Optimal. Leaf size=241 \[ \frac {b^{3/2} (5 a B+2 A b) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {(-b+i a)^{5/2} (A+i B) \tan ^{-1}\left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {b (2 a A+b B) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}-\frac {(b+i a)^{5/2} (A-i B) \tanh ^{-1}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{d \sqrt {\tan (c+d x)}} \]
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Rubi [A] time = 2.35, antiderivative size = 241, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 10, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3605, 3647, 3655, 6725, 63, 217, 206, 93, 205, 208} \[ \frac {b^{3/2} (5 a B+2 A b) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {(-b+i a)^{5/2} (A+i B) \tan ^{-1}\left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {b (2 a A+b B) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}-\frac {(b+i a)^{5/2} (A-i B) \tanh ^{-1}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{d \sqrt {\tan (c+d x)}} \]
Antiderivative was successfully verified.
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Rule 63
Rule 93
Rule 205
Rule 206
Rule 208
Rule 217
Rule 3605
Rule 3647
Rule 3655
Rule 6725
Rubi steps
\begin {align*} \int \frac {(a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\tan ^{\frac {3}{2}}(c+d x)} \, dx &=-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{d \sqrt {\tan (c+d x)}}+2 \int \frac {\sqrt {a+b \tan (c+d x)} \left (\frac {1}{2} a (4 A b+a B)-\frac {1}{2} \left (a^2 A-A b^2-2 a b B\right ) \tan (c+d x)+\frac {1}{2} b (2 a A+b B) \tan ^2(c+d x)\right )}{\sqrt {\tan (c+d x)}} \, dx\\ &=\frac {b (2 a A+b B) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{d \sqrt {\tan (c+d x)}}+2 \int \frac {\frac {1}{4} a \left (6 a A b+2 a^2 B-b^2 B\right )-\frac {1}{2} \left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right ) \tan (c+d x)+\frac {1}{4} b^2 (2 A b+5 a B) \tan ^2(c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}} \, dx\\ &=\frac {b (2 a A+b B) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{d \sqrt {\tan (c+d x)}}+\frac {2 \operatorname {Subst}\left (\int \frac {\frac {1}{4} a \left (6 a A b+2 a^2 B-b^2 B\right )+\frac {1}{2} \left (-a^3 A+3 a A b^2+3 a^2 b B-b^3 B\right ) x+\frac {1}{4} b^2 (2 A b+5 a B) x^2}{\sqrt {x} \sqrt {a+b x} \left (1+x^2\right )} \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac {b (2 a A+b B) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{d \sqrt {\tan (c+d x)}}+\frac {2 \operatorname {Subst}\left (\int \left (\frac {b^2 (2 A b+5 a B)}{4 \sqrt {x} \sqrt {a+b x}}+\frac {3 a^2 A b-A b^3+a^3 B-3 a b^2 B-\left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right ) x}{2 \sqrt {x} \sqrt {a+b x} \left (1+x^2\right )}\right ) \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac {b (2 a A+b B) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{d \sqrt {\tan (c+d x)}}+\frac {\operatorname {Subst}\left (\int \frac {3 a^2 A b-A b^3+a^3 B-3 a b^2 B-\left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right ) x}{\sqrt {x} \sqrt {a+b x} \left (1+x^2\right )} \, dx,x,\tan (c+d x)\right )}{d}+\frac {\left (b^2 (2 A b+5 a B)\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {x} \sqrt {a+b x}} \, dx,x,\tan (c+d x)\right )}{2 d}\\ &=\frac {b (2 a A+b B) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{d \sqrt {\tan (c+d x)}}+\frac {\operatorname {Subst}\left (\int \left (\frac {a^3 A-3 a A b^2-3 a^2 b B+b^3 B+i \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right )}{2 (i-x) \sqrt {x} \sqrt {a+b x}}+\frac {-a^3 A+3 a A b^2+3 a^2 b B-b^3 B+i \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right )}{2 \sqrt {x} (i+x) \sqrt {a+b x}}\right ) \, dx,x,\tan (c+d x)\right )}{d}+\frac {\left (b^2 (2 A b+5 a B)\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+b x^2}} \, dx,x,\sqrt {\tan (c+d x)}\right )}{d}\\ &=\frac {b (2 a A+b B) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{d \sqrt {\tan (c+d x)}}-\frac {\left ((a-i b)^3 (A-i B)\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {x} (i+x) \sqrt {a+b x}} \, dx,x,\tan (c+d x)\right )}{2 d}+\frac {\left ((a+i b)^3 (A+i B)\right ) \operatorname {Subst}\left (\int \frac {1}{(i-x) \sqrt {x} \sqrt {a+b x}} \, dx,x,\tan (c+d x)\right )}{2 d}+\frac {\left (b^2 (2 A b+5 a B)\right ) \operatorname {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}\\ &=\frac {b^{3/2} (2 A b+5 a B) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {b (2 a A+b B) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{d \sqrt {\tan (c+d x)}}-\frac {\left ((a-i b)^3 (A-i B)\right ) \operatorname {Subst}\left (\int \frac {1}{i-(-a+i b) x^2} \, dx,x,\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {\left ((a+i b)^3 (A+i B)\right ) \operatorname {Subst}\left (\int \frac {1}{i-(a+i b) x^2} \, dx,x,\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}\\ &=\frac {(i a-b)^{5/2} (A+i B) \tan ^{-1}\left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {b^{3/2} (2 A b+5 a B) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {(i a+b)^{5/2} (A-i B) \tanh ^{-1}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {b (2 a A+b B) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{d \sqrt {\tan (c+d x)}}\\ \end {align*}
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Mathematica [C] time = 41.18, size = 209298, normalized size = 868.46 \[ \text {Result too large to show} \]
Warning: Unable to verify antiderivative.
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fricas [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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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.
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maple [B] time = 1.10, size = 2651144, normalized size = 11000.60 \[ \text {output too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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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.
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )\,{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{5/2}}{{\mathrm {tan}\left (c+d\,x\right )}^{3/2}} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (A + B \tan {\left (c + d x \right )}\right ) \left (a + b \tan {\left (c + d x \right )}\right )^{\frac {5}{2}}}{\tan ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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