^2+15 A b^4+9 a^3 b B-3 a b^3 B\right ) \sin (c+d x)}{4 a^3 \left (a^2-b^2\right )^2 d \sqrt {\cos (c+d x)}}+\frac {b (A b-a B) \sin (c+d x)}{2 a \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^2}+\frac {b \left (11 a^2 A b-5 A b^3-7 a^3 B+a b^2 B\right ) \sin (c+d x)}{4 a^2 \left (a^2-b^2\right )^2 d \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}-\frac {\int \frac {\frac {1}{8} b \left (24 a^4 A b-33 a^2 A b^3+15 A b^5-8 a^5 B+5 a^3 b^2 B-3 a b^4 B\right )-\frac {1}{8} a b^2 \left (11 a^2 A b-5 A b^3-7 a^3 B+a b^2 B\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{a^3 b \left (a^2-b^2\right )^2}-\frac {\left (8 a^4 A-29 a^2 A b^2+15 A b^4+9 a^3 b B-3 a b^3 B\right ) \int \sqrt {\cos (c+d x)} \, dx}{8 a^3 \left (a^2-b^2\right )^2}\\ &=-\frac {\left (8 a^4 A-29 a^2 A b^2+15 A b^4+9 a^3 b B-3 a b^3 B\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{4 a^3 \left (a^2-b^2\right )^2 d}+\frac {\left (8 a^4 A-29 a^2 A b^2+15 A b^4+9 a^3 b B-3 a b^3 B\right ) \sin (c+d x)}{4 a^3 \left (a^2-b^2\right )^2 d \sqrt {\cos (c+d x)}}+\frac {b (A b-a B) \sin (c+d x)}{2 a \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^2}+\frac {b \left (11 a^2 A b-5 A b^3-7 a^3 B+a b^2 B\right ) \sin (c+d x)}{4 a^2 \left (a^2-b^2\right )^2 d \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}+\frac {\left (11 a^2 A b-5 A b^3-7 a^3 B+a b^2 B\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{8 a^2 \left (a^2-b^2\right )^2}-\frac {\left (35 a^4 A b-38 a^2 A b^3+15 A b^5-15 a^5 B+6 a^3 b^2 B-3 a b^4 B\right ) \int \frac {1}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{8 a^3 \left (a^2-b^2\right )^2}\\ &=-\frac {\left (8 a^4 A-29 a^2 A b^2+15 A b^4+9 a^3 b B-3 a b^3 B\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{4 a^3 \left (a^2-b^2\right )^2 d}+\frac {\left (11 a^2 A b-5 A b^3-7 a^3 B+a b^2 B\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{4 a^2 \left (a^2-b^2\right )^2 d}-\frac {\left (35 a^4 A b-38 a^2 A b^3+15 A b^5-15 a^5 B+6 a^3 b^2 B-3 a b^4 B\right ) \Pi \left (\frac {2 b}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{4 a^3 (a-b)^2 (a+b)^3 d}+\frac {\left (8 a^4 A-29 a^2 A b^2+15 A b^4+9 a^3 b B-3 a b^3 B\right ) \sin (c+d x)}{4 a^3 \left (a^2-b^2\right )^2 d \sqrt {\cos (c+d x)}}+\frac {b (A b-a B) \sin (c+d x)}{2 a \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^2}+\frac {b \left (11 a^2 A b-5 A b^3-7 a^3 B+a b^2 B\right ) \sin (c+d x)}{4 a^2 \left (a^2-b^2\right )^2 d \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}\\ \end {align*}
________________________________________________________________________________________
Mathematica [A]
time = 15.56, size = 458, normalized size = 1.09 \begin {gather*} \frac {-\frac {\frac {\left (56 a^4 A b-95 a^2 A b^3+45 A b^5-16 a^5 B+19 a^3 b^2 B-9 a b^4 B\right ) \Pi \left (\frac {2 b}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{a+b}+\frac {8 a \left (2 a^4 A-10 a^2 A b^2+5 A b^4+4 a^3 b B-a b^3 B\right ) \left ((a+b) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )-a \Pi \left (\frac {2 b}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )\right )}{b (a+b)}+\frac {\left (8 a^4 A-29 a^2 A b^2+15 A b^4+9 a^3 b B-3 a b^3 B\right ) \left (-2 a b E\left (\left .\text {ArcSin}\left (\sqrt {\cos (c+d x)}\right )\right |-1\right )+2 a (a+b) F\left (\left .\text {ArcSin}\left (\sqrt {\cos (c+d x)}\right )\right |-1\right )+\left (-2 a^2+b^2\right ) \Pi \left (-\frac {b}{a};\left .\text {ArcSin}\left (\sqrt {\cos (c+d x)}\right )\right |-1\right )\right ) \sin (c+d x)}{a b \sqrt {\sin ^2(c+d x)}}}{(a-b)^2 (a+b)^2}+\frac {\sqrt {\cos (c+d x)} \left (2 a b \left (16 a^4 A-47 a^2 A b^2+25 A b^4+11 a^3 b B-5 a b^3 B\right ) \sin (c+d x)+b^2 \left (8 a^4 A-29 a^2 A b^2+15 A b^4+9 a^3 b B-3 a b^3 B\right ) \sin (2 (c+d x))+16 A \left (a^3-a b^2\right )^2 \tan (c+d x)\right )}{\left (a^2-b^2\right )^2 (a+b \cos (c+d x))^2}}{8 a^3 d} \end {gather*}
Antiderivative was successfully verified.
[In]
[Out]
________________________________________________________________________________________
Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(1974\) vs.
\(2(480)=960\).
time = 1.55, size = 1975, normalized size = 4.70
method | result | size |
default | \(\text {Expression too large to display}\) | \(1975\) |
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Fricas [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {A+B\,\cos \left (c+d\,x\right )}{{\cos \left (c+d\,x\right )}^{3/2}\,{\left (a+b\,\cos \left (c+d\,x\right )\right )}^3} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________